Below grade 7
This test has the first 120 of the 191 questions that match.
Question 1 of 120
2026 Practice Test A English, question 9Passage The Best Laid Plans of Ravens
The Best Laid Plans of Ravens
In Edgar Allan Poe’s poem “The Raven,” a raven visits a lonely man’s home and responds to the man’s pleading questions with only the word “nevermore.” The poem’s narrator interprets the word as a prediction of doom for his future. A talking, prophetic raven may seem to be the wild imaginings of the poet, but a new study published in the journal Science hints that one particular idea behind the poem might not be as far-fetched as it seems. For most of human history, people assumed that animals do not understand the passage of time in the same way people do. Some people believed that animals might remember events from the past and that instinct might drive them to make preparations in order to guarantee survival, but most people did not think that animals had the ability to plan. At Lund University in Sweden, researchers argue that ravens may be able to think ahead and even plan for the future.
It can be difficult to test an animal’s ability to plan because human observers must be certain they are not mistaking instinctual behavior for intentional planning. For example, many animals hoard food so that they will not run out later, but scientists who study animals would not call hoarding a decision to plan for the future. This action is merely instinctual. Cognitive scientists argue that in order for an animal’s behavior to qualify as preparing for the future, the animal must use specific decision-making skills to solve a problem.
To avoid mistaking instinctual behavior for evidence of decision-making, the Lund University researchers designed two experiments to test ravens’ ability to plan. Ravens belong to the corvid family, a group of birds known for their intelligence. A study in 2007 showed that corvids have the tendency to save only certain types of food, which suggests that they are planning for the future rather than acting on instinct. In order to investigate that theory, the researchers had to design experiments that would achieve results that could not be explained by an instinctual behavior of food hoarding. Therefore, the ravens were taught two behaviors that they do not normally perform in the wild.
For the first experiment, the researchers showed the birds how to use a small stone to open a box and get treats. Once the ravens learned the behavior, the researchers presented the birds with four stones. Only one stone was the right size to open the box. The birds learned to select that stone and set it aside until the researchers presented the box. The second experiment involved bartering. A researcher would trade the ravens a large treat for a bottle cap. Later, the researchers presented the ravens with a group of items, including small treats and the bottle cap. The ravens chose the bottle cap over the treats and waited for the original researcher to trade with them again so that they could get more treats. In both experiments, the ravens waited patiently for up to seventeen hours for the researcher to return.
The results of these experiments are exciting, but more evidence needs to be gathered before scientists can fully conclude that ravens can plan for the future. Some scientists argue that the ravens might be choosing the stone and bottle cap because the ravens have been trained to do so, not necessarily because the ravens are thinking ahead. Regardless, like other recent advances in animal science, these experiments show that ravens could be much smarter than first believed, and scientists now believe that ravens do actually think about their own future.
Question
Read these sentences from paragraph 2.
For example, many animals hoard food so that they will not run out later, but scientists who study animals would not call hoarding a decision to plan for the future. This action is merely instinctual.
Which statement describes the effect of the phrase “merely instinctual” in the passage?
- AIt implies that animals are skilled at finding and saving food for later consumption.
- BIt conveys that many animals will usually prioritize gathering food over other activities.
- CIt suggests that animals often store more food than they will be able to consume.
- DIt emphasizes that many animals collect food automatically rather than with true intention.
Show the answer
Correct answer: D
The question asks which statement describes the effect in the passage of the phrase “merely instinctual” from paragraph 2.
A. Incorrect. The passage does not question animals’ ability to find and save food but rather whether this or other behavior involves thought and planning for the future.
B. Incorrect. While the sentence states that “many animals hoard food,” the passage does not suggest that they give priority of thought to this activity; they hoard food out of instinct.
C. Incorrect. Although the passage discusses animals’ tendency to store a great deal of food, it does not suggest that their hoarding goes beyond their needs.
D. CORRECT. The passage suggests that in order for an animal’s actions to qualify as planning for the future, “the animal must use specific decision-making skills to solve a problem” (paragraph 2). The behavior of hoarding food is based on instinct and done automatically without real thought.
Question analysis
This word-in-context question from “The Best Laid Plans of Ravens” asks about the effect of the phrase “merely instinctual,” which the article applies to food hoarding in paragraph 2. Students read two quoted sentences and decide what the phrase emphasizes about how animals behave. The choices all sound like reasonable facts about animals, so the answer has to come from how the article uses the phrase.
- Grade level
- Grade 7 reading and task
- Compared with grade 7
- Below grade 7 Grade 7 word-meaning questions often need clues gathered from a whole paragraph. Here the phrase is short and familiar, and the sentence just before it supplies a direct contrast, so one careful reading of the two quoted sentences is enough.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- Hard words minor
- The right answer is reworded minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 11, 599 words; question and choices: about grade 10, 108 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Most students know that an instinct is something an animal does without thinking, and the sentence just before the phrase says hoarding is not “a decision to plan.” The only real pitfall is “merely”: a student who skips it can hear praise of a natural talent, and one who thinks only about hoarding may pick a choice about storing too much. Reading the phrase with its contrast makes the answer plain.
Hard words minor Understanding the question
The phrase pairs “merely,” which shrinks what follows, with “instinctual,” and the sentence before it uses “hoard”; a student unsure of any of the three has little to go on. (Perfetti, 2007; Freedle & Kostin, 1993)
The right answer is reworded minor Choosing the answer
The right choice restates the idea as acting “automatically rather than with true intention,” while the wrong ones stay with the sentence’s own subject, finding, gathering and storing food. (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)
Strategy: Ask what the word “merely” adds to “instinctual,” then reread the sentence before the phrase and use its contrast to decide what the author is saying about hoarding.
Traps
- Dictionary meaning, not meaning here: A common meaning of the word that does not fit its use in this text.
- Outside knowledge: A choice is plausible in the real world but not supported by this text.
The wrong answers, classified
A Wrong tone or attitude (Graesser, Singer & Trabasso, 1994)
Hears “instinctual” as a natural gift, as in an instinctive athlete, and turns the phrase into praise of animals’ skill, missing that “merely” makes it a put-down.
B Outside knowledge (King et al., 2004; Recht & Leslie, 1988)
Brings in the idea of instinct as an urge that overrides everything else, but the passage never ranks food gathering above an animal’s other activities.
C Dictionary meaning, not meaning here (Perfetti, 2007)
Gives “hoard” its everyday sense of piling up too much, though the passage says animals store food so they “will not run out later,” not beyond their needs.
Working it out, step by step
Figure out what the question asks
This asks about the effect of a phrase. What do the words “merely instinctual” add to what the passage says about animals and food? Read the phrase together with the sentence before it.
Read the sentences around the phrase
Paragraph 2 says “many animals hoard food so that they will not run out later, but scientists who study animals would not call hoarding a decision to plan for the future. This action is merely instinctual.” To hoard is to collect and store a lot of something.
Work out what the phrase means
An instinct is a natural urge an animal is born with. It acts on it without thinking it through. So instinctual behavior is automatic. Merely means “only” or “just.” Put together, the phrase says hoarding is just automatic. It is not a real decision. The next sentence agrees: to count as planning, “the animal must use specific decision-making skills to solve a problem.”
Check each choice
- A does not fit. The phrase does not praise animals’ skill. “Merely” makes the behavior sound smaller.
- B does not fit. The passage never says animals put food ahead of other activities.
- C does not fit. The passage never says animals store too much. They store food “so that they will not run out later.”
- D fits. “Automatically rather than with true intention” means by instinct, not on purpose. An intention is a plan you have in mind.
Choose the answer
The phrase stresses that animals hoard food without real thought or purpose. So the answer is D.
Why this answer and not the closest wrong one
A is the closest wrong choice, because an instinct can sound like a natural talent. But “merely” turns the phrase into a put-down. The author is saying hoarding is only instinct, not a real choice. D says the same thing: animals collect food automatically, not with true intention.
More questions like this
The original DOE question

The passage “The Best Laid Plans of Ravens” as printed (page 12)

Question 2 of 120
2026 Practice Test A English, question 48Question
Which sentence of the paragraph should be revised to correct a vague pronoun?
(1) Eliza and Brianna have been singing in their school chorus since they were in fourth grade. (2) The girls always sing a duet at the school talent show, and they take turns singing the national anthem before school sporting events. (3) Outside of school, she also sings in a choir made up of young and old members of her community. (4) Both girls hope that they will be able to continue singing for many more years.
- Asentence 1
- Bsentence 2
- Csentence 3
- Dsentence 4
Show the answer
Correct answer: C
The question asks for identification of the sentence with a vague pronoun.
A. Incorrect. There is no vague pronoun in sentence 1, which correctly uses the plural pronouns “their”/“they” to refer to both Eliza and Brianna.
B. Incorrect. There is no vague pronoun in sentence 2, which uses the pronoun “they” to correctly refer to both girls again.
C. CORRECT. Sentence 3 is the only sentence where the pronoun is vague. The sentence uses the pronoun “she” near the beginning, but whether “she” refers to Eliza or Brianna is unclear.
D. Incorrect. There is no vague pronoun in sentence 4, which uses the pronoun “they” to correctly refer to “both girls.”
Question analysis
This editing question gives a short paragraph about Eliza and Brianna, two girls who have sung in their school chorus since fourth grade, and asks which sentence should be revised to fix a vague pronoun. Students must trace each pronoun back to the person or people it stands for and ask whether a reader could tell who is meant.
- NYS standards
- 7L1 grammar and usage conventions
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Recognizing and fixing vague pronouns is taught in grade 6. This paragraph is four short sentences in everyday language with one rule to apply, lighter than the multi-error paragraphs typical of grade 7 editing.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Question and choices: about grade 5, 90 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The paragraph is short, plainly worded and about two named girls, so the one singular pronoun stands out once students look for the person each pronoun stands for. Points are lost mainly by students unsure what “vague” means, who suspect a sentence with several pronouns or a plural “they” instead. Tracing each pronoun back to its noun settles the question in seconds.
Something you must already know minor Working it out
Students must know what makes a pronoun vague: “she” in sentence 3 could mean either Eliza or Brianna, while each plural “they” clearly means both girls. (Anderson & Krathwohl, 2001)
Strategy: Circle every pronoun, draw a line from each to the noun it replaces, and look for one where the line could go to more than one person.
Traps
- Fixes what isn't broken: Changes a part of the sentence that was already correct, while the real error stays; or, where the question asks which sentence has an error or should go, picks one that is correct and belongs.
The wrong answers, classified
A Fixes what isn't broken (Connors & Lunsford, 1988)
Sees two pronouns, “their” and “they,” in the first sentence and suspects confusion, though both are plural and plainly refer to Eliza and Brianna together.
B Fixes what isn't broken (Connors & Lunsford, 1988)
Thinks “The girls” and “they” are too general and wants the names repeated, though both point clearly back to the two girls just named.
D Fixes what isn't broken (Connors & Lunsford, 1988)
Wonders whether “they” could include the choir members of the sentence before, though “Both girls” at the start of sentence 4 fixes who “they” are.
Working it out, step by step
Learn the rule for pronouns
A pronoun is a short word that stands in for a noun, like she, he, they or it. A pronoun is vague when the reader can’t tell who it means. For example: Ana and Lily went to the park, and she fell. Who fell, Ana or Lily? You can’t tell.
Trace each pronoun to its noun
- Sentence 1: “their” and “they” mean both girls, Eliza and Brianna.
- Sentence 2: “they” means “The girls,” both of them.
- Sentence 3: “she” and “her” mean one girl. But which one?
- Sentence 4: “they” means “Both girls.”
Find the vague one
Sentence 3 says, “Outside of school, she also sings in a choir made up of young and old members of her community.” Two girls were just named. “She” could be Eliza or Brianna. The reader can’t tell who sings in the choir.
Show the fix
Replace the pronoun with a name: Outside of school, Brianna also sings in a choir made up of young and old members of her community. It could just as well be Eliza. The point is to name one girl. Now the reader knows who sings in the choir.
Check each choice
- A is wrong. “Their” and “they” are plural, meaning more than one. Both clearly mean the two girls.
- B is wrong. “They” points back to “The girls,” just named.
- C is right. “She” could mean either girl.
- D is wrong. “Both girls” at the start makes “they” clear.
Watch out: Having many pronouns doesn’t make a sentence vague. Sentence 1 has two, but each one is clear.
Choose the answer
Only sentence 3 has a pronoun that could point to two different people. So the answer is C, sentence 3.
Why this answer and not the closest wrong one
Sentence 1 is the closest wrong answer, because it has two pronouns, “their” and “they.” But both are plural, and both mean Eliza and Brianna together. In sentence 3, the singular “she” could mean either girl. So only sentence 3 needs a fix.
More questions like this
The original DOE question

Question 3 of 120
2026 Practice Test B English, question 8Passage A Miracle Mile
A Miracle Mile
In the 1950s people compared running one mile in four minutes to scaling Mount Everest and nicknamed the feat a “dream mile.” Although such an accomplishment was considered humanly impossible, several elite runners aimed to break that supposedly impenetrable barrier. One of them was a twenty-five-year-old medical student named Roger Bannister.
Roger Bannister had tasted failure during the 1952 Olympics. There, he was favored to win the 1,500-meter competition, a distance slightly shorter than a mile, but he finished in a dismal fourth place instead. Bannister’s performance was a disappointment for him and his country, Great Britain. Determined to redeem himself, Bannister postponed his plans to retire from racing and focused on the ultimate prize—breaking the four-minute-mile barrier.
Bannister attacked the elusive milestone with a positive attitude and logical planning. The amateur athlete decided to use intensive interval training to develop endurance and speed. For these workouts, Bannister ran an interval of ten consecutive laps on a quarter-mile track, aiming for sixty seconds each lap. In between intervals, he let his body recover for two minutes.
By early 1954, Bannister had succeeded in lowering his quarter-mile pace to sixty-one seconds, but he had to shave off at least one more second in order to reach his target. Frustrated by the plateau he had reached, Bannister took a break from training and went mountain climbing for three days. The rest from running permitted his muscles to recuperate and left him feeling refreshed. When Bannister returned to the track, he completed ten quarter-mile-long intervals at fifty-nine seconds each. He finally felt prepared to attempt to break the world record.
As a member of the Amateur Athletic Association (AAA), Bannister joined the AAA team for a track meet against Oxford University. The event took place on a cinder track in Oxford on May 6, 1954. Bannister and his two AAA teammates, Chris Chataway and Chris Brasher, were close friends and frequent running partners. Chataway and Brasher agreed to help Bannister accomplish his goal by being his “rabbits.”
In track and field, rabbits are runners who enter the race solely to pace a teammate for a segment of the course. Typically, a runner settles in behind the rabbit and allows the rabbit to set an appropriate tempo. Additionally, by running behind the rabbit, the runner conserves about 15 percent of his or her effort. When the starting pistol fired, Brasher pounced into the lead, and Bannister followed behind his first rabbit.
Propelled by the excitement, Bannister lost his instinctive feel for his pace and shouted “Faster!” at Brasher. Brasher, however, remained composed and maintained his current steady but grueling pace, completing the first two laps in a desirable one minute and fifty-eight seconds. Then Chataway surged forward, leading Bannister at this same punishing rate for another lap and a half. At the beginning of the back straightaway of the track, Bannister bolted past Chataway. Bannister said, “I felt that the moment of a lifetime had come. There was no pain, only a great unity of movement and aim.” Bannister crossed the finish line in 3 minutes 59.4 seconds. The ecstatic crowd erupted the moment the timekeeper announced the word “three.”
Soon after Bannister’s achievement, four other athletes matched his performance. A new mindset had taken root among runners. Over the years, the record continued to fall. However, the current record, 3 minutes 43.13 seconds, has stood unbroken since 1999. Some question whether this record represents the limits of human ability. But perhaps there is another Bannister, an athlete who, with willpower and dedication, will accomplish the miraculous.
Question
The words “feat,” “humanly impossible,” and “impenetrable barrier” in paragraph 1 affect the tone of the paragraph because they
- Ahighlight the idea that only the most skilled runners would be able to run a four-minute mile.
- Bemphasize the idea that running a mile in less than four minutes was a seemingly unattainable goal.
- Cconvey the competitiveness among elite runners to consistently set and break speed records.
- Dshow the intensity of the training programs athletes endure in order to achieve their goals.
Show the answer
Correct answer: B
The question asks how the words “feat,” “humanly impossible,” and “impenetrable barrier” in paragraph 1 affect the tone of the paragraph.
A. Incorrect. Although paragraph 1 states that “several elite runners aimed to break that supposedly impenetrable barrier,” the words “feat,” “humanly impossible,” and “impenetrable barrier” suggest that the goal was unattainable even for the most skilled runners.
B. CORRECT. No one had ever been able to run a mile in less than four minutes, and the words “feat,” “humanly impossible,” and “impenetrable barrier” imply that the goal seemed unattainable. These words also reinforce the comparison of running a four-minute mile to “scaling Mount Everest” (paragraph 1).
C. Incorrect. The words “feat,” “humanly impossible,” and “impenetrable barrier” describe how challenging and almost impossible running a four-minute mile seemed, and while other elite runners “aimed to break that supposedly impenetrable barrier” (paragraph 1), the paragraph focuses on Roger Bannister, not the competition among elite runners seeking to break the record.
D. Incorrect. Bannister’s training program is described in paragraphs 3 and 4, and the words “feat,” “humanly impossible,” and “impenetrable barrier” are used in paragraph 1 to describe the magnitude of Bannister’s goal to run a four-minute mile. Though his training program was intense, these words relate to the intensity of the goal itself, not to the intensity of his training.
Question analysis
This tone question on “A Miracle Mile,” an article about Roger Bannister and the four-minute mile, asks how three word choices in the opening paragraph, “feat,” “humanly impossible” and “impenetrable barrier,” affect its tone. Students must read the connotations of strong words in context and decide what they stress, while choices that echo other parts of the passage compete for attention.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Grade 7 tone questions often ask students to find the telling words themselves. Here the three words are quoted and sit in one short opening paragraph, and their shared meaning is strong and plain, so one careful reading is enough.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- A very close wrong answer minor
- Hard words minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 10, 587 words; question and choices: about grade 11, 80 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The three quoted words all make the same point, and the opening comparison with “scaling Mount Everest” makes it again, so the effect is easy to sense. Points are lost on degree: the sentence holding two of the three words also names “several elite runners,” which makes a choice about only the best runners look supported. Students who hold on to “humanly impossible” rule it out, and the choices about rivalry and training fall away once they check that the words describe the goal, not the runners or their preparation.
A very close wrong answer minor Choosing the answer
The choice about only the most skilled runners and the right one both stress how hard the four-minute mile was; they differ on degree, possible for the best runners or thought impossible for anyone. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Hard words minor Understanding the question
“Impenetrable” and “feat” carry the tone, and a student who does not know that impenetrable means impossible to break through may miss how absolute the barrier sounded. (Perfetti, 2007; Freedle & Kostin, 1993)
Strategy: Decide what feeling the three quoted words share, then reread paragraph 1 to see what they describe, and keep only a choice that fits both the feeling and the thing described.
Traps
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Word match: Repeats striking words or phrases from the text, so it looks supported, but its meaning does not fit the question.
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
The wrong answers, classified
A Twists what the text says (King et al., 2004)
Sees that the words stress difficulty but scales it down to a feat for the best runners only, though “humanly impossible” says no runner at all was thought able to do it.
C Word match (Freedle & Kostin, 1993)
Picks up “elite runners aimed to break” from the same sentence and turns it into a rivalry over records, though the three words describe the barrier, not the runners.
D Wrong part of the text (Kirsch & Mosenthal, 1990)
Carries the intensive interval training of paragraphs 3 and 4 back to paragraph 1, where the words measure the size of the goal, not the effort of preparing for it.
Working it out, step by step
Figure out what the question asks
This is a tone question. Tone is the feeling or attitude that a writer’s words give a passage. The question names three word choices from paragraph 1. You need to decide what feeling they share and what they describe.
Find the words in paragraph 1
Paragraph 1 says people “compared running one mile in four minutes to scaling Mount Everest.” Scaling means climbing. Mount Everest is the tallest mountain on Earth. The paragraph calls running that fast a “feat.” The next sentence says “Although such an accomplishment was considered humanly impossible, several elite runners aimed to break that supposedly impenetrable barrier.”
Work out what the words mean
A feat is an amazing deed that is very hard to do. Humanly impossible means no person could ever do it. Impenetrable means impossible to break through, and a barrier is a wall. All three say the same thing. A mile under four minutes seemed like a goal no one could reach. A word for that is unattainable.
Check each choice
- A does not fit. It says the best runners could do it. But “humanly impossible” means no one could.
- B fits. The words make the four-minute mile sound out of reach.
- C does not fit. The words describe the barrier, not a contest between runners.
- D does not fit. Training comes later, in paragraphs 3 and 4. These words describe the goal itself.
Watch out: A is tempting because the same sentence mentions “several elite runners.” Elite means the very best. But even they were not expected to succeed.
Choose the answer
The three words make the goal sound almost impossible for anyone. So the answer is B, “emphasize the idea that running a mile in less than four minutes was a seemingly unattainable goal.”
Why this answer and not the closest wrong one
A is the closest wrong choice. It also says the four-minute mile was hard. But A says that only the most skilled runners could do it. The passage says it was “considered humanly impossible.” People thought no runner at all could do it. That is why B fits better.
More questions like this
The original DOE question

The passage “A Miracle Mile” as printed (page 13)

Question 4 of 120
2026 Practice Test B English, question 13Uses the passage above: “A Miracle Mile”.
Question
How did interval training affect Bannister’s performance?
- AIt helped him learn how to moderate his pace while running.
- BIt helped him conserve effort when running with teammates.
- CIt helped him improve his pace and stamina while running.
- DIt helped him decrease his recovery time after an intense run.
Show the answer
Correct answer: C
The question asks how interval training affected Bannister’s performance.
A. Incorrect. The moderation of his pace was achieved through the use of “rabbits,” described in paragraph 6, more than through his method of interval training.
B. Incorrect. Bannister worked individually during the interval training period and then paced himself with a two-person team of “rabbits,” which were introduced in paragraph 5 and explained in paragraph 6. It was the use of “rabbits” during the race that helped him conserve some of his effort, not the interval training.
C. CORRECT. As paragraph 3 states, the goal of Bannister’s training program was to “develop endurance and speed,” which are synonyms of the words “stamina” and “pace.”
D. Incorrect. Although Bannister “let his body recover for two minutes” (paragraph 3) during interval training, this was not the primary effect of the training on Bannister’s performance. Instead, interval training allowed him to develop greater endurance and speed.
Question analysis
This question asks how interval training affected Roger Bannister’s performance as he prepared to run a mile in under four minutes, according to “A Miracle Mile.” It tests finding a detail the passage states and recognizing it in a choice that may use different words. The passage describes several parts of his preparation and race, so students must keep track of which one the question is about.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Below typical grade 7 demand: the passage states the purpose of the training in a single sentence. A grade 7 student only needs to locate it and compare the choices with it, without inferring anything.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly understand
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 10, 587 words; question and choices: about grade 7, 48 words.
- Corrected from the DOE handbook
- The DOE’s 2026 explanations appendix skips this question; the explanation and answer key here are the DOE’s own, from its 2021–2022 guide (question 40), relettered.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Paragraph 3 states what the training was for, so the task is to match that sentence and ignore details from elsewhere. The rabbits of paragraph 6, who set Bannister’s tempo and save his effort, supply two tempting effects that belong to the race, not the workouts. The two-minute rests sit in the training paragraph itself, so a student skimming for “recover” can mistake part of the method for its result.
A very close wrong answer minor Choosing the answer
The choice about moderating his pace and the right one both concern pace; only the right one matches the training’s stated aim, to “develop endurance and speed.” (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
The right answer is reworded minor Choosing the answer
The right choice turns “endurance and speed” into stamina and pace, while two wrong ones reuse the passage’s own words, “conserves” effort from paragraph 6 and “recover” from paragraph 3. (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)
Strategy: Find the paragraph that introduces interval training, note what the training was meant to develop, and pick the choice that means the same thing.
Traps
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
- Word match: Repeats striking words or phrases from the text, so it looks supported, but its meaning does not fit the question.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
The wrong answers, classified
A Twists what the text says (King et al., 2004)
Sees the sixty-second laps as practice in holding a pace, but the training aimed to build speed and endurance, and in the race his rabbits had to hold his tempo for him.
B Wrong part of the text (Kirsch & Mosenthal, 1990)
Matches the saving of “about 15 percent of his or her effort,” which paragraph 6 credits to running behind a rabbit in the race, not to interval training.
D Word match (Freedle & Kostin, 1993)
Spots “recover” in paragraph 3 and picks the choice about recovery time, though the two-minute rest is part of the workout, and nothing says it grew shorter.
What was corrected from the DOE handbook, and why
- Explanation and answer key: The 2026 printable test includes this question (Passage Set 2, Item 6), but the explanations appendix skips it: it explains only five of the six questions on “A Miracle Mile.” The question is question 40 of the DOE’s NYC Guide to the SHSAT for 2022 Admissions (the 2021–2022 guide, page 165), with the same stem and options lettered E–H. The explanation is the guide’s (page 218) and its key is G; both are relettered E–H to A–D here, so the key is C.
Working it out, step by step
Figure out what the question asks
This asks what interval training did for Bannister’s running. So find the paragraph that first describes interval training. Look for what it was meant to build.
Find interval training in paragraph 3
Paragraph 3 says “The amateur athlete decided to use intensive interval training to develop endurance and speed.” In each workout he ran ten laps, aiming for sixty seconds a lap. Then he rested for two minutes and ran again. That is what an interval is here: a hard stretch of running between rests.
See what the training did
Endurance means being able to keep going for a long time. Another word for it is stamina. Speed is how fast you run, which is your pace. Paragraph 4 shows the result. By early 1954, he had lowered “his quarter-mile pace to sixty-one seconds.” He was getting faster.
Check each choice
- A does not fit. To moderate your pace means to keep it steady and not too fast. In the race, his rabbits did that for him. He even lost his feel for his pace.
- B does not fit. Saving effort came from running behind rabbits in the race (paragraph 6).
- C fits. Stamina and pace are other words for endurance and speed.
- D does not fit. He rested two minutes between runs. Nothing says that rest got shorter.
Watch out: D is tempting because paragraph 3 mentions rest: “he let his body recover for two minutes.” But the rest is part of the workout, not a result of it.
Choose the answer
Interval training built Bannister’s speed and endurance. So the answer is C, “It helped him improve his pace and stamina while running.”
Why this answer and not the closest wrong one
A is the closest wrong choice, because it is also about pace. But the training aimed “to develop endurance and speed,” not to keep his pace in check. In the race, Bannister “lost his instinctive feel for his pace,” and Brasher had to keep it steady. C matches what the training was for.
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The original DOE question

Question 5 of 120
2026 Practice Test B English, question 23Passage Looking for the Smallest Spark of Everything
Looking for the Smallest Spark of Everything
Located underground near Geneva, Switzerland, the Large Hadron Collider (LHC) helps scientists study and understand how the smallest particles of matter interact with one another. The LHC propels atomic particle beams along a 17-mile-long ring.
There are the things the world is made of,
things we can see and feel, and then
there are the things even smaller, things that seem
to exist only when we are looking right at them.
So beneath the level of molecules are atoms,
and beneath the level of atoms, we find
protons, electrons, neutrons; and beneath that—
what, exactly? Because we are talking about
everything here: your fingernail,
the candy at the back of your mouth,
the coffee your teacher drank this morning,
your little sister, and the stuffed dog
she used to carry around with her everywhere.
And everywhere.
Everything and everywhere
are made up of the same stuff, whatever
it is. How do we find it? We can listen for it
in the wavelengths from deep space,
talking back to us from unimaginable
distances. Or we can build long,
deep circular tunnels beneath the surface
of this Earth and race particles
(like racing cars, only very tiny cars
flashing along at close to the speed of light)
until everything we think we know
bangs against everything else we know.
And there, in the explosion, in the darkness,
briefly incandescent, they appear:
the quarks, the leptons, and the bosons;
the baryons and the mesons. Their names
sound like dinosaurs or maybe bands
playing terrible music in someone’s garage.
The one thing we know for sure
is that they spin. This is how gravity
enters our world, how our world is held
both together and apart, what keeps
together the pencil in your hand right now
as well as separate from, say,
Jupiter. They spin, and it is only down there
in the darkness—in the vast garage
where physicists jot down
what they can, whatever seems most real—
that they let us perceive their wild dancing,
combusting to the music they make.
Question
Read lines 5–8 from the poem.
So beneath the level of molecules are atoms,
and beneath the level of atoms, we find
protons, electrons, neutrons; and beneath that—
what, exactly?
The lines help develop a central idea of the poem by revealing the speaker’s
- Ainterest in determining how the parts of matter work together.
- Bcuriosity about what makes up matter.
- Cdesire to prove that particles make up matter.
- Dquestions about what tools are needed to study matter.
Show the answer
Correct answer: B
The question asks what lines 5–8 reveal about the speaker to help develop a central idea of the poem.
A. Incorrect. In lines 5–8, the speaker is demonstrating knowledge about the structure of matter, not demonstrating interest in how matter’s parts work together; for example, “beneath the level of molecules are atoms, / and beneath the level of atoms, we find / protons, electrons, neutrons” (lines 5–7).
B. CORRECT. Lines 5–8 show the connection between the particles by using the words “beneath the level” repeatedly. The progression of particles of matter, from atoms to subatomic particles, “protons, electrons, neutrons” (line 7), ends with a question. By asking, “what, exactly?” (line 8), the speaker questions the mysteries that are still unsolved about the nature of matter, which is a central idea of the poem.
C. Incorrect. The speaker expresses curiosity in line 8 (“what, exactly?”) but does not use language in lines 5–8 that expresses a desire to prove a point. The poem does not show any conflict, either from within or from other scientists, about the concept that particles make up matter.
D. Incorrect. While the poem is about the Large Hadron Collider, which, according to the introductory text before the poem, is a tool that “helps scientists study and understand how the smallest particles of matter interact with one another,” lines 5–8 do not suggest the use of tools but rather question what comes after the “protons, electrons, neutrons” (line 7).
Question analysis
This question on the poem “Looking for the Smallest Spark of Everything,” about the Large Hadron Collider, quotes lines 5–8, which move from molecules down to atoms and smaller particles. Students decide what the lines reveal about the speaker and how that develops a central idea of the poem. The choices name related attitudes toward matter and science, so precise reading matters.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Below typical grade 7 demand: the four lines are quoted in the question and use everyday words, so a single step of interpretation links them to the speaker’s attitude and to the poem’s larger subject.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: 346 words (a poem; reading level not measured); question and choices: about grade 7, 76 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The lines move down level by level, molecules, atoms, protons, and then stop at a question, so the speaker’s curiosity is close to the surface. The choice about how the parts work together comes nearest to the right one, helped by the head-note, which says the collider studies how particles “interact.” Students who notice that the lines list what lies beneath what, and never how things interact, settle on curiosity about what matter is made of.
Reading between the lines minor Working it out
The speaker never names a feeling; the curiosity shows in the dash and the question “what, exactly?” that cut off a confident list of layers. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
A very close wrong answer minor Choosing the answer
The choice about how the parts of matter work together and the right one both give the speaker an interest in matter; they part on its object, how parts interact or what matter is made of. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Strategy: Read lines 5–8 as one sentence, describe in a word or two what the speaker is doing in them, and pick the choice that matches those exact lines.
Traps
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
- Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
The wrong answers, classified
A Twists what the text says (King et al., 2004)
Keeps the speaker’s interest in matter but shifts it from what lies beneath atoms to how parts interact, the subject the head-note gives the collider, not these lines.
C Reads too much in (Graesser, Singer & Trabasso, 1994)
Reads the confident list of layers as a case being argued, though the speaker treats them as known and ends on a question, seeking rather than proving.
D Wrong part of the text (Kirsch & Mosenthal, 1990)
Jumps ahead to stanza 3, where the speaker asks “How do we find it?” and describes tunnels, but lines 5–8 say nothing about tools or ways of studying matter.
Working it out, step by step
Figure out what the question asks
This asks what lines 5–8 show about the speaker, the voice talking in the poem. It also asks how that fits a central idea, a main point of the whole poem. Decide what the speaker is doing in these lines.
Read lines 5–8
The lines say “So beneath the level of molecules are atoms, / and beneath the level of atoms, we find / protons, electrons, neutrons; and beneath that— / what, exactly?” Matter is the stuff everything is made of. A molecule is a tiny group of atoms joined together. Protons, electrons and neutrons are the even tinier parts of an atom.
See what the speaker is doing
The speaker goes down layer by layer: molecules, then atoms, then the parts of atoms. Then the list stops with a dash and a question: “what, exactly?” The speaker does not know what lies beneath, and wants to know. That is curiosity. The whole poem keeps asking what everything is made of.
Check each choice
- A does not fit. The lines ask what is underneath, not how the parts work together.
- B fits. The speaker wonders what matter is made of at its smallest.
- C does not fit. The speaker is not trying to prove a point. The lines end with a question.
- D does not fit. Ways to search, like tunnels, come later, in lines 17–22. Lines 5–8 do not mention them.
Watch out: A is tempting because the note before the poem says scientists study how particles “interact,” or act on one another. But lines 5–8 are only about what lies beneath.
Choose the answer
The question “what, exactly?” shows a speaker who wants to know what matter is made of. So the answer is B, “curiosity about what makes up matter.”
Why this answer and not the closest wrong one
A is the closest wrong choice, since it also shows interest in matter. But A is about how the parts work together. Lines 5–8 never talk about that. They go down layer by layer and end with “what, exactly?” That question asks what matter is made of, so B fits.
More questions like this
The original DOE question

The passage “Looking for the Smallest Spark of Everything” as printed (pages 32–33)


Question 6 of 120
2026 Practice Test B English, question 24Uses the passage above: “Looking for the Smallest Spark of Everything”.
Question
The main purpose of the comparison in lines 23–24 of the poem is to show
- Athat the particles’ size makes them difficult to see.
- Bhow the particles move within the tunnel.
- Cthat the particles have a familiar shape.
- Dhow the particles can be seen only in darkness.
Show the answer
Correct answer: B
The question asks what the comparison in lines 23–24 of the poem is mainly meant to show.
A. Incorrect. While the particles are moving “close to the speed of light” (line 24), which would make them difficult to see, the comparison in lines 23–24 focuses on the movement of the particles, not how visible they are.
B. CORRECT. Lines 23–24 describe “tiny cars / flashing along at close to the speed of light,” which describes how the particles move.
C. Incorrect. Lines 23–24 do not describe the shape of the particles but rather their movement, which is “flashing along at close to the speed of light” (line 24).
D. Incorrect. The word “flashing” in line 24 does not refer to light in darkness but rather to speed, and the word “light” is used to reference a measurement of speed, not brightness in darkness.
Question analysis
This figurative language question asks for the main purpose of a comparison in lines 23–24 of the poem “Looking for the Smallest Spark of Everything,” where particles in an underground tunnel are likened to tiny racing cars. Students must work out what the image is mainly meant to show about the particles, judging the comparison as a whole rather than word by word.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Explaining what a simple comparison shows is grade 6 work, and this one likens particles to racing cars in two short lines of familiar words, so a single step of interpretation is enough.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- Evidence
- Answered from one sentence or line
- Reading level
- Passage: 346 words (a poem; reading level not measured); question and choices: about grade 5, 46 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The image is built from words about motion, racing cars “flashing along at close to the speed of light,” so most students see its purpose quickly. Each wrong choice seizes one piece of the picture, the cars’ tininess, their shape or the word “light,” and makes it the point. Students who ask what the whole comparison makes them picture, rather than what one word suggests, avoid all three.
Figurative or literary language minor Understanding the question
The comparison sits in parentheses, “like racing cars, only very tiny cars / flashing along at close to the speed of light,” and students must see that racing cars stand for speed. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Strategy: Name the two things being compared in lines 23–24 and ask what a reader is meant to picture, then choose the purpose that the comparison’s own words support.
Traps
- Taking figurative language literally: A choice reads a figure of speech or image at face value.
- Word match: Repeats striking words or phrases from the text, so it looks supported, but its meaning does not fit the question.
- Dictionary meaning, not meaning here: A common meaning of the word that does not fit its use in this text.
The wrong answers, classified
A Word match (Freedle & Kostin, 1993)
Fixes on “very tiny cars” and decides the point is that the particles are too small to see, though racing cars are chosen for their speed, not their size.
C Taking figurative language literally (Graesser, Singer & Trabasso, 1994)
Takes the comparison at face value, as if particles looked like little cars, when the speaker uses racing to picture how they move, not what shape they have.
D Dictionary meaning, not meaning here (Perfetti, 2007)
Reads “flashing” and “light” as a glow, like the one in the darkness of lines 27–28, but “the speed of light” measures how fast the particles travel.
Working it out, step by step
Figure out what the question asks
This asks why the poet makes a comparison in lines 23–24. A comparison shows how two different things are alike. First find the two things. Then ask what the reader is meant to picture.
Read the lines around it
Lines 20–22 describe building “deep circular tunnels beneath the surface / of this Earth” to race particles. Then lines 23–24 add “(like racing cars, only very tiny cars / flashing along at close to the speed of light)”. The speed of light is the fastest speed there is.
Work out the comparison
The particles are compared to racing cars. Racing cars zoom around a track, lap after lap, very fast. The particles zoom around the round tunnel the same way. “Flashing along” means moving so fast they are a blur. So the comparison helps you picture how the particles move.
Check each choice
- A does not fit. The cars are “very tiny,” but the point is racing, not being hard to see.
- B fits. Like racing cars, the particles speed around the tunnel.
- C does not fit. The particles do not look like cars. The comparison is about motion, not shape.
- D does not fit. “Light” here measures speed. It is not about glowing in the dark.
Watch out: A is tempting because of the word “tiny.” But a poet picks racing cars to show speed, not size.
Choose the answer
The racing cars help you picture particles speeding around the tunnel. So the answer is B, “how the particles move within the tunnel.”
Why this answer and not the closest wrong one
A is the closest wrong choice, since the poem does call them “very tiny cars.” But the comparison is to racing cars “flashing along at close to the speed of light.” Every word there is about speed and motion. So B, how the particles move, fits better.
More questions like this
The original DOE question

Question 7 of 120
2026 Practice Test B English, question 48Question
In September 2016 the National Museum of African American History and Culture opened as part of the Smithsonian Institution, the museum is already the Smithsonian’s third most popular site. Experts say that they expect this newest Smithsonian facility to welcome nearly 4 million visitors a year. The museum features more than 30,000 objects, including Muhammad Ali’s boxing gloves and a dress sewn by Rosa Parks. A commemorative copy of the Emancipation Proclamation, written in 1863 during the presidency of Abraham Lincoln, is also on display at the museum.
Which part of the paragraph contains an error in sentence structure?
Select the one error.
On the test you click the words in the text; here, choose them below.
- AInstitution, the
- Byear. The
- Cobjects, including
- DProclamation, written
Show the answer
Correct answer: A
The question asks which part of the paragraph contains an error in sentence structure.
Choice 1: “Institution, the” contains an error in sentence structure. The comma between these two words creates a type of run-on sentence known as a comma splice because both clauses—“In September 2016 the National Museum of African American History and Culture opened as part of the Smithsonian Institution” and “the museum is already the Smithsonian’s third most popular site”—are independent clauses, and must be separated with either a period or semicolon. If a comma is used to join the clauses together, a conjunction such as and would need to be added after the comma [converting the comma splice into a correct compound sentence].
Choice 2: “year. The” is correctly constructed, with independent clauses on both sides of the period (“Experts say that they expect this newest Smithsonian facility to welcome nearly 4 million visitors a year” and “The museum features more than 30,000 objects, including Muhammad Ali’s boxing gloves and a dress sewn by Rosa Parks”).
Choice 3: “objects, including” is correctly constructed. The sentence is correct as written with the nonrestrictive prepositional phrase (“including Muhammad Ali’s boxing gloves and a dress sewn by Rosa Parks”) properly preceded by a comma.
Choice 4: “Proclamation, written” is correctly constructed. The sentence is correct, with the nonrestrictive participial phrase (“written in 1863 during the presidency of Abraham Lincoln”) properly offset by a comma on either side.
Question analysis
This sentence-structure question shows a short paragraph about the 2016 opening of the National Museum of African American History and Culture, with four short phrases underlined. On the test students click the one part that contains an error in sentence structure, and the long names and the commas in the paragraph can make every spot look suspicious.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Finding run-ons and fragments is taught before grade 7, and here students check four short underlined spots in one paragraph for a single error in sentence structure. Long institution names add reading load but no extra grammar.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- A rule that doesn't sound wrong moderate
- Complex sentences minor
- Every choice must be checked minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Question and choices: about grade 10, 107 words.
- Corrected from the DOE handbook
- Two grammar terms in the explanation were corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Students no longer have to supply the fix, only find the faulty spot, so the question rests on recognizing a comma splice. The error sits in the first sentence behind a long museum name, while the phrases “including Muhammad Ali’s boxing gloves” and “written in 1863” look more suspicious to anyone unsure about phrases set off by commas. Testing each spot for a full sentence on both sides finds it quickly.
A rule that doesn't sound wrong moderate Working it out
A comma alone joins two full clauses at “Institution, the,” and because the long museum name delays the break, the comma reads as a pause rather than a missing period. (Connors & Lunsford, 1988)
Complex sentences minor Understanding the question
The first clause runs 19 words, most of them the museum’s name, so by the time “the museum is already” arrives it sounds like a continuation, not a new sentence. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Every choice must be checked minor Choosing the answer
Each of the four clickable phrases straddles a comma or a period, so students must check, at every one, whether the words on each side could stand as a sentence. (Embretson & Wetzel, 1987)
Strategy: Find the subject and main verb of each sentence and count how many complete thoughts it holds, then check whether the punctuation at each underlined spot suits what comes on either side.
Traps
- Fixes what isn't broken: Changes a part of the sentence that was already correct, while the real error stays; or, where the question asks which sentence has an error or should go, picks one that is correct and belongs.
The wrong answers, classified
B Fixes what isn't broken (Connors & Lunsford, 1988)
Suspects the break after “year” because two sentences meet there, though a period already separates two complete sentences correctly, each with its own subject and verb.
C Fixes what isn't broken (Connors & Lunsford, 1988)
Believes a phrase opening with “including” after a comma is a fragment, though it correctly attaches examples to “more than 30,000 objects” within a complete sentence.
D Fixes what isn't broken (Connors & Lunsford, 1988)
Thinks “written in 1863” cannot describe the copy without “which was,” though the sentence already has its verb “is” and the participial phrase is correctly set off.
What was corrected from the DOE handbook, and why
- Explanation: The DOE’s explanation of choice 3 calls “including Muhammad Ali’s boxing gloves and a dress sewn by Rosa Parks” a “nonrestrictive appositive phrase.” An appositive renames a noun; this phrase gives examples of the objects, and “including” followed by a noun phrase works as a preposition, so it is a prepositional phrase. The explanation of choice 4 calls “written in 1863 during the presidency of Abraham Lincoln” a “nonrestrictive clause”; it has no subject or finite verb of its own, so it is a participial phrase. The site now says “nonrestrictive prepositional phrase” and “nonrestrictive participial phrase.” Both parts are still correctly constructed, and the answer, “Institution, the,” is unchanged.
Working it out, step by step
Figure out what the question asks
This question asks you to find the one error in sentence structure. Sentence structure is how a sentence is built. On the test, you click the part with the mistake. At each part, ask: could the words on each side stand as a sentence?
Learn the rule: no comma splice
An independent clause is a group of words with a subject and a verb that can stand alone as a sentence. You can’t join two of them with just a comma. That mistake is called a comma splice. Example: I ran home, the rain started. Fix it with a period: I ran home. The rain started. You could also add and after the comma.
Test “Institution, the”
Before the comma: “In September 2016 the National Museum of African American History and Culture opened as part of the Smithsonian Institution.” That is a full sentence. After the comma: “the museum is already the Smithsonian’s third most popular site.” That is a full sentence too. Only a comma joins them, so this is a comma splice. Fix: put a period after Institution and start The museum with a capital.
Check “year. The” and “objects, including”
- “year. The” is correct. A period already separates two full sentences.
- “objects, including” is correct. A preposition is a linking word like with or for, and “including” works like one here. A prepositional phrase is that word plus the words after it. “Including Muhammad Ali’s boxing gloves and a dress sewn by Rosa Parks” is a prepositional phrase. It gives examples of the objects, and a comma correctly attaches it.
Check “Proclamation, written”
“Written in 1863 during the presidency of Abraham Lincoln” is a participial phrase. That is a group of words built on a verb form, here written, with no subject of its own. It tells when the Proclamation was written. Commas correctly set it off, and the sentence keeps its main verb, “is.”
Watch out: Phrases starting with “including” and “written” can’t stand alone, so they may look like errors. But each one is attached to a full sentence. That is correct.
Choose the answer
Only the first part joins two full sentences with just a comma. So the answer is A: “Institution, the” contains the error. On the test, you would click that phrase.
Why this answer and not the closest wrong one
“Objects, including” is the closest wrong answer, because the words after its comma can’t stand alone. But they don’t need to. That phrase only adds examples to a full sentence. At “Institution, the,” both sides are full sentences, so a comma alone is not enough.
More questions like this
The original DOE question

Question 8 of 120
2026 Practice Test B English, question 51Question
How should the paragraph be revised?
(1) Danielle spent several hours preparing for an upcoming audition for a play at the community theater. (2) First she did vocal exercises to practice her diction and projection so that her words would carry clearly throughout the large auditorium. (3) Then she studies the text of the monologue to better understand the emotions, and motivations of the character she plans to portray. (4) Finally she recited her monologue in front of a mirror many times, making slight adjustments and improvements to her performance each time.
- ASentence 1: Change spent to had spent, AND insert a comma after play.
- BSentence 2: Change did to does, AND insert a comma after projection.
- CSentence 3: Change studies to studied, AND delete the comma after emotions.
- DSentence 4: Change recited to recites, AND delete the comma after times.
Show the answer
Correct answer: C
The question asks how the paragraph should be revised.
A. Incorrect. The revisions in this option introduce new errors. The word “spent” is correct in the simple past tense because it encompasses the actions described throughout the rest of the paragraph, all of which happened during those “several hours.” Additionally, adding a comma after the word “play” would be incorrect because it would unnecessarily separate the prepositional phrase “at the community theater” from the rest of the sentence.
B. Incorrect. The revisions in this option introduce new errors. The word “did” is correct as written in the past tense because the past tense is used throughout the paragraph. There is no comma needed after the word “projection” because “so” is functioning not as a conjunction but rather as part of the conjunction phrase “so that,” which does not take a comma.
C. CORRECT. Changing the word “studies” from the present tense to the past tense “studied” is necessary to match the past tense established in the paragraph (“spent,” “recited”). Additionally, the comma after the word “emotions” needs to be removed because the words “emotions and motivations” are part of a group (series) of two elements, and when there are only two elements in a series, a comma is not used.
D. Incorrect. The revisions in this option introduce new errors. The word “recited” is correct as written in the past tense because the past tense is used throughout the paragraph. Also, removing the comma after “times” is incorrect because the comma is needed in order to separate the modifying phrase “making slight adjustments and improvements to her performance each time” from the main clause.
Question analysis
This grammar and punctuation question gives a four-sentence paragraph about Danielle preparing for an audition at the community theater, with eight spots underlined. Each answer choice pairs a verb change with a comma change in a single sentence, so students must judge verb tense and comma use together and accept a choice only if both of its edits are needed.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Keeping verb tense consistent and using commas correctly are taught before grade 7, and this paragraph is short and plain. Each choice bundles two edits, so a student who spots one real error can confirm the answer with the other.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Question and choices: about grade 10, 133 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The shift to “Then she studies” stands out in a story told in the past tense with “spent,” “did” and “recited,” and finding either half of the right option is enough to choose it. That makes this one of the easiest editing questions. Students who miss it usually trust a half-remembered comma rule, such as a comma before “so,” and stop checking the verb in the same option.
A rule that doesn't sound wrong minor Working it out
Only two nouns, “emotions” and “motivations,” follow “the,” so the comma between them has no job, yet a habit of putting commas before “and” in lists hides it. (Connors & Lunsford, 1988)
Every choice must be checked minor Choosing the answer
The four options each pair a tense change with a comma change, one option per sentence, and an option is right only when both of its edits repair real errors. (Embretson & Wetzel, 1987)
Strategy: Find the tense of the paragraph's first verb, check every underlined verb against it, and then confirm that the comma change in the same choice is needed too.
The wrong answers, classified
A Verb tense or form (Connors & Lunsford, 1988)
Thinks “had spent” is needed because the preparing came before the audition, and accepts a comma after “play” that would cut off “at the community theater.”
B Punctuation (Connors & Lunsford, 1988)
Applies the rule of a comma before “so” joining two clauses, missing that “so that” opens a purpose clause with no comma, and accepts the present-tense “does.”
D Verb tense or form (Connors & Lunsford, 1988)
Notices that the tenses clash but takes the present-tense “studies” as the standard, shifts “recited” to match, and drops the comma that sets off “making slight adjustments.”
Working it out, step by step
Figure out what the question asks
This question asks how to fix the paragraph. Each choice makes two changes in one sentence: a verb change and a comma change. A choice is right only if both changes fix real errors. So check both halves.
Rule 1: keep one tense
Tense is when an action happens: past, present or future. A story should stay in one tense. Example: I woke up and eat breakfast mixes tenses. It should be I woke up and ate breakfast. This paragraph is in the past: “spent,” “did,” “recited.” But sentence 3 says “Then she studies.” That is present tense. It should be “studied.”
Rule 2: no comma between two
When only two things are joined by and, don’t put a comma between them. Example: I like apples and pears. Sentence 3 says “the emotions, and motivations.” There are only two things, so the comma should go. Fixed: the emotions and motivations of the character.
Check each choice
- A is wrong. “Spent” is already correct past tense. A comma after “play” would cut off “at the community theater.”
- B is wrong. “Does” is present tense, so it would create an error. And no comma goes before “so that.”
- C is right. “Studied” fixes the tense, and dropping the comma fixes the pair.
- D is wrong. “Recites” would create an error. The comma after “times” is needed to set off “making slight adjustments and improvements to her performance each time.”
Watch out: B tempts students who remember a comma before “so.” But “so that” tells a purpose, why she practiced, and takes no comma.
Choose the answer
So the answer is C: “Sentence 3: Change studies to studied, AND delete the comma after emotions.”
Why this answer and not the closest wrong one
B is the closest wrong answer, because many students put a comma before “so.” But “so that” explains a purpose, and it takes no comma. Also, B’s change to “does” would break the past tense. Both of C’s changes fix real errors.
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The original DOE question

Question 9 of 120
2026 Practice Test B Math, question 13Question
A basket contains 3 apples, 4 oranges, 5 bananas, and 8 pears.
If one piece of fruit is randomly pulled from the basket, what is the probability that it will be a pear?
Select the place on the number line to plot the point.
On the test you click a point on the number line; here, choose it from the menu.
Show the answer
Correct answer: the point 0.4
The total number of fruits is:
The probability of picking a pear is:
Question analysis
A basket holds apples, oranges, bananas and pears, and students find the probability that a piece of fruit pulled at random is a pear, then plot it as a point on a number line from 0 to 1. It tests simple probability as a part of a whole. With no choices to pick from, the fraction has to be turned into tenths to find its place.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding the probability of a simple event from counts is grade 6 work in New York. This one adds four whole numbers, forms one fraction and plots it among marked tenths, a shorter and more routine task than grade 7 probability.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 6, 40 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Adding the four counts to 20 and writing 8 over 20 is routine, so most students succeed. The losses come from the number line: its decimal labels make it easy to read the 8 pears as 0.8, and a part-to-part fraction such as 8/12 falls between marks and has to be rounded, which should prompt a second look. Simplifying 8/20 to 2/5 and then to 4/10 makes the mark plain.
Something you must already know minor Working it out
The question does not say that a probability is the favorable count over the whole count, here the 8 pears over all 20 pieces of fruit rather than over the other 12. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The number line is labeled only in tenths from 0 to 1, so 8/20 has to be rewritten as a decimal before the right mark can be chosen. (Enright, Morley & Sheehan, 2002)
Strategy: Add up all the fruit for the total, write the number of pears over that total, rewrite the fraction in tenths, and select that mark on the line.
Traps
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
0.8 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Takes the 8 pears straight to the mark labeled 0.8, never dividing by the 20 pieces of fruit in the basket.
0.7 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Compares the 8 pears with the 12 other pieces of fruit, 8/12 or about 0.67, a part-to-part ratio, and plots the nearest tenth.
0.6 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Puts the 12 apples, oranges and bananas over the 20 pieces of fruit, plotting the chance of not getting a pear instead of getting one.
Worked solution, step by step
Formulas used
- Probability of an event
- P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want, here pears. The bottom counts every possible result, here every piece of fruit. It works when each result is equally likely, like grabbing fruit without looking.
- Equal fractions
- a is the top of a fraction, and b is the bottom. n is any number except 0. Dividing the top and the bottom by the same number gives an equal fraction. Use it to get 10 on the bottom, which is easy to write as a decimal.
Sort out the question
The basket has 3 apples, 4 oranges, 5 bananas and 8 pears. You pull out one piece without looking. You need the probability that it is a pear. Then you plot that number as a point. The number line runs from 0 to 1, with a mark at every tenth.
Count all the fruit
The bottom of the fraction is every piece of fruit you could pull out. Add all four kinds:
There are 20 pieces of fruit in all.
Write the probability as a fraction
Use the probability formula. There are 8 pears out of 20 pieces of fruit:
Watch out: the bottom is all the fruit, not just the other fruit. compares the pears with the 12 fruits that are not pears. It is about 0.67, and it leads you to the mark at 0.7. And 12 over 20 is the chance of not getting a pear.
Change it to tenths
The number line is marked in tenths. So rewrite the fraction with 10 on the bottom. Divide the top and the bottom by 2:
Four tenths is written 0.4.
Watch out: the 8 pears do not make 0.8. That skips dividing by the 20 pieces of fruit.
Plot the point
Start at 0 and count four marks to the right: 0.1, 0.2, 0.3, 0.4. On the test, you click the mark labeled 0.4. On this site, you choose 0.4 from the menu.
So the answer is the point at 0.4.
Check your answer
The chance of not getting a pear is . A pear and not a pear cover every result, so the two must add to 1: . Also, pears are less than half of the fruit, so the point should be left of 0.5. It is.
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The original DOE question

Question 10 of 120
2026 Practice Test B Math, question 17Question
How many positive integers satisfy the inequality ?
Enter your answer in the space.
Fill-in question: work out the answer and type it as a number.
Show the answer
Correct answer: 15
Solve the inequality for x.
There are 15 positive integers that are less than 16.
Question analysis
This inequality question asks how many positive integers make x + 7 < 23 true, and the count is typed into a box. Students solve a one-step inequality, then count its whole-number solutions. The strict less-than sign and the word positive decide where the count starts and stops, so both ends need checking.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 inequality problems usually take two steps, such as 2x − 5 > 9. This one needs a single subtraction; the care goes into counting the whole-number solutions and deciding whether the boundary itself counts.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 12, 14 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One subtraction solves the inequality, so the points are lost in the count that follows. The boundary 16 is the last number the solving produces and is easy to type in, or to count up to; checking that 16 + 7 is 23, which is not less than 23, rules it out. Listing 1 through 15, the positive integers below the boundary, settles the count.
An answer in an unexpected form minor Choosing the answer
Solving gives x < 16, but the box asks how many positive integers make the inequality true, a count that still has to be made after the solving is done. (Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Two conditions set the ends of the count at once: the word positive starts it at 1 rather than 0, and the strict < stops it below 16. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Strategy: Undo the addition to find the boundary for x, then count the positive whole numbers below it, checking the largest one and the boundary in the original inequality.
Traps
- Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
The wrong answers, classified
16 Counting the ends (Brown & Burton, 1978)
Solves to x < 16 and then enters 16, either the boundary itself or a count of 1 through 16, though x = 16 makes x + 7 equal 23, not less than it.
29 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Moves the 7 to the right side without changing its sign, 23 + 7 = 30, and counts the 29 positive integers below 30.
Worked solution, step by step
Formulas used
- Doing the same to both sides
- a and b are the two sides of an inequality, and c is any number. Taking the same number away from both sides keeps the inequality true. Think of two people who each walk down the same number of stairs: the lower one stays lower. Use it to get x by itself.
Sort out the question
An inequality compares two amounts. The sign < means is less than. A positive integer is a whole number bigger than 0: 1, 2, 3, and so on. You need to count how many of them make true. The answer is a count, not a value of x.
Get x by itself
The 7 is added to x. Subtracting 7 undoes adding 7, so subtract 7 from both sides:
So x must be less than 16.
Watch out: adding 7 to 23 instead gives x < 30, and a count of 29.
Check the boundary number
The number 16 is the boundary, where the solutions stop. Does 16 itself work? Try it:
But 23 is not less than 23. It is equal. So 16 does not count. The largest number that works is 15, since , and 22 is less than 23.
Watch out: entering 16 counts the boundary, which fails.
Count the positive integers
The numbers that work start at 1, the smallest positive integer. Zero is not positive. They stop at 15:
Counting from 1 up to 15 gives 15 numbers.
Enter the count
There are 15 positive integers that make the inequality true. Type the count in the box as a plain whole number. So you would enter 15.
Check your answer
Test both ends. For 1: , and 8 is less than 23, so 1 works. For 15: 22 is less than 23, so 15 works. For 16: 23 is not less than 23, so 16 fails. That leaves exactly the numbers 1 through 15.
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The original DOE question

Question 11 of 120
2026 Practice Test C English, question 28Passage Serpent Mound
Serpent Mound
The Great Serpent Mound, located in Adams County, Ohio, is a human-made mound of earth that researchers believe was created between 300 B.C. and A.D. 1100 by an indigenous culture.
Ohio, 1846
Brush Creek stood low when the museum men came
with their measuring tapes and sketchbooks.
It was winter. Fringed with ice,
the creek doubled back on itself
as if it had forgotten something.
Pa was in Cincinnati, or else on his way home,
so Ma told me to lead the men
into the marshy low grounds. It being winter,
there was little underbrush to speak of—
in the summer there would have been
briars, poison ivy, biting flies. I listened
for the swish of a beaver’s heavy tail,
the chitter of a chickadee, or the cry of a hawk,
but the winter silence of the creek pressed
down on all of us like a weight.
The humps in the ground were all but
invisible until you were right up on them. The figure
was even less obvious: the sinuous body,
the tail coiled three times around,
and at the other end, the mouth wide open.
In the summer the creek bottom was crowded
with so much life that you could trip over
the ridges of earth before you saw
anything at all. In winter you could climb
a tree and get some idea of the whole thing:
the serpent’s body undulating, slithering
silently across the ancient
earth. At the mouth end, there was an oval mound
as if the snake were about to swallow an egg—
as snakes sometimes did in our rickety
henhouse—my Pa always said, or
as if swallowing the sun, one of the museum
men suggested, taking notes with his quill pen,
an old-style inkhorn slung at his side.
I liked that: swallowing the sun,
just the sort of thing a snake might do,
might want to do. When, later, I told my sister Ruth,
she disagreed. It is singing to the sun,
she insisted. That is why its mouth
is wide open. She said, “Sometimes I think
I hear it on summer nights. Not swallowing, singing.”
Question
Lines 1–2 contribute to the development of ideas in the poem by
- Asuggesting that the men are too busy with their work to talk to the speaker.
- Bhelping establish the reason for the men’s visit and purposeful behavior.
- Chinting that the men have hidden motives for studying the mound.
- Dindicating why the speaker is fascinated by the men and wants to help them.
Show the answer
Correct answer: B
The question asks how lines 1–2 contribute to the development of ideas in the poem.
A. Incorrect. The idea that the men are too busy with their work to talk to the speaker is not conveyed in the poem, and, in fact, one of the museum men discusses the oval mound with the speaker (“as if swallowing the sun, one of the museum / men suggested” [lines 32–33]).
B. CORRECT. The lines, with their mention of measuring tapes and sketchbooks, indicate that the reasons the museum men came were to measure the mound and to draw sketches of it.
C. Incorrect. There is little indication in the poem that the museum men have a hidden motive for studying the mound. Lines 1–2 present a straightforward description of the men and their equipment, which emphasizes the men’s academic interest in studying the mound.
D. Incorrect. While the speaker displays some curiosity about the work the museum men are doing (“one of the museum / men suggested, taking notes with his quill pen, / an old-style inkhorn slung at his side. / I liked that” [lines 32–35]), lines 1–2 focus on introducing the idea that the mound is a formation that is worth formal study and examination by experts.
Question analysis
This question asks how the opening lines of “Serpent Mound,” in which museum men arrive at a low winter creek with measuring tapes and sketchbooks, contribute to the development of ideas in the poem. The lines are short and plain, but the choices credit the men or the speaker with motives and feelings, and students must check each against the rest of the poem.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Below typical grade 7 demand: the two opening lines are short and concrete, and checking the choices takes one reading of the poem, much like grade 6 questions on how a single sentence fits into a text.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- How, not what minor
- Evidence in several places minor
- Evidence
- Answered from one sentence or line
- Reading level
- Passage: 354 words (a poem; reading level not measured); question and choices: about grade 7, 62 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The opening lines are short and concrete, and their tools, measuring tapes and sketchbooks, plainly say why the men came. The wrong choices each add something: men too busy to talk, a hidden motive, or a fascinated speaker, and the last borrows the speaker’s later delight in “swallowing the sun.” Checking each against the rest of the poem, rather than against first impressions of strangers arriving, is what the question rewards.
How, not what minor Working it out
The stem asks what lines 1–2 do in the poem, so students must see the measuring tapes and sketchbooks as setting up the visit’s purpose, not only as a description of the men. (Anderson & Krathwohl, 2001)
Evidence in several places minor Working it out
Ruling out two wrong choices takes later lines: a museum man offers the speaker an idea in lines 32–33, and the speaker leads the men in line 7 because Ma says to. (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)
Strategy: Say in your own words what lines 1–2 tell you about the men, then reject any choice that depends on something the poem never shows.
Traps
- Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
The wrong answers, classified
A Reads too much in (Graesser, Singer & Trabasso, 1994)
Pictures men absorbed in measuring who ignore the child, though nothing says so and one of them later suggests the mound looks “as if swallowing the sun.”
C Reads too much in (Graesser, Singer & Trabasso, 1994)
Reads a secret aim into the visit, but lines 1–2 show the men carrying measuring tools in the open, and the poem never hints at any other purpose.
D Wrong part of the text (Kirsch & Mosenthal, 1990)
Jumps to line 35, where the speaker likes a museum man’s phrase; in lines 1–2 the men have only arrived, and the speaker guides them because Ma says to.
Working it out, step by step
Figure out what the question asks
This asks what lines 1–2 do for the poem’s ideas. They are the poem’s first lines, so they set things up. You need to decide what they tell you about the museum men, and why that matters.
Read lines 1–2
Lines 1–2 say “Brush Creek stood low when the museum men came / with their measuring tapes and sketchbooks.” A measuring tape is for measuring lengths. A sketchbook is for drawing. So the men came to measure the mound and draw it. They have a job to do. Purposeful means acting with a clear goal. Their tools show that goal from the very start.
Check each choice
- A does not fit. Nothing says the men were too busy to talk. Later, one of them shares an idea: “as if swallowing the sun.”
- B fits. The tools show why the men came and that they came to work.
- C does not fit. The men carry their tools in plain view. The poem never hints at a hidden motive, or secret reason.
- D does not fit. The speaker leads the men because “Ma told me to lead the men.” The speaker’s liking for a man’s idea comes much later, in line 35.
Watch out: D borrows a feeling from later in the poem, “I liked that.” In lines 1–2, the men have only just arrived.
Choose the answer
Lines 1–2 use the men’s tools to show why they came and what they plan to do. So the answer is B, “helping establish the reason for the men’s visit and purposeful behavior.”
Why this answer and not the closest wrong one
D is the closest wrong choice. The speaker does later like a museum man’s phrase, “swallowing the sun.” But that comes in line 35. Lines 1–2 say nothing about the speaker’s feelings. They name only the men’s tools. Tapes and sketchbooks tell you why the men came: to measure and draw the mound.
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The original DOE question

The passage “Serpent Mound” as printed (page 40)

Question 12 of 120
2026 Practice Test C English, question 30Uses the passage above: “Serpent Mound”.
Question
What impact do the phrases “all but / invisible” and “even less obvious” in lines 16–18 have on the meaning of the poem?
- AThey indicate that viewing the full size and shape of the mound is difficult.
- BThey suggest that the location of the mound is unknown to most people.
- CThey imply that the speaker wants the location of the mound to remain a secret.
- DThey reveal that the speaker is unfamiliar with the significance of the mound.
Show the answer
Correct answer: A
The question asks what impact the phrases “all but / invisible” and “even less obvious” in lines 16–18 have on the meaning of the poem.
A. CORRECT. The words “all but / invisible” indicate that a person must be close to see the humps in the ground, while the phrase “even less obvious” indicates that a person must be far away to see the overall snake shape of the mound.
B. Incorrect. While the speaker has to lead the museum men to the mound because they did not know the exact location, the words in lines 16–18 are related to the difficulty in viewing the mound from the immediate surrounding area.
C. Incorrect. The speaker is willing to take the museum men out to study the mound, so there is no indication in the poem that the speaker wants the location of the mound to remain a secret. The words in lines 16–18 convey the idea that the mound is difficult to see.
D. Incorrect. While the speaker contemplates the significance of certain elements of the mound, the phrases in lines 16–18 relate to a viewer’s ability to see the physical shape of the mound, not its deeper meaning.
Question analysis
This word-meaning question asks what impact the phrases “all but / invisible” and “even less obvious” in lines 16–18 of “Serpent Mound” have on the poem’s meaning. Students must understand the idiom “all but” and then judge which of four claims about the mound and the speaker the two phrases actually support.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Below typical grade 7 demand: both phrases are quoted with their line numbers, in plain modern language, and one careful reading settles the question. The one hurdle is the idiom “all but,” which some readers take literally.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- Hard words moderate
- Reading between the lines minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: 354 words (a poem; reading level not measured); question and choices: about grade 7, 76 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Both phrases are quoted with their lines, and the sentence that holds them goes on to list the body, the coiled tail and the open mouth, so its subject is what can be seen. The wrong choices slide from seeing to knowing: who knows where the mound is, who wants it hidden, and what it means. “Less obvious” is the slippery phrase, since obvious can also mean easy to understand.
Hard words moderate Understanding the question
“All but” is an idiom meaning almost, so “all but / invisible” means nearly unseen; a student who takes it as the opposite, plainly visible, loses the sense that the humps are hard to see. (Perfetti, 2007; Freedle & Kostin, 1993)
Reading between the lines minor Working it out
Students must keep both phrases about sight and connect them to lines 24–25, where only climbing a tree gives “some idea of the whole thing.” (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Strategy: Put both phrases into plain words, starting with the idiom “all but,” then test each choice against what lines 16–25 actually describe.
Traps
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
- Dictionary meaning, not meaning here: A common meaning of the word that does not fit its use in this text.
The wrong answers, classified
B Twists what the text says (King et al., 2004)
Turns “invisible” into “unknown,” helped by Ma sending the speaker to guide the men, but the phrases say how hard the mound is to see, not who knows where it is.
C Reads too much in (Graesser, Singer & Trabasso, 1994)
Finds a wish for secrecy in the hidden humps, though the speaker leads the museum men straight to the mound and never tries to keep it from anyone.
D Dictionary meaning, not meaning here (Perfetti, 2007)
Takes “less obvious” in its sense of harder to understand, but the sentence goes straight on to list the body, the coiled tail and the open mouth, so the word is about sight.
Working it out, step by step
Figure out what the question asks
This asks what two phrases add to the poem’s meaning. They are “all but / invisible” and “even less obvious,” in lines 16–18. Put each phrase in plain words first. Then see what they tell you about the mound.
Read lines 16–18
Lines 16–18 say “The humps in the ground were all but / invisible until you were right up on them. The figure / was even less obvious.” “All but” means almost. So the humps were almost invisible until you got very close. The figure is the whole snake shape. It was even harder to see than the humps. Here, obvious means easy to see.
Connect the phrases to later lines
The same sentence goes on to list what you would see: “the sinuous body, / the tail coiled three times around, / and at the other end, the mouth wide open.” So the phrases are about seeing the shape. Later, lines 24–25 say “In winter you could climb / a tree and get some idea of the whole thing.” Only from up high could you begin to see it. The mound is huge, low and hard to take in all at once.
Check each choice
- A fits. Both phrases show that seeing the whole mound is hard.
- B does not fit. The phrases are about how hard the mound is to see, not about who knows where it is.
- C does not fit. The speaker leads the museum men right to the mound. There is no wish to keep it secret.
- D does not fit. Here “less obvious” is about sight, not about understanding what the mound means.
Watch out: Obvious can also mean easy to understand. That reading leads to D. But the sentence goes straight on to list parts of the snake you can see.
Choose the answer
The phrases show how hard it is to see the mound’s full size and shape. So the answer is A, “They indicate that viewing the full size and shape of the mound is difficult.”
Why this answer and not the closest wrong one
B is a close wrong choice, because “invisible” can make you think nobody knows the mound is there. But the phrases describe looking, not knowing. The speaker knows exactly where the mound is. The problem is seeing it. The humps vanish until “you were right up on them,” and the whole shape shows only from a tree.
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The original DOE question

Question 13 of 120
2026 Practice Test C Math, question 1Question
What is the prime factorization of 756?
- A
- B
- C
- D
Show the answer
Correct answer: D
Determine the prime factors of 756 using a factor tree:

The prime factors are 2, 2, 3, 3, 3, and 7, or , which can be expressed using exponents as .
Question analysis
This number question asks for the prime factorization of 756, written with exponents. Students break the number into prime factors, with a factor tree or repeated division, and match the result to one of four products. Every choice multiplies back to 756, so the real test is whether each factor shown is actually prime.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Factor trees and exponent notation are grade 6 skills, and 756 splits quickly by small divisors. What adds a little is that every choice multiplies out to 756, so checking the product alone settles nothing.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Something you must already know minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 11, 10 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The factoring itself is quick: halving twice and dividing by 3 three times leaves 7. The question is built so that multiplying back proves nothing, since every choice equals 756, and three of them simply stop the factor tree at a composite number. Students who check only the product, or who take an odd factor such as 21 or 27 for a prime, pick a factorization that is unfinished.
A very close wrong answer moderate Choosing the answer
2² · 3² · 21 shares the right answer's 2² and differs only in 3² · 21 against 3³ · 7, and both equal 756, so neither an estimate nor a product check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that a prime factorization may hold only primes, and must spot that 21, 9 and 27 are composite even though none of them is even. (Anderson & Krathwohl, 2001)
Strategy: Divide 756 by 2 while the result is even, then by 3 while it divides evenly, and keep going until every factor left is prime; then compare factor by factor, not by product.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Divides out 2 twice and 3 twice, then stops at 21, which is 3 × 7, so one composite factor remains in a product that still equals 756.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Splits off a 3 and a 7 but keeps 9 as if it were prime, perhaps because it is odd, though 9 = 3 × 3.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Splits 756 into 4 × 189 and 189 into 7 × 27, then stops, not seeing that 27 is 3 × 3 × 3.
Worked solution, step by step
Formulas used
- Writing a repeated factor with an exponent
- a is the number being multiplied, called the base. n is the exponent, the small raised number. It tells how many times a appears in the product. So 2² means 2 × 2, and 3³ means 3 × 3 × 3. Exponents are a short way to write a factor that repeats.
Sort out the question
You need the prime factorization of 756. That means writing 756 as a product of prime numbers only. A product is the answer to a multiplication. A prime number has exactly two factors, 1 and itself. Examples are 2, 3, 5 and 7.
All four choices multiply to 756. So multiplying a choice back does not pick the answer. You must check that every factor is prime.
Divide by 2 while you can
A factor tree splits a number into two factors, again and again. Start with the smallest prime, 2. 756 is even, so 2 goes into it:
378 is even too:
189 is odd, so 2 does not go into it. You have pulled out two 2s.
Divide by 3 while you can
Add the digits of 189: . A number can be divided by 3 when its digits add up to a number that 3 goes into. 3 goes into 18, so it goes into 189:
7 is prime, so the tree stops. You have pulled out three 3s and one 7.
Write the factors with exponents
Collect every prime you pulled out:
Use the exponent rule. Two 2s make , and three 3s make . The choices use a raised dot, ·, for times. So 756 is .
Choose the answer
Watch out: each wrong choice is a factor tree that stopped too soon. A ends in 21, which is 3 × 7. B has 9, which is 3 × 3. C has 27, which is 3 × 3 × 3. Being odd does not make a number prime.
Only D uses primes alone. So the answer is D, .
Check your answer
Multiply your answer back: , and . Then look at each factor. 2, 3 and 7 are all prime. D passes both tests.
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Question 14 of 120
2026 Practice Test C Math, question 4Question
A person buys a used car with a down payment of $1,200 and makes monthly payments of $275 for 3 years. What is the total amount the person pays for the car?
- A$2,025
- B$9,900
- C$11,100
- D$13,500
Show the answer
Correct answer: C
Determine the total cost of the car by adding the down payment to the total monthly payments over 3 years.
There are 12 months in a year, so multiply (12 months/year × 3 years) to get a total of 36 months. Determine the total monthly payments by multiplying the monthly payment ($275) × the total number of months (36):
Determine the total cost of the car by adding the down payment ($1,200) to the total monthly payments ($9,900):
The total cost of the car is $11,100.
Question analysis
A used car is bought with a down payment of $1,200 followed by monthly payments of $275 for 3 years, and students find the total paid. The work is to count the payments, multiply, and add the down payment. Mixing up years and months, or forgetting the money paid up front, leads to the wrong totals among the choices.
- NYS standards
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 A total made of one payment up front plus equal monthly payments is a short whole-number calculation, simpler than grade 7 word problems with percents or negatives. The only conversion is from years to months.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Much to keep track of minor
- Error-prone arithmetic minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 29 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The situation is familiar and the method direct, so this is one of the easier questions on the test. Points are lost in two places: turning 3 years into 36 payments, and remembering the money paid up front after the long multiplication. Each slip, like paying the down payment once a year, leads to a listed total, so checking that the answer holds exactly one $1,200 and 36 payments of $275 is worth the seconds.
Much to keep track of minor Working it out
The $1,200 down payment is paid once while the $275 repeats every month, so it has to be set aside while 36 payments are totaled and then added at the end. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Error-prone arithmetic minor Working it out
The product 275 × 36 = 9,900 is a two-digit multiplication done by hand; splitting it as 275 × 4 = 1,100 and then 1,100 × 9 makes it quick and safe. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Turn 3 years into a number of months, multiply the monthly payment by it, and add the down payment once at the end.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
The wrong answers, classified
A Units or scale (King et al., 2004)
Multiplies the $275 by 3, the number of years, instead of by 36 months, then adds the $1,200 down payment: 825 + 1,200 = 2,025.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the 36 monthly payments, 36 × $275 = $9,900, and stops without adding the $1,200 down payment made at the start.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Works out one year as $1,200 + 12 × $275 = $4,500 and multiplies by 3, so the down payment is paid three times instead of once.
Worked solution, step by step
Formulas used
- Changing years into months
- Each year has 12 months. Multiply the number of years by 12 to get the number of months. You need months here because a payment is made every month.
- Total cost with a down payment
- The down payment is money paid once, at the start. The monthly payment is paid every month. Multiply it by the number of months to get all the monthly payments. Then add the down payment only once, because it is paid only once.
Sort out the question
The person pays $1,200 once, at the start. This is the down payment. Then the person pays $275 every month for 3 years. You need the total amount paid for the car.
The payments come once a month, but the time is given in years. So change the years into months first.
Change 3 years into months
Use the months formula:
So the person makes 36 monthly payments.
Watch out: multiplying $275 by 3 counts only 3 payments. That leads to choice A, $2,025.
Add up the monthly payments
Multiply $275 by 36. A quick way is to split 36 into 4 × 9:
So the 36 payments add up to $9,900.
Watch out: stopping here gives choice B, $9,900. The down payment is still missing.
Add the down payment once
Use the total-cost formula:
The person pays $11,100 in all. Adding $1,200 for every year counts it three times. That gives choice D, $13,500.
So the answer is C, $11,100.
Check your answer
Estimate. $275 is a little less than $300, and . Add the $1,200: . The real total should be a little less than $12,000. $11,100 fits.
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Question 15 of 120
2026 Practice Test C Math, question 7Question
For what values of x is the above inequality true?
- A
- B
- C
- D
Show the answer
Correct answer: C
Divide to solve for x:
Rewriting with the x term first requires reversing the inequality symbol in order to keep the meaning “x is less than ”.
Question analysis
This inequality question asks which values of x make −12 > 6x true. Students solve a one-step inequality by dividing and then state the solution with x written first. A negative number on one side and the variable on the other make the direction of the symbol the main thing to get right.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 inequalities usually take two steps, such as 3x + 4 > 19. This one needs a single division, and the care goes into reading the result correctly when the variable ends up on the right of the symbol.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Something you must already know minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 10, 15 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One division finds the boundary, so the question is really about direction. Two things pull students the wrong way: the −12 on the left makes the inequality look like one that must be flipped, and the result −2 > x has x on the right, where it is easy to read backward. Testing x = −3, which gives −12 > −18, shows that values below −2 work.
A very close wrong answer moderate Choosing the answer
x < −2 and x > −2 share the boundary −2 and differ only in direction, so finding the right number settles nothing until the direction is read correctly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that the symbol flips only when both sides are multiplied or divided by a negative number, and that −2 > x says the same as x < −2. (Anderson & Krathwohl, 2001)
Strategy: Divide both sides by 6, asking whether you divided by a negative number, then rewrite the result with x on the left so that the symbol still opens toward the larger side.
Traps
- Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Adds 6 to −12 instead of dividing by 6, getting −6, and also reverses the direction, so both the boundary and the symbol are wrong.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Undoes the 6 by adding it to −12 rather than dividing, as if 6x meant x − 6, getting −6 > x and so x < −6.
D Negatives and direction (Brown & Burton, 1978)
Finds −2 > x but flips the symbol because −12 is negative, or reads −2 > x as x > −2, though dividing by a positive 6 keeps the direction.
Worked solution, step by step
Formulas used
- Dividing both sides of an inequality
- a and b are the two sides of an inequality, and c is the number you divide by. If c is positive, the symbol stays the same. It flips only when you multiply or divide both sides by a negative number. Here c is 6, which is positive.
- Reading an inequality the other way
- a and b are any two numbers. Saying a is greater than b is the same as saying b is less than a. When you swap the two sides, turn the symbol around. Its point still aims at the smaller number.
Sort out the question
The inequality is . The symbol > means is greater than. So it says −12 is greater than 6 times x. You need every value of x that makes this true.
Divide both sides by 6
To get x alone, divide both sides by 6:
You divided by 6, which is positive. So the symbol does not flip.
Watch out: −12 is negative, but that does not matter. Only dividing by a negative number flips the symbol.
Put x first
says −2 is greater than x. That means x is less than −2. Swap the sides and turn the symbol around:
Watch out: keeping > when you swap the sides gives choice D, . That is the opposite meaning.
Choose the answer
Choices A and B use −6. That comes from adding 6 to −12 instead of dividing by 6. The right boundary is −2, and x is below it.
So the answer is C, .
Check your answer
Test a number below −2, such as −3. Then , and is true. Now test 0, which is above −2. Then , and is false. So only numbers below −2 work.
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Question 16 of 120
2026 Practice Test C Math, question 8Question
Jordan is helping plan snacks for a school dance. He knows that 0.75 of the students voted for pizza, and he wants to record the same amount using a fraction.
Which fractions are equivalent to the decimal 0.75?
Select the four correct answers.
Choose every answer that is correct, then check. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.
- A
- B
- C
- D
- E
- F
Show the answer
Correct answer: A, B, C, and E
Check each fraction to determine whether it is equal to 0.75 when simplified or converted to a decimal.
This is the most common fraction equal to 0.75.
Simplifies to 3/4, because 15 and 20 can both be divided by 5.
This is a direct match to 0.75 when written as a decimal.
, which is less than 0.75.
Simplifies to because both 6 and 8 can be divided by 2.
, which is more than 0.75.
Question analysis
Jordan wants to record 0.75 of the students' pizza votes as a fraction, and students select the four fractions, out of six, that equal 0.75. It tests converting between decimals and fractions and recognizing equivalent fractions. Each choice has to be checked by its value, not by how its numbers look.
- NYS standards
- 7.NS.2d fractions to decimals by long division
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Matching fractions to a decimal such as 0.75 builds on grade 4 to 6 work, and the denominators here divide easily. The added demand is the format: four of the six must be chosen, and one wrong or missing pick loses the item.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Every choice must be checked moderate
- A very close wrong answer minor
- Much to keep track of minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 47 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Three of the four right fractions, 3/4, 75/100 and 6/8, are familiar forms of 0.75, so the item comes down to the fourth. Students who stop after the familiar ones must decide between 15/20, which needs dividing by 5 to recognize, and 8/10, which sits close to 0.75 on a number line. Dividing each numerator by its denominator, or scaling 3/4 up by 5, settles it.
Every choice must be checked moderate Choosing the answer
Six fractions with six different denominators have to be tested one by one, by dividing or by simplifying, since 15/20 and 8/10 cannot be judged by how they look. (Embretson & Wetzel, 1987)
A very close wrong answer minor Choosing the answer
8/10 is 0.8, only 0.05 above 0.75, and it follows 6/8 by adding 2 to the top and the bottom, so it looks like one more member of the family. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
Four of the six must be chosen and all four must be right for the item to count, so each verdict has to be kept while the remaining fractions are tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Strategy: Turn each choice into a decimal by dividing its numerator by its denominator, or reduce it to lowest terms, and keep only the ones whose value is exactly the decimal given.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No usual working links 2/5 to 0.75, since 2 ÷ 5 = 0.4, so it mostly catches students filling the fourth slot by guessing.
F Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Builds 8/10 from 6/8 by adding 2 to the top and the bottom, as if equal additions kept a fraction's value, or accepts 0.8 as close enough.
Worked solution, step by step
Formulas used
- Changing a fraction to a decimal
- a is the numerator, the top. b is the denominator, the bottom. A fraction bar means divide. So dividing the top by the bottom gives the fraction’s value as a decimal.
- Equivalent fractions
- a is the top and b is the bottom. n is a number that goes evenly into both. Dividing the top and the bottom by the same number gives an equal fraction. Equivalent fractions name the same amount in different ways.
Sort out the question
Six fractions are given, A to F. You need the four that equal the decimal 0.75. Equivalent means equal in value. Test every choice, because some fractions look alike but are not equal.
Write 0.75 as a fraction
0.75 means 75 hundredths:
So choice C works. Now simplify it. 25 goes into both 75 and 100:
So choice A, 3/4, works too.
Test choices B and E
Simplify each one to see if it becomes 3/4:
- B: 5 goes into 15 and 20. . It works.
- E: 2 goes into 6 and 8. . It works.
Test choices D and F
Divide the top by the bottom:
- D: . That is less than 0.75.
- F: . That is more than 0.75.
Watch out: F looks like 6/8 with 2 added to the top and the bottom. Adding the same number to both changes the value. Only multiplying or dividing both keeps it.
Name the four answers
A, B, C and E all equal 0.75. On the test, you check the box beside each of them. On this site, you choose all four, then check your answer.
So the answers are A, B, C and E.
Check your answer
Turn each answer into a decimal. , , and all give 0.75. Exactly four choices passed, and the question asks for four. So none were missed.
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Question 17 of 120
2026 Practice Test C Math, question 9Question
Maria rides her bicycle to school at a constant speed of 15 miles per hour. If the distance to school is 6 miles, how many minutes will it take Maria to get to school?
- A10
- B15
- C24
- D30
Show the answer
Correct answer: C
Maria’s rate of speed is given as 15 miles per hour. Since 1 hour is equivalent to 60 minutes, her rate of speed, in miles per minute, can be determined.
, or mile per minute.
Therefore, it takes Maria 4 minutes to ride 1 mile.
The distance to school is 6 miles, so it will take minutes to ride to school.
Question analysis
Maria bikes 6 miles to school at a constant 15 miles per hour, and students find how many minutes the ride takes. It tests using a rate to find a time and converting hours to minutes. With the speed given per hour and the question asking, in bold, for minutes, keeping track of units is the main hurdle.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Distance, speed and time with whole numbers is grade 6 rate work. The only extra demand is an answer in minutes when the speed is per hour, a conversion grade 7 students make routinely.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 6, 32 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The numbers are small, but the speed is per hour and the answer is wanted in minutes, so students have to choose a route through the units. Finding the minutes per mile first, 4 at 15 miles an hour, turns the rest into a single multiplication. The wrong choices come from using a number in the stem as the answer, from mixing the 60 minutes with the 6 miles, or from rounding 0.4 hour to a half hour.
Something you must already know minor Working it out
Nothing on the page says that time is distance divided by speed or that an hour has 60 minutes, and the bold word minutes requires both. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
6 ÷ 15 is less than 1, 0.4 hour, and turning it into minutes means 0.4 × 60; the cleaner route, 60 ÷ 15 = 4 minutes a mile, avoids decimals. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Find how many minutes Maria needs to ride one mile at 15 miles per hour, then multiply by the number of miles to school.
Traps
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
- Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Divides the 60 minutes in an hour by the 6 miles, 60 ÷ 6 = 10, as if the ride took a full hour, and ignores the 15 mph speed.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Takes the 15 in 15 miles per hour as the number of minutes, a number copied from the question without any calculation.
D Rounds too early (King et al., 2004)
Sees that 6 miles is a little under half of 15, rounds the time of 0.4 hour up to half an hour, and gives 30 minutes.
Worked solution, step by step
Formulas used
- Minutes for each mile
- Miles per hour tells how many miles you ride in one hour, which is 60 minutes. Sharing the 60 minutes evenly among those miles gives the minutes for each mile. This works when the speed stays the same the whole time.
- Distance, speed and time
- Distance is how far you go, here in miles. Speed is how far you go in one hour. Dividing the distance by the speed gives the time in hours.
Sort out the question
Maria rides at a constant speed of 15 miles per hour. Constant means it never changes. School is 6 miles away. You need the time in minutes, not hours. You must know that one hour has 60 minutes.
Find the minutes for one mile
15 miles per hour means Maria rides 15 miles in 60 minutes. Use the minutes-per-mile formula:
So each mile takes Maria 4 minutes.
Find the minutes for 6 miles
Each of the 6 miles takes 4 minutes:
So the ride takes 24 minutes.
Watch out: 15 is the speed, not a time, so choice B is a trap. And , choice A, ignores the speed.
Choose the answer
Another way is the time formula. , so the ride takes 0.4 of an hour. In minutes, that is .
Watch out: rounding 0.4 hour up to half an hour gives choice D, 30.
So the answer is C, 24.
Check your answer
Check with the speed. 24 minutes is of an hour. In a full hour Maria rides 15 miles. So in 2/5 of an hour she rides miles. That is exactly the distance to school.
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Question 18 of 120
2026 Practice Test C Math, question 11Question
Allison has 5 stamp albums with 576 stamps in each album. If she transfers her stamp collection to 6 albums, each holding 378 stamps, how many stamps will be left over?
- A198
- B612
- C620
- D632
Show the answer
Correct answer: B
Determine the total number of stamps in 5 albums holding 576 stamps:
Determine the total number of stamps in 6 albums holding 378 stamps:
Subtract 2,268 from 2,880 to find the number of stamps that will be left over:
Question analysis
Allison moves a stamp collection from 5 albums of 576 stamps into 6 albums that hold 378 each, and students find how many stamps are left over. It tests multiplying and subtracting large whole numbers in a word problem. The question gives no total, so the collection has to be counted before the new albums' capacity is taken away.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Whole-number multiplication and subtraction in a three-step word problem is easier than grade 7 work with fractions, decimals or negatives. Because the products are large, a slip or a rounded estimate easily gives another choice.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Error-prone arithmetic minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 6, 27 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The plan is clear from the stem: count the stamps, then take away what the six new albums hold, so the question tests careful arithmetic. Its choices punish the estimating that saves time elsewhere, since rounding either factor shifts the result by only a few tens, onto another listed number. Students who skip the album counts altogether and compare one album with one album find 198 waiting.
A very close wrong answer moderate Choosing the answer
612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Multiply to find the total number of stamps, multiply again to find how many the six new albums hold, subtract, and check the last digit of each product.
Traps
- Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Subtracts 378 from 576, comparing one old album with one new album, and ignores that there are 5 old albums and 6 new ones.
C Rounds too early (King et al., 2004)
Rounds 576 to 580 and 378 to 380 before multiplying, so 2,900 − 2,280 = 620, an estimate rather than the exact number left over.
D Rounds too early (King et al., 2004)
Rounds only 576 up to 580, getting 2,900 stamps, then subtracts the exact 2,268, so the count comes out 20 too high.
Worked solution, step by step
Formulas used
- Total in equal groups
- Here the groups are albums. The amount in each group is how many stamps one album holds. When every group is the same size, multiplying is a fast way to add them all up.
Sort out the question
Allison has 5 albums with 576 stamps in each. She moves all her stamps into 6 new albums. Each new album holds 378 stamps. Some stamps will not fit. You need the number left over.
Count all her stamps
Use the equal-groups formula. Split 576 into 500 + 76 to make it easy:
Allison has 2,880 stamps.
Count what the new albums hold
Do the same for the 6 new albums. Split 378 into 300 + 78:
The new albums hold 2,268 stamps in all.
Subtract to find the leftovers
The stamps left over are the ones she has minus the ones that fit:
Watch out: , choice A, compares just one old album with one new one. Rounding to 580 and 380 first gives 620, choice C. Rounding only 576 gives 632, choice D. Use the exact numbers.
So the answer is B, 612.
Check your answer
Add the leftovers to the stamps that fit: . That is exactly the number Allison started with. So no stamps were lost or counted twice.
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Question 19 of 120
2026 Practice Test C Math, question 13Question
Start with the number 135.28 and perform the following steps in order:
- Step 1: Multiply by 10.
- Step 2: Add 0.5 to the result of Step 1.
- Step 3: Drop the digits after the decimal point.
- Step 4: Divide by 10.
What is the result?
- A13.5
- B135
- C135.2
- D135.3
Show the answer
Correct answer: D
Begin by multiplying 135.28 by 10:
Add 0.5:
Drop the digits after the decimal point, so becomes .
Divide by 10:
Question analysis
Students start with 135.28 and carry out four listed steps: multiply by 10, add 0.5, drop the digits after the decimal point, and divide by 10. It tests place value and operations with decimals. Skipping a step, or doing two of them in the wrong order, changes the result, and the choices are built from those slips.
- NYS standards
- 6.NS.3 multi-digit decimal operations
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Multiplying and dividing a decimal by 10 and adding 0.5 are grade 5 and 6 skills. The four steps are spelled out, so the demand is following them exactly and in order, not choosing a method.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Many steps minor
- Error-prone arithmetic minor
- Solution
- 4 steps
- Reading level
- Question and choices: about grade 3, 36 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The steps spell out how to round 135.28 to the nearest tenth, but nothing says so, and they have to be followed exactly. The decisive moment is step 2, where adding 0.5 pushes 1,352.8 past 1,353; skip it and the digits are simply cut, not rounded. Students who recognize the routine can check the result directly, since 135.28 lies closer to 135.3 than to 135.2.
A very close wrong answer moderate Choosing the answer
135.3 and 135.2 differ only in the tenths digit, and 135 shares their whole-number part, so a rough sense of the result's size rules out nothing. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
Four listed steps move the decimal point twice, and doing them in another order, or skipping one, still produces a believable number near 135. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Adding 0.5 to 1,352.8 carries into the ones place, since 8 tenths and 5 tenths make 1.3, which turns 1,352 into 1,353 before the decimals are dropped. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Write down the number after each step instead of doing them in your head, and carry out the steps exactly in the order listed.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
The wrong answers, classified
A Misreads the question (Hegarty, Mayer & Monk, 1995)
Skips step 1, so adds 0.5 to 135.28, drops the decimals to get 135, and then divides by 10 to reach 13.5.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Divides by 10 before dropping the digits after the decimal point, so 1,353.3 becomes 135.33 and then 135, doing steps 3 and 4 in reverse order.
C Rounds too early (King et al., 2004)
Leaves out step 2, or adds 0.5 without carrying, so 1,352 is kept and 135.2 results: 135.28 cut down instead of rounded up.
Worked solution, step by step
Formulas used
- Multiplying and dividing by 10
- a is any decimal number. Multiplying by 10 moves its decimal point one place to the right, so each digit is worth ten times as much. Dividing by 10 moves the point one place to the left. The two steps undo each other.
Sort out the question
You start with 135.28 and do four steps, in order. Write down the number after each step. Doing them in your head, or in another order, leads to a wrong choice. You need the number at the end.
Multiply by 10
Step 1. Multiplying by 10 moves the decimal point one place to the right:
The digits stay the same. Only the decimal point moves.
Add 0.5
Step 2. Line up the decimal points. 8 tenths plus 5 tenths is 13 tenths. That is 1 whole and 3 tenths, so carry the 1 into the ones place:
Watch out: skipping this step, or forgetting to carry, keeps 1,352. That leads to choice C, 135.2.
Drop the digits after the point
Step 3. Cut off the digits after the decimal point. Do not round. So 1,353.3 becomes 1,353.
Watch out: do step 3 before step 4. Dividing first gives 135.33. Dropping its decimals then gives 135, choice B.
Divide by 10
Step 4. Dividing by 10 moves the decimal point one place to the left:
Skipping step 1 gives 13.5, choice A. So the answer is D, 135.3.
Check your answer
These four steps are a recipe for rounding to the nearest tenth. 135.28 is between 135.2 and 135.3. Its hundredths digit is 8, which is 5 or more, so it rounds up to 135.3. The recipe and the rounding agree.
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The original DOE question

Question 20 of 120
2026 Practice Test C Math, question 16Question
- A−1
- B0
- C
- D1
Show the answer
Correct answer: B
Simplify the expression.
is the equivalent of subtracting 3 from 3, and is the equivalent of adding 3 to 3:
The zero property of division states that zero divided by any nonzero number equals zero, so .
Question analysis
This fraction has 3 plus negative 3 on top and 3 minus negative 3 on the bottom, and students find its value. It tests adding and subtracting negative numbers and dividing integers. The top and the bottom look almost alike, so each sign has to be read carefully before anything is divided.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Adding and subtracting a negative integer and dividing are grade 7 basics, and here the numbers are small. It is a short routine calculation, easier than typical grade 7 work, though the two signs invite a quick misreading.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 16+, 2 words.
- Corrected from the DOE handbook
- A grammatical error in the explanation was corrected.
- A rule stated too broadly in the explanation was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The arithmetic is two one-step sign calculations and a division by 6. The danger is reading the top and the bottom as alike: they differ in a single sign, and treating them the same way gives 1 or −1. Students who rewrite 3 − (−3) as 3 + 3 and 3 + (−3) as 3 − 3 see 0 over 6 at once, and zero divided by a nonzero number is zero.
Error-prone arithmetic minor Working it out
The top adds −3 and the bottom subtracts −3, signs that look nearly the same but give 0 and 6, so each has to be rewritten before anything is divided. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Simplify the top and the bottom separately, rewriting the subtraction of a negative as an addition, and only then divide the top by the bottom.
Traps
- Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
The wrong answers, classified
A Negatives and direction (Brown & Burton, 1978)
Adds the sizes on top and keeps the minus, taking 3 + (−3) as −6, so the fraction becomes −6/6 = −1.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No sign rule gives a top of 1, so 1/6 mostly draws students who find the bottom, 6, and are unsure what a top of 0 means.
D Negatives and direction (Brown & Burton, 1978)
Treats the top like the bottom, adding 3 and 3 in both places, so the fraction becomes 6/6 = 1.
What was corrected from the DOE handbook, and why
- Explanation: The DOE’s explanation says “3 − (−3) is equivalent of adding 3 to 3”, without the article that the same sentence uses in “3 + (−3) is the equivalent of subtracting 3 from 3”. The site now says “is the equivalent of adding 3 to 3”.
- Explanation: The DOE’s explanation says “The zero property of division states that zero divided by any number equals zero.” Zero divided by zero is undefined, not zero, so the rule holds only when the divisor is not zero. The site now says “zero divided by any nonzero number equals zero”. Here the divisor is 6, so 0/6 = 0, as before.
Worked solution, step by step
Formulas used
- Adding a negative
- a and b are numbers. Adding a negative is the same as subtracting. On a number line, adding −3 moves you 3 steps left, just like subtracting 3.
- Subtracting a negative
- a and b are numbers. Subtracting a negative is the same as adding. Taking away a debt of 3 dollars leaves you 3 dollars better off. On a number line, you move right.
- Zero divided by a number
- b is any number except 0. Zero split into b equal parts leaves 0 in each part. The rule needs b not to be 0, because dividing by 0 has no answer.
Sort out the question
The top of the fraction is , and the bottom is . A fraction bar means divide. Work out the top and the bottom separately first. Then divide the top by the bottom.
Work out the top
Adding −3 is the same as subtracting 3:
A number plus its opposite is always 0.
Watch out: do not add the 3s and keep the minus sign. That makes the top −6 and leads to choice A, −1.
Work out the bottom
Subtracting −3 is the same as adding 3:
Watch out: the top and bottom look alike, but they are not. Making both of them 6 gives 6/6 = 1, choice D.
Divide the top by the bottom
Now the fraction is 0 over 6. Zero divided by any number other than zero is zero:
A top of 0 does not make the fraction 1/6, choice C. So the answer is B, 0.
Check your answer
Check by multiplying. A division is right when the answer times the bottom gives the top. Here , which is the top. So 0/6 really is 0.
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The original DOE question

Question 21 of 120
2026 Practice Test C Math, question 21Question
In a fruit basket, the ratio of apples to oranges is the same as the ratio of cherries to walnuts. If there are 6 oranges, 16 cherries, and 48 walnuts, how many apples are there?
- A2
- B3
- C18
- D38
Show the answer
Correct answer: A
The ratios of the items are equal, so they can be expressed using a proportion.
Let a represent the number of apples, o the number of oranges, c the number of cherries, and w the number of walnuts. Write a proportion showing apples:oranges is equal to cherries:walnuts:
Substitute the given values for oranges (6), cherries (16), and walnuts (48):
Cross multiply to form an equation to solve for the number of apples (a):
Solve for the value of a:
Question analysis
In a fruit basket, the ratio of apples to oranges equals the ratio of cherries to walnuts, and students find the number of apples from the counts of oranges, cherries and walnuts. It tests setting up and solving a proportion. Matching each quantity to its place in the ratio matters, since the wrong order gives a very different count.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Equal ratios with whole numbers are grade 6 work, and the numbers divide evenly. Grade 7 proportion problems are usually longer; here the only care is keeping apples with oranges in the same order as cherries with walnuts.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Turning words into math minor
- Much to keep track of minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 7, 32 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The numbers divide evenly, since 16/48 is 1/3, so the work is in the setup. Writing the two ratios in the order the sentence gives them makes the apples a third of the oranges; writing either one upside down makes them three times the oranges. A quick sense check helps: cherries are fewer than walnuts, so apples must be fewer than oranges.
Turning words into math minor Understanding the question
The sentence pairs apples with oranges and cherries with walnuts in the same order, so a/6 = 16/48; reading either pair the other way turns one ratio upside down. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Four kinds of fruit are named and three counts given, and the unknown apples must be matched with the smaller count, the cherries, not the larger, the walnuts. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Strategy: Write apples over oranges equal to cherries over walnuts, simplify the known ratio, and scale it to match the 6 oranges.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Stops one step early: A choice is a number from partway through the solution.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Divides the 48 walnuts by the 16 cherries to get the factor 3 and stops before using it to divide the 6 oranges.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Turns one ratio upside down, setting apples over oranges equal to 48 over 16, so the apples come out as 3 × 6 = 18.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reasons with differences instead of ratios, taking the gap of 32 between walnuts and cherries and adding it to the 6 oranges.
Worked solution, step by step
Formulas used
- Equal ratios
- A ratio compares two amounts. Writing it as a fraction puts the first thing named on top. The question says the two ratios are the same, so the two fractions are equal. Keep the same order in both.
Sort out the question
A ratio compares two amounts. The ratio of apples to oranges equals the ratio of cherries to walnuts. You know there are 6 oranges, 16 cherries and 48 walnuts. You need the number of apples.
Write the proportion
Put the first thing of each pair on top. Apples go with oranges, and cherries go with walnuts. Let a be the number of apples:
Watch out: keep the order the same. Writing 48 over 16 flips the second ratio. That gives 18, choice C.
Simplify the known ratio
16 goes into both 16 and 48. Divide the top and the bottom by 16:
So there is 1 cherry for every 3 walnuts. There must also be 1 apple for every 3 oranges.
Find the apples
Now the proportion is:
The bottom 6 is 3 × 2, so the top is 1 × 2. That makes .
Watch out: is choice B, but that is not the number of apples. And 38 comes from using differences instead of ratios.
So the answer is A, 2.
Check your answer
Compare the ratios. Apples to oranges is 2 to 6. Cherries to walnuts is 16 to 48. In both, the second number is 3 times the first: and . So the ratios are equal.
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The original DOE question

Question 22 of 120
2026 Practice Test C Math, question 23Question

On parallelogram PQRS above, the length of is 18 centimeters. What is the area of the parallelogram?
- A112 sq cm
- B396 sq cm
- C612 sq cm
- D748 sq cm
Show the answer
Correct answer: C
PQRS is a parallelogram in which PS is a base and QT is an altitude. The base is 34 cm in length, and the height is the length of the altitude, so the height is 18 cm.
To determine the area of a parallelogram, multiply the base × the height:
Question analysis
Parallelogram PQRS is drawn with a 34 cm base PS, a 22 cm slanted side and an 18 cm segment QT at a right angle to the base, and students find its area. It tests the area formula for a parallelogram. One labeled length is not needed, so the base and height that go together have to be chosen.
- NYS standards
- 6.G.1 area of polygons
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 The area of a parallelogram is grade 6 content, used here once with the height already drawn and labeled. Grade 7 area tasks usually involve circles, composite shapes or scale drawings, so this is easier.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 5, 25 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One multiplication answers the question once the right pair of lengths is chosen. Each wrong choice pairs or combines the labeled lengths in another way: the two sides as for a rectangle, the height with the slanted side, or all four sides as a perimeter. Following the right-angle mark at T from QT down to PS identifies the base and height that belong together.
Reading a figure, table or graph minor Understanding the question
Three lengths are labeled, 34 cm, 18 cm and 22 cm, but only the base PS and the height QT meet at the right-angle mark; the slanted side RS is not needed. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The formula, base times height, is not given, and the height must be the perpendicular distance between two sides, not the length of a slanted side. (Anderson & Krathwohl, 2001)
Strategy: Multiply the base by the height measured at a right angle to it, and check that the length you use as the height meets that base at a right-angle mark.
Traps
- Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
- Information you don't need: The question gives numbers or facts that play no part in the answer.
The wrong answers, classified
A Wrong formula (King et al., 2004)
Adds all four sides, 2 × (34 + 22) = 112, finding the perimeter of the parallelogram instead of its area.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Multiplies the 18 cm height by the 22 cm slanted side, a pair that does not meet at a right angle, getting 396.
D Wrong formula (King et al., 2004)
Multiplies the two side lengths, 34 × 22 = 748, as if a parallelogram's area were side times side like a rectangle's.
Worked solution, step by step
Formulas used
- Area of a parallelogram
- A is the area. b is the base, the side the shape sits on. h is the height, the straight-up distance from the base to the opposite side, at a right angle to the base. It works because you can cut a triangle off one end, move it to the other end, and make a rectangle.
Sort out the question
PQRS is a parallelogram, a four-sided shape whose opposite sides are parallel. The figure shows three lengths. Side PS, the bottom, is 34 cm. Segment QT goes straight down from Q to PS and is 18 cm. Side RS is slanted and is 22 cm. You need the area, in square centimeters.
Pick the base and the height
The base is PS, 34 cm. The height must meet the base at a right angle. The small square at T shows that QT does. So the height is QT, 18 cm.
Watch out: the 22 cm side is slanted, so it is not the height. Multiplying 22 × 18 gives 396, choice B.
Multiply the base by the height
Use the area formula:
Break it up to multiply:
The area is 612 square centimeters.
Choose the answer
Watch out: 34 × 22 = 748, choice D, treats the shape like a rectangle. Adding all four sides gives 112, choice A. That is the perimeter, the distance around.
So the answer is C, 612 sq cm.
Check your answer
Estimate: 34 is about 30 and 18 is about 20, and . 612 is close. Also, the area must be less than , because the height is shorter than the slanted side.
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The original DOE question

Question 23 of 120
2026 Practice Test C Math, question 30Question
Out of 2,000 high school seniors, 191 were enrolled in the Honor Society. What percentage of the entire group of 2,000 seniors were in the Honor Society?
- A0.955%
- B1.91%
- C9.55%
- D19.1%
Show the answer
Correct answer: C
The fraction represents the 191 high school seniors out of 2,000 enrolled in the Honor Society.
To determine the percentage, convert the fraction to a decimal. Then, rewrite the decimal as a percentage.
Question analysis
Out of 2,000 high school seniors, 191 belong to the Honor Society, and students find what percent of the whole group that is. It tests writing a part of a whole as a percent. Several choices share the same digits, so the decimal point has to be placed with care when the fraction becomes a percent.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Writing a part of a whole as a percent is grade 6 work, done here with one division and one conversion. The awkward numbers, 191 out of 2,000, make placing the decimal point the only real demand.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 7, 24 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The method is a single fraction, 191 over 2,000, so the question tests decimal places. The four choices share the digits 955 or 191 and differ only in where the point falls, which means a slip of one place still finds a choice waiting. A rough check settles it: 191 is a little under 200, and 200 is 10% of 2,000, so the share must be a little under 10%.
Error-prone arithmetic moderate Working it out
Dividing 191 by 2,000 gives 0.0955, with a zero right after the decimal point, and writing it as a percent moves the point two more places; each move is a chance to land one place off. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Write the Honor Society members as a fraction of all 2,000 seniors, divide, and move the decimal point two places to the right to turn the decimal into a percent.
Traps
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
A An arithmetic slip (Junker & Sijtsma, 2001)
Finds 0.0955 but moves the decimal point only one place when writing the percent, giving 0.955%, ten times too small.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Divides 191 by 100, as if a percent were always out of 100, and never uses the 2,000 seniors, so 191 ÷ 100 = 1.91 becomes 1.91%.
D An arithmetic slip (Junker & Sijtsma, 2001)
Divides 191 by 1,000 instead of 2,000, dropping the 2 while moving the decimal point, so 0.191 becomes 19.1%, twice the true share.
Worked solution, step by step
Formulas used
- Percent of a whole
- The part is the smaller group you care about. The whole is the entire group. The fraction tells what share of the whole the part is. Multiplying by 100 turns that share into a percent, which means out of 100.
Sort out the question
There are 2,000 seniors in all. Of them, 191 are in the Honor Society. You need the percent of all 2,000 seniors who are in it.
So 191 is the part, and 2,000 is the whole.
Write the share as a fraction
Put the part over the whole:
This fraction says 191 out of every 2,000 seniors. Next, turn it into a decimal by dividing.
Divide to get a decimal
Dividing by 2,000 is the same as dividing by 2 and then by 1,000:
Dividing by 1,000 moves the decimal point 3 places to the left. So the share is 0.0955.
Change the decimal to a percent
Multiply by 100. That moves the decimal point 2 places to the right:
So the share is 9.55%.
Watch out: moving the point only 1 place gives 0.955%, choice A. Dividing by 1,000 instead of 2,000 gives 19.1%, choice D. Dividing 191 by 100 gives 1.91%, choice B.
Give the answer
The 191 Honor Society members are 9.55% of all 2,000 seniors. That is a little less than one tenth of them.
So the answer is C, 9.55%.
Check your answer
Estimate. 191 is a little less than 200. And 200 out of 2,000 is one tenth, which is 10%. So the answer should be a little less than 10%. Only 9.55% fits.
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The original DOE question

Question 24 of 120
2026 Practice Test C Math, question 31Question
NUMBER OF VISITORS
TO BROOKSIDE ZOO
| Male | Female | |
|---|---|---|
| Children | 360 | 340 |
| Adults | 210 | 290 |
The table above shows the number of people who visited Brookside Zoo in one day. The largest of the four groups made up what percentage of all visitors on that day?
- A30%
- B28%
- C24%
- D17%
Show the answer
Correct answer: A
First, determine the total number of visitors.
The largest group is 360 (Male Children).
Find the percent that represents .
Question analysis
A two-way table counts one day's visitors to Brookside Zoo by age, children or adults, and by male or female. Students find what percent of all visitors the largest of the four groups made up. They must total every cell first, then make sure the count they divide is really the largest one.
- NYS standards
- 6.RP.3c percent as a rate per 100
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding what percent one count is of a total is grade 6 work. A two-way table adds two quick steps, totaling the four groups and spotting the largest, but the numbers divide evenly, so it stays below grade 7 demand.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Reading a figure, table or graph minor
- Many steps minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 39 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The arithmetic is friendly, since 360 over 1,200 reduces to 3 over 10, so the work is in reading the table. Two of its cells are close, 340 just below 360, and every wrong choice is the share of one of the other groups. Adding all four counts first, and circling the largest before dividing, keeps the right cell in play.
Reading a figure, table or graph minor Understanding the question
The two-way table gives four separate counts and no totals, so the 1,200 visitors must be added up from the cells and the largest group picked out of a grid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three short steps run in order: total the four cells, pick the largest, and divide; using the wrong cell in the last step still gives a believable percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Strategy: Add all four cells for the day's total, pick the biggest single cell, and divide it by the total; simplifying the fraction before dividing keeps the arithmetic short.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
- Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
The wrong answers, classified
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Takes 340, the female children, as the largest cell, misjudging it against the 360 male children beside it, so 340 ÷ 1,200 ≈ 28%.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Answers for the larger adult group, the 290 female adults, as if the question were about adults, so 290 ÷ 1,200 ≈ 24%.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reads largest as smallest and uses the 210 male adults, so 210 ÷ 1,200 = 17.5%, and takes the nearest choice.
Worked solution, step by step
Formulas used
- Percent of a whole
- The part is the one group you care about. The whole is everyone. The fraction tells what share of everyone the group is. Multiplying by 100 turns that share into a percent, which means out of 100.
Sort out the question
The table splits one day’s visitors into four groups:
Male Female Children 360 340 Adults 210 290 You need the largest group as a percent of all the visitors. The table gives no total, so you must find it.
Add up all the visitors
Add all four groups to get the whole:
So 1,200 people visited the zoo that day.
Find the largest group
Compare the four numbers. The biggest is 360, the male children.
Watch out: 340 is close to 360, but it is smaller. Using 340 gives about 28%, choice B. Using the 290 female adults gives about 24%, choice C. And 210 is the smallest group, which gives 17.5%, near choice D.
Find the percent
Put the part over the whole. Then simplify by dividing the top and bottom by 120:
Now multiply by 100 to get a percent:
So the male children made up 30% of the visitors.
Give the answer
The largest group, 360 male children, is 30% of all 1,200 visitors.
So the answer is A, 30%.
Check your answer
Work backward. 30% of 1,200 is , the male children. Also, with four groups an even split would be 25% each. The largest group should be a bit more than 25%, and 30% is.
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The original DOE question

Question 25 of 120
2026 Practice Test C Math, question 37Question
The table shows the probability of randomly picking each flavor of candy from a bowl.
PICKING CANDY FROM
A BOWL
| Flavor | Probability |
|---|---|
| Butterscotch | 0.57 |
| Lemon | 0.095 |
| Maple | 0.125 |
| Strawberry | 0.21 |
Which flavor is most likely to be randomly picked?
- Abutterscotch
- Blemon
- Cmaple
- Dstrawberry
Show the answer
Correct answer: A
The flavor most likely to be picked at random is butterscotch, because its probability is the greatest of the probabilities in the list. More than half the candies in the bowl are butterscotch.
Question analysis
A table lists the probability of picking butterscotch, lemon, maple or strawberry candy at random from a bowl, each as a decimal. Students decide which flavor is most likely to be picked. It tests the meaning of probability and comparing decimals, which have different numbers of digits here and must be compared by place value.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly understand
- What makes it hard
- Error-prone arithmetic minor
- Solution
- 1 step
- Reading level
- Question and choices: about grade 9, 37 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One idea answers the question: the most likely flavor has the greatest probability. The only obstacle is comparing decimals of different lengths, where 0.125 and 0.095 look bigger than 0.57 to students who count digits. Writing each as thousandths, 570, 95, 125 and 210, or noticing that 0.57 alone is more than one half, makes the order plain.
Error-prone arithmetic minor Working it out
The probabilities have two and three decimal places, 0.57 beside 0.095 and 0.125, so comparing them by length, or by their digits as whole numbers, ranks them wrongly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Line the four decimals up by place value, adding zeros so each has three decimal places, and compare them digit by digit starting from the tenths place.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reads lemon’s 0.095 as 0.95, overlooking the 0 in the tenths place, so it looks almost certain and larger than 0.57.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Treats the decimal with more digits as the larger, or compares 125 with 57 as whole numbers, so 0.125 seems bigger than 0.57.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common misreading makes 0.21 the largest, since it is neither the longest decimal nor the one with the biggest digits, so it mostly draws guesses.
Worked solution, step by step
Formulas used
- Probability
- P is the probability of picking a flavor. It is a number from 0 to 1. The more candies of a flavor there are, the bigger its probability. So the flavor with the greatest probability is the most likely to be picked.
Sort out the question
The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.
The greater the probability, the more likely the pick. So you need the greatest of the four decimals.
Line up the decimals
The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:
Flavor Probability Butterscotch 0.570 Lemon 0.095 Maple 0.125 Strawberry 0.210 Compare the tenths digits
Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.
Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.
Give the answer
Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.
So the answer is A, butterscotch.
Check your answer
0.57 is more than 0.5, which is one half. So more than half the candies are butterscotch. The other three flavors add up to , less than half. So butterscotch must be the most likely.
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The original DOE question

Question 26 of 120
2026 Practice Test C Math, question 43Question
A softball player bought bottles of water for her team to have at practice. The total number of bottles she bought, b, is proportional to n, the number of players at practice. Which equation represents the relationship between b and n? The constant of proportionality is p.
- A
- B
- C
- D
Show the answer
Correct answer: A
In this context, the constant of proportionality, p, represents the relationship between b, the number of bottles of water, and n, the number of players on the team.
The number of bottles of water is the product of the constant of proportionality and the number of players.
The equation that represents this relationship is .
Question analysis
A softball player buys bottles of water for practice, and the number of bottles, b, is proportional to the number of players, n, with constant of proportionality p. Students choose the equation that represents the relationship. It tests writing an equation for a proportional relationship using letters only, so the meaning of proportional has to be clear.
- NYS standards
- 7.RP.2c proportional relationship equations
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 students write equations for proportional relationships from tables, graphs and descriptions, usually with a number to find. Here the constant is already given as a letter, so only the general form must be recalled, a short task.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly remember
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 7, 51 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is a recall question dressed in a short story: the softball team adds nothing but names for the letters. The one idea needed is that proportional means multiplied by a constant, so doubling the players doubles the bottles. Students who think of a constant as something added, or as a fixed total, are drawn to the two sums that use the same letters.
Something you must already know minor Working it out
The general form of a proportional relationship, one amount equal to the constant times the other, has to be recalled; the stem names p but never says it multiplies. (Anderson & Krathwohl, 2001)
Variables instead of numbers minor Understanding the question
No numbers are given, only b, n and p, so a choice cannot be checked by computing unless students make up a case, such as 2 bottles per player for 10 players. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Strategy: Recall how two proportional quantities are linked through the constant of proportionality, then test each choice with sample numbers: if the players double, the bottles should double too.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Treats proportional as adding a fixed amount, so the bottles would be the players plus p, rather than the players multiplied by a constant.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Takes the constant of proportionality as the total of bottles and players, confusing a constant sum with a constant ratio.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Puts all three letters into one sum equal to 1, an equation that uses the right letters but matches no idea of proportion, so it mostly draws guesses.
Worked solution, step by step
Formulas used
- Proportional relationship
- y and x are two amounts that change together. k is the constant of proportionality, the number you always multiply x by to get y. Because it is a multiplication, doubling x doubles y.
Sort out the question
The letter b is the number of bottles, and n is the number of players. The bottles are proportional to the players. The constant of proportionality is p.
You need the equation that links b and n. There are no numbers, only letters.
Recall what proportional means
Proportional means one amount is always a fixed number times the other. For example, say there are 2 bottles for each player. Then 10 players need 20 bottles, and 20 players need 40 bottles. Doubling the players doubles the bottles.
The fixed number, here 2, is the constant of proportionality.
Write it with these letters
Start with the rule . The bottles take the place of y, the players take the place of x, and p is the constant:
Here pn means p times n.
Check the other choices
Watch out: choice B, , adds instead of multiplying. With p = 2, 10 players give 12 bottles. Then 20 players give 22, which is not double.
Choice C makes p a total of bottles and players, not a multiplier. Choice D makes all three letters add up to 1, which does not fit at all.
Give the answer
Only multiplies the players by a constant. So it is the only choice where doubling the players doubles the bottles.
So the answer is A, .
Check your answer
Try p = 3 bottles per player. With 5 players, . With 10 players, . Twice the players, twice the bottles. Also, , so b ÷ n gives back the constant.
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The original DOE question

Question 27 of 120
2026 Practice Test C Math, question 46Question
A rain gauge shows that 10 centimeters of rain fell yesterday. If 1 inch is equal to approximately 2.5 centimeters, about how many inches of rain fell yesterday?
- A1
- B2.5
- C4
- D10
Show the answer
Correct answer: C
Set up a proportion to solve:
10 centimeters is approximately 4 inches.
Question analysis
A rain gauge shows that 10 centimeters of rain fell, and 1 inch is about 2.5 centimeters. Students find about how many inches of rain fell. It tests converting between metric and customary units with a given conversion fact, and the main risk is converting in the wrong direction or reusing a number from the question.
- NYS standards
- 6.RP.3d converting measurement units
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Converting units with a given conversion fact is grade 6 content, and this needs a single division with friendly numbers, lighter than grade 7 work with rates and percents.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Turning words into math minor
- Solution
- 1 step
- Reading level
- Question and choices: about grade 10, 25 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
A single division answers the question, and its numbers are friendly: four lengths of 2.5 centimeters make 10. Every wrong choice is simply a number from the stem, so the only real risk is reaching for a given value instead of converting. Inches are longer than centimeters, so the rainfall in inches has to be less than 10 but more than the 1 inch that 2.5 centimeters make.
Turning words into math minor Understanding the question
The fact “1 inch is equal to approximately 2.5 centimeters” has to be turned into a division, how many 2.5-centimeter lengths fit into 10, rather than a multiplication. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Ask how many groups of 2.5 centimeters fit into the 10 centimeters of rain, since each group of 2.5 centimeters makes one inch.
Traps
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
- Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Picks the 1 from the conversion fact, 1 inch for 2.5 centimeters, reading the given fact as the answer instead of applying it to 10 centimeters.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Picks 2.5, the number of centimeters in one inch, straight from the conversion fact rather than finding how many inches 10 centimeters make.
D Units or scale (King et al., 2004)
Keeps the 10 from the rain gauge, as if centimeters and inches were the same size, so no conversion is done at all.
Worked solution, step by step
Formulas used
- Converting centimeters to inches
- About 2.5 centimeters make 1 inch. So each group of 2.5 centimeters is 1 inch. Dividing the centimeters by 2.5 counts how many of those groups fit. That count is the number of inches.
Sort out the question
The rain gauge shows 10 centimeters (cm) of rain. You know that 1 inch is about 2.5 cm. You need the rain in inches.
Decide whether to multiply or divide
An inch is longer than a centimeter. So it takes fewer inches than centimeters to measure the same rain. The answer must be less than 10. That tells you to divide, not multiply.
Watch out: choice D, 10, keeps the same number, as if inches and centimeters were the same size.
Divide by 2.5
Use the conversion formula:
You can see it by adding: . Four groups of 2.5 cm fit into 10 cm. So 10 cm is about 4 inches.
Give the answer
Watch out: choices A and B, 1 and 2.5, are just the numbers in the conversion fact. They do not use the 10 cm at all.
About 4 inches of rain fell. So the answer is C, 4.
Check your answer
Change back. Each inch is 2.5 cm, so 4 inches is cm. That is what the gauge showed. Also, 4 inches is less than 10, as it should be, because inches are longer.
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The original DOE question

Question 28 of 120
2026 Practice Test C Math, question 47Question
The height of a plant, in millimeters, is proportional to the number of days the plant has grown. The plant has grown 20 millimeters in 8 days. What is the constant of proportionality, in millimeters per day?
- A0.4
- B2.5
- C12
- D20
Show the answer
Correct answer: B
The constant of proportionality is determined by dividing the number of millimeters by the number of days:
Question analysis
A plant's height in millimeters is proportional to the number of days it has grown, and it grew 20 millimeters in 8 days. Students find the constant of proportionality in millimeters per day. It tests finding a unit rate, and the order of the division decides whether the rate comes out in the units asked for.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a constant of proportionality from one pair of values is a basic grade 7 skill. With one division and the units named in the question, it asks less than most grade 7 problems.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 7, 35 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One division gives the constant, and the question itself names the order by asking for millimeters per day. The tempting slip is dividing the smaller number by the larger, 8 by 20, out of habit, which gives a rate in days per millimeter. A plant that grows 20 millimeters in 8 days grows more than 2 millimeters a day, a quick sense check on any result.
Something you must already know minor Working it out
Students must know that the constant is the height divided by the days; the units in the question, millimeters per day, name the order of the division. (Anderson & Krathwohl, 2001)
Strategy: Use the units in the question, millimeters per day, as the instruction: divide the millimeters grown by the number of days.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Divides the days by the millimeters, 8 ÷ 20 = 0.4, which gives days per millimeter, the rate turned upside down.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Subtracts the days from the millimeters, 20 − 8 = 12, combining the two numbers without thinking about a rate.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Gives the 20 millimeters the plant grew over all 8 days, the total growth, rather than the growth for each single day.
Worked solution, step by step
Formulas used
- Constant of proportionality
- Here y is the height in millimeters, and x is the number of days. k is the growth in one day, in millimeters per day. In a proportional relationship, y = kx, so dividing y by x gives k.
Sort out the question
The plant’s height is proportional to the number of days it grows. It grew 20 millimeters (mm) in 8 days.
You need the constant of proportionality in millimeters per day. Per day means for each day. So you need how much the plant grows in one day.
Divide millimeters by days
The units tell you the order: millimeters per day means millimeters divided by days. Use the formula:
So the plant grows 2.5 mm each day.
Avoid the tempting slips
Watch out: , choice A, is days per millimeter. That is the rate upside down.
Choice C, 12, comes from . Subtracting does not make a rate. Choice D, 20, is the growth over all 8 days, not one day.
Give the answer
The constant of proportionality is 2.5 mm per day.
So the answer is B, 2.5.
Check your answer
Multiply back. At 2.5 mm a day for 8 days, the plant grows mm. That matches. Also, 2 mm a day would give only 16 mm, and 3 mm a day would give 24 mm. So the rate must be between 2 and 3.
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The original DOE question

Question 29 of 120
2026 Practice Test D English, question 14Passage Excerpt from “How We Made the First Flight”
Excerpt from “How We Made the First Flight”
by Orville Wright
In 1903 brothers Wilbur and Orville Wright conducted experiments related to flying machines. These experiments would eventually lead to air travel becoming a reliable form of transportation.
During the night of December 16, 1903, a strong cold wind blew from the north. When we arose on the morning of the 17th, the puddles of water, which had been standing about camp since the recent rains, were covered with ice. The wind had a velocity of 10 to 12 meters per second (22 to 27 miles an hour). We thought it would die down before long, and so remained indoors the early part of the morning. But when ten o’clock arrived, and the wind was as brisk as ever, we decided that we had better get the machine out and attempt a flight. We hung out the signal for the men of the Life Saving Station.1 We thought that by facing the flyer into a strong wind, there ought to be no trouble in launching it from the level ground about camp. We realized the difficulties of flying in so high a wind, but estimated that the added dangers in flight would be partly compensated for by the slower speed in landing.
Final Preparations
We laid the track on a smooth stretch of ground about one hundred feet north of the new building. The biting cold wind made work difficult, and we had to warm up frequently in our living room, where we had a good fire in an improvised stove made of a large carbide2 can. By the time all was ready, J. T. Daniels, W. S. Dough and A. D. Etheridge, members of the Kill Devil3 Life Saving Station; W. C. Brinkley of Manteo, and Johnny Moore, a boy from Nags Head,4 had arrived.
We had a “Richard” hand anemometer with which we measured the velocity of the wind. Measurements made just before starting the first flight showed velocities of 11 to 12 meters per second, or 24 to 27 miles per hour. . . .
Audacity—and Calculation
Wilbur having used his turn in the unsuccessful attempt on the 14th, the right to the first trial now belonged to me. After running the motor a few minutes to heat it up, I released the wire that held the machine to the track, and the machine started forward in the wind. Wilbur ran at the side of the machine, holding the wing to balance it on the track. Unlike the start on the 14th, made in a calm, the machine, facing a 27-mile wind, started very slowly. Wilbur was able to stay with it till it lifted from the track after a forty-foot run. One of the Life Saving men snapped the camera for us, taking a picture just as the machine had reached the end of the track and had risen to a height of about two feet. The slow forward speed of the machine over the ground is clearly shown in the picture by Wilbur’s attitude. He stayed along beside the machine without any effort.
Flight
The course of the flight up and down was exceedingly erratic, partly due to the irregularity of the air, and partly to lack of experience in handling this machine. The control of the front rudder was difficult on account of its being balanced too near the center. This gave it a tendency to turn itself when started; so that it turned too far on one side and then too far on the other. As a result the machine would rise suddenly to about ten feet, and then as suddenly dart for the ground. A sudden dart when a little over a hundred feet from the end of the track, or a little over 120 feet from the point at which it rose into the air, ended the flight. As the velocity of the wind was over 35 feet per second and the speed of the machine over the ground against this wind ten feet per second, the speed of the machine relative to the air was over 45 feet per second, and the length of the flight was equivalent to a flight of 540 feet made in calm air. This flight lasted only 12 seconds, but it was nevertheless the first in the history of the world in which a machine carrying a man had raised itself by its own power into the air in full flight, had sailed forward without reduction of speed and had finally landed at a point as high as that from which it started.
From “How We Made the First Flight” by Orville Wright—Public Domain/Federal Aviation Administration
1Life Saving Station: one of the rescue stations along the Atlantic coastline that provided assistance to mariners in distress
2carbide: a very hard material composed of carbon and other heavy metals
3Kill Devil: the town of Kill Devil Hills in eastern North Carolina
4Nags Head: a town in eastern North Carolina
Question
The use of chronological structure contributes to the development of ideas in the excerpt by
- Aoutlining the actions that the Wright brothers took to prepare for and successfully complete the first flight.
- Bidentifying the primary factors that allowed the Wright brothers to overcome obstacles and achieve the first flight.
- Cshowing how the Wright brothers applied lessons learned from their previous flight attempts to accomplish the first flight.
- Ddemonstrating how the Wright brothers analyzed the impact of wind velocity to identify the ideal conditions for the first flight.
Show the answer
Correct answer: A
The question asks how the use of chronological structure contributes to the development of ideas in the excerpt.
A. CORRECT. Through the chronological structure, Orville addresses all stages of the flight in a way that builds from flight concerns and preparations for takeoff to his experiences during the flight and his successful landing.
B. Incorrect. While obstacles are presented within the excerpt, the chronological structure emphasizes the events that ended in the successful achievement of flight on that day rather than how the Wright brothers overcame obstacles.
C. Incorrect. The chronological structure of the excerpt does not show that the Wright brothers applied lessons learned from their previous attempts at flight; instead, the structure describes Orville’s December 17 attempt.
D. Incorrect. While Orville’s narrative certainly shows a strong need to analyze wind speed, this idea is not conveyed through the chronological structure of the excerpt. Additionally, the description of the first flight on December 17 indicates that it did not take place in ideal conditions (“We realized the difficulties of flying in so high a wind” [paragraph 1] and “The course of the flight up and down was exceedingly erratic, partly due to the irregularity of the air” [paragraph 5]).
Question analysis
This text-structure question asks how the chronological structure of Orville Wright’s “How We Made the First Flight,” an account that begins on the night of December 16, 1903, contributes to the development of ideas in the excerpt. Students must see what telling events in time order does for the account. Each choice names a purpose that could suit a story of the first airplane flight.
- Grade level
- Grade 7 reading and task
- Compared with grade 7
- Below grade 7 Grade 7 structure questions often ask how a less obvious pattern, such as problem and solution, shapes ideas. This excerpt is plainly told in time order, with dates and section headings, so recognizing what that order does is a lighter task.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- How, not what minor
- A very close wrong answer minor
- Evidence
- Answered from the whole text
- Reading level
- Passage: about grade 7, 816 words; question and choices: about grade 11, 87 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The excerpt is plainly chronological, with a date in its first sentence and headings that move from “Final Preparations” to “Flight,” so the structure is easy to recognize. The difficulty is in saying what that order does. The strongest pull sounds like the excerpt’s central idea, success over obstacles, but ranking the causes of success is analysis, not sequence, and the other choices invent lessons from the failed attempt on the 14th or a search for ideal winds.
How, not what minor Working it out
The question asks what telling events in time order does, and the excerpt moves from the night of December 16 through laying the track and the launch to the flight, so the structure’s job is to trace those steps. (Anderson & Krathwohl, 2001)
A very close wrong answer minor Choosing the answer
The choice about the primary factors shares the key’s ending, achieving the first flight; it differs in its first half, an analysis of what made success possible, which a time order does not set out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Strategy: Skim the section headings and the first sentence of each paragraph to see how the excerpt moves through time, then test each choice against what a time order can do.
Traps
- Wrong purpose or structure: Names a purpose, structure or effect that the part does not serve, or describes what a part says when the question asks what it does.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
The wrong answers, classified
B Wrong purpose or structure (Anderson & Krathwohl, 2001)
Expects the structure to explain why the flight succeeded, a fair summary of the excerpt’s message, but time order tells events as they happened and never ranks which factors mattered most.
C Reads too much in (Graesser, Singer & Trabasso, 1994)
Builds on the failed attempt of the 14th and assumes the brothers learned from it, though the excerpt only contrasts that calm start with this windy one.
D Twists what the text says (King et al., 2004)
Recalls the anemometer readings and the risk estimate and turns them into a search for ideal conditions, though the brothers flew in a wind whose difficulties and added dangers they admitted.
Working it out, step by step
Figure out what the question asks
This is a structure question. Chronological structure means telling events in time order, from first to last. The question asks what that order does for the excerpt. So first check how the excerpt moves through time.
Trace the order of events
Look at the section headings and the start of each paragraph.
- Paragraph 1 starts “During the night of December 16, 1903.” The brothers decide to fly.
- Under “Final Preparations,” they lay the track and measure the wind.
- Under “Audacity—and Calculation,” Orville takes off.
- Under “Flight,” the machine flies and lands.
Say what the time order does
The story moves step by step. It goes from getting ready, to taking off, to flying, to landing. Each step leads to the next. By the end, you have followed everything the brothers did to make the first flight happen. The last sentence says the machine “had finally landed at a point as high as that from which it started.”
Check each choice
- A fits. Time order lays out what the brothers did to prepare and then to fly.
- B does not fit. Primary factors are the main causes. Time order tells events as they happen. It does not rank which causes mattered most.
- C does not fit. The excerpt mentions the failed try on the 14th. But it never says what they learned from it.
- D does not fit. They did not wait for ideal, or perfect, conditions. They flew in a strong wind and knew its dangers.
Watch out: B is tempting because the excerpt does show success over problems. But naming the main causes of success is a different job from telling events in order.
Choose the answer
The time order walks you through the preparations, the takeoff and the flight. So the answer is A, “outlining the actions that the Wright brothers took to prepare for and successfully complete the first flight.”
Why this answer and not the closest wrong one
B is the closest wrong choice. The excerpt does show the brothers succeeding despite problems. But time order cannot explain which factors mattered most. It just tells events in the order they happened. That is what A says: the time order outlines the actions, from getting ready to completing the flight.
More questions like this
The original DOE question

The passage “Excerpt from “How We Made the First Flight”” as printed (pages 13–14)


Question 30 of 120
2026 Practice Test D English, question 21Passage Ode to Fireworks
Ode to Fireworks
In autumn my mother drove us to the edge of the field
where the fair was set up year after year:
the carousel, the bumper cars, the long, low sheds
filled with prizewinning animals.
We—my sister, my cousin, and I—were ready for bed,
already in our pajamas. This was a treat we waited
all year for. We waited in the darkness
for the first low, dull thwumps, like someone
beating an old, filthy rug hung on a wash line.
Then we counted the seconds between the lightning
and thunder, as we also used to do, until the sky
lit up: red, blue, green, gold. In my mind’s eye
I can still see the straggly, ancient oak whose branches
reached up past the exhibition halls, silhouetted
against the spectrum of stars that cascaded behind it.
It was one thing to look up into the sky
and imagine yourself in it or to make out pictures
among the clouds, which my sister liked to do.
No, I would tell her, that cloud
does not look like an elephant, a hat, an umbrella.
But it was another thing to see
the sky at night written upon
with those jewels. (We lived in the country:
night was night.) All around us, crickets
stridulated in the stubble of what had been
somebody’s cornfield, their song rising and falling.
You could smell winter on the air’s edge.
Now, in the city, when the sky dips into shadow
at New Year’s or on the Fourth of July, I find myself
craning my neck upward at odd moments.
The city sky is always lit up. This is where we live now,
and it is how we live now, awash in light
of every hue. Everything is a constant celebration:
picking up washing at the cleaner’s or stopping by
the corner market for a loaf of heavy bread.
And the music around me is the music of people,
their voices rising and falling in a hundred languages.
But beneath the yellowish glow deep in the sky
of all our city lights pelting out into the universe,
I remember the feel of the pickup truck bumping
across the ridged field, as I kept waiting for those
childhood bursts, watching as they escorted us home.
Question
The comparison in lines 8–9 of the poem is used to convey
- Athe muffled pounding of explosions in the distance.
- Bthe way lightning streaks through the clouds.
- Cthe echoes of thunder on an autumn night.
- Dthe glow of sparks falling from the sky.
Show the answer
Correct answer: A
The question asks what the comparison in lines 8–9 of the poem is used to convey.
A. CORRECT. In lines 8–9 the speaker compares the deep thumping noises at the start of a fireworks show to the muffled thumping sound made when beating a rug to clean it. The imagery of “low, dull thwumps” (line 8) (onomatopoeia) describes a sound that is not clear or powerful. To the speaker, these low, distant explosions are the signal that the fireworks display is starting and that the loud, cracking sound of fireworks will be heard soon (“Then we counted the seconds between the lightning / and thunder” [lines 10–11]).
B. Incorrect. The comparison in lines 8–9 deals with the low sound of the first fireworks shooting off rather than the streaks of light they emit. To the speaker, the low thwumps (onomatopoeia) are the signal that fireworks are about to explode overhead, much like the way a streak of lightning during a storm indicates that a crash of thunder will follow in a few seconds.
C. Incorrect. The thwump (onomatopoeia) sounds are the start of the fireworks show, not thunder. As a comparison, the speaker says the initial thwumps of the fireworks signal anticipation for the full explosion that will come, much like the way the speaker would watch for lightning and count the seconds before an impending thunderclap (lines 10–11).
D. Incorrect. The muffled thwump (onomatopoeia) sounds occur before the dazzling explosion of lights and before sparks start to fall from the sky. The comparison focuses on the sounds of the experience, not the sight of the experience.
Question analysis
This figurative language question asks what the comparison in lines 8–9 of “Ode to Fireworks” conveys. In the poem an adult speaker remembers waiting with family, already in pajamas, for the fireworks at a country fair. Students must interpret an old-fashioned household image and decide what part of the show it describes, since the nearby lines mention sights as well as sounds.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Interpreting a simile is taught from grade 6, and this one takes up two lines of concrete words. The household chore it pictures may be unfamiliar, but the task is a single step, lighter than typical grade 7 poetry work.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- Evidence
- Answered from one sentence or line
- Reading level
- Passage: 379 words (a poem; reading level not measured); question and choices: about grade 3, 42 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The comparison is about sound, and once students notice that, the sight-based choices about lightning and sparks fall away. The pull comes from the next lines, where the children count “the seconds between the lightning / and thunder,” a storm game that makes thunder seem like the thing being described. The question rewards staying inside lines 8–9, where “low, dull” and the beaten rug describe the first faint booms of the show.
Figurative or literary language moderate Understanding the question
Lines 8–9 pair the made-up sound word “thwumps” with a simile, someone beating an old rug on a wash line, so students must hear a dull, distant thud in an image about housework. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Something you must already know minor Working it out
The simile works only for readers who can imagine the flat whack of a rug being beaten clean on a line, a chore many students have never seen or heard. (Anderson & Krathwohl, 2001)
Strategy: Reread lines 8–9 slowly and ask which sense the comparison appeals to, sight, sound or touch, then pick the choice that describes that same experience.
Traps
- Taking figurative language literally: A choice reads a figure of speech or image at face value.
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
The wrong answers, classified
B Wrong part of the text (Kirsch & Mosenthal, 1990)
Jumps to the “lightning” of line 10, but lines 8–9 come before the sky lights up and describe something heard in the darkness, not a streak of light.
C Taking figurative language literally (Graesser, Singer & Trabasso, 1994)
Takes the “lightning / and thunder” of lines 10–11 as a real storm and hears the thwumps as thunder, though the counting game only likens the fireworks to one.
D Wrong part of the text (Kirsch & Mosenthal, 1990)
Pictures the colors of lines 12–15, “red, blue, green, gold,” but the rug-beating image describes a dull sound heard before any sparks appear in the sky.
Working it out, step by step
Figure out what the question asks
This asks what a comparison helps you picture. Lines 8–9 compare one thing to another using “like.” That kind of comparison is called a simile. First ask which sense it describes: sight, sound or touch.
Find the comparison
The family waits in the dark. Lines 8–9 say they wait “for the first low, dull thwumps, like someone / beating an old, filthy rug hung on a wash line.” Thwumps is a made-up word that sounds like the noise it names. So the comparison is about a sound. It comes before the sky lights up.
Picture the rug
People used to hang a dusty rug on a line and hit it with a stick. Each hit makes a flat, soft thud, not a sharp bang. The words “low” and “dull” say the same thing. So the first fireworks sound like soft booms far away. Muffled means soft, as if wrapped in cloth.
Check each choice
- A is right. Dull thuds heard in the dark are the muffled pounding of explosions far off.
- B is wrong. Lightning is something you see. Lines 8–9 describe a sound heard before anything lights up.
- C is wrong. The thunder in lines 10–11 is part of a counting game. It compares the show to a storm. The rug image is about the fireworks.
- D is wrong. Sparks are something you see. The colors don’t come until line 12.
Watch out: Lines 10–11 mention “lightning” and “thunder.” That can make C look right. But the rug comparison comes first and is about the fireworks.
Choose the answer
The rug-beating simile makes the first fireworks sound like soft, faraway thuds. So the answer is A, “the muffled pounding of explosions in the distance.”
Why this answer and not the closest wrong one
C is the closest wrong answer, because thunder is a low boom too. But lines 10–11 only compare the show to a storm in a counting game. The rug image in lines 8–9 describes the fireworks themselves: dull, muffled thuds from far away. That is A.
More questions like this
The original DOE question

The passage “Ode to Fireworks” as printed (page 32)

Question 31 of 120
2026 Practice Test D English, question 37Passage Martial Arts for the Mind and Body
Martial Arts for the Mind and Body
(1) The martial arts blend a series of physical movements with strategic mental discipline so that practitioners can defend themselves, physically defeat an opponent, or both. (2) Historians are unsure of exactly when and where martial arts were first used. (3) Martial arts have been practiced by several different societies for many centuries. (4) Martial arts such as karate, kung fu, tae kwon do, and judo are still taught and practiced as methods of self-defense, but they offer students more than that. (5) The study of martial arts can provide students with a way to enhance their mental discipline as well as their physical fitness.
(6) Discipline, focus, and respect are important qualities for everyone to have. (7) However, for most people these qualities are not innate; they must be learned and practiced. (8) The study of martial arts can provide an opportunity to develop these skills. (9) Students are rewarded for their dedication by passing tests and advancing to higher ranks or levels. (10) For example, in a typical tae kwon do class, students learn discipline by diligently practicing moves, improve focus by listening carefully, and demonstrate respect by bowing to the instructor and following directions.
(11) For teenagers, martial arts classes provide a safe and structured environment for gaining physical skills, building confidence, and enjoying a sense of community. (12) A lot of teens go through hard situations as they try to do well in school and in life. (13) A martial arts class can provide teens with a physical outlet for stress while also challenging them mentally. (14) Participating in a martial arts program also helps children and teens focus on self-improvement rather than on competition. (15) Progressing through levels of achievement involves mastering more physically demanding techniques. (16) It requires students to take responsibility and be accountable for achieving set goals. (17) Students gain confidence and experience companionship with other students who are progressing through the ranks.
(18) Adults who practice martial arts can experience many of the same benefits that younger people do, but perhaps the greatest of these is health and fitness. (19) Adult martial arts students often see changes in their body within weeks of beginning a program.
(20) For people interested in studying a martial art, there are many ways to learn and practice. (21) In addition to private studios, community recreation centers often offer low-cost or free martial arts classes. (22) There are even online videos that introduce students to the basic concepts. (23) People should study martial arts.
Question
Which revision of sentence 12 best maintains the formal style established in the passage?
- AA lot of teens have to put up with difficult things while trying to do well in school and in life.
- BMany teenagers deal with tough situations as they try to stay on top of their studies and do well in life.
- CMany teenagers encounter challenges as they work to succeed both academically and personally.
- DA lot of teens face problems as they work to keep up with their schoolwork and find personal success.
Show the answer
Correct answer: C
The question asks for the revision of sentence 12 that best maintains the formal style established in the passage.
A. Incorrect. The phrases “A lot,” “put up with,” “difficult things,” and “do well in school” are worded informally; therefore, this sentence does not consistently maintain the formal style established in the passage.
B. Incorrect. The phrases “deal with,” “tough situations,” “stay on top of,” and “do well in life” are worded informally; therefore, this sentence does not consistently maintain the formal style established in the passage.
C. CORRECT. This option is correct because it uses clear and scholarly wording throughout the whole sentence. The phrases “many teenagers,” “encounter challenges,” and “succeed both academically and personally” make the sentence’s style more consistently formal than the other options.
D. Incorrect. The phrases “A lot,” “face problems,” and “keep up with” are worded informally; therefore, this sentence does not consistently maintain the formal style established in the passage.
Question analysis
This style question asks which revision of sentence 12, “A lot of teens go through hard situations as they try to do well in school and in life,” best keeps the formal style of the martial arts passage. Students compare four rewrites phrase by phrase, listening for casual expressions in each one.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Keeping a formal style is practiced from grade 6, and here it is tested in one sentence with familiar casual phrases. Typical grade 7 style work asks for subtler choices of precise, content-specific words across a passage.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly evaluate
- What makes it hard
- Evidence
- Answered from one sentence or line
- Reading level
- Passage: about grade 11, 402 words; question and choices: about grade 8, 87 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Most students hear the casual phrases at once, and only one version avoids all of them, so this is one of the easier revising questions. The risk is stopping at a formal start: one version opens with “Many teenagers” and then slides into “tough situations,” and another sounds serious with “face problems” but begins with “A lot of teens.” Checking every phrase, not the first few words, is all the question asks.
Every choice must be checked minor Choosing the answer
Each rewrite mixes formal and casual wording, “Many teenagers” beside “deal with” and “stay on top of,” so every phrase of every version has to be checked, not just the opening words. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
Students must recognize idioms such as “put up with” and “keep up with” as spoken English, a sense of register that the question expects them to bring. (Anderson & Krathwohl, 2001)
Strategy: Underline every casual phrase in each rewrite, such as everyday idioms or vague words, and keep only the version with none; one formal phrase is not enough.
Traps
- Wordy, vague or informal: A revision is grammatical but wordy, imprecise or too casual for the passage.
The wrong answers, classified
A Wordy, vague or informal (Connors & Lunsford, 1988)
Trades “go through hard situations” for “put up with difficult things” and keeps “A lot of teens,” swapping one set of casual phrases for another.
B Wordy, vague or informal (Connors & Lunsford, 1988)
Is reassured by the formal opening “Many teenagers,” but the rest, “deal with,” “tough situations” and “stay on top of,” is everyday speech.
D Wordy, vague or informal (Connors & Lunsford, 1988)
Likes the more serious “face problems” and “personal success,” but keeps the casual “A lot of teens” and “keep up with their schoolwork.”
Working it out, step by step
Figure out what the question asks
This asks which rewrite of sentence 12 matches the passage’s formal style. Formal style is the careful, serious way of writing you use in a school report. It skips the casual words you might use with friends.
Spot the casual words
Sentence 12 says, “A lot of teens go through hard situations as they try to do well in school and in life.” “A lot of,” “teens” and “go through” sound like talking with a friend. Compare sentence 11, which speaks of “a safe and structured environment.” The rest of the passage uses careful words like those.
Name the rule
Formal writing uses exact, serious words and skips everyday sayings. For example, Kids have a ton of homework is casual. Students have a large amount of homework is formal. Check every phrase in each choice, not just the first few words.
Check each choice
- A is wrong. “A lot of teens,” “put up with” and “difficult things” are casual.
- B is wrong. It starts well with “Many teenagers.” But “deal with,” “tough situations” and “stay on top of” are casual.
- C is right. Every phrase is formal. To encounter challenges is to meet hard problems. Academically means in school. Personally means in your own life.
- D is wrong. “Face problems” sounds serious. But “A lot of teens” and “keep up with” are casual.
Watch out: B and D each have a formal phrase or two. One formal phrase is not enough.
Choose the answer
Only C is formal from start to finish. So the answer is C, “Many teenagers encounter challenges as they work to succeed both academically and personally.”
Why this answer and not the closest wrong one
B is the closest wrong answer, because it opens with the formal “Many teenagers.” But it then slides into everyday phrases like “deal with” and “stay on top of.” C keeps the same meaning and stays formal from start to end. Words like “encounter challenges” and “academically” match the careful style of the passage.
More questions like this
The original DOE question

The passage “Martial Arts for the Mind and Body” as printed (page 52)

Question 32 of 120
2026 Practice Test D English, question 48Question
Read this sentence.
In the Whispering Gallery of Grand Central Terminal, people whispering a message in one corner can be clearly heard in the opposite corner, regardless of surrounding noise, because the tiled archways send the sound through the environment.
Which revision to the words send the sound through the environment uses the most precise language?
Drag the correct phrase into the box.
In the Whispering Gallery of Grand Central Terminal, people whispering a message in one corner can be clearly heard in the opposite corner, regardless of surrounding noise, because the tiled archways ____.
On the test you drag one of the tiles into the box; here, choose it.
- Aguide the sound waves in a specific direction
- Bfocus the sound waves to travel through the space
- Cdirect the sound waves to move along the curves of the arches
- Dtransfer the sound waves to a variety of different locations
Show the answer
Correct answer: C
The question asks for the revision to the words send the sound through the environment that uses the most precise language.
Incorrect. The phrase “guide the sound waves in a specific direction” is not the most precise revision. “A specific direction” does not explain precisely where the sound waves travel.
Incorrect. The phrase “focus the sound waves to travel through the space” is not the most precise revision, because it does not precisely describe where the waves move. The phrase “through the space” does not provide the reader with more specific information as to where in the space the sound waves travel.
CORRECT. The phrase “direct the sound waves to move along the curves of the arches” uses precise language to explain how the waves travel. The wording “direct the sound waves” is more specific than “send the sound” as in the original sentence, and the action “move along the curves” conveys more detail than the phrase “through the environment.”
Incorrect. The phrase “transfer the sound waves to a variety of different locations” is not the most precise revision, because it does not precisely convey where the waves move to. The phrase “variety of different locations” is vague and does not add clarity.
Question analysis
This precise-language question gives a sentence about the Whispering Gallery in Grand Central Terminal, where a whisper in one corner is heard in the opposite corner, and asks which revision of “send the sound through the environment” is most precise. On the test students drag one of four phrase tiles into the box, and every tile sounds more technical than the original.
- NYS standards
- 7L3a precise and concise language
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Replacing vague wording with exact wording is taught by grade 6. Here it is one sentence about a familiar place and four phrase tiles, a single judgment, which is lighter than typical grade 7 revision of a whole paragraph.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly evaluate
- What makes it hard
- A very close wrong answer moderate
- Every choice must be checked minor
- Evidence
- Answered from one sentence or line
- Reading level
- Question and choices: about grade 8, 133 words.
- Corrected from the DOE handbook
- A misquotation of the sentence in the explanation was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
All four tiles upgrade “send the sound” to “the sound waves” with a sharper verb, so the original’s vagueness survives only in the endings. The tile that promises “a specific direction” sounds precise because of that word, yet it never names the direction; the key is the only tile that says where the sound travels. Dragging a tile into the box adds nothing to the task beyond picking one choice.
A very close wrong answer moderate Choosing the answer
The “specific direction” tile is built like the key, a precise verb and “the sound waves”; the two differ only in the ending, a direction never named versus “along the curves of the arches.” (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
Every tile sounds more exact than the original, so students must ask of each one whether a reader could tell how the sound gets from one corner to the opposite one. (Embretson & Wetzel, 1987)
Strategy: Reread the whole sentence, then ask of each tile whether a reader could picture exactly how the sound gets from one corner of the gallery to the other.
Traps
- Wordy, vague or informal: A revision is grammatical but wordy, imprecise or too casual for the passage.
- Changes the meaning: A revision is correct English but no longer says what the writer meant.
The wrong answers, classified
A Wordy, vague or informal (Connors & Lunsford, 1988)
Trusts the word “specific,” which sounds exact, but “a specific direction” never says which direction, so the phrase stays as vague as the original.
B Wordy, vague or informal (Connors & Lunsford, 1988)
Is drawn by the technical “focus the sound waves,” but “through the space” just restates “through the environment” and says nothing about the archways.
D Changes the meaning (Connors & Lunsford, 1988)
Picks the scientific-sounding “transfer,” but sending sound to “a variety of different locations” contradicts a whisper heard in the one opposite corner, and the phrase repeats itself.
What was corrected from the DOE handbook, and why
- Explanation: The DOE’s explanation of the correct phrase says that “move along the curves” conveys more detail than “the phrase ‘moving through the environment.’” The sentence in the question says the archways “send the sound through the environment”; the word “moving” is not in it. The site now quotes the sentence’s own words: “the phrase ‘through the environment.’”
Working it out, step by step
Figure out what the question asks
This asks which phrase is most precise. Precise language is exact. It tells the reader just what happens, so they can picture it. The phrase you pick replaces the underlined words.
See what the sentence needs
The sentence says a whisper in one corner “can be clearly heard in the opposite corner.” Why? “because the tiled archways send the sound through the environment.” That ending is vague, or unclear. “Through the environment” could mean anywhere. A better phrase would say how the sound gets from one corner to the other.
Name the rule
Precise writing swaps general words for exact ones. For example, The dog went over there is vague. The dog ran under the porch is precise. Now you know just where the dog went.
Check each choice
- A is wrong. “A specific direction” sounds exact. But it never says which direction.
- B is wrong. “Through the space” is just another way to say “through the environment.”
- C is right. “Along the curves of the arches” tells exactly where the sound travels. It follows the curved arches to the far corner.
- D is wrong. “A variety of different locations” is vague. It also clashes with a whisper heard in one corner. And “variety” and “different” say the same thing twice.
Watch out: The word “specific” makes A sound precise. Ask whether you can picture the path. With A, you can’t.
Choose the answer
With the tile in place, the sentence ends because the tiled archways direct the sound waves to move along the curves of the arches. So the answer is C. On the test, you drag the tile “direct the sound waves to move along the curves of the arches” into the box.
Why this answer and not the closest wrong one
A is the closest wrong answer. It has a strong verb, “guide,” and the word “specific.” But it never names the direction, so the reader still can’t picture the path. C says exactly where the sound goes: along the curves of the arches. That explains how a whisper reaches the opposite corner.
More questions like this
The original DOE question

Question 33 of 120
2026 Practice Test D Math, question 1Question
What is the prime factorization of 3,575?
- A
- B
- C
- D
Show the answer
Correct answer: B
By observation, 3,575 ends in 75; therefore, 3,575 is divisible by 25.
Test 143 for factors and eliminate 3, 5, and 7, but 11 is a factor.
; both 11 and 13 are prime numbers.
The first factor, 25, is not prime but is the product of primes, .
The prime factorization of 3,575 is .
Question analysis
This number question asks for the prime factorization of 3,575, chosen from four products of factors. Students divide by primes, using divisibility tests, until only primes remain. The four products look much alike, so the work must be carried all the way through, keeping count of every prime divided out along the way.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Breaking a number into primes is grade 6 number work that relies on divisibility tests. The number here is larger than in most class examples, but the work is still a few divisions, lighter than typical grade 7 problems.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Something you must already know minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 11, 10 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Three of the four choices multiply back to 3,575, so checking a product rules out only the one that equals 715. The question turns on whether 143 is prime: it survives the quick tests for 2, 3 and 5, and the factor 11 is easy to miss. Students who keep dividing until every factor is prime, and who write a prime as often as it divides, finish in under a minute.
A very close wrong answer moderate Choosing the answer
5² · 143 matches the right factorization in its first factor and multiplies back to 3,575 just as well, so a product check cannot separate them; only splitting 143 into 11 × 13 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Nothing on the page defines a prime factorization or gives a test for 11, and 143 passes the familiar quick tests for 2, 3 and 5, so it looks prime. (Anderson & Krathwohl, 2001)
Strategy: Divide by the smallest prime that goes in evenly, again and again, writing down every prime you use until the quotient reaches 1, then compare that list with each choice.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Lists each prime that divides 3,575 only once, 5, 11 and 13, as if a factorization named which primes divide it, though their product is only 715.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Divides out one 5 and stops at 5 · 715, a true factorization but not a prime one, though 715 ends in 5 and splits further.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Divides by 25 and stops at 143, which passes the tests for 2, 3 and 5 and looks prime, though 143 = 11 × 13.
Worked solution, step by step
Formulas used
- Divisibility by 25
- A number is divisible by 25 when 25 goes into it with no remainder. You only need to look at the last two digits. If they are 00, 25, 50 or 75, the number is divisible by 25. This works because 100 is 4 × 25, so the hundreds never leave a remainder.
- Writing a repeated factor with an exponent
- a is any number. The small raised 2 is an exponent. It tells you how many times a is used as a factor, so 5² means 5 × 5 = 25. Use an exponent in a prime factorization when the same prime shows up more than once.
Sort out the question
A prime number is a whole number greater than 1 whose only factors are 1 and itself. Examples are 2, 3, 5, 7, 11 and 13. The prime factorization of 3,575 writes it as a product (the answer to a multiplication) of prime numbers only. So you must break 3,575 down until every factor is prime.
Divide by 25 first
3,575 ends in 75. So it is divisible by 25:
That means 3,575 = 25 × 143. But 25 is not prime. It is 5 × 5, which you can write as 5².
Test 143 for prime factors
Now see if 143 splits. Try the primes in order. 143 is odd, so 2 does not work. Its digits add to 1 + 4 + 3 = 8, which is not a multiple of 3, so 3 does not work. It does not end in 0 or 5, so 5 does not work. Since 7 × 20 = 140, 143 ÷ 7 leaves a remainder of 3, so 7 does not work. Now try 11:
So 143 = 11 × 13, and both 11 and 13 are prime.
Watch out: 143 looks prime, but it is not. Stopping at 5² · 143 leads to choice D.
Put all the primes together
Collect every prime you found. The 5 appears twice, then 11 and 13 once each. The dot · means times, just like ×.
Watch out: write the 5 twice, since it divides 3,575 twice. Writing it once gives 5 · 11 · 13, choice A, which equals only 715. Choice C, 5 · 715, stops too early, because 715 ends in 5 and splits further.
So the answer is B, .
Check your answer
Multiply the primes back together: 25 × 11 = 275, and 275 × 13 = 3,575. That is the number you started with. Also, 5, 11 and 13 are all prime, so nothing can be split further.
More questions like this
The original DOE question

Question 34 of 120
2026 Practice Test D Math, question 4Question
If , what is the value of ?
- A
- B
- C
- D
Show the answer
Correct answer: C
The first step is to substitute the value for x into the expression to get .
Then, evaluate the denominator and finally divide the numerator by the denominator.
Question analysis
This substitution question sets x equal to one fourth and asks for the value of 2 divided by (2 − x). Students substitute, subtract a fraction from a whole number, and divide by the result. The expression is a stacked fraction, so treating the whole denominator as one quantity and dividing rather than multiplying are where students slip.
- NYS standards
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Substituting a value and dividing by a fraction are grade 6 skills, and here they take three short steps with friendly numbers. The stacked fraction asks for care, but the demand is below typical grade 7 work with signed rational numbers.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- An answer in an unexpected form minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 10, 12 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Substitution is one step, so the question is about handling a fraction inside a fraction. The bar over 2 − x groups the whole denominator, and the division by 7/4 must flip it; both errors lead to listed choices. A size check helps: 2 − 1/4 is a little less than 2, so the value must be a little more than 1, which rules out every wrong choice.
Error-prone arithmetic moderate Working it out
The denominator 2 − 1/4 has to become 7/4 before anything is divided, and dividing 2 by 7/4 means multiplying by 4/7; multiplying by 7/4 instead is the usual slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The work ends at the improper fraction 8/7, but the choices show it only as a mixed number, listed among 3/4, 8/9 and 3 1/2. (Enright, Morley & Sheehan, 2002)
Strategy: Simplify the denominator to a single fraction first, then divide 2 by that fraction by multiplying by its reciprocal; treating the fraction bar as a grouping symbol avoids the common slip.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
- Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
The wrong answers, classified
A Order of operations (Brown & Burton, 1978)
Lets the fraction bar cover only the first 2, working 2/2 − 1/4 = 1 − 1/4 = 3/4 instead of dividing 2 by the whole denominator.
B Negatives and direction (Brown & Burton, 1978)
Adds 1/4 to 2 instead of subtracting it, so the denominator becomes 9/4, and 2 ÷ 9/4 = 8/9.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Finds the denominator 7/4 correctly but multiplies 2 by it instead of dividing, 2 × 7/4 = 14/4 = 3 1/2.
Worked solution, step by step
Formulas used
- Dividing by a fraction
- a is any number, and b/c is a fraction. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing asks how many copies of b/c fit into a.
- Changing an improper fraction to a mixed number
- Use it when the top, a, is bigger than the bottom, b. q is how many whole times b fits into a. r is the remainder, what is left over. For 8/7, the 7 fits once with 1 left, so 8/7 = 1 1/7.
Sort out the question
You know . You need the value of 2 divided by (2 − x). A fraction bar means divide. It also groups, so you work out the whole bottom, 2 − x, before you divide.
Work out the bottom
Put in 1/4 for x. To subtract, write 2 as fourths, since 2 = 8/4:
The bottom is 7/4.
Watch out: the sign is minus. Adding 1/4 gives 9/4 and then 8/9, choice B.
Divide 2 by 7/4
Now the problem is 2 ÷ 7/4. Use the rule for dividing by a fraction. Flip 7/4 to get 4/7, then multiply:
Watch out: multiplying by 7/4 instead of 4/7 gives 14/4 = 3 1/2, choice D.
Write it as a mixed number
The choices use mixed numbers. 7 fits into 8 once, with 1 left over:
Watch out: the fraction bar covers all of 2 − 1/4. Working 2/2 = 1 first and then taking away 1/4 gives 3/4, choice A.
So the answer is C, .
Check your answer
Estimate. The bottom, 1 3/4, is a bit less than 2. So 2 divided by it should be a bit more than 1, and 1 1/7 fits. Also check by multiplying: 7/4 × 8/7 = 56/28 = 2. That is the top, so the division is right.
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The original DOE question

Question 35 of 120
2026 Practice Test D Math, question 7Question
A middle school library has a shelf with 40 books, each labeled with a different number from 201 to 240. One book is selected at random. What is the probability, as a percentage, of selecting book number 223?
Enter your answer in the space provided.
Fill-in question: work out the answer and type it as a number.
Show the answer
Correct answer: 2.5%
The probability that book number 223 will be selected at random from 40 books is 1 out of 40. To find the probability as a percentage, convert the fraction to a decimal.
The chance of selecting one specific book is .
Convert the decimal to a percentage:
Question analysis
A library shelf holds 40 books numbered 201 to 240, and one is picked at random. Students find the probability of picking one particular book and type it as a percentage. It tests simple probability and converting a fraction to a percent; the book numbers play no part, and the decimal point needs care.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding the probability of one equally likely outcome is grade 6 content. The only further step is rewriting the fraction as a percent, which needs care with place value, so it is lighter than grade 7 compound probability.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- An answer in an unexpected form moderate
- Error-prone arithmetic minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 6, 41 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The book numbers, 201 to 240, and the chosen number 223 play no part, since every book is equally likely and the shelf holds 40. The chance is 1 in 40, and the real demand is the form: changing it to a percent means a decimal with a zero in the tenths place, then a shift of two places. Students who know that 1/4 is 25% can check that 1/40 must be a tenth of that.
An answer in an unexpected form moderate Choosing the answer
Only the bold word percentage sets the form, since the empty box shows no % sign, so the probability 1/40 must be turned into 2.5, not entered as the fraction or as 0.025. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Dividing 1 by 40 gives 0.025, with a zero right after the decimal point, and losing that zero turns the percent into 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Count the books, write the chance of picking one particular book as a fraction, then turn the fraction into a decimal and multiply by 100 to get the percent.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Information you don't need: The question gives numbers or facts that play no part in the answer.
The wrong answers, classified
0.025 Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the probability as a decimal, 1 ÷ 40 = 0.025, and enters it without multiplying by 100, though the box asks for a percentage.
25 An arithmetic slip (Junker & Sijtsma, 2001)
Loses the zero after the decimal point when dividing, taking 1 ÷ 40 as 0.25, as 1 ÷ 4 would be, and so enters 25.
1/40 Misreads the question (Hegarty, Mayer & Monk, 1995)
Enters the probability as the fraction 1/40, the form it was found in, overlooking the bold word percentage in the question.
Worked solution, step by step
Formulas used
- Probability of one outcome
- P is the probability, the chance of an event. The top counts the outcomes you want. The bottom counts all the equally likely outcomes. Use it when each outcome has the same chance, like picking one book at random.
- Changing a fraction to a percent
- a/b is the fraction. Percent means out of 100. Multiplying by 100 tells you how many out of 100 the fraction is. For example, 1/2 × 100 = 50, so 1/2 is 50%.
Sort out the question
The shelf has 40 books, numbered 201 to 240. One book is picked at random, so every book has the same chance. You need the chance of picking book 223, written as a percentage. Book 223 is on the shelf, since 223 is between 201 and 240.
Write the probability as a fraction
Only 1 book is number 223, and there are 40 books in all:
The numbers on the labels do not matter beyond this. They just give each book its own name.
Change the fraction to a decimal
Divide 1 by 40. It helps to start from 1/4 = 0.25. Since 40 is ten times 4, 1/40 is one tenth of 1/4:
Watch out: there is a 0 right after the decimal point. Losing it gives 0.25 and then 25, which is wrong.
Change the decimal to a percent
Multiply by 100 to get a percent. That moves the decimal point two places to the right:
So the chance is 2.5%.
Watch out: 0.025 is the probability as a decimal, and 1/40 is the same chance as a fraction. The question asks for a percentage, so neither one is the answer here.
Enter the number
The box shows no % sign. Type 2.5. Adding the sign, as 2.5%, also counts.
So you would enter 2.5.
Check your answer
2.5% means 2.5 out of every 100. Out of 40 books, that is 40 × 0.025 = 1 book, the one numbered 223. Also, one book out of 40 should be a small chance, and 2.5% is small.
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The original DOE question

Question 36 of 120
2026 Practice Test D Math, question 15Question
What is the value of the expression shown above?
- A
- B
- C
- D
Show the answer
Correct answer: C
To add fractions, find a common denominator and rename the fractions using the common denominator. The least common denominator for 11, 22, and 33 is 66. After adding the fractions, the final fraction can be reduced to lowest terms.
Question analysis
This fractions question asks for the value of 1/11 + 1/22 + 1/33. Students find a common denominator for three fractions, rename each one, add and simplify. The denominators are all multiples of 11, and adding the tops and bottoms straight across, or forgetting to rename a numerator, leads to tempting wrong choices.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Adding fractions with unlike denominators is taught before grade 7. Finding a common denominator for three related numbers and simplifying is short, routine work, lighter than typical grade 7 problems with signed rational numbers.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- An answer in an unexpected form minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 9, 14 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The question checks whether students rename fractions before adding, and each wrong choice comes from a shortcut that skips that step or stops early. Seeing that 22 and 33 are multiples of 11 points to 66 as the common denominator. A size check helps too: the sum must be more than 1/11, its largest term, which rules out two of the wrong choices at once.
Error-prone arithmetic moderate Working it out
Three unlike denominators must be renamed over 66, not over the product of all three, and the numerators 6, 3 and 2 have to be found one by one before adding. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The sum comes out as 11/66, which must be reduced to match a choice; its simplest form has a denominator, 6, unlike any in the expression. (Enright, Morley & Sheehan, 2002)
Strategy: Notice that 22 and 33 are multiples of 11, rename all three fractions over their least common multiple, add the numerators, and simplify the result.
Traps
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Stops one step early: A choice is a number from partway through the solution.
The wrong answers, classified
A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Adds the numerators and the denominators straight across, 3 over 11 + 22 + 33 = 66, and simplifies 3/66 to 1/22.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Puts the three 1s over the largest denominator, 33, without renaming the other two fractions, so 3/33 simplifies to 1/11.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Adds the first two fractions correctly, 2/22 + 1/22 = 3/22, and stops before adding the third fraction, 1/33.
Worked solution, step by step
Formulas used
- Equivalent fractions
- a/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction. It works because n/n is 1, and multiplying by 1 changes nothing. Use it to give fractions the same bottom.
- Adding fractions with the same denominator
- d is the denominator, the bottom number, which tells the size of each piece. a, b and c are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.
Sort out the question
You need to add three fractions: 1/11, 1/22 and 1/33. Their denominators (bottom numbers) are all different. You can only add fractions when the pieces are the same size. So first give all three the same denominator.
Find a common denominator
A common denominator is a number that 11, 22 and 33 all divide into. Notice that 22 = 2 × 11 and 33 = 3 × 11. So the smallest number that all three divide into is:
Rename each fraction
Multiply the top and bottom of each fraction by the number that makes the bottom 66:
Add the tops
Now every piece is a sixty-sixth. Add the tops and keep the bottom:
Watch out: never add the bottoms. Adding straight across gives 3/66 = 1/22, choice A. Putting the three 1s over 33 gives 3/33 = 1/11, choice B.
Simplify the fraction
To simplify, divide the top and bottom by a number that goes into both. Here 11 goes into 11 and into 66:
Watch out: add all three fractions. Stopping after the first two gives 3/22, choice D.
So the answer is C, .
Check your answer
Estimate with decimals. 1/11 is about 0.091, 1/22 is about 0.045 and 1/33 is about 0.030. They add to about 0.166. And 1/6 is 0.1666…, a match. The other choices are all 0.136 or less.
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The original DOE question

Question 37 of 120
2026 Practice Test D Math, question 16Question
The table shows the proportional relationship between x cups of oatmeal and y ounces of raisins in a bread recipe.
BREAD RECIPE
| Cups of Oatmeal (x) | Ounces of Raisins (y) |
|---|---|
| 2 | 1.0 |
| 3 | 1.5 |
| 6 | 3.0 |
What is the constant of proportionality of the number of ounces of raisins to the number of cups of oatmeal?
Enter your answer in the space.
Fill-in question: work out the answer and type it as a number.
Show the answer
Correct answer: 0.5
To find p, the constant of proportionality of ounces of raisins, y, to cups of oatmeal, x, use the equation .
The table shows that there is 1 ounce of raisins for 2 cups of oatmeal.
The constant of proportionality is 0.5.
Question analysis
A table for a bread recipe lists cups of oatmeal and the ounces of raisins that go with them, in a proportional relationship. Students type the constant of proportionality of raisins to oatmeal. It tests reading a unit rate from a table, and the order in which the question names the two quantities matters.
- NYS standards
- 7.RP.2b constant of proportionality
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a constant of proportionality from a table is a grade 7 standard, and here a single division in one row gives it. The demand is lighter than typical grade 7 work; the order of the ratio in the wording needs the most care.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Turning words into math moderate
- An answer in an unexpected form minor
- Solution
- 1 step
- Reading level
- Question and choices: about grade 6, 56 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Every row gives the same ratio, so the division itself is easy; the question tests which quantity goes on top. Because oatmeal comes first in the table and its numbers are the larger ones, dividing cups by ounces feels natural and produces a clean whole number. Writing the relationship as y = kx, raisins equal k times oatmeal, and checking it on a second row such as 3 cups and 1.5 ounces fixes the order.
Turning words into math moderate Understanding the question
The phrase “of the number of ounces of raisins to the number of cups of oatmeal” fixes the order y ÷ x, while the table lists oatmeal first, inviting 2 ÷ 1.0. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed in with no choices, so an upside-down ratio of 2, a whole number, looks as finished as the decimal the table actually gives. (Enright, Morley & Sheehan, 2002)
Strategy: Divide an ounces-of-raisins value by the cups of oatmeal in the same row, in the order the question names them, and check that every row gives the same value.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
2 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Divides cups of oatmeal by ounces of raisins, 2 ÷ 1.0, turning the ratio upside down: cups per ounce instead of ounces per cup.
1 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reads the first raisin amount, 1.0 ounce, as the constant without dividing it by the 2 cups of oatmeal in the same row.
Worked solution, step by step
Formulas used
- Constant of proportionality
- x and y are two amounts that grow together at the same rate. k is the constant of proportionality, the number you always multiply x by to get y. Here y is ounces of raisins and x is cups of oatmeal, so k is ounces of raisins per cup of oatmeal.
Sort out the question
The table shows cups of oatmeal, x, and ounces of raisins, y. Proportional means the raisins are always the same multiple of the oatmeal. That multiple is the constant of proportionality. The question asks for raisins to oatmeal. So you want ounces of raisins for each cup of oatmeal: y ÷ x.
Divide y by x in one row
Use the first row of the table: 2 cups of oatmeal go with 1.0 ounce of raisins.
So there is 0.5 ounce of raisins for each cup of oatmeal.
Check the other rows
A constant must be the same in every row:
Every row gives 0.5, so the rule is .
Watch out: the table lists oatmeal first, which tempts you to divide 2 by 1.0. That gives 2, which is cups per ounce, the ratio upside down. And 1.0 alone is just one raisin amount, not the constant.
Enter the answer
Type the constant in the box as a number, with no units. On this site, 0.5, .5 and 1/2 all count.
So you would enter 0.5.
Check your answer
Multiply each oatmeal amount by 0.5: 2 × 0.5 = 1.0, 3 × 0.5 = 1.5 and 6 × 0.5 = 3.0. Each result matches the raisins in its row. Also, every raisin number is smaller than its oatmeal number, so the constant must be less than 1.
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The original DOE question

Question 38 of 120
2026 Practice Test D Math, question 21Question
Maxie borrowed $7.75 from her mother, $11.00 from her father, and $4.50 from her brother to purchase a video game. Her grandmother gave her $25.00 as a gift. How much money will Maxie have left or still owe if she uses the money her grandmother gave her to pay back the money she borrowed?
- AMaxie will still owe $2.25.
- BMaxie will still owe $1.75.
- CMaxie will have $1.75 left.
- DMaxie will have $2.25 left.
Show the answer
Correct answer: C
First, determine the total amount Maxie borrowed, in dollars:
Then use subtraction to determine the difference between the amount her grandmother gave her and the amount she borrowed:
The difference is positive, which means that Maxie will still have $1.75 left from the amount her grandmother gave her.
Question analysis
Maxie borrowed money from her mother, father and brother to buy a video game, then received $25.00 from her grandmother to pay them back. Students decide how much money she will have left over or still owe. It tests adding and subtracting decimals with money, and the choices also test which way the difference goes.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic minor
- Turning words into math minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 4, 66 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Few students miss this one, since it needs only one addition and one subtraction with money. The choices come in pairs, so the right amount must also be matched with the right outcome. Lining up the decimal points, and asking whether $25.00 is more or less than the $23.25 borrowed, settles both the amount and whether Maxie owes or keeps money.
Error-prone arithmetic minor Working it out
The loans add to $23.25, and subtracting that from $25.00 means borrowing across two zeros, a classic place for a digit to go wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Each amount appears twice among the choices, once as money left and once as money owed, so the result must also be read for direction by asking which is larger, the gift or the loans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Add the three amounts borrowed, compare the total with the $25.00 gift, and decide from which is larger whether Maxie ends up with money left or still owes some.
Traps
- Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
The wrong answers, classified
A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Needs two errors at once, a wrong difference such as $2.25 and the reversed outcome, so no single line of working reaches it and it mostly draws guesses.
B Negatives and direction (Brown & Burton, 1978)
Finds the $1.75 difference correctly but reverses its meaning, as if the $25.00 gift fell short of the $23.25 borrowed instead of covering it.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Subtracts $23.25 from $25.00 column by column, taking the smaller digit from the larger instead of borrowing, so 5 − 0 and 2 − 0 give $2.25.
Worked solution, step by step
Formulas used
- Money left over
- Money she has is the $25.00 gift. Money she owes is the total she borrowed. If the result is positive, she has that much left. If it were negative, she would still owe that much. Comparing the two amounts first tells you which.
Sort out the question
Maxie borrowed from three people: $7.75, $11.00 and $4.50. Her grandmother gave her $25.00. She uses the $25.00 to pay everyone back. You need to know if money is left over or if she still owes some, and how much.
Add up what she borrowed
Line up the decimal points and add:
She owes $23.25 in all. That is the amount she must pay back.
Compare the gift with the debt
The gift, $25.00, is more than the $23.25 she owes. So she can pay everyone back and still have money left. She will not owe anything.
Watch out: once you find $1.75, keep its meaning straight. Saying she still owes $1.75, choice B, gets the direction backwards.
Subtract to find what is left
Take the debt away from the gift:
You need to regroup (borrow) across the zeros. Think of $25.00 as 24 dollars and 100 cents. Then 100 − 25 = 75 cents, and 24 − 23 = 1 dollar.
Watch out: taking the smaller digit from the larger one in each column gives $2.25, choice D.
So the answer is C: Maxie will have $1.75 left.
Check your answer
Add back. The $23.25 she pays plus the $1.75 she keeps is 23.25 + 1.75 = 25.00. That is the whole gift, so the subtraction is right.
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The original DOE question

Question 39 of 120
2026 Practice Test D Math, question 22Question
Seth grows strawberries in his garden.
- He picked cups of strawberries on Monday.
- He picked 4 cups of strawberries on Tuesday.
- He ate cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.
How many cups of strawberries did Seth have left?
- A
- B
- C6
- D9
Show the answer
Correct answer: B
When Seth picks strawberries, the value represents a positive number, and when he eats the strawberries, the value represents a negative number. Write an expression and then evaluate the expression, where the negative value is multiplied by 3 because he ate the same amount on 3 different days.
If the multiplication is confusing, evaluate the expression below which represents the sum for the 5 days with each day listed separately.
Question analysis
Seth picked strawberries on Monday and Tuesday, in amounts given as a mixed number and a whole number, then ate the same fraction of a cup at breakfast on three days. Students find how many cups he has left. It tests adding, multiplying and subtracting fractions in a short word problem, and the eaten amount repeats each day.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Adding mixed numbers and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so the demand is below typical grade 7 work with signed rational numbers.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- Turning words into math minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 6, 47 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The bulleted list lays out the story day by day, so setting up the problem is easy; the points are lost in the fractions. The number of breakfast days, 3, sits right beside the 3/4 cup, and subtracting 3 cups instead of three quarters of a cup each day is the tempting shortcut. Writing every amount in fourths, 26/4 picked and 9/4 eaten, makes the last step a whole-number subtraction.
Error-prone arithmetic moderate Working it out
Three breakfasts of 3/4 cup make 9/4 = 2 1/4 cups, and taking that from 6 1/2 means renaming the half as 2/4, two fraction steps where a slip is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Picking adds and eating subtracts, and “each day on Wednesday, Thursday, and Friday” means the 3/4 cup counts three times, not once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Add the two amounts picked, multiply the breakfast amount by the number of days, and subtract that from the total; writing every amount in fourths keeps the arithmetic simple.
Traps
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Subtracts 3 cups, the number of breakfast days, instead of 3 × 3/4 = 2 1/4 cups, so 6 1/2 − 3 = 3 1/2.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives exactly 6; it is near the 6 1/2 cups picked and mostly catches students who drop the fractions or guess.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Adds the 2 1/4 cups eaten to the 6 1/2 cups picked instead of subtracting, 8 3/4 cups, and takes the nearest choice, 9.
Worked solution, step by step
Formulas used
- Total of a repeated amount
- Use it when the same amount happens again and again. Here Seth eats 3/4 cup on each of 3 days, so the number of times is 3 and the amount each time is 3/4 cup.
- A whole number times a fraction
- n is a whole number, and a/b is a fraction. Multiply the whole number by the top only, and keep the bottom. It works because n groups of a pieces make n × a pieces, and each piece is still size 1/b.
Sort out the question
Seth picked 2 1/2 cups on Monday and 4 cups on Tuesday. Then he ate 3/4 cup each day on Wednesday, Thursday and Friday. That is 3 days of eating. Picking adds strawberries, and eating takes them away. You need the number of cups he has left.
Add what he picked
Add the two days of picking:
He picked 6 1/2 cups in all, before he ate any.
Find how much he ate
He ate 3/4 cup on each of 3 days. Multiply:
To change 9/4 to a mixed number, note that 4 fits into 9 twice, with 1 left over. He ate 2 1/4 cups.
Watch out: he ate 3/4 cup three times, not 3 cups. Subtracting 3 gives 3 1/2, choice A.
Take away what he ate
Write the half as fourths, so both fractions have the same bottom: 1/2 = 2/4. Then subtract the wholes and the fourths:
Watch out: eating takes away. Adding the 2 1/4 cups gives 8 3/4, and the nearest choice is 9, choice D.
So the answer is B, cups.
Check your answer
Use decimals. 2.5 + 4 = 6.5 cups picked. 3 × 0.75 = 2.25 cups eaten. Then 6.5 − 2.25 = 4.25 cups, which is 4 1/4.
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The original DOE question

Question 40 of 120
2026 Practice Test D Math, question 23Question
Which situation involves quantities that combine to equal zero?
- Areceiving $5 as a gift and then giving $5 to a friend
- Bbuying a book for $10 and then buying lunch for $10
- Cgetting on an elevator at the ground floor, rising 3 floors, and then rising 3 more floors
- Dstarting in a cave 20 feet below the ground, climbing up 20 feet, and then climbing up 20 feet more
Show the answer
Correct answer: A
When looking for a situation where quantities combine to make zero, find two values with the same absolute value where one value is positive and one value is negative.
In the situation in option A, receiving a gift of $5 is represented by , and giving a friend $5 is represented by . These values combine to equal zero because .
Question analysis
This question gives four everyday situations involving money, an elevator and a cave, and asks which one involves quantities that combine to make zero. It tests representing gains and losses, or ups and downs, with positive and negative numbers. Equal numbers alone do not make zero, so the direction of each amount must be read carefully.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Describing situations where opposite quantities make zero is named in the grade 7 standards, and this is its most direct form: one reading of four everyday situations, easier than typical grade 7 work with signed numbers.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly understand
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 5, 60 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The idea that a quantity and its opposite add to zero is familiar, and the right situation states it plainly. The wrong choices are built from equal amounts: two of them repeat a number but move in one direction, and the cave climb reaches zero partway before going 20 feet past it. Writing each situation as a sum of signed numbers takes a few seconds and settles it.
Turning words into math minor Understanding the question
Each situation must be turned into signed numbers, receiving or rising as plus and giving or buying as minus, before its total can be judged. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
Every choice repeats one amount, $5, $10, 3 floors or 20 feet, so equal numbers alone do not identify the answer; the direction of each move has to be checked. (Embretson & Wetzel, 1987)
Strategy: Write each situation as signed numbers, using + for gains or upward moves and − for losses or downward moves, add them, and see which total is zero.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Treats the two equal $10 amounts as opposites that cancel, though buying a book and buying lunch both take money away, for a total of −$20.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Pairs the two 3s as amounts that cancel, missing that rising and then rising again both go up, ending 6 floors above the ground floor.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Sees that starting 20 feet below ground and climbing 20 feet reaches zero, then overlooks the last 20 feet climbed, which ends 20 feet above ground.
Worked solution, step by step
Formulas used
- Opposites add to zero
- a is any number, and −a is its opposite: the same size, but the other sign. A number and its opposite always add to 0. On a number line, they sit the same distance from 0 on opposite sides. Equal moves in opposite directions cancel out.
Sort out the question
You need the situation whose amounts add up to zero. Give each amount a sign. Use + for getting money or going up. Use − for spending money or going down. Then add the amounts in each choice. The direction matters, not just the numbers.
Write each choice with signs
Turn each situation into signed numbers and add them up:
Choice Amounts Total A +5, then −5 0 B −10, then −10 −20 C +3, then +3 +6 D start at −20, then +20, then +20 +20 Look at the tricky choices
In B, buying a book and buying lunch both take money away. Equal amounts cancel only when they go in opposite directions. In C, rising and rising again both go up, so you end 6 floors up.
Watch out: in D, starting at −20 and climbing 20 feet does reach 0, ground level. But then you climb 20 feet more. You end at +20, 20 feet above the ground, so D is wrong.
Choose the zero total
Only choice A has two opposite amounts of the same size. Getting $5 is +5, and giving $5 away is −5:
So the answer is A: receiving $5 as a gift and then giving $5 to a friend.
Check your answer
Think about where you end up. In A, you end with exactly the money you started with, so the total change is zero. In every other choice, you end with less money, higher up, or above the ground.
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The original DOE question

Question 41 of 120
2026 Practice Test D Math, question 25Question
A bus trip takes 9 hours if the mean speed is 50 miles per hour. How many hours would this trip take if the mean speed was 45 miles per hour?
- A8
- B10
- C12
- D14
Show the answer
Correct answer: B
First, find the length of the bus trip in miles by multiplying the time, 9 hours, by the average (mean) speed, 50 miles per hour, to get 450 miles.
Then, divide the total distance, 450 miles, by the average (mean) speed, 45 miles per hour, to get the total time:
Question analysis
A bus trip takes 9 hours at a mean speed of 50 miles per hour, and students find how long the same trip takes at 45 miles per hour. It tests speed, distance and time. Working out the fixed distance first is the key, since guessing how the time should change leads many students astray.
- NYS standards
- 6.RP.3b unit rate problems
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- No obvious method moderate
- Something you must already know minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 3, 28 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Two short calculations solve it, but only after noticing that speed and time move in opposite directions: a slower bus takes longer. Students who scale the 9 hours by 45/50, as in a direct proportion, get a little over 8 hours, a result that should feel wrong for a slower trip. Finding the distance first turns the question into one division with no direction to decide.
No obvious method moderate Working it out
The question never mentions the trip's length, yet the 450-mile distance is what stays fixed; setting up a proportion between hours and speeds, the usual reflex, points the wrong way. (Schoenfeld, 1985)
Something you must already know minor Working it out
Students must bring distance = speed × time themselves, to turn 9 hours at 50 miles per hour into miles before dividing by the new speed. (Anderson & Krathwohl, 2001)
Strategy: Find the trip's length in miles from the first speed and time, since that length stays the same, then divide it by the new mean speed.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Treats time as proportional to speed, scaling 9 hours by 45/50 to 8.1, about 8, though a slower bus needs more time, not less.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 12 hours; it needs a slip such as 540 miles for 9 × 50, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Adds the 5-mile-per-hour drop in speed to the 9 hours, 9 + 5 = 14, combining numbers from the question without a model of the trip.
Worked solution, step by step
Formulas used
- Distance, speed and time
- d is the distance traveled. r is the rate, or speed, such as miles per hour. t is the time. It works because at 50 miles each hour, 9 hours covers 9 groups of 50 miles.
- Finding the time
- This is the same rule turned around. d is the distance and r is the speed. Divide the distance by the speed to find t, the number of hours. Use it when you know how far and how fast.
Sort out the question
A bus trip takes 9 hours at a mean speed of 50 miles per hour. The mean speed is the average speed over the whole trip. You need the time for the same trip at 45 miles per hour. The length of the trip does not change, so find it first.
Find the length of the trip
Use distance equals speed times time, with a speed of 50 miles per hour and a time of 9 hours:
The trip is 450 miles long.
Find the new time
Now the bus goes 45 miles per hour over the same 450 miles. Divide the distance by the speed:
The trip takes 10 hours.
Check that it makes sense
A slower bus needs more time for the same trip. 10 hours is more than 9 hours, so it fits.
Watch out: time does not shrink with speed. Scaling 9 hours by 45/50 gives about 8 hours, choice A, which would make the slower bus arrive sooner. Adding the drop of 5 miles per hour to 9 hours gives 14, choice D, which has no reason behind it.
So the answer is B, 10 hours.
Check your answer
Check the distance both ways. At 50 miles per hour for 9 hours, the bus goes 50 × 9 = 450 miles. At 45 miles per hour for 10 hours, it goes 45 × 10 = 450 miles. Same trip, same distance.
More questions like this
The original DOE question

Question 42 of 120
2026 Practice Test D Math, question 32Question
There were 36 people at Emily’s party. If each person at the party drank 2 cups of juice, how many gallons of juice did they drink altogether?
Note: 1 gallon = 4 quarts; 1 quart = 4 cups.
- A
- B
- C9
- D72
Show the answer
Correct answer: B
Since 36 people each drank 2 cups of juice, the total amount of juice they drank was cups.
Convert 72 cups to gallons:
Question analysis
At Emily's party 36 people each drank 2 cups of juice, and this unit conversion question asks how many gallons they drank in all, given 4 quarts to a gallon and 4 cups to a quart. Students multiply to get the total in cups, then convert through quarts to gallons, and the bold word gallons sets the unit of the answer.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Converting units with given factors is grade 6 work. This adds a first step, the total number of cups, and a two-stage conversion through quarts, but the numbers are friendly and the path is routine for grade 7 students.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Many steps minor
- Turning words into math minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 32 words.
- Corrected from the DOE handbook
- A stray word at the end of the explanation was removed.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
With both conversion facts printed, the question is mainly about keeping the units straight. Students who skip the word each convert 36 cups instead of 72, and those who stop at cups find a total that is right in the wrong unit. Working out that a gallon is 4 × 4 = 16 cups before dividing turns two conversions into one and avoids adding the 4s.
Many steps minor Working it out
A multiplication, 36 × 2, comes before two divisions by 4, cups to quarts and quarts to gallons, and stopping after the first step leaves a believable 72. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The word “each” turns 36 people into 72 cups, and the bold word gallons sets the unit of the answer, a change the note supplies in two steps. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Find the total number of cups first, then divide by 4 to get quarts and by 4 again to get gallons, or work out how many cups make a gallon and divide once.
Traps
- Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
The wrong answers, classified
A Misreads the question (Hegarty, Mayer & Monk, 1995)
Converts 36 cups to gallons, 36 ÷ 16 = 2 1/4, missing the word each, so every person’s second cup of juice is left out.
C Units or scale (King et al., 2004)
Uses 8 cups to a gallon, adding the two 4s in the note instead of multiplying them, so 72 cups become 72 ÷ 8 = 9 gallons.
D Units or scale (King et al., 2004)
Finds the 72 cups of juice and stops, giving the amount in cups though the bold word gallons asks for it in gallons.
What was corrected from the DOE handbook, and why
- Explanation: The DOE’s explanation ends with a line that reads only “lineExtend”, a formatting code left in the document by mistake. The site leaves it out.
Worked solution, step by step
Formulas used
- Total amount
- When every person has the same amount, multiply the number of people by the amount each one has. The word each tells you to multiply.
- Cups in a gallon
- A gallon holds 4 quarts, and each quart holds 4 cups. So a gallon holds 4 groups of 4 cups. You multiply, not add, because each of the 4 quarts has its own 4 cups.
- Changing cups to gallons
- It takes 16 cups to make 1 gallon. Dividing by 16 finds how many groups of 16 cups there are. A gallon is bigger than a cup, so the number of gallons is smaller.
Sort out the question
You know:
- 36 people were at the party.
- Each person drank 2 cups of juice.
- 1 gallon = 4 quarts, and 1 quart = 4 cups.
You need the total amount of juice in gallons. The bold word tells you the unit of the answer.
Find the total in cups
Each of the 36 people drank 2 cups:
They drank 72 cups in all.
Watch out: skipping the word each uses only 36 cups. Then gallons, choice A.
Find the cups in a gallon
A gallon is 4 quarts, and each quart is 4 cups:
So 1 gallon holds 16 cups.
Watch out: adding the two 4s gives 8 cups in a gallon. Then , choice C.
Change cups to gallons
Divide the cups by 16. Simplify the fraction by dividing the top and bottom by 8:
They drank gallons of juice.
Watch out: 72 is the amount in cups. The question asks for gallons, so 72, choice D, is in the wrong unit.
So the answer is B, .
Check your answer
Work backward. gallons is quarts. 18 quarts is cups. Shared among 36 people, that is cups each. It matches.
More questions like this
The original DOE question

Question 43 of 120
2026 Practice Test D Math, question 37Question
What 3-digit number between 630 and 650 is divisible by both 5 and 8?
Enter your answer in the space.
Fill-in question: work out the answer and type it as a number.
Show the answer
Correct answer: 640
The final digit of the three-digit number is 0. Each of the digits 0 through 9, inclusive, must be considered.
To narrow the choices, first consider that the number between 630 and 650 must be divisible by both 5 and 8.
A number divisible by 5 must end in 0 or 5. Therefore, the only possibilities are 0 and 5.
-
If the number ended in 0, the whole number 640 must also be divisible by 8:
Both result in a remainder of 0.
-
If the number ended in 5, the whole numbers 635 and 645 must also be divisible by 8:
Dividing 635 by 5 results in a remainder of 0, but dividing 635 by 8 results in a remainder of 3.
Dividing 645 by 5 results in a remainder of 0, but dividing 645 by 8 results in a remainder of 5.
Therefore, 640 is the only whole number between 630 and 650 that is divisible by both 5 and 8.
Question analysis
This number question asks for the 3-digit number between 630 and 650 that is divisible by both 5 and 8, typed into a box. Students use divisibility rules or common multiples to find it. The work is short, but the number has to pass both tests, so checking only one divisor is the easy slip.
- NYS standards
- 6.NS.4 common factors and multiples
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Common multiples are grade 6 content. Divisibility by 5 limits the last digit, and only a few numbers lie between 630 and 650, so this is a quick check, shorter than typical grade 7 work.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 16 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Few numbers need checking, so most students succeed; the risk is testing only one divisor. Students who start from multiples of 8 meet 632 first, and those who start from multiples of 5 can slip on 645 ÷ 8. Using the least common multiple, 40, shows at once that the number must be a multiple of 40, which leaves one candidate in the range.
Much to keep track of minor Working it out
Each candidate must pass two tests, divisible by 5 and by 8, and both ends of the range, 630 and 650, pass the first test but fail the second. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The shortcut that a multiple of 5 ends in 0 or 5 is assumed, and it cuts the search to 635, 640 and 645 before any division by 8. (Anderson & Krathwohl, 2001)
Strategy: Use divisibility by 5 to limit the last digit, then check only those few numbers in the range for divisibility by 8.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Stops one step early: A choice is a number from partway through the solution.
The wrong answers, classified
632 Misreads the question (Hegarty, Mayer & Monk, 1995)
Takes the first multiple of 8 after 630, 8 × 79 = 632, and stops without checking divisibility by 5, which a number ending in 2 fails.
645 An arithmetic slip (Junker & Sijtsma, 2001)
Picks a number ending in 5 so that 5 divides it, then slips on the check for 8, which leaves a remainder of 5.
40 Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the least common multiple of 5 and 8, 40, and enters it, one step before finding its multiple between 630 and 650.
Worked solution, step by step
Formulas used
- Divisible by 5
- Divisible means it divides evenly, with no remainder. A whole number is divisible by 5 only when its last digit is 0 or 5. That is because counting by 5s always lands on a number ending in 0 or 5.
- Least common multiple
- The least common multiple, or LCM, is the smallest number that both 5 and 8 divide evenly. 5 and 8 share no factor except 1, so their LCM is their product. Any number divisible by both is a multiple of 40.
Sort out the question
You need a 3-digit number between 630 and 650. It must be divisible by 5 and also by 8. Between means 630 and 650 themselves do not count.
Use the rule for 5
A number divisible by 5 ends in 0 or 5. Between 630 and 650, only three numbers do:
Only these need checking for 8.
Test each number for 8
Divide each one by 8 and look at the remainder, the amount left over:
- 635 ÷ 8 = 79, with 3 left over.
- 640 ÷ 8 = 80, with nothing left over.
- 645 ÷ 8 = 80, with 5 left over.
Only 640 divides evenly by 8.
Watch out: 645 passes the test for 5 but fails for 8. And 632, which is , passes for 8 but ends in 2, so it fails for 5.
Enter the number
640 passes both tests. Type the whole number in the box.
Watch out: 40 is the smallest number divisible by 5 and 8, but it is not between 630 and 650.
So you would enter 640.
Check your answer
Use the LCM. A number divisible by both 5 and 8 must be a multiple of 40. . The multiples next to it are 600 and 680, which are outside the range. So 640 is the only one.
More questions like this
The original DOE question

Question 44 of 120
2026 Practice Test D Math, question 39Question
The table shows the probabilities of a hockey team scoring different numbers of goals in a game.
HOCKEY GOAL PROBABILITIES
| Number of Goals | Probability |
|---|---|
| 0 | 0.16 |
| 1 | 0.21 |
| 2 | 0.29 |
| 3 | 0.21 |
| 4 | 0.09 |
| 5 | 0.04 |
What is the probability that the team will score three or more goals in the next game?
- A0.13
- B0.21
- C0.34
- D0.50
Show the answer
Correct answer: C
To find the probability that any one of several mutually exclusive events occurs, use the addition rule, and add the probabilities of each event:
Using the data in the table, the probability that the team will score 3 or more goals in the next game is represented by
Question analysis
A table gives the probabilities of a hockey team scoring 0 to 5 goals in a game, and students find the chance that it scores three or more goals in the next one. This probability question tests adding the probabilities of separate outcomes. Reading exactly which rows the phrase covers is the main point of care.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 8, 40 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Adding three decimals is easy, so the question turns on which rows the words cover. Including the 3-goal row, and not stopping at it, are the two decisions, and each misreading gives a listed choice. Checking the complement, 1 − (0.16 + 0.21 + 0.29), confirms both the sum and the choice of rows.
Turning words into math minor Understanding the question
The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.
Traps
- Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
A Counting the ends (Brown & Burton, 1978)
Adds only the rows for 4 and 5 goals, 0.09 + 0.04 = 0.13, reading three or more as more than three and leaving out the end row.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reads only the row for exactly 3 goals, 0.21, overlooking the words “or more” and the two rows below it.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Counts 3 of the 6 rows, for 3, 4 and 5 goals, as half the outcomes, ignoring that the table gives them different probabilities.
Worked solution, step by step
Formulas used
- Adding probabilities
- A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
- The chance it does not happen
- The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.
Sort out the question
The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.
Pick the right rows
Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:
- 3 goals: 0.21
- 4 goals: 0.09
- 5 goals: 0.04
Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.
Add the chances
A game can end with only one number of goals. So add the three chances:
Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.
So the answer is C, 0.34.
Check your answer
Find the chance of fewer than 3 goals, the other rows: . Then the chance of 3 or more is . Both ways give 0.34.
More questions like this
The original DOE question

Question 45 of 120
2026 Practice Test D Math, question 41Question
The distance an eagle flies is proportional to the time the eagle spends flying. An eagle flies 90 miles in 3 hours. What is the constant of proportionality for this relationship?
- A3
- B30
- C60
- D90
Show the answer
Correct answer: B
The constant of proportionality is the ratio of y (90) to x (3). . So the constant of proportionality is 30.
Question analysis
An eagle flies 90 miles in 3 hours, and the distance it flies is proportional to its flying time. This proportional relationship question asks for the constant of proportionality. Students need to know that the constant is the unit rate, here miles per hour, and find it from the two values given.
- NYS standards
- 7.RP.2b constant of proportionality
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 students find constants of proportionality in tables, graphs and equations. This one is a single division of miles by hours given in a sentence, the same unit rate work students do in grade 6.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 7, 29 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One division answers the question, so it is among the easiest on the test. The wrong choices are the two numbers in the stem and one unrelated value, so a student who knows what the constant means is not tempted. Checking that the constant times 3 hours gives back the 90 miles confirms it.
Something you must already know minor Working it out
Students must know that the constant of proportionality is the unit rate, distance over time, here the miles flown in a single hour; the stem gives only the term. (Anderson & Krathwohl, 2001)
Strategy: Divide the distance by the time, because the constant of proportionality is the distance flown in one hour, then check by multiplying it by the 3 hours.
Traps
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Gives the 3 hours printed in the stem as the constant, instead of the miles flown in each hour.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No working from 90 miles and 3 hours leads to 60, which may echo the 60 minutes in an hour, so it mostly attracts guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Gives the 90 miles, the distance for the whole 3-hour flight, as if the eagle covered it in a single hour.
Worked solution, step by step
Formulas used
- Proportional relationship
- x is one amount, here the time in hours, and y is the other, here the distance in miles. k is the constant of proportionality. It is the same for every pair, and it equals y divided by x. Here it is the miles flown in 1 hour.
Sort out the question
The eagle flies 90 miles in 3 hours. Proportional means the distance is always the same number times the time. That number is the constant of proportionality. You need to find it.
Divide the distance by the time
Use the formula. The distance is y and the time is x:
The constant is 30. It means the eagle flies 30 miles in each hour.
Avoid the numbers in the question
Watch out: 3 is the number of hours, choice A. And 90 is the distance for all 3 hours, choice D. Neither is the distance for 1 hour.
So the answer is B, 30.
Check your answer
Multiply back. At 30 miles each hour, the eagle flies miles in 3 hours. That matches the question. Hour by hour, it goes 30, 60, then 90 miles.
More questions like this
The original DOE question

Question 46 of 120
2026 Practice Test D Math, question 43Question
Two student clubs are selling snacks as part of a fundraiser. The Art Club sells a 24-pack of granola bars for $12.00. The Science Club sells a 10-pack of granola bars for $6.50. To the nearest cent, what is the difference in cost per bar between the two clubs?
Select the correct answer from each drop-down to complete the sentences.
The cost for the Art Club’s bar is per bar.
The cost for the Science Club’s bar is per bar.
The difference between the cost of the bars is .
Choose an answer from each menu, then check. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.
Show the answer
Correct answer: blank 1: $0.50; blank 2: $0.65; blank 3: $0.15
To determine the cost per bar from the Art Club, divide the total cost by the total number of bars:
To determine the cost per bar from the Science Club, divide the total cost by the total number of bars:
To find the difference in price per bar, subtract the cost per bar for the Art Club from the cost per bar for the Science Club:
Question analysis
Two clubs sell granola bars for a fundraiser: the Art Club a 24-pack for $12.00 and the Science Club a 10-pack for $6.50. Students complete three sentences with drop-down menus, giving each club's cost per bar and the difference between them. It tests unit prices with money, and all three menus must be right.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Comparing unit prices is grade 6 rate work. The menus split it into three short steps, two divisions of money amounts and one subtraction, which is lighter than typical grade 7 work.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Many steps minor
- Error-prone arithmetic minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 3, 87 words.
- Corrected from the DOE handbook
- The order of the answers in the DOE’s key was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The menus guide the work, one sentence per step, and every option is a reasonable price, so the question rewards accurate division more than planning. No common misconception leads to the wrong options; they catch slips, such as misreading a pack size or subtracting carelessly. Checking that 24 bars at the first price cost $12.00 and 10 bars at the second cost $6.50 confirms both menus before the subtraction.
Many steps minor Working it out
Three menus must all be right to score: two divisions for the costs per bar, then a subtraction that uses both, so a slip in either division spoils the third menu. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Each division is with money, $12.00 ÷ 24 and $6.50 ÷ 10, so the decimal point must be placed carefully, though the menus offer only sensible amounts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Divide each pack's price by the number of bars in it, then subtract the smaller cost per bar from the larger; checking that 24 bars at your first price make $12.00 catches slips.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
The wrong answers, classified
CBA Misreads the question (Hegarty, Mayer & Monk, 1995)
Divides $12.00 by 20 instead of 24, misreading the size of the Art Club’s pack, so $0.60 a bar leaves a gap of only $0.05.
BBB An arithmetic slip (Junker & Sijtsma, 2001)
Finds both costs per bar correctly, then slips in the subtraction, taking $0.65 − $0.50 as $0.10, the option beside the right one.
What was corrected from the DOE handbook, and why
- Answer key: The DOE’s explanation opens with the key “($0.65; $0.50; $0.15)”, which puts $0.65 in the first drop-down and $0.50 in the second. Neither is possible: the first menu (the Art Club’s cost per bar) offers $0.40, $0.50, $0.60 and $0.75, and the second (the Science Club’s) offers $0.55, $0.65, $0.75 and $0.85. The explanation’s own working gives the Art Club $12.00 ÷ 24 = $0.50 and the Science Club $6.50 ÷ 10 = $0.65, so the key is $0.50, $0.65, $0.15. The site’s key was already in that order; the DOE’s key line is not shown on the site.
Worked solution, step by step
Formulas used
- Cost of one item
- The price of the pack is what you pay for all the bars together. Dividing shares that price equally among the bars. The result is the price of one bar, called the unit price.
Sort out the question
You know:
- The Art Club sells 24 bars for $12.00.
- The Science Club sells 10 bars for $6.50.
There are three menus. The first is the Art Club’s cost per bar. The second is the Science Club’s. The third is the difference, how much more one bar costs than the other.
Find the Art Club’s cost
Share $12.00 among 24 bars:
Each bar costs $0.50. That makes sense: 24 half-dollars make $12. Pick $0.50 in the first menu.
Watch out: dividing by 20 instead of 24 gives $0.60, which is also in the menu.
Find the Science Club’s cost
Share $6.50 among 10 bars. Dividing by 10 moves the decimal point one place to the left:
Each bar costs $0.65. Pick $0.65 in the second menu.
Find the difference
Subtract the smaller cost from the larger one:
The bars differ by $0.15. Pick $0.15 in the third menu.
Watch out: $0.10 sits right next to $0.15 in the menu. A slip in the subtraction lands there.
Fill in the three menus
On the test and on this site, you choose each answer from a drop-down menu.
So the menus read $0.50, $0.65 and $0.15.
Check your answer
Multiply back. and , so both costs per bar match the pack prices. And , so the difference fits too.
More questions like this
The original DOE question

Question 47 of 120
2026 Practice Test D Math, question 49Question
Anise has a piece of cloth measuring 45 inches by 75 inches. She wants to cut it into squares of equal size, without any cloth left over. What is the greatest possible side length, in inches, of one of those squares?
- A3
- B5
- C15
- D45
Show the answer
Correct answer: C
Given that each piece will be a square, the length and width will be the same. The greatest possible length can be determined by finding the greatest common factor of the length and width of the original piece of cloth.
Begin by finding the prime factorization of the length (75 in.) and width (45 in.).
The common factors can be seen as .
The greatest possible side length, in inches, will be .
Question analysis
Anise wants to cut a 45-inch by 75-inch piece of cloth into equal squares with nothing left over. This factors question asks for the greatest possible side length of each square, in inches. Students must see that the side has to divide both lengths evenly, then find the largest number that does.
- NYS standards
- 6.NS.4 common factors and multiples
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Greatest common factors of two numbers up to 100 are grade 6 content. The cloth setting asks students to recognize that this is what is needed, but the work is one short factor search, below typical grade 7 demand.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 4, 39 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Once the side is seen as a common factor, the work is short, since 45 and 75 share both 3 and 5. Students who spot one shared factor and stop reach a listed choice, and those who think only of the largest square that fits use the full 45-inch width. Checking that 45 and 75 both divide by the chosen side, and that no larger number does, closes it.
Turning words into math minor Understanding the question
The words “squares of equal size, without any cloth left over” mean the side must divide both 45 and 75 exactly; nothing names the greatest common factor. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding the greatest common factor of 45 and 75 calls for prime factorization or a list of factors, a grade 6 skill the question assumes. (Anderson & Krathwohl, 2001)
Strategy: Write each length as a product of primes, multiply the primes they share, and check that both 45 and 75 divide evenly by the result.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds a factor that 45 and 75 share, 3, and stops without combining it with the 5 they also share.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Spots the common factor 5, since both lengths end in 5, and stops before multiplying it by the 3 they also share.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Uses the cloth’s full width, 45 inches, as the side, the largest square that fits, but 45 does not divide 75, leaving a 45-by-30 strip.
Worked solution, step by step
Formulas used
- Greatest common factor
- A factor divides a number evenly. The greatest common factor, or GCF, is the biggest number that divides both. Write each number as a product of primes, numbers whose only factors are 1 and themselves. Then multiply the primes they share.
Sort out the question
The cloth is 45 inches by 75 inches. It is cut into equal squares, with no cloth left over. You need the greatest side the squares can have.
No cloth left over means the side must fit evenly into 45 and into 75. So the side is a common factor of both. The greatest one is the GCF.
Break each number into primes
Write each length as a product of prime numbers:
The two lists share one 3 and one 5.
Multiply the shared primes
Multiply the primes the two numbers share:
The squares can be 15 inches on a side.
Watch out: 3 and 5 are each common factors, choices A and B. But they are not the greatest. Use every prime the numbers share.
Rule out the full width
Watch out: a 45-inch square, choice D, fits the width. But 75 ÷ 45 = 1 with 30 left over. That leaves a strip of cloth 45 inches by 30 inches.
So the answer is C, 15.
Check your answer
Divide each length by 15: and . So the cloth cuts into 3 rows of 5 squares, with nothing left. Both divisions come out even.
More questions like this
The original DOE question

Question 48 of 120
2025 Practice Test A English, question 23Passage Excerpt from Winter Wheat
Excerpt from Winter Wheat
by Mildred Walker
Ellen, the narrator, is preparing to leave her parents and the family farm for college the next day.
I love Dad’s way of talking that makes him seem different from other ranchers. He’s lived here twenty-three years, but he still says “back East where I come from.” He’s the one who gets excited when I do about spring coming or a serial running in the magazine we’re both reading, but it’s what Mom says that I depend on. When Mom used to say “Don’t worry” about my pet chicken or dog or new calf, it always got well. Dad is always talking of going some place, not now, but next year, maybe. Mom seems to think of nothing farther away than today or perhaps yesterday or tomorrow morning.
Mom folded the ironing board and put it inside their bedroom that was just off the kitchen. She carried in the freshly ironed clothes. Dad went back to his paper. When Mom came back she took beans from the cupboard to soak for tomorrow. Dad always said Mom could make all the dishes he’d had back in Vermont as well as though she were a New Englander herself, instead of a Russian. All of a sudden, I realized that tomorrow when those beans would be ready to eat I’d be going away. It gave me a funny feeling.
“I’ll be taking the train tomorrow night,” I said aloud, more to hear it myself.
“We can drive you into town in the afternoon,” Dad said, dropping his paper on the floor.
“There’s no need to go to town; she can catch the train at Gotham just as well. We haven’t nothing to take us into town for,” Mom said.
“Well, we don’t have to decide tonight,” Dad said, but I knew he wanted to go into Clark City. It wouldn’t be so flat as just seeing me go off on the train from Gotham. My going away was hard on both of them; they were so different—and I was part of them both. It made me uncomfortable to think of leaving them.
While I was getting ready for bed in my room that’s off the front room, I saw how it would be if I left from town. We’d go in right after dinner and go around to the stores, Dad going one way and Mom and I another. Dad would probably have his hair cut at the barbershop and stop in the bank and meet someone he knew to talk to. Then we’d meet at the big store on the corner and go to the cafeteria for supper. The train stops ten minutes or so at the station in town and there are other people and excitement and you have time to wave from the platform and then again from your window by your seat. We went to the station in Clark City to see the Goodals off when they went back to Iowa.
If I left from Gotham, we’d just drive down in the truck and wait till the train came. It only stops long enough for you to get on and you hardly have time to taste the flavor of going away.
I sat on the bed in my pyjamas with my arms around my knees. I couldn’t keep from thinking of that time Dad went back East. I tried to, and then I just sat still and looked straight at it. Sometimes that’s better than working so hard to keep from looking at what’s in your mind.
Dad went all the way back to Vermont. . . . It was in November and it was already dark when the train came through Gotham. Even now, I could feel how cold and dark it was. I held Mom’s hand. Dad was so dressed-up he seemed strange. . . . We stood there without saying anything until Dad told Mom to remember to call Mr. Bardich, our neighbor, if the cow didn’t calve tomorrow.
“I’ll manage,” Mom snapped back.
“I wish you could go, Anna,” Dad said to Mom, “and we could take Ellen.” . . .
“Good-by, Anna Petrovna,” he said, looking at Mom. I had never heard him call her by two names before.
“Good-by,” Mom said, standing still, without smiling.
Then he was gone and the crossroads were darker than ever. The train light shone on the high window in the top of the grain elevator for a moment and then that too was dark. We got into our old Ford and Mom drove back to the house. My throat ached all the way. The name Dad had called Mom kept saying itself in my ears: “Anna Petrovna, Anna Petrovna.” . . .
Our house seemed lonely when we came back to it. It seemed to be hiding under the coulee. I went with Mom to put the truck in the barn that was bigger than the house. I think Mom was prouder of our barn than the house, anyway. We walked back to look at the cow that was going to calve. She was just a big light blob in the dark, waiting. I had thought she was exciting this morning, but now she seemed sad, too.
The wind blew when we walked across the open space to the house and I couldn’t help shivering with the cold. Inside the house it was warm, but empty.
“Bring your nightgown in here and I heat you some milk,” Mom said.
I drank the milk sitting on a stool in front of the stove. It tasted good, but the lonely ache in my throat was still there. I picked up my clothes and hung them neatly behind the stove and put my cup on the sink board. Mom was fixing oatmeal for tomorrow morning.
“Good night, Mom,” I said almost timidly, standing beside her. She seemed wrapped around in a kind of strangeness. Then she turned around and drew me to her. The front of her dress was warm from the stove. I felt the comfortable heat through my gown. She laid her hand against my face and it felt rough and hard but firm. I dared ask her something I wanted to know.
“Mom, was that really your name—what Dad called you?”
Her voice sounded surprised. “Why, Yeléna, you know that; Anna Petrovna. You know I am born in Russia, in Seletskoe.”
“Yes, but I didn’t know your other name,” I said.
“Anna Petrovna Webb.” She pronounced it slowly. “Once I think what a funny name Ben Webb is!” She laughed. Her laugh was warm and low like our kitchen, and comfortable. The house seemed natural and right again. . . .
But now that I am grown, I feel the wall of strangeness between them, more than when I was a child. I wondered how they would get along without me.
From WINTER WHEAT by Mildred Walker, published by University of Nebraska Press. Copyright © 1944 by Harcourt, Brace and Company, Inc. Copyright renewed 1971 by Mildred Walker. All rights reserved.
serial: story published in short segments at regular intervals
coulee: small gulch or ravine
Question
Which sentence from the excerpt provides evidence that Mom wants Ellen to understand the family’s heritage?
- A“Mom folded the ironing board and put it inside their bedroom that was just off the kitchen.” (paragraph 2)
- B“ ‘There’s no need to go to town; she can catch the train at Gotham just as well.’ ” (paragraph 5)
- C“We’d go in right after dinner and go around to the stores, Dad going one way and Mom and I another.” (paragraph 7)
- D“ ‘Why, Yeléna, you know that; Anna Petrovna.’ ” (paragraph 22)
Show the answer
Correct answer: D
The question asks which sentence from the excerpt provides evidence that Mom wants Ellen to understand the family’s heritage.
A. Incorrect. Although this sentence from paragraph 2 describes a domestic moment that takes place within the family home, it does not relate to the heritage of either parent and therefore does not provide evidence that her mother wants Ellen to understand the family’s heritage.
B. Incorrect. Although this sentence from paragraph 5 provides the name of the train station closest to Ellen’s family’s farm, the train stop at Gotham is not related to the family’s heritage. Therefore, this sentence does not provide clear evidence that her mother wants Ellen to understand the family’s heritage.
C. Incorrect. Although this sentence from paragraph 7 describes an aspect of the family dynamic (Ellen predicts that while browsing in stores in town, “Dad [would go] one way and Mom and I another”), their shopping habits do not provide clear evidence that her mother wants Ellen to understand the family’s heritage.
D. CORRECT. In this sentence from paragraph 22, Ellen’s mother addresses her by the Russian version of her name (Yeléna) and repeats her own Russian name (Anna Petrovna). The choice to call her daughter Yeléna instead of Ellen follows immediately after Ellen asks, “Mom, was that really your name—what Dad called you?” (paragraph 21), and the mother’s surprised response provides evidence that she both wants and expects Ellen to understand the family’s Russian heritage.
Question analysis
This evidence question asks which quoted sentence from the Winter Wheat excerpt shows that Mom wants Ellen to understand the family’s heritage. The four sentences, from paragraphs 2, 5, 7 and 22, all involve Mom, so students must reread each one in its paragraph to see what it really shows about her.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Finding the sentence that supports a stated idea is everyday grade 7 work. The quotations are short and plainly worded and the idea is concrete, so one careful rereading of each in context is enough, a lighter task than usual.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly evaluate
- What makes it hard
- Evidence
- Answered from several paragraphs or stanzas
- Reading level
- Passage: about grade 4, 1179 words; question and choices: about grade 8, 83 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Only one of the four sentences touches the family’s origins at all, so a student who knows what heritage means can rule out the others at a glance. The small risk lies with students who match “Mom wants” to any line where Mom states a preference, or who do not recognize the Russian names, since the quotation stops before Mom mentions Russia.
Reading between the lines minor Working it out
The quoted line never mentions heritage; students must see that Mom calling Ellen “Yeléna” and repeating “Anna Petrovna,” as things Ellen should already know, shows her wish to pass on their Russian roots. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Evidence in several places minor Working it out
The names’ Russian origin is clearest outside the quotation, in the rest of paragraph 22, where Mom names her birthplace, and in paragraph 2’s “instead of a Russian.” (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)
Strategy: Define heritage in your own words, then reread each quoted sentence in its paragraph and ask whether it shows Mom handing something about the family on to Ellen.
Traps
- True but not the answer: A choice says something accurate about the text that does not answer this question.
- Supports only part of the claim: In a question about evidence: a line that backs one half of the claim but not the other, or a related point rather than the claim itself.
The wrong answers, classified
A True but not the answer (King et al., 2004; Embretson & Wetzel, 1987)
Picks an everyday moment of Mom putting away the ironing board, perhaps remembering that paragraph 2 also mentions her Russian roots, though this sentence shows only housework.
B Supports only part of the claim (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Matches “Mom wants” to Mom’s spoken preference that Ellen catch the train at Gotham, which backs half the idea, a wish of Mom’s, but concerns travel, not heritage.
C True but not the answer (King et al., 2004; Embretson & Wetzel, 1987)
Notices Mom and Ellen shopping together in Ellen’s imagined trip to town, a sign of closeness, but going around the stores says nothing about where the family comes from.
Working it out, step by step
Figure out what the question asks
This asks which sentence shows that Mom wants Ellen to understand the family’s heritage. Heritage is the history and traditions a family passes down, such as where it came from. Look for a sentence where Mom shares something about the family’s roots.
Look at the conversation
At the train, Dad called Mom Anna Petrovna, a name Ellen had never heard him use. In paragraph 21, Ellen asks, “Mom, was that really your name—what Dad called you?” In paragraph 22, Mom answers, “Why, Yeléna, you know that; Anna Petrovna.” Then she reminds Ellen that she was born in Russia.
Work out what Mom’s answer shows
Mom calls her daughter Yeléna, the Russian form of Ellen. She repeats her own Russian name. She also says “you know that,” as if Ellen should already know. Mom is surprised that Ellen doesn’t. That shows Mom wants and expects Ellen to know where the family comes from.
Check each choice
- A does not fit. Folding the ironing board is housework. It says nothing about the family’s past.
- B does not fit. Mom does want something here, the train at Gotham. But that is about travel, not heritage.
- C does not fit. Shopping in town shows family habits, not family roots.
- D fits. Mom uses Russian names and expects Ellen to know them.
Watch out: B tempts because it shows what Mom wants. But the question asks what she wants Ellen to understand about the family.
Choose the answer
Only one sentence has Mom passing on the family’s Russian roots. So the answer is D.
Why this answer and not the closest wrong one
B is the closest wrong choice, because Mom states a wish in it: Ellen can catch the train at Gotham. But that wish is about the trip, not the family’s past. In D, Mom uses the Russian names Yeléna and Anna Petrovna. She is surprised that Ellen doesn’t know them. That shows she wants Ellen to understand her heritage.
More questions like this
The original DOE question

The passage “Excerpt from Winter Wheat” as shown on the test







Question 49 of 120
2025 Practice Test A English, question 55Question
Which revision corrects the error in sentence structure in the paragraph?
The land on Earth has not always been separated into the seven continents, at one time a massive supercontinent, known as Pangaea, covered one-third of Earth’s surface. Additionally, the supercontinent was surrounded by ocean waters called Panthalassa, much of which were in Earth’s Southern Hemisphere. Geologists believe that the supercontinent split apart over millions of years because of the movement of the tectonic plates that form Earth’s crust. In fact, experts predict that over the next 250 million years the movement of the plates will cause the seven continents to merge into a supercontinent again.
- Acontinents. At
- Bsurface; additionally,
- CPanthalassa. Much
- Dcrust, in fact,
Show the answer
Correct answer: A
The question asks for the revision that corrects the error in sentence structure in the paragraph.
A. CORRECT. The first sentence of the paragraph is a run-on because both clauses—“The land on Earth has not always been separated into the seven continents” and “at one time a massive supercontinent, known as Pangaea, covered one-third of Earth’s surface”—are independent clauses. Independent clauses are complete sentences, and two complete sentences should not be combined with just a comma. The revision is correct because it separates the two sentences by placing a period where it is needed after “continents” and by capitalizing the word “At.”
B. Incorrect. While adding a semicolon after “surface” can be considered a correct way to separate the independent clauses “at one time a massive supercontinent, known as Pangaea, covered one-third of Earth’s surface” and “Additionally, the supercontinent was surrounded by ocean waters called Panthalassa, much of which were in Earth’s Southern Hemisphere,” this change does not address the sentence structure error in the paragraph (the run-on sentence). The use of a period separating the independent clauses (complete sentences) is correct as is.
C. Incorrect. Adding a period after “Panthalassa” would separate the dependent clause “much of which were in Earth’s Southern Hemisphere” from its antecedent: “ocean waters called Panthalassa.” The dependent clause cannot stand on its own, because it shares elements with the main clause. A dependent clause is not a complete sentence and must be joined with an independent clause. The clause beginning with “much of which” is an adjective clause that modifies “ocean waters.” Changing the comma to a period creates a new error (a sentence fragment) and does not address the existing sentence structure error at the beginning of the paragraph (the run-on sentence).
D. Incorrect. Removing the period after “crust” would create a run-on sentence because the period is needed to separate two complete sentences—“Geologists believe that the supercontinent split apart over millions of years because of the movement of the tectonic plates that form Earth’s crust” and “In fact, experts predict that over the next 250 million years the movement of the plates will cause the seven continents to merge into a supercontinent again.” A comma after “crust” would combine these sentences into a run-on, creating a new error in the sentence without fixing the existing sentence structure error (the run-on sentence) at the beginning of the paragraph.
Question analysis
This editing question shows a short paragraph about Pangaea, the supercontinent that once covered a third of Earth’s surface, and the ocean around it, with four underlined spots, and asks which revision corrects the error in sentence structure. Each revision changes only punctuation and capitals at one spot, so the decision rests on how the clauses around each spot fit together.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Fixing run-ons and fragments is taught well before grade 7, and this item tests that basic skill in one short paragraph. The sentences are long, with an appositive and a clause beginning “much of which,” but the grammar involved stays simple.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- A rule that doesn't sound wrong moderate
- Every choice must be checked minor
- Complex sentences minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Question and choices: about grade 9, 114 words.
- Corrected from the DOE handbook
- Two grammar terms in the explanation were corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The error is in the first sentence, but it hides behind a comma that sounds like an ordinary pause, while the later spots carry features that look riskier: a transition word, a clause beginning “much of which,” the phrase “in fact.” Students who test each spot for a complete sentence on both sides find the splice at once. Those who look for a revision that merely sounds polished are drawn to the semicolon, which is correct but fixes nothing.
A rule that doesn't sound wrong moderate Working it out
The comma after “continents” joins two complete sentences, and because “at one time” sounds like an introductory phrase continuing the thought, the splice reads as a natural pause. (Connors & Lunsford, 1988)
Every choice must be checked minor Choosing the answer
Each revision changes only punctuation at one underlined spot, and the semicolon one is correct English, so students must ask whether each spot was wrong to begin with, not only whether the change looks right. (Embretson & Wetzel, 1987)
Complex sentences minor Understanding the question
The paragraph’s sentences are long, with the appositive “known as Pangaea” and the clause “much of which were,” whose odd-sounding verb draws suspicion to a spot that is correctly punctuated. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Strategy: At each underlined spot, read the words on both sides and ask whether each side could stand as a sentence, then pick the revision that leaves every sentence correctly joined or separated.
Traps
- Fixes what isn't broken: Changes a part of the sentence that was already correct, while the real error stays; or, where the question asks which sentence has an error or should go, picks one that is correct and belongs.
- Fragment or run-on: Creates or keeps a sentence fragment, a run-on sentence or a comma splice.
The wrong answers, classified
B Fixes what isn't broken (Connors & Lunsford, 1988)
Knows a semicolon can join two clauses before “additionally” and picks that tidy-looking change, though the period it replaces was already correct and the comma splice up front remains.
C Fragment or run-on (Connors & Lunsford, 1988)
Treats the long second sentence as a run-on and splits it after “Panthalassa,” leaving “Much of which were in Earth’s Southern Hemisphere” as a fragment with no main clause.
D Fragment or run-on (Connors & Lunsford, 1988)
Thinks “in fact” belongs inside the sentence before it, set off by commas, which joins two complete sentences with only a comma and creates the error the paragraph already has.
What was corrected from the DOE handbook, and why
- Explanation: The DOE’s explanation of Option C says a period after “Panthalassa” would separate the dependent clause “much of which were in Earth’s Southern Hemisphere” “from its subject: ‘ocean waters called Panthalassa.’” The clause has a subject of its own, “much”; “ocean waters called Panthalassa” is its antecedent, the noun phrase that “which” refers to. The explanation also calls the clause “an adjective for ‘ocean waters’”; a clause is not an adjective but an adjective (relative) clause, which modifies the noun. The site now says “from its antecedent” and “an adjective clause that modifies ‘ocean waters.’” The reasoning and the answer, A, are unchanged.
Working it out, step by step
Learn the rule for joining sentences
An independent clause is a group of words with a subject and a verb that can stand alone as a sentence. Two of them can’t be joined with only a comma. That mistake is a run-on sentence, also called a comma splice. For example: I love soccer, my brother loves chess. Fix it with a period: I love soccer. My brother loves chess.
Test each underlined spot
At each spot, read the words on both sides. Ask: could each side stand alone as a sentence?
- After “continents”: both sides are complete. A comma joins them, so this is an error.
- After “surface”: a period already separates two sentences. That is correct.
- After “Panthalassa”: the words after the comma can’t stand alone. The comma is correct.
- After “crust”: a period already separates two sentences. That is correct.
Look closely at the first sentence
The first part says “The land on Earth has not always been separated into the seven continents.” That is a full sentence. The second part says “at one time a massive supercontinent, known as Pangaea, covered one-third of Earth’s surface.” Its subject is “supercontinent,” and its verb is “covered.” So it is a full sentence too. The words “At one time” start a new sentence.
Show the fix
Change the comma to a period, and capitalize the next word: The land on Earth has not always been separated into the seven continents. At one time a massive supercontinent, known as Pangaea, covered one-third of Earth’s surface. Now each sentence stands on its own. That is the revision continents. At.
Check each choice
- A is right. It splits the run-on into two sentences.
- B is wrong. A semicolon could join those two sentences, but the period there was already correct. It fixes nothing.
- C is wrong. “Much of which were in Earth’s Southern Hemisphere” is a dependent clause, a group of words with a verb that can’t be a sentence by itself. “Which” points back to “ocean waters called Panthalassa,” its antecedent. So the clause must stay attached.
- D is wrong. A comma after “crust” would join two full sentences. That makes a new run-on.
Choose the answer
Only the comma after “continents” joins two complete sentences. So the answer is A, continents. At.
Why this answer and not the closest wrong one
B is the closest wrong choice, because a semicolon can correctly join two sentences. But the spot after “surface” already had a period, which was correct. Changing it does not fix the real error. The comma splice after “continents” would still be there. Only A fixes it.
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Question 50 of 120
2025 Practice Test B English, question 15Passage A Miracle Mile
A Miracle Mile
In the 1950s people compared running one mile in four minutes to scaling Mount Everest and nicknamed the feat a “dream mile.” Although such an accomplishment was considered humanly impossible, several elite runners aimed to break that supposedly impenetrable barrier. One of them was a twenty-five-year-old medical student named Roger Bannister.
Roger Bannister had tasted failure during the 1952 Olympics. There, he was favored to win the 1,500-meter competition, a distance slightly shorter than a mile, but he finished in a dismal fourth place instead. Bannister’s performance was a disappointment for him and his country, Great Britain. Determined to redeem himself, Bannister postponed his plans to retire from racing and focused on the ultimate prize—breaking the four-minute-mile barrier.
Bannister attacked the elusive milestone with a positive attitude and logical planning. The amateur athlete decided to use intensive interval training to develop endurance and speed. For these workouts, Bannister ran an interval of ten consecutive laps on a quarter-mile track, aiming for sixty seconds each lap. In between intervals, he let his body recover for two minutes.
By early 1954, Bannister had succeeded in lowering his quarter-mile pace to sixty-one seconds, but he had to shave off at least one more second in order to reach his target. Frustrated by the plateau he had reached, Bannister took a break from training and went mountain climbing for three days. The rest from running permitted his muscles to recuperate and left him feeling refreshed. When Bannister returned to the track, he completed ten quarter-mile-long intervals at fifty-nine seconds each. He finally felt prepared to attempt to break the world record.
As a member of the Amateur Athletic Association (AAA), Bannister joined the AAA team for a track meet against Oxford University. The event took place on a cinder track in Oxford on May 6, 1954. Bannister and his two AAA teammates, Chris Chataway and Chris Brasher, were close friends and frequent running partners. Chataway and Brasher agreed to help Bannister accomplish his goal by being his “rabbits.”
In track and field, rabbits are runners who enter the race solely to pace a teammate for a segment of the course. Typically, a runner settles in behind the rabbit and allows the rabbit to set an appropriate tempo. Additionally, by running behind the rabbit, the runner conserves about 15 percent of his or her effort. When the starting pistol fired, Brasher pounced into the lead, and Bannister followed behind his first rabbit.
Propelled by the excitement, Bannister lost his instinctive feel for his pace and shouted “Faster!” at Brasher. Brasher, however, remained composed and maintained his current steady but grueling pace, completing the first two laps in a desirable one minute and fifty-eight seconds. Then Chataway surged forward, leading Bannister at this same punishing rate for another lap and a half. At the beginning of the back straightaway of the track, Bannister bolted past Chataway. Bannister said, “I felt that the moment of a lifetime had come. There was no pain, only a great unity of movement and aim.” Bannister crossed the finish line in 3 minutes 59.4 seconds. The ecstatic crowd erupted the moment the timekeeper announced the word “three.”
Soon after Bannister’s achievement, four other athletes matched his performance. A new mindset had taken root among runners. Over the years, the record continued to fall. However, the current record, 3 minutes 43.13 seconds, has stood unbroken since 1999. Some question whether this record represents the limits of human ability. But perhaps there is another Bannister, an athlete who, with willpower and dedication, will accomplish the miraculous.
Question
How does the author’s use of chronological structure contribute to the development of ideas in the passage?
- AIt presents the increasing physical effects of Bannister’s training methods as he prepared to break the four-minute-mile barrier.
- BIt shows the increase in Bannister’s confidence in his ability to break the four-minute-mile barrier.
- CIt emphasizes the key events in Bannister’s life that inspired him to break the four-minute-mile barrier.
- DIt highlights the progression of Bannister’s training and details about his successful attempt to break the four-minute-mile barrier.
Show the answer
Correct answer: D
The question asks how the author’s use of chronological structure contributes to the development of ideas in the passage.
A. Incorrect. While paragraph 4 mentions the “plateau” Bannister dealt with and the break that “permitted his muscles to recuperate,” the chronological structure of the passage focuses mainly on the progression of Bannister’s process over time.
B. Incorrect. Even though the passage suggests that Bannister’s ability to break the four-minute-mile barrier improved and that he felt “prepared to attempt to break the world record” (paragraph 4), the chronological structure of the passage primarily serves to show how Bannister’s progress over time led to his success.
C. Incorrect. The chronological structure of the passage begins with an acknowledgment that Bannister was inspired to break the four-minute-mile barrier by his loss at the 1952 Olympics (paragraph 2); however, the overall organization of the passage highlights his dedicated efforts over time to break the four-minute-mile barrier.
D. CORRECT. The chronological structure of the passage follows the progression of Bannister’s training from his loss at the 1952 Olympics to the race where he ran the first sub-four-minute mile on May 6, 1954 (paragraph 5).
Question analysis
This text-structure question asks how the chronological structure of “A Miracle Mile,” an article about Roger Bannister’s attempt to run a mile in under four minutes, contributes to the development of its ideas. The article moves from the 1950s to the present day, and each choice names a thread a reader might trace through it, so students must decide which one the time order really follows.
- Grade level
- Grade 7 reading and task
- Compared with grade 7
- Below grade 7 Grade 7 students analyze how an author organizes a text. Here the stem names the structure, and the article moves through dated events in plain prose, so seeing what the order brings out is lighter than most grade 7 structure questions.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- How, not what minor
- Every choice must be checked minor
- Evidence in several places minor
- Evidence
- Answered from the whole text
- Reading level
- Passage: about grade 10, 587 words; question and choices: about grade 13, 84 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The article tells Bannister’s story in order, and the right choice simply names what that order follows: training that improves, then the race itself. Each wrong choice picks a thread the text touches without tracing, a confidence present from the start, a physical toll mentioned once, an inspiration that comes only from 1952. Students who sum up each paragraph in a few words see that only the training and the race run all the way through.
How, not what minor Working it out
The stem names the structure and asks what it does, so students must say which thread the time order follows from the 1952 Olympics to May 6, 1954, not retell the events. (Anderson & Krathwohl, 2001)
Every choice must be checked minor Choosing the answer
Each wrong thread has a foothold: “a positive attitude” and “finally felt prepared” for confidence, muscles that needed to “recuperate” for physical effects, and the 1952 loss for inspiration. (Embretson & Wetzel, 1987)
Evidence in several places minor Working it out
A thread counts only if it runs from paragraph 2 to paragraph 7, so it must cover both the training of paragraphs 3 and 4 and the race told in paragraphs 5 to 7. (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)
Strategy: Note in a few words what each paragraph covers, in order, then choose the description that matches what the whole sequence follows rather than one detail along the way.
Traps
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
- Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
The wrong answers, classified
A Twists what the text says (King et al., 2004)
Recalls the plateau and the muscles that needed to “recuperate” in paragraph 4, but the article tracks lap times falling from sixty-one to fifty-nine seconds, not a physical toll that grows.
B Reads too much in (Graesser, Singer & Trabasso, 1994)
Follows Bannister from “Frustrated by the plateau” to “finally felt prepared” and “the moment of a lifetime,” but he began with “a positive attitude,” and the order traces training, not confidence.
C Too narrow (King et al., 2004)
Starts from the 1952 Olympic loss that left him “Determined to redeem himself,” but that single event belongs to paragraph 2, and the later paragraphs follow training and a race, not further inspirations.
Working it out, step by step
Figure out what the question asks
This is a structure question. Chronological structure means telling events in time order, from first to last. The question asks what that order helps the passage show. So sum up each paragraph in a few words. Then see which thread runs through all of them.
Sum up each paragraph in order
- Paragraph 1: In the 1950s, a four-minute mile seemed impossible.
- Paragraph 2: Bannister loses at the 1952 Olympics and sets his goal.
- Paragraph 3: He makes a training plan.
- Paragraph 4: By early 1954, his laps get faster.
- Paragraph 5: A race is set for May 6, 1954. Two teammates will set his pace.
- Paragraphs 6–7: The race is run. He finishes in 3 minutes 59.4 seconds.
- Paragraph 8: Other runners match him, and over the years the record is lowered.
See what the order follows
Most of the passage follows two things. First, his training improves step by step. The track is a quarter mile, so four laps make a mile. Sixty seconds a lap would be exactly four minutes. He goes from sixty-one seconds a lap to fifty-nine, so the goal comes within reach. Second, the passage gives details of the race. It is set up in paragraph 5 and run in paragraphs 6 and 7. Progression means step-by-step progress.
Check each choice
- A does not fit. His body comes up only now and then, as when a rest “permitted his muscles to recuperate” in paragraph 4. Recuperate means recover. The passage never tracks effects that grow, and A leaves out the race.
- B does not fit. He had “a positive attitude” from the start, in paragraph 3. The order follows what he did, not how sure he felt.
- C does not fit. Only one event inspired him, the 1952 loss in paragraph 2. After that come training and the race.
- D fits. The order traces his training and then the record-breaking race.
Watch out: B is tempting because he “finally felt prepared” in paragraph 4. But that is one moment, not the thread of the whole passage.
Choose the answer
The time order follows Bannister’s training as it improves, then the race where he broke the barrier. So the answer is D, “It highlights the progression of Bannister’s training and details about his successful attempt to break the four-minute-mile barrier.”
Why this answer and not the closest wrong one
B is the closest wrong choice. Bannister does end up ready, and he “finally felt prepared” in paragraph 4. But he began with “a positive attitude,” so his confidence does not build from low to high. The paragraphs follow his training and then the race itself. Only D names both.
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The passage “A Miracle Mile” as printed (page 13)

Question 51 of 120
2024 Test A English, question 1Question
Read this sentence.
The engineers tried some other things in the hope of finding a more effective insulation for the compartment.
What is the most precise revision for the words The engineers tried some other things?
- AThe engineers did experiments with several new materials
- BThe engineers tested foam and fiberglass
- CThe engineers examined two new materials
- DThe engineers worked with foam and fiberglass
Show the answer
Correct answer: B
The question asks for the most precise revision for the words “The engineers tried some other things.”
A. Incorrect. This revision uses the word “materials” rather than precisely identifying what the engineers used.
B. CORRECT. This option revises the words to be more precise by using the specific words “tested” for the verb and “foam and fiberglass” for the materials.
C. Incorrect. This revision uses the word “examined” rather than a word indicating a process of experimental trial and error. The sentence also uses the word “materials” rather than precisely identifying what the engineers used.
D. Incorrect. This revision identifies the materials, but the imprecise verb “worked with” does not specify what the engineers were doing.
Question analysis
This revising question gives one sentence about engineers looking for better insulation for a compartment and asks which rewrite of its vague opening words is the most precise. The tricky part is that some choices sharpen one half of the phrase while leaving the other half as loose as the original.
- NYS standards
- 7L3a precise and concise language
- Grade level
- Grade 7 reading and task
- Compared with grade 7
- Below grade 7 Choosing precise wording is a grade 7 language standard, but this item is a single sentence about an engineering test with everyday words. Typical grade 7 revision asks for the same judgment across a paragraph or a full draft.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly evaluate
- What makes it hard
- A very close wrong answer moderate
- Every choice must be checked minor
- Evidence
- Answered from one sentence or line
- Reading level
- Question and choices: about grade 7, 63 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Precision here has two halves: a verb that names the actual activity and an object that names the actual materials. Most students who miss it stop at the first concrete detail they see, the named materials or the count of two, and never test the verb. Only one rewrite names both an exact action and the exact materials; each of the others leaves one half, or both, nearly as loose as the original.
A very close wrong answer moderate Choosing the answer
The rewrite with “worked with foam and fiberglass” matches the right one word for word except the verb, so only asking whether “worked with” says what the engineers did separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
The vague words have two parts, the action “tried” and the object “some other things,” and each rewrite sharpens a different part, so every one must be tested for both. (Embretson & Wetzel, 1987)
Strategy: Split the underlined words into the action and the things acted on, then ask of each revision whether a reader would know exactly what the engineers did and with what.
Traps
- Half right: One half of a two-part choice is right and the other is not.
- Wordy, vague or informal: A revision is grammatical but wordy, imprecise or too casual for the passage.
The wrong answers, classified
A Wordy, vague or informal (Connors & Lunsford, 1988)
Takes “did experiments with” as an exact verb, but the wordier phrase still ends in “several new materials,” which leaves the reader guessing what was tried.
C Wordy, vague or informal (Connors & Lunsford, 1988)
Treats the count “two” as precise detail, though the materials stay unnamed and “examined” suggests looking them over rather than trying them as insulation.
D Half right (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Sees that naming foam and fiberglass makes the object exact and stops there, missing that “worked with” is as vague as the original “tried.”
Working it out, step by step
Figure out what the question asks
This question is about precise language. Precise means exact and clear. The words “The engineers tried some other things” are vague. You can’t tell what the engineers did. You also can’t tell what they used. You need the rewrite that tells both.
Learn what makes words precise
Precise writing names the exact action and the exact thing. Vague: She did stuff with food. Precise: She baked bread.
The underlined words have two vague parts. The verb (the action word) is “tried.” The things are “some other things.” A good rewrite must fix both parts.
Think about what the sentence needs
The engineers want “a more effective insulation.” Insulation is material that keeps heat or cold from getting through. To find the best one, they would try out real materials and see how well each works. So the best verb says they tested. The best object names the materials.
Check each choice
- A is wrong. “Did experiments with” is wordy, and “several new materials” still doesn’t say which ones.
- B is right. “Tested” names the exact action. “Foam and fiberglass” names the exact materials.
- C is wrong. “Two new materials” gives a number but no names. “Examined” means looked over, not tried out.
- D is wrong. It names foam and fiberglass. But “worked with” is as vague as “tried.”
Watch out: D is tempting because it names the materials. Always check the verb too.
Choose the answer
Only B fixes both vague parts. The reader learns what the engineers did and what they used. So the answer is B: The engineers tested foam and fiberglass.
Why this answer and not the closest wrong one
D is the closest wrong answer. It matches B word for word except for the verb. “Worked with” could mean cut, built or studied. “Tested” tells you the engineers tried each material to see how well it insulates. That is what the sentence is about, so B is more precise.
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Question 52 of 120
2024 Test A English, question 3Question
Which edit should be made to correct the sentence?
The Colosseum in Rome, Italy which is considered one of the “new” Seven Wonders of the World, is the largest amphitheater ever built and once could hold up to 50,000 spectators.
- ADelete the comma after Rome.
- BInsert a comma after Italy.
- CDelete the comma after World.
- DInsert a comma after built.
Show the answer
Correct answer: B
The question asks for the correction needed for an error in the sentence.
A. Incorrect. When a place name has more than one element, a comma is used to separate the elements for clarity.
B. CORRECT. The sentence needs a comma to set off the clause “which is considered one of the ‘new’ Seven Wonders of the World,” because the clause is not essential to the meaning of the sentence (nonrestrictive clause). This option is correct because it places a comma where it is needed, after “Italy,” to set off the nonrestrictive clause that follows.
C. Incorrect. This edit would remove the comma at the end of the nonrestrictive clause, which should be set off by commas at the beginning and the end.
D. Incorrect. A comma after “built” would separate the predicate “and once could hold up to 50,000 spectators” from the subject of the main clause. The predicates “is the largest amphitheater ever built” and “once could hold up to 50,000 spectators” both relate to the same subject (“The Colosseum in Rome, Italy”), so a comma should not be used to separate the second predicate from that subject.
Question analysis
This editing question gives one long sentence about the Colosseum in Rome and asks which comma edit corrects it. Students must know how commas work with a city and its country, with a descriptive clause beginning with which, and with two verbs that share one subject, because every choice adds or removes a comma.
- NYS standards
- 7L2 capitalization, punctuation, and spelling
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Comma rules for place names and for extra information set off in a sentence are taught by grade 6. Here they are tested in one sentence with four marked spots, lighter than typical grade 7 editing of a whole paragraph.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- A rule that doesn't sound wrong moderate
- Complex sentences minor
- Evidence
- Answered from one sentence or line
- Reading level
- Question and choices: about grade 6, 59 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The which-clause has a closing comma after “World” but no opening one, and the sentence still sounds natural, so many students see the lone comma as the mistake instead of noticing its missing partner. Bracketing the extra description first, from “which” to “World,” exposes the gap quickly. The remaining choices fall to two rules students usually know: city and country are separated, and a second verb sharing the subject takes no comma.
A rule that doesn't sound wrong moderate Working it out
“Rome, Italy which is considered” reads smoothly aloud, yet the comma after “Italy” is needed twice over: to close the city-and-country pair and to open the nonessential which-clause. (Connors & Lunsford, 1988)
Complex sentences minor Understanding the question
Thirty-one words hold a two-part place name, a which-clause wedged between subject and verb, and two verbs, “is” and “once could hold,” sharing one subject, so each comma must be judged against that frame. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Strategy: Strip the sentence down to its main subject and verb, then check every marked spot against that core: what is the comma, or the missing comma, doing there?
Traps
- Punctuation: A comma, semicolon, colon, dash or apostrophe used where it does not belong, or missing where it does.
The wrong answers, classified
A Punctuation (Connors & Lunsford, 1988)
Treats “Rome, Italy” as a single name that should not be broken, not knowing that a comma always separates a city from the country that follows it.
C Punctuation (Connors & Lunsford, 1988)
Applies the sound rule against a comma between a subject and its verb to the comma before “is,” missing that it closes the which-clause and must stay.
D Punctuation (Connors & Lunsford, 1988)
Adds a pause before “and once could hold” because the sentence is long, though that second verb shares the subject, the Colosseum, and takes no comma.
Working it out, step by step
Figure out what the question asks
This is a punctuation question. Punctuation means marks like commas and periods. Four spots are underlined, at “Rome,” “Italy,” “World” and “built.” One edit fixes the sentence. Two comma rules decide it.
Learn the rule for extra information
Some groups of words just add an extra fact. If you take them out, the sentence still works. This is called a nonessential clause. A clause is a group of words with its own verb. A nonessential clause needs a comma before it and a comma after it, like a pair of brackets.
Example: Our school, which is very old, has a new gym. Without the middle part: Our school has a new gym.
Find the extra part
The extra part here starts at “which is considered” and ends at “of the World.” It adds a fun fact. Take it out and the sentence still works: The Colosseum in Rome, Italy, is the largest amphitheater ever built.
The comma after “World” closes the extra part. But nothing opens it. A comma is missing after “Italy.”
Learn the rule for places
A comma goes between a city and its country, as in “Rome, Italy.” If the sentence keeps going, put a comma after the country too. Example: She flew from Lima, Peru, to Miami.
So the comma after “Italy” is needed for two reasons. One comma does both jobs.
Check each choice
- A is wrong. The comma after “Rome” splits the city from the country. It must stay.
- B is right. It adds the missing comma that opens the extra part.
- C is wrong. The comma after “World” closes the extra part. Without it, neither end would be marked.
- D is wrong. The Colosseum “is the largest amphitheater” and “once could hold” many people. These are two verbs with one subject. No comma goes before the second verb. Example: Max swam fast and won a medal.
Watch out: C is tempting. A comma between a subject (who or what the sentence is about) and its verb is usually wrong. Here it is fine, because it is the second comma of a pair.
Choose the answer
With a comma after “Italy,” the extra part has a comma at each end. So the answer is B: insert a comma after Italy.
Why this answer and not the closest wrong one
C is the closest wrong answer. It treats the comma before “is” as the mistake. But that comma ends the extra part about the Seven Wonders. Delete it and nothing marks either end. B adds the missing opening comma, so both ends are marked.
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The original DOE question

Question 53 of 120
2024 Test A English, question 12Passage Excerpt from “Impressions of an Indian Childhood”
Excerpt from “Impressions of an Indian Childhood”
by Zitkala-Sa
Zitkala-Sa (Gertrude Simmons Bonnin) was a Native American writer, musician, teacher, and political activist who was raised on the Yankton Sioux Reservation in South Dakota. In 1900 she published “Impressions of an Indian Childhood” (the term Indian was commonly used at the time to refer to Native American people) to expose readers to what life is like on a reservation.
Soon after breakfast Mother sometimes began her beadwork. On a bright, clear day, she pulled out the wooden pegs that pinned the skirt of our wigwam1 to the ground, and rolled the canvas part way up on its frame of slender poles. Then the cool morning breezes swept freely through our dwelling, now and then wafting the perfume of sweet grasses from newly burnt prairie.
Untying the long tasseled strings that bound a small brown buckskin2 bag, my mother spread upon a mat beside her bunches of colored beads, just as an artist arranges the paints upon his palette. On a lapboard she smoothed out a double sheet of soft white buckskin; and drawing from a beaded case that hung on the left of her wide belt a long, narrow blade, she trimmed the buckskin into shape. Often she worked upon small moccasins for her small daughter. Then I became intensely interested in her designing. With a proud, beaming face, I watched her work. In [my] imagination, I saw myself walking in a new pair of snugly fitting moccasins. I felt the envious eyes of my playmates upon the pretty red beads decorating my feet.
Close beside my mother I sat on a rug, with a scrap of buckskin in one hand and an awl in the other. This was the beginning of my practical observation lessons in the art of beadwork. From a skein3 of finely twisted threads of silvery sinews my mother pulled out a single one. With an awl she pierced the buckskin, and skillfully threaded it with the white sinew. Picking up the tiny beads one by one, she strung them with the point of her thread, always twisting it carefully after every stitch.
It took many trials before I learned how to knot my sinew thread on the point of my finger, as I saw her do. Then the next difficulty was in keeping my thread stiffly twisted, so that I could easily string my beads upon it. My mother required of me original designs for my lessons in beading. At first I frequently ensnared many a sunny hour into working a long design. Soon I learned from self-inflicted punishment to refrain from drawing complex patterns, for I had to finish whatever I began.
After some experience I usually drew easy and simple crosses and squares. These were some of the set forms. My original designs were not always symmetrical nor sufficiently characteristic, two faults with which my mother had little patience. The quietness of her oversight made me feel strongly responsible and dependent upon my own judgment. She treated me as a dignified little individual as long as I was on my good behavior; and how humiliated I was when some boldness of mine drew forth a rebuke from her!
In the choice of colors she left me to my own taste. I was pleased with an outline of yellow upon a background of dark blue, or a combination of red and myrtle-green. There was another of red with a bluish-gray that was more conventionally used. When I became a little familiar with designing and the various pleasing combinations of color, a harder lesson was given me. It was the sewing on, instead of beads, some tinted porcupine quills, moistened and flattened between the nails of the thumb and forefinger. My mother cut off the prickly ends and burned them at once in the centre fire. These sharp points were poisonous, and worked into the flesh wherever they lodged. For this reason, my mother said, I should not do much alone in quills until I was as tall as my cousin Warca-Ziwin.
Always after these confining lessons I was wild with surplus spirits, and found joyous relief in running loose in the open again. Many a summer afternoon a party of four or five of my playmates roamed over the hills with me. We each carried a light sharpened rod about four feet long, with which we pried up certain sweet roots. When we had eaten all the choice roots we chanced upon, we shouldered our rods and strayed off into patches of a stalky plant under whose yellow blossoms we found little crystal drops of gum. Drop by drop we gathered this nature’s rock-candy, until each of us could boast of a lump the size of a small bird’s egg. Soon satiated with its woody flavor, we tossed away our gum, to return again to the sweet roots.
From “Impressions of an Indian Childhood” by Zitkala-Sa—Public Domain
A BRIEF HISTORY OF BEADWORK IN SOUTH DAKOTA
| Date | Event |
|---|---|
| 1500s | The Dakota use beads made from bones, shells, stones, and animal teeth. European traders bring glass beads to North America. |
| Late 1600s | The Dakota begin trading with the French in Minnesota. |
| 1830s | The Dakota begin using glass beads in clothing, artwork, and decorations in place of Native-made beads. |
| 1862–1865 | The Dakota are expelled from their homelands in Minnesota as a result of the U.S.-Dakota War. |
| 1900s–1920s | The Dakota create items to sell outside their community. |
Source: MNopedia
1wigwam: hut with an arched framework of poles covered with bark, mats, or animal hides
2buckskin: leather made from the skin of a male deer
3skein: coiled length of yarn or other thread loosely wound on a reel
Question
The phrase “just as an artist arranges the paints upon his palette” in paragraph 2 suggests that
- Abeadwork is a true form of art.
- Bcolor is a source of artistic inspiration.
- Call artistic activities begin with a series of steps.
- Dthe beadworker tries to imitate art.
Show the answer
Correct answer: A
The question asks what the phrase “just as an artist arranges the paints upon his palette” in paragraph 2 suggests in the excerpt.
A. CORRECT. In paragraph 2, by comparing the way her mother arranges the beads to the way a painter “arranges the paints upon his palette,” the author emphasizes that her mother approaches her craft just as a painter does, thus suggesting that beadwork is a true form of art.
B. Incorrect. Although the author mentions “bunches of colored beads” in paragraph 2, this detail does not provide enough evidence for the idea that color inspires beadworkers. The phrase “just as an artist arranges the paints upon his palette” indicates the connection of beadwork to other forms of art in general, not necessarily to the specifics of color as a source of inspiration.
C. Incorrect. In paragraph 2, the author describes her mother “untying the long tasseled strings” and then spreading “upon a mat beside her bunches of colored beads.” However, the phrase “just as an artist arranges the paints upon his palette” does not suggest that all artistic activities begin with a series of steps. There is no mention in the passage of any other artistic activities involving multiple initial steps.
D. Incorrect. In paragraph 2, the author describes how her mother “spread upon a mat beside her bunches of colored beads” as the first step in the artistic process described in the passage. Therefore, the comparison between the mother’s process and a painter arranging paints on a palette suggests that the mother is creating art, not imitating it. There is no indication that the mother is copying a technique she has seen in other art forms.
Question analysis
This reading question asks what the comparison between Zitkala-Sa's mother laying out her colored beads and an artist arranging paints on a palette suggests. It tests reading a simile in context, and the choices include readings that fix on a single word or go further than the comparison itself.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Interpreting a simile is taught by grade 6. The phrase is short and quoted in the question, and one step of interpretation is enough, so this asks less than typical grade 7 work on figurative language.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- Evidence
- Answered from one sentence or line
- Reading level
- Passage: about grade 8, 946 words; question and choices: about grade 6, 45 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The comparison is quoted in the stem and makes one point: the mother lays out her beads the way a painter lays out paints, so her craft counts as art. Wrong answers fix on the word “paints,” stretch one comparison into a rule about all art, or read the comparison as copying. Asking what the two things being compared have in common settles it.
Figurative or literary language minor Understanding the question
The simile sets the mother laying out bunches of colored beads beside a painter arranging paints, and students must see what that comparison says about beadwork itself. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Strategy: Name the two things being compared and what they share, then choose the idea that comparison supports without adding anything the sentence does not say.
Traps
- Word match: Repeats striking words or phrases from the text, so it looks supported, but its meaning does not fit the question.
- Too broad or too strong: A choice overstates, generalizes, or uses absolute words the text does not support.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
The wrong answers, classified
B Word match (Freedle & Kostin, 1993)
Links “paints” with the “bunches of colored beads” and decides color is the point, though the simile compares how materials are arranged, not what inspires the artist.
C Too broad or too strong (King et al., 2004)
Turns one comparison about laying out beads into a rule about all artistic activities, a generalization the sentence does not make.
D Twists what the text says (King et al., 2004)
Sees that the simile links beadwork with painting but twists its point, reading it as the mother imitating art rather than practicing it.
Working it out, step by step
Figure out what the question asks
The phrase is a simile. A simile compares two things using “like” or “as.” You need to know what this comparison suggests about the mother’s beadwork.
Read the phrase in paragraph 2
The mother “spread upon a mat beside her bunches of colored beads, just as an artist arranges the paints upon his palette.” The word order is old-fashioned. It means she spread her bunches of colored beads on a mat beside her. A palette is the board a painter uses to hold and mix paints.
Work out what the comparison means
The two things being compared are the mother and a painter. Both lay out their colors before they start to make something. So the author treats her mother like a painter. That means beadwork is real art, just like painting.
Check each choice
- A is right. Comparing the mother to a painter shows that beadwork is a true art.
- B is wrong. The simile is about laying out materials, not about what gives an artist ideas.
- C is wrong. One comparison can’t prove a rule about all art.
- D is wrong. The mother is not copying art. She works the way an artist does, so she is making art.
Watch out: B is tempting because paints and colored beads both bring color to mind. But the simile is about how she arranges them.
Choose the answer
The simile puts the beadworker side by side with a painter. So the answer is A.
Why this answer and not the closest wrong one
B is the closest wrong answer. Paints and colored beads do share color. But the simile compares what the two people do: both arrange their materials before they create. Nothing is said about inspiration, which means where ideas come from. A matches the point that the mother works as an artist does.
More questions like this
The original DOE question

The passage “Excerpt from “Impressions of an Indian Childhood”” as printed (pages 42–43)


Question 54 of 120
2024 Test A English, question 32Passage Champion of the Channel
Champion of the Channel
In 1926 an editor at the London Daily News predicted that Gertrude Ederle, an American swimmer with eighteen world records and three Olympic medals, would fail in her attempt to swim across the English Channel. He claimed that “even the most uncompromising champion of the rights and capacities of women must admit that in contests of physical skill, speed and endurance they must remain forever the weaker sex.” Yet, at only nineteen years old, Ederle not only became the first woman to accomplish this feat, she also broke the men’s record by two hours. Gertrude Ederle’s triumphant swim across the English Channel was a testimony to her determination, innovative spirit, and passion for swimming.
Crossing the English Channel is a daunting task for any swimmer. At its narrowest point, the channel measures twenty-one miles across. Its icy waters hover around sixty degrees Fahrenheit, and its unruly tides and currents toss swimmers about like bobbing corks. Stinging jellyfish, seaweed, and floating debris from shipwrecks and lost cargoes present added hazards.
For decades the channel’s perils have defeated countless swimmers. Ederle, too, failed in her first attempt to cross the channel in 1925. Just six miles short of finishing, she became ill, and her coach had to haul her out of the water. Undeterred, Ederle decided to try again. Ederle knew that if she did not complete the challenge this time, she might never get the opportunity to set this record, because a rival female swimmer was preparing to make her second attempt at the crossing as well.
To prepare for the marathon swim, Ederle found ways to improve her equipment. She and her sister Meg discovered that melted candle wax perfectly sealed goggle edges, effectively waterproofing Ederle’s goggles against hammering waves. The sisters also designed a two-piece silk swimsuit for Ederle. During her first channel-crossing attempt she had worn a standard one-piece swimsuit that, after the lengthy hours of swimming across the channel, had stretched out, filling with water and creating drag, making an already challenging task almost insurmountable. Unlike the cumbersome typical bathing suit, this silk invention weighed little and allowed for easy movement.
On August 6, 1926, Ederle waded into the channel near Cape Gris-Nez, France. At first she shivered in the bone-chilling water even though she had covered her body in eight layers of grease for insulation. Her limbs felt stiff. Her strokes were irregular. Driving forward, she fought to clear her mind and find what she called her “sphere,” a place where the sea became her only companion and the shrieks of gulls and the humming of boat engines faded away. Using a new overhand stroke called the American crawl, Ederle eventually settled into a steady pace, briskly breaking through waves.
Throughout Ederle’s swim, two tugboats accompanied her. One carried newspaper reporters who wired dispatches of her progress to shore. The other, displaying a sign that read “This way, ole kid!” with an arrow pointing forward, transported her coach, family, and friends. Her coach played songs, such as “Yes, We Have No Bananas,” on a phonograph so that Ederle could time her strokes to the rhythm. Using a net, her coach also passed her baby bottles of broth for nourishment.
For hours Ederle swam, dodging debris with an amused smile. However, as she neared the English shore, a sudden fierce storm erupted. The tides and waves forced Ederle backward, and she fought the stubborn swells for several hours. The salty water caused her tongue to swell and inflamed her ears. Yet Ederle felt indescribably happy as she churned through the sea. Finally, as she neared the English shore, the storm abated, and the tide turned. No longer fighting against her, the sea pushed her toward the shore and victory.
After fourteen hours and thirty-one minutes, Ederle, on wobbly legs, stepped onto the English shore. The waiting crowd roared, honked their automobile horns, blasted their tugboat whistles, and set off flares that flashed in the sky. Ederle had swum into history.
When Ederle returned to New York, she received a parade, where thousands of people shouted “Trudy!” Not only were everyday American citizens proud of Ederle, but she also inspired them to be more active. Over the next few years, more than 60,000 people credited her with motivating them to earn their American Red Cross swimming certificates. Gertrude Ederle’s accomplishment proved to the world that with determination and passion, it was possible for a person to achieve his or her goals.
SOLO SWIMS ACROSS THE ENGLISH CHANNEL
|
Earliest Speed Records |
||
| Year | Swimmer | Time |
| 1875 | Matthew Webb (M) | 21 hours, 45 minutes |
| 1923 | Enrico Tiraboschi (M) | 16 hours, 33 minutes |
| 1926 | Gertrude Ederle (F) | 14 hours, 31 minutes |
| 1926 | Arnst Vierkotter (M) | 12 hours, 38 minutes |
| Current Speed Records | ||
| Year | Swimmer | Time |
| 2012 | Trent Grimsey (M) | 6 hours, 55 minutes |
| 2006 | Yvetta Hlavacova (F) | 7 hours, 25 minutes |
Question
Which sentence is the best summary of the steps that Ederle took to prepare for her second attempt to swim across the English Channel?
- AWorking with her sister, Ederle waterproofed her goggles using melted candle wax to seal the edges and designed a two-piece silk swimsuit that was lightweight and would not stretch out during the long swim.
- BEderle covered her body in numerous layers of grease for insulation and focused on finding her “sphere” during her swim.
- CEderle began training with her coach, who played music while she swam to help her time her strokes to the music.
- DEderle focused on developing better equipment than the standard bathing suit that proved cumbersome during her first attempt to cross the channel.
Show the answer
Correct answer: A
The question asks for the best summary of Ederle’s steps to prepare for her second attempt to swim across the English Channel.
A. CORRECT. Ederle’s preparation is outlined in paragraph 4. This option is correct because it acknowledges the idea that Ederle involved her sister in this process, and it details the efforts the two took to improve Ederle’s equipment, including sealing her goggles with wax and designing a better swimsuit.
B. Incorrect. The option focuses on Ederle’s actions in the moments before her swim (covering her body with grease for insulation) and during her swim (finding her “sphere”), not her overall preparation methods.
C. Incorrect. The option refers to a way that Ederle’s coach helped Ederle keep her strokes in rhythm during her swim. While this option suggests that Ederle and her coach had made thoughtful preparations, it does not address the many other steps that Ederle took to prepare for the swim.
D. Incorrect. The option does not refer to the series of steps that Ederle took to prepare for her swim, but rather to just one of the efforts made (improving equipment), without acknowledging her sister’s contributions.
Question analysis
This summary question asks which sentence best sums up how Gertrude Ederle got ready for her second try at swimming the English Channel. Because the passage also describes what she and her team did during the crossing, students must separate preparation from the swim itself and judge which summary is complete.
- NYS standards
- 7R2 theme or central idea
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 This is a one-paragraph summary, the kind of task grade 6 standards describe; grade 7 central-idea work usually follows an idea through a whole text. The real demand is telling preparation apart from the swim.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- A very close wrong answer moderate
- Every choice must be checked minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 9, 790 words; question and choices: about grade 12, 121 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Paragraph 4 holds all of the preparation, but the passage keeps describing things Ederle and her team did, so getting ready must be told apart from the swim. The grease is the hardest to place, because she put it on before wading in, on the day of the crossing itself. Between the two equipment choices, the one that lists the goggles, the silk suit and her sister's help is the fuller summary of several steps.
A very close wrong answer moderate Choosing the answer
The right summary and the one about better equipment than the standard bathing suit describe the same work; only the waxed goggles and Meg's part, missing from the shorter one, separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
Two choices borrow real details from the crossing, the grease she wore and the songs played from the tugboat, so each has to be placed before or during the swim. (Embretson & Wetzel, 1987)
Strategy: Locate the paragraph about getting ready, list each preparation step and who took part, and choose the sentence that covers that list most fully without adding events from another part of the passage.
Traps
- Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
The wrong answers, classified
B Wrong part of the text (Kirsch & Mosenthal, 1990)
Takes the grease she wore on the day and her search for her “sphere” in the water, both from paragraph 5's account of the swim, as advance preparation.
C Twists what the text says (King et al., 2004)
Turns the phonograph her coach played from a tugboat during the crossing into training sessions before it, which the passage never describes.
D Too narrow (King et al., 2004)
Captures the idea of better equipment but names only the bathing suit, leaving out the wax-sealed goggles and her sister Meg, so it summarizes one step of several.
Working it out, step by step
Know what a summary needs
A summary retells the main points in a few words. Here you need the steps Ederle took to get ready before her second try. A good summary includes all of those steps. It leaves out things that happened during the swim itself.
Find the getting-ready paragraph
Paragraph 4 opens with “To prepare for the marathon swim, Ederle found ways to improve her equipment.” The word “prepare” signals that this paragraph holds the steps. Equipment means the gear she used.
List each step
- With her sister Meg, she found that “melted candle wax perfectly sealed goggle edges,” so water stayed out of her goggles.
- Together the sisters “designed a two-piece silk swimsuit” to replace her old one.
- The new suit “weighed little and allowed for easy movement,” unlike her old suit, which had stretched and filled with water.
Check each choice
- A fits: it names her sister, the waxed goggles and the light silk suit.
- B is wrong because the grease and her “sphere” belong to paragraph 5, the day of the swim.
- C is wrong because her coach played music from a tugboat during the swim (paragraph 6), not in training.
- D is wrong because it leaves out the goggles and Meg.
Watch out: D is true as far as it goes. But it covers only one step, so it is too narrow to be the best summary.
Choose the answer
Choice A lists all the preparation in paragraph 4, and nothing from the swim itself. So the answer is A.
Why this answer and not the closest wrong one
D comes closest, since improving her equipment is what paragraph 4 is about. But D mentions only the swimsuit. It skips the wax-sealed goggles and her sister’s help. A summary of the steps should include every step, and only A does.
More questions like this
The original DOE question

The passage “Champion of the Channel” as printed (pages 56–57)


Question 55 of 120
2024 Test A English, question 35Passage 5 Reasons Physical Books Might Be Better Than E-Books
5 Reasons Physical Books Might Be Better Than E-Books
by Shaunacy Ferro
Though e-book1 readers have become a more common sight around town, traditional books still have their evangelists.2 According to The New York Times, e-book sales have been falling in 2015. Print definitely isn’t dead. In fact, according to some research, it may actually be a better choice for some readers. While scientists are still trying to tease out exactly how digital reading affects us differently, here are five ways e-books might be inferior to their dead-tree cousins.
E-books can reduce reading comprehension.
In a study of middle schoolers, West Chester University researchers found that students who read on iPads3 had lower reading comprehension than when they read traditional printed books. They discovered that the kids sometimes skipped text in favor of interactive features in the e-books, suggesting that certain multimedia in children’s e-books can be detrimental to the practice of reading itself. However, the researchers noted that some interactive features in e-books are designed to enhance comprehension, and that those might be more helpful than game-type interactive graphics.
Young kids can get distracted by e-books.
Similar results were found by a small study by the Joan Ganz Cooney Center that consisted of 32 kids reading e-books and print books with their parents. It found that “enhanced” e-books might be distracting. Kids who read enhanced e-books—ones with interactive, multimedia experiences—were more engaged with them physically, but in the end they remembered fewer narrative details than those who read print books or basic e-books.
You remember less about a book’s timeline.
Another study of adults also found that e-books can be hard to absorb. The researchers asked 25 people to read a 28-page story on a Kindle4 and 25 to read the story in paperback, then asked the readers to put 14 events from the story in chronological order. Those who read the story on a Kindle performed worse on the chronology test than the book readers, though they performed about the same as print readers in other tests. Earlier research by the same scholars, from Stavanger University in Norway, found that Norwegian 10th graders also remembered more about texts if they read them in print rather than on a computer screen.
They’re not great as textbooks.
While e-book textbooks are often cheaper (and easier to carry) than traditional door-stop5 textbooks, college students often don’t prefer them. In some surveys of college kids, the majority of students have reported preferring print books. However, a 2012 study from the UK’s National Literacy Trust of kids ages 8 to 16 found that more than 50 percent of children reported preferring screen reading.
They’re tiring.
Staring at a lit screen can be tiring for the eyes and the brain. A 2005 study from Sweden found that reading digitally required a higher cognitive workload than reading on paper. Furthermore, staring at LED6 screens at night can disrupt sleep patterns. A 2014 Harvard study found that people who used e-readers with LED screens at night slept worse and were more tired the next day. So, if you’re going to go for an e-book, go for one without the backlight.
The take-away message? If you’re really trying to absorb material, you might want to go for a physical book. And if you’re going to be up all night studying, turn off the backlight.
However, all this may not mean that reading on a Kindle is really going to melt your brain. For instance, reading an e-book on a computer is a much different experience than reading on a Kindle, which is specifically designed for consuming books. So, too, is playing with an interactive e-book on an iPad, compared to using a simpler e-book device that only presents the text, with no opportunities to click away into digital distractions.
And some studies have found that part of the difference between the way people absorb information from e-books versus paper might be due to approaching e-books differently—in one test, participants didn’t regulate their study time with digital books like they did with paper texts, leading to worse performances. It’s possible that our expectations of e-book reading—as well as the different designs of the digital reading experience on a computer or iPad or Kindle—might affect how we approach the text and how much effort we put into studying them. As generations of e-book readers evolve, and people become more accustomed to the idea of sitting down with a digital textbook, these factors could change—for better or for worse.
“5 Reasons Physical Books Might Be Better Than E-Books” by Shaunacy Ferro from MENTAL FLOSS, October 2, 2015. Copyright © 2015 Mental Floss, Inc.
1e-book: an electronic book, a publication made available digitally for reading on a computer or other device
2evangelists: advocates or supporters
3iPads: tablet computers
4Kindle: device for reading e-books
5door-stop: a reference to the idea that some textbooks are large and heavy enough to hold a door open
6LED: Light-emitting diode; LEDs are often used as a backlight for electronic devices such as smart phones and tablets.
Question
The details in paragraph 1 convey a central idea of the passage by noting that
- Afewer people are buying e-books than printed books.
- Bprinted books have some advantages over e-books.
- Cprinted books rely on natural materials more than e-books.
- Dsome people recommend buying printed books over e-books.
Show the answer
Correct answer: B
The question asks how the details in paragraph 1 convey a central idea of the passage. A specific central idea that needs to be identified in order to correctly answer this question is that reading a digital book “might be inferior” (paragraph 1) to reading a traditional book.
A. Incorrect. In paragraph 1, the author notes that despite the popularity of e-books, their sales fell in 2015, leading to the author’s statement that “print definitely isn’t dead.” While these statements suggest that readers still purchase and enjoy printed books, sales information between print and e-books is not compared. Therefore, these details do not convey a central idea of the passage.
B. CORRECT. The author states in paragraph 1 that “according to some research,” reading a traditional book “may actually be a better choice for some readers” because “digital reading affects us differently.” The author then supports this statement throughout the passage, detailing five advantages that printed books have over e-books. This detail helps convey a central idea of the passage.
C. Incorrect. In paragraph 1, the author refers to printed books as “traditional books” and the “dead-tree cousins” of e-books. Although these references suggest that printed books rely on natural materials more than e-books do, the author does not provide additional details in the passage to convey this idea. Therefore, this detail does not help convey a central idea of the passage.
D. Incorrect. Although the author states in paragraph 1 that “traditional books still have their evangelists,” the author does not provide additional statistics or details in the passage to convey the idea that some people recommend buying printed books over e-books. Therefore, this detail does not help convey a central idea of the passage.
Question analysis
This central-idea question asks how the details in the opening paragraph of a magazine article comparing printed books and e-books convey one of the passage's central ideas. Students must connect the introduction to what the rest of the article goes on to develop, while each choice seizes on something the first paragraph mentions.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Connecting an opening paragraph to a central idea is a grade 6 and 7 skill. This article is plainly organized, with a short introduction and bold headings, so the task is lighter than typical grade 7 central-idea work.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 10, 829 words; question and choices: about grade 7, 46 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Paragraph 1 packs four separate details into five sentences, and three of them become wrong choices: the falling e-book sales, print's “evangelists” and the joke about dead trees. The question rewards checking which detail the rest of the article develops, and the five bold headings that follow each name a way print beats the screen. Students who judge each choice against paragraph 1 alone can find support for all four.
The right answer is reworded minor Choosing the answer
The right choice recasts “may actually be a better choice” and “e-books might be inferior” in new words, while the wrong choices echo “sales”, “evangelists” and “dead-tree cousins”. (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)
Figurative or literary language minor Understanding the question
Paragraph 1 is playful, calling print's supporters “evangelists” and paper books “dead-tree cousins”, and both phrases must be read as jokes rather than as claims. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Strategy: Read paragraph 1, then skim the bold headings to see which of its statements the rest of the article develops; a central idea is one the passage keeps returning to.
Traps
- Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Taking figurative language literally: A choice reads a figure of speech or image at face value.
The wrong answers, classified
A Twists what the text says (King et al., 2004)
Reads “e-book sales have been falling” as print outselling e-books, a comparison of sales the paragraph never makes and the rest of the article never mentions.
C Taking figurative language literally (Graesser, Singer & Trabasso, 1994)
Takes “dead-tree cousins” as a point about what books are made of, when it is a joking name for paper books and the article never discusses materials.
D Too narrow (King et al., 2004)
Rests on the opening remark that print still has its “evangelists”, a single detail about supporters that the article drops in favor of research findings.
Working it out, step by step
Know what a central idea is
A central idea is a main point that the whole passage keeps coming back to. So you need a detail in paragraph 1 that the rest of the article develops. A detail that shows up once and is then dropped doesn’t count.
Read paragraph 1 closely
Paragraph 1 makes several points. It says e-book “sales have been falling.” It says print still has its “evangelists,” which the footnote explains means supporters. Then comes the key line: print “may actually be a better choice for some readers.” The paragraph ends by promising “five ways e-books might be inferior,” or worse, compared with paper books.
Use the headings as a map
The bold headings show what the rest of the article is about. Two of them are “E-books can reduce reading comprehension” and “They’re tiring.” Each heading names one way paper books may beat e-books. So the point the article develops is that print has some advantages.
Check each choice
- B fits: the five sections that follow each give one advantage of printed books.
- A is wrong because the paragraph says e-book sales fell, not that fewer e-books than print books are sold.
- C is wrong because “dead-tree cousins” is a joke name for paper books, and the article never discusses materials.
- D is wrong because the supporters are mentioned once, and the article reports research instead of advice.
Watch out: A starts from a real detail but adds a comparison the paragraph never makes.
Choose the answer
The article keeps returning to the ways print may be better than e-books. So the answer is B.
Why this answer and not the closest wrong one
D comes close, because paragraph 1 does say print has supporters. But the article never returns to them. Instead, every section after paragraph 1 gives research showing one way print may beat e-books. That is why B, “printed books have some advantages over e-books,” is the central idea.
More questions like this
The original DOE question

The passage “5 Reasons Physical Books Might Be Better Than E-Books” as printed (pages 58–59)


Question 56 of 120
2024 Test A English, question 47Passage Excerpt from The Highest Tide
Excerpt from The Highest Tide
by Jim Lynch
The narrator’s summer job is collecting marine specimens in Puget Sound, Washington. In this excerpt, he is looking for specimens in the mudflats before daybreak. Mudflats, or flats, are coastal landforms created when sediment and silt from tides are deposited as the tides recede.
I rounded the oyster beds, to the far side. . . . It was low tide by then, and I saw the water hesitating at its apex, neither leaving nor returning, patiently waiting for the gravitational gears to shift. Dozens of anxious clams started squirting in unison like they did whenever vibrating grains of sand warned them predators were approaching. I stopped and waited with them, to actually see the moment when the tide started returning with its invisible buffet of plankton for the clams, oysters, mussels and other filter feeders. It was right then, ankle deep in the Sound, feet numbing, eyes relaxed, that I saw the nudibranch.1
In all my time on the flats I’d never seen one before. I’d read about them, sure. I’d handled them at aquariums but never in the wild, and I’d never even seen a photo of one this stunning. It was just three inches long but with dozens of fluorescent, orange-tipped hornlike plumes jutting from the back of its see-through body that appeared to be lit from within.
Nudibranchs are often called the butterflies of the sea, but even that understates their dazzle. Almost everything else in the northern Pacific is dressed to blend with pale surroundings. Nudibranchs don’t bother, in part because they taste so lousy they don’t need camouflage to survive. But also, I decided right then, because their beauty is so startling it earns them a free pass, the same way everyday life brakes for peacocks, parade floats. . . .
The dark mudflats loomed like wet, flattened dunes stretching deep into Skookumchuck Bay in front of our house. From a distance, they looked too barren to support sea life. Up close, they still did, unless you knew where to find the hearty clams, worms and tiny creatures that flourish in mud. . . . I’m not sure why I decided to take a look. It was still an hour before sunrise, and I knew exactly what the bars2 looked like in the moonlight, but for some reason, I couldn’t resist.
I heard it long before I saw it. It was an exhale, a release of sorts, and I instantly wondered if a whale was stranded again. We had a young minke3 stuck out there two summers prior, and it made similar noises until the tide rose high enough for rescuers to help free it. . . . I looked for a hulking silhouette but couldn’t find one. I waited, but there were no more sounds. Still, I went toward what I thought I’d heard, avoiding stepping into the mud until I had to. I knew the flats well enough to know I could get stuck just about anywhere. The general rule was you didn’t venture out past the shells and gravel with an incoming tide. I sank up to my knees twice, and numbing water filled my boots.
. . . I kept stepping toward the one sound I’d heard, a growing part of me hoping I’d find nothing at all. When I stopped to rest and yank up my socks, my headlamp crossed it. My first thought? A giant octopus.
Puget Sound has some of the biggest octopi in the world. They often balloon to a hundred pounds. Even the great Jacques Cousteau4 himself came to study them. But when I saw the long tubular shape of its upper body and the tangle of tentacles below it, I knew it was more than an octopus. I came closer, within fifty feet, close enough to see its large cylindrical siphon5 quiver. I couldn’t tell if it was making any sounds at that point, because it was impossible to hear anything over the blood in my ears. . . .
The creature’s body came to a triangular point above narrow fins that lay flat on the mud like wings, but it was hard to be sure exactly where it all began or ended, or how long its tentacles truly were, because I was afraid to pry my eyes off its jumble of arms for more than half a second. I didn’t know whether I was within reach, and its arms were as big around as my ankle and lined with suckers the size of half-dollars. If they even twitched I would have run. So, I was looking at it and not looking at it while my heart spangled my vision. I saw fragments, pieces, and tried to fuse them in my mind but couldn’t be certain of the whole. I knew what it had to be, but I wouldn’t allow myself to even think the two words. Then I gradually realized the dark shiny disc in the middle of the rubbery mass was too perfectly round to be mud or a reflection.
It was too late to smother my scream. Its eye was the size of a hubcap.6
From THE HIGHEST TIDE by Jim Lynch. Copyright © 2005 by Jim Lynch. Used by permission of Bloomsbury USA. All rights reserved.
1nudibranch: type of sea slug
2bars: sandbars
3minke: species of small whale
4Jacques Cousteau: renowned French oceanographer
5siphon: tube-like organ that is used for drawing in or removing fluids
6hubcap: removable cap over the end of a wheel axle
Question
Which sentence from paragraph 5 supports the idea that the narrator is taking a risk?
- A“It was an exhale, a release of sorts, and I instantly wondered if a whale was stranded again.”
- B“I looked for a hulking silhouette but couldn’t find one.”
- C“I waited, but there were no more sounds.”
- D“I knew the flats well enough to know I could get stuck just about anywhere.”
Show the answer
Correct answer: D
The question asks which sentence from paragraph 5 supports the idea that the narrator is taking a risk.
A. Incorrect. In this sentence, the narrator is describing the exhale of a creature and his concern that the sound might be coming from a whale in danger. At no point does the narrator suggest that this is a dangerous situation.
B. Incorrect. This sentence describes a “hulking silhouette” because the narrator is searching for a whale that might be stranded. This expression refers to the size of the whale and not to any danger the narrator might face.
C. Incorrect. The narrator’s words “no more sounds” are used to illustrate his listening for what may be a whale. At this point, the narrator is still concerned about the possibility that a whale may be stranded, not the possibility of being in danger.
D. CORRECT. The idea that the narrator could become stuck in the mud of the flats means that the surroundings pose some danger or difficulty.
Question analysis
Students choose which sentence from one tense paragraph of The Highest Tide shows that the narrator is taking a risk as he follows a strange sound across the mudflats. In this evidence question, all four sentences come from the same scene, so students must tell real danger to the narrator apart from a mood of suspense.
- NYS standards
- 7R1 textual evidence and inferences
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Choosing the sentence that supports an idea is a grade 6 and 7 skill. Here all four choices come from one paragraph and only the idea of danger needs matching, a lighter task than typical grade 7 evidence work.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly evaluate
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 6, 902 words; question and choices: about grade 5, 65 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Only one of the four sentences mentions anything that could happen to the narrator himself, so this is among the easiest questions on the test. The other three build suspense, a strange exhale, a missing silhouette, a silence, and a student who confuses suspense with danger may pick one. Asking who is at risk in each sentence settles it at once.
Reading between the lines minor Working it out
The right sentence states only what the narrator knows, that he could get stuck; it shows a risk because the sentences around it have him walking out onto the mud anyway. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Strategy: Decide what taking a risk would mean for the narrator in this place, then test each sentence for a danger to him rather than a feeling of suspense.
Traps
- Supports only part of the claim: In a question about evidence: a line that backs one half of the claim but not the other, or a related point rather than the claim itself.
- Word match: Repeats striking words or phrases from the text, so it looks supported, but its meaning does not fit the question.
The wrong answers, classified
A Supports only part of the claim (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Senses danger in a possibly stranded whale, but the danger would be the whale's, and the sentence shows the narrator only wondering, not exposing himself to anything.
B Word match (Freedle & Kostin, 1993)
Hears menace in “hulking silhouette”, but the phrase describes the size of the whale he expected, and failing to find one puts him in no danger.
C Reads too much in (Graesser, Singer & Trabasso, 1994)
Reads the silence after the sound as ominous, but waiting and listening is a pause in his search, not something that puts him at risk.
Working it out, step by step
Know what the question asks
This asks you to find evidence. Taking a risk means doing something that could hurt you. So look for a sentence that shows a danger to the narrator himself.
Think about the danger on the flats
Mudflats are soft and wet. A person can sink into the mud and get stuck as the water rises. Paragraph 5 even gives a rule: “The general rule was you didn’t venture out past the shells and gravel with an incoming tide.”
Find the sentence about danger
Choice D says, “I knew the flats well enough to know I could get stuck just about anywhere.” So the narrator knows he could get stuck. Yet he keeps walking out toward the sound. That means he is taking a risk.
Check each choice
- D fits: getting stuck is a danger to him, and he goes on anyway.
- A is wrong because a stranded whale would be in danger, not the narrator.
- B is wrong because “hulking silhouette,” a huge dark shape, describes the whale he expected, not a threat.
- C is wrong because waiting in silence builds suspense but puts him in no danger.
Watch out: A, B and C create a spooky mood. But a spooky mood is not the same as a risk.
Choose the answer
Only one sentence names something that could happen to the narrator himself. So the answer is D.
Why this answer and not the closest wrong one
A comes closest, because a stranded whale sounds serious. But the danger in that sentence belongs to the whale. Only D names something that could happen to the narrator: getting stuck in the mud.
More questions like this
The original DOE question

The passage “Excerpt from The Highest Tide” as printed (pages 64–65)


Question 57 of 120
2024 Test A English, question 52Passage Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”
Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”
by Emma Marris
The author of this opinion article attended a symposium, or conference, in Missoula, Montana, at which biologists and other scientists discussed species of plants and animals that are not native to the areas in which they live.
Invasive species are scary. It was ecologist Charles Elton, back in the 1950s, who introduced the militaristic “invasion” metaphor to describe exotic plants and animals—but there’s no question some can be extremely destructive.
The brown tree snake has eaten a dozen kinds of forest birds in Guam to extinction; zebra mussels clog pipes around the Great Lakes; the common house cat turns out to be, in Australia, a mercilessly effective killer of cute, fluffy marsupials like the bilby and the numbat.
As scientists have sounded the alarm about these pests, the public has gotten the message. Citizen groups rip out non-native plants. Native gardens have become increasingly popular, both as ways to celebrate the unique flora of each region and as tiny hot spots of diversity. Native trees provide food for native bugs, which feed native birds. Food chains developed over thousands of years of co-evolution unfold in our backyards. . . .
So we’ve learned, scientists and laypeople alike, that native species are good and non-natives are bad.
Julian Olden, a biologist at the University of Washington, Seattle, who co-organized the symposium, recently polled nearly 2,000 ecologists. Among his findings: a substantial number of them said they would immediately eradicate a hypothetical non-native forest plant, even if it were shown to have no effect on the forest. Olden calls this the “guilty even when proven innocent” approach.
That kind of approach is not very useful on a rapidly changing planet.
Exotics Are Everywhere
Climate change is making it harder even to decide who the invaders are.
How, scientists at the symposium wondered, do you define “native” on a warming planet, when plants and animals are already moving toward the poles or up mountainsides in search of climate conditions they can tolerate? Should we consider them “invasive” in their new homes? Regardless of what we label them, conservationists will be reluctant to remove them from their new environs—to do so would stymie1 their chances of adapting to the warmer future we’re creating.
And then there are the non-natives that we actually like. Most domestic crops are exotic in most of the places they’re grown, but there are even wild exotics that “do good,” forming useful relationships with native species.
Edwin Grosholz of the University of California, Davis, told the recent symposium about one such relationship. On beaches in his state, non-native spartina grass has become important habitat for the endangered California clapper rail, a plump shorebird with a downward curving bill more at home on land than in the air. A project to rip out and poison the spartina—which grows in dense swaths that exclude many other shorebirds—saw clapper rail numbers go tumbling downward.
There are other examples like that. The endangered southwestern willow flycatcher nests in “invasive” tamarisk shrubs. Many native (and beautiful) Hawaiian flowers are now pollinated by the Japanese white-eye bird—because the native pollinators have been driven extinct by other non-native species.
Should we impose further risk on already endangered natives by severing these relationships? Or should we admire the resilience of nature and let such “well-behaved” exotics stay? . . .
Leave them alone, more and more conservationists are arguing, and stop focusing obsessively on categorizing species as native or non-native. Mark Davis, an ecologist at Macalester College in St. Paul, Minnesota, once considered himself an “invasion biologist”—but not anymore. “I am actively trying to get the field to retire the invader narrative,” he said in Missoula.
A Good Thing, Not the Only Thing
After all, nativeness is just one environmental value, and arguably not as important as preventing extinctions and preserving biodiversity. In some cases we can best serve biodiversity by leaving the non-natives alone or even—brace yourself, now—introducing them on purpose.
This is the thinking behind, for example, installing the Aldabra tortoise on the islands of Mauritius. The islands lost their own large tortoises, and the fruiting plants that formerly had their seeds moved around by these fruit-loving reptiles have been on the decline. A tortoise that’s related to the island’s large tortoises—a non-native from the Seychelles in the Indian Ocean that was intentionally introduced in 2004—is now handling some of that work.
Most of the time, for the time being, conserving species still means focusing on supporting them in their historical habitats, planting natives and removing non-natives. We can and should do that in places where it is feasible and important to us.
From “It’s Time to Stop Thinking That All Non-Native Species Are Evil” by Emma Marris from NATIONAL GEOGRAPHIC MAGAZINE, July 24, 2014. Copyright © 2014 by National Geographic Society.

1stymie: hinder or interfere with
Question
Which detail provides the most relevant support for the claim that “climate change is making it harder even to decide who the invaders are” (paragraph 7)?
- Athe mention in paragraph 8 of non-native species that are forced to seek new habitats
- Bthe acknowledgment in paragraph 9 that non-native species include many domestic crops
- Cthe description in paragraph 11 of the interactions between native and non-native species
- Dthe account in paragraph 13 of an ecologist who changed his beliefs about non-native species
Show the answer
Correct answer: A
The question asks which detail provides the most relevant support for the claim that “climate change is making it harder even to decide who the invaders are” (paragraph 7).
A. CORRECT. The fact that many plants and animals have no choice but to seek out “climate conditions they can tolerate” (paragraph 8) makes it much harder to decide whether to call these species invaders.
B. Incorrect. The fact that many domestic crops are non-native species has nothing to do with climate change affecting our discernment of which species are invaders.
C. Incorrect. The description in paragraph 11 of native and non-native species interacting never suggests that climate change is a factor in these interactions and never suggests that climate change muddies our understanding of which species is native and which is non-native.
D. Incorrect. While the example of Mark Davis so thoroughly changing his beliefs about non-native species is a powerful one, the role that climate change might have played in this reversal is never discussed.
Question analysis
Here students choose the detail that gives the most relevant support to the claim that climate change is making it harder even to decide which species are invaders, in an article rethinking non-native species. Each choice names a paragraph, and the article gives several different reasons to question the label, so every detail must be checked against the whole claim.
- Grade level
- Grade 7 reading and task
- Compared with grade 7
- Below grade 7 Judging which detail supports a claim is grade 7 argument work, but here the choices name their paragraphs, the claim is short and specific, and only one key idea must be matched, so the task is lighter than typical.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly evaluate
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 11, 815 words; question and choices: about grade 11, 76 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The claim has two parts, climate change as the cause and blurred labels as the effect, and every wrong detail muddies the word “invader” without any help from the climate. Crops, helpful partnerships and a converted invasion biologist all complicate the label, so a student who checks only the effect finds four plausible answers. Only paragraph 8 talks about a warming planet.
Every choice must be checked minor Choosing the answer
The choices describe paragraphs instead of quoting them, so each of paragraphs 8, 9, 11 and 13 must be reread for a link to climate. (Embretson & Wetzel, 1987)
The right answer is reworded minor Choosing the answer
The right choice says “forced to seek new habitats” where paragraph 8 says species are moving “toward the poles or up mountainsides in search of climate conditions they can tolerate”. (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)
Strategy: Name the cause and the effect in the claim, then test each detail for both; a detail that fits only one part is not the most relevant support.
Traps
- Supports only part of the claim: In a question about evidence: a line that backs one half of the claim but not the other, or a related point rather than the claim itself.
The wrong answers, classified
B Supports only part of the claim (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Points to crops as non-natives people like, which complicates the label “invader”, but paragraph 9 never connects them to a changing climate.
C Supports only part of the claim (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Picks the helpful pairings of native and non-native species in paragraph 11, which blur good and bad but have nothing to do with a warming planet.
D Supports only part of the claim (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Chooses Mark Davis's turn away from invasion biology, a striking change of mind that nothing in paragraph 13 ties to climate change.
Working it out, step by step
Split the claim into two parts
This asks for the detail that best supports a claim. The claim has a cause and an effect. The cause is climate change. The effect is that it is harder to tell which species are invaders. The best support must connect both parts.
Find where climate comes up
Paragraph 8 comes right after the claim, and it talks about “a warming planet.” Plants and animals “are already moving toward the poles or up mountainsides in search of climate conditions they can tolerate.” To tolerate means to be able to live with. As places get hotter, many species move to cooler ones.
See why the labels get harder
When a species moves to a new place, it is not native there. But it didn’t really invade. Heat pushed it out of its old home. So scientists ask whether such species should be called invasive “in their new homes.” Climate change blurs the line between native and invader.
Check each choice
- A fits: species forced to move by the climate link the cause to the effect.
- B is wrong because crops are non-natives people like, but paragraph 9 never mentions climate.
- C is wrong because the helpful pairings in paragraph 11 have nothing to do with climate change.
- D is wrong because nothing ties Mark Davis’s change of mind to a warming planet.
Watch out: B, C and D all make the word invader less clear. But only A shows climate change doing it.
Choose the answer
Only A shows climate change making the word invader hard to apply. So the answer is A.
Why this answer and not the closest wrong one
B comes closest, since crops are non-natives we welcome, and that does complicate the word invader. But paragraph 9 says nothing about climate. A has both parts of the claim: a warming planet, and species that end up in new homes where the old labels don’t fit.
More questions like this
The original DOE question

The passage “Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”” as printed (pages 68–70)



Question 58 of 120
2024 Test A Math, question 11Question
Jar 1 and Jar 2 each contain cup of water. If of the water in Jar 1 is poured into Jar 2, how much water is now in Jar 2?
- A cup
- B cup
- C cup
- D cup
Show the answer
Correct answer: C
Jar 1 starts with cup of water. If of its water is poured into Jar 2, then the amount of water added to Jar 2 is cup.
Jar 2 starts with cup of water. If cup of water is added to it, then the amount of water now in Jar 2 is cup.
Question analysis
A fraction word problem about two jars that each hold half a cup of water; a quarter of the water in the first jar is poured into the second. Students find how much water the second jar now holds. Reading exactly what the quarter is a quarter of is the detail that matters.
- NYS standards
- 7.NS.3 four operations with rational numbers
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Multiplying and adding positive fractions is review by grade 7, and the problem takes two short steps. It is easier than typical grade 7 work with negative numbers or longer rational-number problems, apart from careful reading.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Turning words into math moderate
- Error-prone arithmetic minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 4, 35 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The numbers are friendly halves, quarters and eighths, and the work is two steps. The question turns on one phrase: the quarter is a fraction of Jar 1's water, not a measured quarter cup, and reading it the second way gives a choice that looks right. Students who also remember that Jar 2 already held water finish correctly.
Turning words into math moderate Understanding the question
The phrase "1/4 of the water in Jar 1" means a quarter of Jar 1's half cup, an eighth of a cup, not a quarter cup; the word of calls for multiplying. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Adding the 1/8 cup to Jar 2's 1/2 cup needs the common denominator 8, so 1/2 becomes 4/8 before the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Find the amount poured by taking one quarter of Jar 1's half cup, then add that amount to the half cup already in Jar 2.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the amount poured, 1/4 of 1/2 cup = 1/8 cup, and stops there instead of adding it to the half cup already in Jar 2.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Picks the fraction 1/4 straight from the question, as if the share poured were the amount of water now in Jar 2.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reads 1/4 of the water as 1/4 cup and adds it to Jar 2's half cup, 1/2 + 1/4 = 3/4.
Worked solution, step by step
Formulas used
- Fraction of an amount
- N is the whole amount, and a over b is the part of it you want. In math, the word of means multiply. For example, half of 10 is one half times 10, which is 5.
- Multiplying fractions
- a and c are the tops, and b and d are the bottoms. Multiply the tops together and the bottoms together. Taking a part of a part makes the pieces smaller, so the bottom gets bigger.
- Adding fractions with the same bottom
- a and b are the tops, and d is the shared bottom, called the denominator. If the bottoms match, add the tops and keep the bottom. If they do not match, rewrite one fraction first.
Sort out the question
Each jar starts with cup of water. You pour “ of the water in Jar 1” into Jar 2. You need how much water is in Jar 2 after that, in cups.
Find how much water is poured
Jar 1 has cup. You pour of it. Of means multiply.
You pour cup. Watch out: a quarter of the water is not a quarter cup. It is a quarter of a half cup. Adding cup gives cup, choice D.
Make the bottoms match
Now add the poured water to the cup already in Jar 2. The bottoms are 2 and 8, so rewrite the half with 8 on the bottom. Multiply its top and bottom by 4.
Add the water in Jar 2
Jar 2 now holds cup. Watch out: cup is only the water poured in. Stopping there gives choice A. So the answer is C, cup.
Check your answer
Jar 2 started with a half cup and got a little more. So it should hold just over a half cup, and is. Also, the water in both jars still adds up to 1 cup: Jar 1 has cup left, and .
More questions like this
The original DOE question

Question 59 of 120
2024 Test A Math, question 19Question
8.9, 8.2, 8.5, 9.0, 8.4, 8.6, 8.8
At a skating championship, there are seven judges who each award a score for each skater’s performance. The highest and lowest scores given to each skater are discarded, and the mean of the remaining scores is then calculated and reported as the skater’s final score. What is the final score for the skater who received the scores shown above from the judges?
- A8.60
- B8.62
- C8.64
- D8.70
Show the answer
Correct answer: C
The highest and lowest scores are discarded: 8.2 and 9.0.
To find the mean, add the remaining scores and then divide that sum by the total number of remaining scores.
Question analysis
A statistics question about a skating championship with seven judges: the highest and lowest of a skater's seven scores are dropped, and the rest are averaged. Students apply that rule to the listed scores, decimals close together. Following the dropping rule exactly and adding the decimals carefully both matter, since the choices are very close.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a mean is grade 6 content, and this adds only a rule about which scores to drop. Grade 7 statistics usually compares whole data sets, so this is easier, though the close choices demand exact decimal arithmetic.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Error-prone arithmetic minor
- Turning words into math minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 9, 62 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The arithmetic is short, but the choices are spaced so closely that a small slip or a skipped rule lands on a neighbor. Averaging all seven scores, or picking the middle score, gives values within a few hundredths of the answer. The question rewards following the stated rule exactly and adding the five remaining scores with care.
A very close wrong answer moderate Choosing the answer
The choices 8.60, 8.62 and 8.64 lie within 0.04 of one another, so no estimate separates them; only the exact sum of the five kept scores, 43.2, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Five one-decimal scores must be added and divided by 5; subtracting 8 from each first leaves 0.9, 0.5, 0.4, 0.6 and 0.8, which are easier to add. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The scoring rule takes two sentences to state: discard the single highest and the single lowest score, then average the five that remain. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Cross out the single highest and the single lowest score, add the five that remain, and divide that sum by 5.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Gives the median, the middle score 8.6, instead of the mean of the five scores left after the highest and lowest are dropped.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Averages all seven scores without discarding any, 60.4 ÷ 7 = 8.628…, and takes 8.62, the nearest choice to that mean.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Drops only the lowest score, 8.2, and averages the other six, (60.4 − 8.2) ÷ 6 = 8.7, keeping the top score of 9.0.
Worked solution, step by step
Formulas used
- Mean, or average
- The mean is the average. The sum is what you get when you add all the scores. Divide it by how many scores you added. It shares the total out evenly, as if every score were the same.
Sort out the rule
Seven judges gave the seven scores shown. The rule has two parts. First, throw out the highest score and the lowest score. Then find the mean of the scores that are left. The mean is the average. That is the final score you need.
Cross out the highest and lowest
The highest score is 9.0. The lowest is 8.2. Cross both out. Five scores are left: 8.9, 8.5, 8.4, 8.6 and 8.8. Watch out: drop both. Dropping only 8.2 gives 8.70, choice D. Keeping all seven gives about 8.63, which is closest to 8.62, choice B.
Add the five scores
A shortcut: each score is 8 plus a bit. The bits are 0.9, 0.5, 0.4, 0.6 and 0.8, and they add to 3.2. Five 8s make 40, and .
Divide by the number of scores
Watch out: do not just take the middle score, 8.6. That is the median, choice A. The question asks for the mean. So the answer is C, 8.64.
Check your answer
Multiply back: , the sum. Also, the mean should sit between the lowest kept score, 8.4, and the highest kept score, 8.9. 8.64 does.
More questions like this
The original DOE question

Question 60 of 120
2024 Test A Math, question 24Question
In 1991, the total public debt of the United States was about $3,600,000,000,000. In that year, there were about 250,000,000 people in the United States. Which amount is the best estimate of the public debt per person for that year?
- A$1,440
- B$14,400
- C$144,000
- D$14,400,000,000
Show the answer
Correct answer: B
The question asks for the amount of the public debt per person, so divide the total public debt by the number of people.
The amount of the public debt per person was about $14,400.
Question analysis
An estimation question with very large numbers: the 1991 United States public debt of about 3.6 trillion dollars and a population of about 250 million. Students find the debt per person. The choices differ by powers of ten, so cancelling zeros and keeping track of place value is the whole challenge.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Dividing large whole numbers and using place value are grade 5 and 6 skills. Grade 7 work goes further with rational numbers; this asks only for one careful division, with counting zeros as the main risk.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- A very close wrong answer minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 37 words.
- Corrected from the DOE handbook
- A misprinted number in the explanation was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The division itself is easy once the zeros are handled, and the question is built to punish miscounting them. Three choices carry the same digits, 144, at different sizes, so a student who loses or gains one zero still finds a match. A quick estimate in powers of ten, 3.6 trillion over a quarter of a billion, confirms that the answer is in the tens of thousands.
Error-prone arithmetic moderate Working it out
The debt has 13 digits and the population 9; cancelling seven zeros from each leaves 360,000 ÷ 25, and losing or adding one zero shifts the answer tenfold. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
Three choices share the digits 144 and differ only by powers of ten, so getting the digits right does not settle it; only exact place value does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Strategy: Cancel the same number of zeros from the debt and the population, then divide the smaller numbers that remain; check the size by estimating with powers of ten.
Traps
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
The wrong answers, classified
A An arithmetic slip (Junker & Sijtsma, 2001)
Cancels one zero too many while dividing and ends with $1,440, a per-person amount ten times too small.
C An arithmetic slip (Junker & Sijtsma, 2001)
Cancels one zero too few from the debt and ends with $144,000, a per-person amount ten times too large.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Divides the debt by 250 alone, dropping the six zeros that follow 250 in the population without removing any from the debt.
What was corrected from the DOE handbook, and why
- Explanation: The handbook’s first division writes the debt as “3,6000,000,000,000”, with an extra zero. The question gives $3,600,000,000,000, and the handbook’s next line (where the zeros are cancelled) uses 3,600,000,000,000, so the site now reads 3,600,000,000,000 ÷ 250,000,000. The answer, $14,400, is unchanged.
Worked solution, step by step
Formulas used
- Amount per person
- The total is the whole amount, here the debt in dollars. Dividing by the number of people shares it out equally. The result is how much goes with each one person.
- Canceling zeros
- a and b are any numbers, and b is not zero. Dividing the top and bottom of a fraction by 10 does not change its value. So you can cross off a zero at the end of the top and one at the end of the bottom, as a pair.
Sort out the question
The debt was about $3,600,000,000,000, which is 3.6 trillion dollars. There were about 250,000,000 people, which is 250 million. You need the debt per person. Per person means the amount for each one person.
Set up the division
Share the debt equally by dividing it by the number of people. The debt goes on top, and the people go on the bottom.
Cancel zeros in pairs
Count the zeros at the end of each number. The debt ends in 11 zeros. The population ends in 7. Cross off 7 zeros from each. Each pair you cross off divides the top and bottom by 10.
Watch out: cross off the same number from each. One too many from the debt gives $1,440, choice A. One too few gives $144,000, choice C.
Divide what is left
Dividing by 25 is the same as dividing by 100 and then multiplying by 4. That is because 4 groups of 25 make 100.
The debt was about $14,400 per person. Watch out: choice D divides by 250 only and forgets the zeros after it. So the answer is B, $14,400.
Check your answer
Multiply back. First, . Then put back the 7 zeros you crossed off: 3,600,000,000,000. That is the debt, so $14,400 is right.
More questions like this
The original DOE question

Question 61 of 120
2024 Test A Math, question 26Question

What is the area of the parallelogram shown above?
- A750 sq ft
- B600 sq ft
- C500 sq ft
- D300 sq ft
Show the answer
Correct answer: B
To find the area of a parallelogram, multiply the length of a side by the altitude for that side.
One property of parallelograms is that opposite sides are the same length.
So the bottom side of the parallelogram is 30 feet long and has an altitude of 20 feet.
Therefore, the area of the parallelogram, in square feet, is .
Question analysis
A geometry question with a figure: a parallelogram is labeled with a 30-foot top side, a 25-foot slanted side and a 20-foot height drawn at a right angle. Students find its area. The figure gives one length that is not needed, so students must pick the base and height that belong together.
- NYS standards
- 6.G.1 area of polygons
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Area of a parallelogram is grade 6 content, and this is a one-step use of the formula. It is easier than typical grade 7 area work, which involves composite figures, circles or scale drawings.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 2, 17 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is a one-step formula question with whole numbers, made slightly harder by one length too many. A student who multiplies the two side lengths, as for a rectangle, or brings in a triangle's one half, finds a listed choice. Matching the base to the height marked with a right angle is all it takes.
Reading a figure, table or graph minor Understanding the question
The figure labels three lengths, but only the 30-foot side and the 20-foot height drawn at a right angle to it belong together; the slanted 25-foot side is extra. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The parallelogram area formula, base times height, is not given, and its height must be the perpendicular distance, not the slanted side. (Anderson & Krathwohl, 2001)
Strategy: Use base times height, where the height is the perpendicular distance between the two parallel sides chosen as bases.
Traps
- Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
- Information you don't need: The question gives numbers or facts that play no part in the answer.
The wrong answers, classified
A Wrong formula (King et al., 2004)
Multiplies the two side lengths, 30 × 25 = 750, as if a parallelogram's area were side times side like a rectangle's.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Pairs the 20-foot height with the slanted 25-foot side, 25 × 20 = 500, though the height is drawn at a right angle to the 30-foot sides.
D Wrong formula (King et al., 2004)
Uses the triangle formula, 1/2 × 30 × 20 = 300, halving the base times the height.
Worked solution, step by step
Formulas used
- Area of a parallelogram
- A is the area. b is the base, one of the two parallel sides. h is the height: the straight distance between the two parallel sides, at a right angle to the base. It works because you can cut a triangle off one end, move it to the other end and make a rectangle.
Sort out the figure
A parallelogram is a four-sided shape whose opposite sides are parallel. Parallel means they never meet. The figure shows three lengths. The top side is 30 ft. The slanted left side is 25 ft. The arrow from top to bottom is 20 ft, and the small square shows a right angle. You need the area, in square feet.
Choose the base and height
The height must make a right angle with the base. The 20 ft arrow meets the bottom side at a right angle. Opposite sides of a parallelogram are equal, so the bottom side is 30 ft too. So the base is 30 ft and the height is 20 ft. Watch out: the slanted 25 ft side is not needed. Using gives choice C.
Find the area
The area is 600 sq ft. Watch out: do not multiply the two sides, . That is choice A. Do not take half either, which is for triangles and gives 300, choice D. So the answer is B, 600 sq ft.
Check your answer
Picture cutting the triangle off the left end and sliding it to the right end. The shape becomes a rectangle 30 ft long and 20 ft tall, with an area of 600 sq ft. 750 is too big, because the shape is only 20 ft tall, not 25 ft.
More questions like this
The original DOE question

Question 62 of 120
2024 Test A Math, question 29Question
Which graph represents the solution to the inequality ?
- A

- B

- C

- D

Show the answer
Correct answer: A
Use the properties of equations to solve the inequality for x.
Subtract 4 from both sides of the inequality.
The graph of will have a point at −1 and will include all values that are greater than, or to the right of, −1. The greater than or equal sign is used, which indicates that the solution includes −1. So the point at −1 will be solid. Option A is the correct answer.
Question analysis
This inequality question asks which number line graph shows the solution set of x + 4 ≥ 3. Students solve a one-step inequality by undoing the addition, then match the result to a graph. All four graphs use a solid dot and an arrow pointing right, so the whole question rests on where the shaded ray begins.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 inequality work is usually two-step, such as 3x − 5 ≥ 10, and asks students to choose an open or closed dot and a direction. This one needs a single subtraction, and all four graphs share a closed dot and a rightward arrow.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 13, 9 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Solving takes a single subtraction, and the graphs differ only in where the solid dot sits, so the question comes down to 3 − 4. Students who lose the negative sign land on the graph that starts at 1, while those who never isolate x pick a graph that starts at a number printed in the inequality. Testing 0, which satisfies 0 + 4 ≥ 3, shows that the shading must include 0.
Reading a figure, table or graph minor Understanding the question
All four graphs share a closed dot and a rightward arrow, so the only thing to read is where the dot sits, against tick labels printed with raised minus signs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Undoing the + 4 means computing 3 − 4, the one step that crosses zero; a dropped minus sign moves the dot from −1 to 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Subtract 4 from both sides to get x alone, then find the graph whose solid dot sits at that number; test one shaded value in the original inequality to confirm.
Traps
- Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
B Negatives and direction (Brown & Burton, 1978)
Computes 4 − 3 instead of 3 − 4, or drops the minus sign of −1, and starts the shaded ray at positive 1.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Starts the ray at 3, the number on the right side of the inequality, as if x ≥ 3, without subtracting the 4.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Starts the ray at 4, the number added to x, reading it off the inequality instead of solving for x.
Worked solution, step by step
Formulas used
- Keeping an inequality balanced
- a and b are the two sides of an inequality, and c is any number. The sign ≥ stands for greater than or equal to. Taking the same number away from both sides keeps the comparison true, like taking the same weight off both pans of a scale. Use it to get x by itself.
Sort out the question
An inequality is a math sentence that compares two amounts. The sign means is greater than or equal to. So says that x plus 4 is 3 or more.
You need the graph that shows every number x that makes this true. All four graphs have a solid dot and an arrow to the right. Only the place where the dot sits is different.
Get x by itself
Right now 4 is added to x. To undo adding 4, subtract 4 from both sides.
So x is −1 or any number greater than −1. Watch out: 3 − 4 is −1, not 1. Dropping the minus sign puts the dot at 1, choice B. Putting the dot at 3 or 4 means you read a number off the inequality without solving. That leads to C or D.
Match the solution to a graph
On a number line, greater numbers are to the right. So the shading starts at −1 and goes right. The arrow shows that it keeps going.
The dot is solid, or filled in, because of the or equal to part. A solid dot means −1 itself is a solution, a number that makes the inequality true.
Graph A has a solid dot at −1 and is shaded to the right. So the answer is A, the graph with a solid dot at −1.
Check your answer
Test a number from the shaded part, such as 0. , and 4 is more than 3. Now test the dot: , and 3 is equal to 3, so −1 works. A number left of the dot, −2, gives . That is less than 3, so −2 should not be shaded, and it isn’t.
More questions like this
The original DOE question

Question 63 of 120
2024 Test A Math, question 31Question
Nura made a square poster with a side length of 13 inches. Latrice made a square poster with a side length of 15 inches. What is the difference, in square inches, between the area of Latrice’s poster and the area of Nura’s poster?
- A56
- B8
- C4
- D2
Show the answer
Correct answer: A
Nura’s poster is square, so all the side lengths are 13 inches. The area, in square inches, is
Latrice’s poster is square, so all the side lengths are 15 inches.
Subtract to determine the difference between the areas.
The difference, in square inches, is 56.
Question analysis
This area word problem compares two square posters, one 13 inches on a side and one 15 inches on a side, and asks for the difference between their areas in square inches. Students square each side length and subtract. Working with the difference in the sides, or with perimeters, instead of the areas is the easy mistake.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Area of a square is elementary content, and squaring 13 and 15 is grade 6 exponent work. Typical grade 7 area problems involve circles, composite shapes or scale drawings; this one has two familiar areas and a subtraction.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 41 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The method is short, but the question invites a shortcut: the sides differ by 2, so it feels as if the areas should differ by 2 squared. Two choices reward working with the side lengths and one rewards perimeters. Squaring each side before subtracting avoids all three, and a sketch of the 13-inch square inside the 15-inch one shows the gap is an L-shaped strip, far more than 4 square inches.
Something you must already know minor Working it out
The area of a square, side times side, is not given; students must square 13 and 15 themselves rather than work with the side lengths. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
Squaring 13 and 15 by hand gives 169 and 225, and the subtraction 225 − 169 needs a borrow, where an arithmetic slip creeps in. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Find each poster's area by squaring its side length, then subtract the smaller area from the larger; work with the areas, not the side lengths.
Traps
- Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
B Wrong formula (King et al., 2004)
Compares perimeters instead of areas, 4 × 15 − 4 × 13 = 60 − 52 = 8, a difference in inches, not square inches.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Subtracts the sides first and squares the difference, (15 − 13)² = 4, as if 15² − 13² equaled (15 − 13)².
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Gives the difference in side lengths, 15 − 13 = 2, a length rather than a difference in area.
Worked solution, step by step
Formulas used
- Area of a square
- A is the area, the flat space inside the shape. s is the length of one side. All four sides of a square are equal, so the area is side times side. The small 2 means squared, or times itself. Area is measured in square units, here square inches.
Write down what you know
Nura’s poster is a square with sides of 13 inches. Latrice’s poster is a square with sides of 15 inches. You need the difference between their areas, in square inches. The difference is how much bigger one area is than the other. So find each area, then subtract.
Find the area of Nura’s poster
Use the area formula with the side of 13 inches.
Nura’s poster has an area of 169 square inches.
Find the area of Latrice’s poster
Use the same formula with the side of 15 inches.
Latrice’s poster has an area of 225 square inches.
Subtract the two areas
Take the smaller area away from the larger one.
The difference is 56 square inches.
Watch out: don’t subtract the sides first and then square. , choice C, is not the same as . The side difference, 2 inches, is choice D, a length. Comparing perimeters, the distance around, gives 8, choice B.So the answer is A, 56 square inches.
Check your answer
Picture Nura’s poster laid in a corner of Latrice’s. The extra part is an L-shaped strip 2 inches wide. One arm is 15 by 2 and the other is 13 by 2. , so the strip is 56 square inches. That matches.
More questions like this
The original DOE question

Question 64 of 120
2024 Test A Math, question 37Question

On the number line above, the distance between X and Y is 1 unit, and the distance between Y and Z is 2 units. What is the distance, in units, between Y and the midpoint of X and Z?
- A
- B1
- C
- D3
Show the answer
Correct answer: A
First, find the distance between X and Z by adding the distance between X and Y to the distance between Y and Z.
Since the distance between X and Z is 3, the midpoint between X and Z is units from X.
The midpoint between X and Z is units from X, and the distance between X and Y is 1. The distance between Y and the midpoint of X and Z is .
Question analysis
This number line question shows points X, Y and Z, with 1 unit from X to Y and 2 units from Y to Z, and asks how far Y is from the midpoint of X and Z. Students find the length of XZ, locate its midpoint, and compare it with Y. Each step produces a tempting number, so stopping early is the usual mistake.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 number line problems usually involve negative numbers and fractions as coordinates. This one has no coordinates, only two whole-number distances, and needs three short steps: a total length, a midpoint and a difference.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Reading a figure, table or graph minor
- Many steps minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 39 words.
- Corrected from the DOE handbook
- An imprecise statement about the midpoint in the explanation was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The arithmetic is light, but every intermediate result is waiting among the choices, so the risk is stopping before the last subtraction. Students also have to keep two midpoints apart: the one asked about lies between X and Z, not halfway along the 2-unit stretch from Y. Marking X at 0 and the midpoint at 1.5 on the drawing makes the half-unit gap visible.
Reading a figure, table or graph minor Understanding the question
The number line has no coordinates, only dimension arrows labeled 1 and 2, so students must assign positions themselves, such as X at 0, Y at 1 and Z at 3. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three short results come in sequence, the length XZ, the midpoint's distance from X, then its distance from Y, and each is also a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Strategy: Place X at 0 so that Y and Z get positions from the given distances, find the midpoint of X and Z as half their distance, then compare it with Y's position.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Halves the 2-unit segment from Y to Z, the midpoint of the wrong pair of points, which is 1 unit from Y; it also matches the labeled XY.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds that the midpoint is 1 1/2 units from X and stops before subtracting the 1 unit between X and Y.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Adds the two labeled distances, 1 + 2 = 3, and gives the whole length XZ rather than any distance to the midpoint.
What was corrected from the DOE handbook, and why
- Explanation: The handbook says “the midpoint between X and Z is 1½” (twice). A midpoint is a point, and this number line has no numbers on it, so the 1½ is the midpoint’s distance from X (half of XZ = 3). Both sentences now say the midpoint is “1½ units from X.” The rest is unchanged: Y is 1 unit from X, so the distance between Y and the midpoint is 1½ − 1 = ½.
Worked solution, step by step
Formulas used
- Adding lengths on a line
- X, Y and Z are points on one line, with Y between X and Z. XY and YZ are the lengths of the two short pieces. XZ is the length of the whole. The two pieces join end to end to make the whole.
- Midpoint of a segment
- A segment is the straight piece of a line between two points, its ends. The midpoint is the point exactly halfway between the ends. It splits the segment into two equal halves. So it is half the length away from each end.
Sort out the picture
The points go X, then Y, then Z, from left to right. X to Y is 1 unit, and Y to Z is 2 units. You need the midpoint of X and Z, the point halfway between them. Then you need how far Y is from that midpoint. The line has no numbers, so measure everything from X.
Find the length from X to Z
X to Z is 3 units. Watch out: 3 is the whole length, choice D. It is not a distance to the midpoint.
Place the midpoint
The midpoint of X and Z is half of 3 units from X.
So the midpoint is units to the right of X. Watch out: this is its distance from X, not from Y. Stopping here gives choice C.
Measure from Y to the midpoint
Y is 1 unit to the right of X. The midpoint is units to the right of X. Subtract to find the gap between them.
Watch out: use the midpoint of X and Z, not of Y and Z. Halving the 2 units from Y to Z gives 1, choice B.So the answer is A, unit.
Check your answer
Give each point a number, with X at 0. Then Y is at 1 and Z is at 3. Halfway between 0 and 3 is . From Y at 1 to is . This fits the picture too: Y sits a little left of the middle of XZ.
More questions like this
The original DOE question

Question 65 of 120
2024 Test A Math, question 42Question
Lian bought enough oranges to fill 4 bags. Each bag contains 8 oranges. The total cost was $11.52. At that rate, how much would Lian pay for 42 oranges?
- A$17.28
- B$15.12
- C$15.02
- D$12.52
Show the answer
Correct answer: B
First, find the amount, in dollars, that Lian paid for each orange. There are 4 bags with 8 oranges each.
Divide the total cost by the total number of oranges to get the cost per orange:
Multiply to find the cost of 42 oranges:
Question analysis
This rate word problem says Lian paid $11.52 for 4 bags of 8 oranges each and asks how much 42 oranges would cost at the same rate. Students find the price of one orange and multiply. Since 42 oranges is not a whole number of bags, working by the bag instead of by the orange invites mistakes.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a unit price and scaling it up is grade 6 rate work. The decimals are multi-digit, but the plan is routine, with less to manage than grade 7 problems built on percents or negative numbers.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- A very close wrong answer minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 2, 25 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The plan is familiar, total oranges, price per orange, then the new cost, so the work lies in the decimals. Because 42 oranges is five bags and two oranges, students who price by the bag either round up to six bags or must handle the leftover. Estimating, since 42 is about a third more than 32, gives roughly $15 and rules out the far choices but not the near one.
Error-prone arithmetic moderate Working it out
Two decimal operations are done by hand: 11.52 ÷ 32 = 0.36 for one orange, then 0.36 × 42 = 15.12, where a dropped carry is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The choices $15.12 and $15.02 differ by 10 cents, so a rough estimate of about $15 cannot choose between them; only the exact product can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Strategy: Multiply to find how many oranges were bought, divide the total cost by that number to get the price of one orange, then multiply by 42.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
The wrong answers, classified
A Misreads the question (Hegarty, Mayer & Monk, 1995)
Prices whole bags, rounding 42 oranges up to 6 bags of 8, and pays 6 × $2.88 = $17.28, missing that "at that rate" means per orange.
C An arithmetic slip (Junker & Sijtsma, 2001)
Uses the right method but slips one digit in multiplying 0.36 by 42, a tens-place error that lands 10 cents short of the total.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives $12.52, exactly one dollar more than $11.52; no price per orange or per bag leads there, so it mostly draws guesses.
Worked solution, step by step
Formulas used
- Unit price
- The unit price is what one item costs. Divide the total cost by how many items you got. It works because the total is shared equally among the items.
- Cost
- The unit price is the cost of one item, here one orange. Multiply it by how many items you buy to get the total cost. This works when every item costs the same.
Sort out the question
Lian bought 4 bags with 8 oranges in each bag. She paid $11.52 in all. You need the cost of 42 oranges. “At that rate” means each orange costs the same as before. So find the price of one orange first.
Count the oranges
Each of the 4 bags holds 8 oranges, so multiply to find how many she bought.
Lian bought 32 oranges for $11.52.
Find the price of one orange
This is easier in cents: . So one orange costs $0.36, or 36 cents.
Find the cost of 42 oranges
Break 42 into 40 and 2 to make it easier.
The 42 oranges cost $15.12.
Avoid the traps and answer
Watch out: $15.02, choice C, is only 10 cents away. Multiply and add with care. Watch out: Lian pays for each orange, not for full bags. Buying 6 whole bags would cost $17.28, choice A.So the answer is B, $15.12.
Check your answer
42 oranges is 10 more than 32. Those 10 oranges cost $3.60, since . Then . Same answer, so $15.12 is right.
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The original DOE question

Question 66 of 120
2024 Test A Math, question 43Question
- A0.005
- B0.648
- C20
- D200
Show the answer
Correct answer: D
Question analysis
This decimal division question asks for 3.6 divided by 0.018, with no context. Students rewrite the division so the divisor is a whole number and then divide. Because the divisor has three decimal places, moving the decimal point the right number of places in both numbers is where students go wrong.
- NYS standards
- 6.NS.3 multi-digit decimal operations
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Dividing multi-digit decimals is a grade 6 fluency standard, and this is a single division with no context. It is easier than typical grade 7 work, though shifting three decimal places invites a place-value slip.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 16+, 1 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is a single division, and the only real hazard is the decimal point. Shifting it three places in the divisor but only two in the dividend gives a listed answer ten times too small. An estimate settles it: 0.018 is about 0.02, and 3.6 ÷ 0.02 = 180, so the answer must be in the hundreds.
Error-prone arithmetic moderate Working it out
The divisor 0.018 has three decimal places, so both numbers must be multiplied by 1,000, turning 3.6 into 3,600, not 360, before dividing by 18. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Multiply both numbers by 1,000 so the divisor becomes a whole number, then divide; estimate first by asking roughly how many 0.02s fit into 3.6.
Traps
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Divides the wrong way round, 0.018 ÷ 3.6 = 0.005, putting the smaller number on top as if division always shrinks.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Multiplies instead of dividing, 36 × 18 = 648, and misplaces the decimal point too, since 3.6 × 0.018 is really 0.0648.
C An arithmetic slip (Junker & Sijtsma, 2001)
Moves the decimal point two places instead of three, dividing 360 by 18 to get 20, a result ten times too small.
Worked solution, step by step
Formulas used
- Scaling a division
- a is the number being divided, and b is the number you divide by. n is any number except 0. Multiplying both numbers by the same n keeps the answer the same, just like an equal fraction. Use it to turn a decimal you divide by into a whole number.
Sort out the problem
You need . The number you divide by, 0.018, has three decimal places. That means three digits after the decimal point. Dividing by a decimal is hard, so first turn 0.018 into a whole number.
Multiply both numbers by 1,000
Multiplying by 1,000 moves the decimal point 3 places to the right. Do it to both numbers, so the answer does not change.
Watch out: 3.6 has only one decimal place, so fill in zeros: 3.6 becomes 3,600. Moving its point only two places gives , choice C.Divide the whole numbers
This works because , so . Watch out: divide 3.6 by 0.018, not the other way around. Dividing 0.018 by 3.6 gives 0.005, choice A.
So the answer is D, 200.
Check your answer
Multiply back: . Also estimate. 0.018 is close to 0.02, and . The answer should be near that, and 200 is. Dividing by a number less than 1 gives a bigger answer, so 200 makes sense.
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The original DOE question

Question 67 of 120
2024 Test A Math, question 45Question
Misha wants to use ribbon to make 2 straps for a backpack. The ribbon costs $5.00 a yard. If each strap requires yards of ribbon, how much will Misha pay for the ribbon (not including tax)?
- A$4.00
- B$6.25
- C$11.25
- D$12.50
Show the answer
Correct answer: D
Each strap requires yards of ribbon, and ribbon costs $5.00 a yard.
, so the ribbon for each strap costs $6.25.
Misha wants to make 2 straps.
, so the cost of the ribbon for 2 straps is $12.50.
Question analysis
Misha needs ribbon for 2 backpack straps, each using 1 1/4 yards, and ribbon costs $5.00 a yard. This cost word problem asks how much she will pay for the ribbon before tax. Students multiply a mixed number by a price per yard and by the number of straps; it is short, but both the mixed number and the two straps need care.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Multiplying a mixed number by a whole number and finding a total cost are grade 5 and 6 skills. With two short steps and friendly numbers, this is easier than typical grade 7 problems with percents or negatives.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic minor
- Much to keep track of minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 35 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The question is short and the numbers are friendly; points are lost by stopping at one strap or by mishandling the mixed number. Pricing a single strap gives a listed choice, as does doubling only the whole-number part of 1 1/4. Finding the total ribbon first, 2 1/2 yards, before multiplying by the price keeps both errors away.
Error-prone arithmetic minor Working it out
The length per strap is a mixed number, 1 1/4 yards, so the arithmetic is easiest as 2 × 1.25 = 2.5 yards, then 2.5 × $5.00. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
The number of straps comes in the first sentence and the length of each in the last, so the 2 has to be carried through to the cost. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Strategy: Find the total ribbon first by doubling the length for one strap, then multiply the yards by $5.00; writing 1 1/4 as 1.25 keeps the arithmetic simple.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Divides the price by the length, $5.00 ÷ 1.25 = 4, instead of multiplying yards by dollars per yard.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Prices the ribbon for one strap, 1 1/4 × $5.00 = $6.25, and forgets that Misha is making 2 straps.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Doubles only the whole-number part of 1 1/4, getting 2 1/4 yards instead of 2 1/2, and pays 2.25 × $5.00 = $11.25.
Worked solution, step by step
Formulas used
- Cost
- Price per unit is the cost of one unit, here $5.00 for one yard. Number of units is how many you buy, here yards of ribbon. Multiply them to get the total before tax.
Sort out the question
Misha makes 2 straps. Each strap needs yards of ribbon. A yard is a length of 3 feet. Ribbon costs $5.00 a yard. You need the total cost for both straps, with no tax.
Find the total ribbon
First change the mixed number to a decimal. One quarter is 0.25, so is 1.25. Then double it for 2 straps.
Misha needs 2.5 yards of ribbon. Watch out: double the whole mixed number, the 1 and the quarter. Doubling only the 1 gives yards and $11.25, choice C.
Find the cost
The ribbon costs $12.50. Watch out: this is for both straps. One strap’s ribbon costs $6.25, choice B. Don’t stop there.
So the answer is D, $12.50.
Check your answer
Estimate. Each strap needs a bit more than 1 yard, so both need a bit more than 2 yards. Two yards cost $10, so the answer should be a bit more than $10. $12.50 fits. Also, 2 yards cost $10 and the extra half yard costs $2.50, which makes $12.50.
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The original DOE question

Question 68 of 120
2024 Test A Math, question 53Question
Carlos has $350 in a savings account that earns 5% simple interest each year. How much will he have in the account after 1 year, if there is no money withdrawn?
- A$17.50
- B$175.00
- C$367.50
- D$525.00
Show the answer
Correct answer: C
Find 5% of 350.
Question analysis
Carlos has $350 in a savings account earning 5% simple interest a year, and this interest question asks how much will be in the account after 1 year with no withdrawals. Students find 5 percent of the balance and use it. Reading whether the question wants the interest or the account's new balance matters most.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Simple interest is a grade 7 percent topic, but one year of interest is just 5 percent of the balance added on. Typical grade 7 problems run for several years or ask for the rate, so this is lighter.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 6, 28 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is a one-year interest question with friendly numbers, so most students who know what interest is will finish it. The two ways to lose the point are stopping at the interest, which is listed, and writing 5% as 0.5, which gives a listed interest and a listed balance of its own. A sensible balance is only a little more than $350, which rules out every wrong choice.
Something you must already know minor Working it out
Simple interest for one year is the rate times the balance, 5% of $350, added to what is already there; the stem does not spell this out. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
Writing 5% as 0.05, not 0.5, is the one place-value decision, and 0.05 × 350 = 17.5 must then be read as $17.50. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Find 5 percent of $350 by multiplying by 0.05, then decide whether the question wants the interest alone or the amount in the account.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the interest, 0.05 × $350 = $17.50, and stops without adding it to the $350 already in the account.
B An arithmetic slip (Junker & Sijtsma, 2001)
Writes 5% as 0.5 instead of 0.05, a decimal slip that makes the interest $175.00, and then also stops at the interest.
D An arithmetic slip (Junker & Sijtsma, 2001)
Writes 5% as 0.5, so the interest comes out as $175.00, and adds it to the $350 to reach $525.00.
Worked solution, step by step
Formulas used
- Simple interest
- I is the interest, the extra money the bank pays you. P is the principal, the money you put in. r is the yearly rate as a decimal. t is the time in years. Simple means you earn interest only on the money you started with.
Sort out the question
Carlos starts with $350. The account earns 5% simple interest each year. Interest is extra money the bank adds to your account. You need the total in the account after 1 year. That is the $350 plus the interest.
Change the percent to a decimal
Percent means out of 100, so 5% is 5 out of 100.
Watch out: 5% is 0.05, not 0.5. Using 0.5 makes the interest $175, which leads to B or D.Find the interest
Put in P = 350, r = 0.05 and t = 1 year.
The interest for one year is $17.50. That is the money added to the account.
Add the interest to the savings
After 1 year, Carlos has $367.50. Watch out: $17.50 is only the interest, choice A. The question asks how much is in the account.
So the answer is C, $367.50.
Check your answer
10% of $350 is $35. 5% is half of that, $17.50, so the interest is right. The total should be a little more than $350, since 5% is small. $367.50 fits. $525.00 would mean the account grew by half.
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The original DOE question

Question 69 of 120
2024 Test A Math, question 54Question
The probability of an event occurring is 0.05. What is the chance that the event will occur?
- Alikely
- Bunlikely
- Cimpossible
- Dneither likely nor unlikely
Show the answer
Correct answer: B
The probability of the event is 0.05, which equals 5%, which is less than 50%.
Option A:
An event that is likely has a probability over 50%.
Option B:
An event that is unlikely has a probability less than 50% but greater than 0%.
Option C:
An event that is impossible has a 0% probability.
Option D:
An event that is neither likely nor unlikely has a 50% probability.
Question analysis
This probability question gives an event a probability of 0.05 and asks how likely the event is to occur, described in words. Students place the decimal on the scale from 0, impossible, to 1, certain. Reading the place value of the decimal correctly is the only real hurdle.
- NYS standards
- 6.SP.6 probability from 0 to 1
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Describing in words how likely an event is from its probability between 0 and 1 is a grade 6 standard. This is one quick judgment with no computation, well below grade 7 probability work with compound events.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly understand
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 10, 23 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One judgment answers the question: 0.05 is a small probability, but it is not zero. The two tempting errors are treating a small chance as no chance at all and misreading 0.05 as 0.5, an even chance. Writing the probability as 5% or as 1 in 20 makes its size plain.
Something you must already know minor Working it out
The probability scale, 0 for impossible, 1/2 for an even chance and 1 for certain, is assumed, and 0.05 has to be placed near its low end. (Anderson & Krathwohl, 2001)
Strategy: Picture the probability scale from 0 to 1, with 0 for impossible, 1/2 for an even chance and 1 for certain, and see where 0.05 falls.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
The wrong answers, classified
A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common reading makes 0.05 likely, which would need a probability above one half, so it mostly draws guesses.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Treats a probability close to 0 as exactly 0, as if an event with a 1 in 20 chance could never happen.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reads 0.05 as 0.5, a place-value slip, which would mean an even chance of happening or not.
Worked solution, step by step
Formulas used
- Probability scale
- P is the probability of an event, a number that tells how likely it is to happen. It is always from 0 to 1. 0 means impossible, and 1 means certain. One half, or 0.5, means an even chance. The closer P is to 0, the less likely the event.
Sort out the question
Probability is a number that tells how likely an event is. An event is something that might happen. Here the probability is 0.05. You need the word that best describes that chance.
Change 0.05 to a percent
So 0.05 is 5 out of 100, or 5%. That is the same as 1 chance in 20. Watch out: 0.05 is not 0.5. A probability of 0.5 would be an even chance, choice D.
Place it on the scale
Probability Chance 0 impossible more than 0, less than 0.5 unlikely 0.5 neither likely nor unlikely more than 0.5, less than 1 likely 1 certain 0.05 is more than 0 but much less than 0.5. So the event is unlikely.
Rule out the other choices
Watch out: a small chance is still a chance. 0.05 is more than 0, so the event can happen. It is not impossible, choice C. Likely, choice A, would need more than 0.5.So the answer is B, unlikely.
Check your answer
Picture 20 tries. With a 5% chance, the event would happen about once, since 5% of 20 is 1. It would not happen about 19 times. That is rare but possible, so unlikely fits.
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The original DOE question

Question 70 of 120
2024 Test A Math, question 55Question
The table below shows the number of cups of red paint and blue paint used to make a purple paint mixture.
PURPLE PAINT
| Cups of Red Paint | Cups of Blue Paint |
|---|---|
| 1 | 1.5 |
| 4 | 6 |
| 11 | y |
Based on the relationship between the number of cups of red paint and the number of cups of blue paint, what is the value of y?
- A10.5
- B13
- C16.5
- D24
Show the answer
Correct answer: C
Divide the number of cups of blue paint by the number of cups of red paint to find the constant of proportionality.
Multiply to find the value of y.
Question analysis
A table gives cups of red and blue paint for a purple mixture: 1 and 1.5, 4 and 6, and 11 and an unknown y. This ratio table question asks for y based on the relationship between the two colors. The tricky part is noticing whether the relationship works by multiplying or by adding.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a constant of proportionality from a table is grade 7 work, but here one row already pairs 1 cup of red with its cups of blue, so a single multiplication finishes it, lighter than typical grade 7 tasks.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 4, 57 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The relationship is multiplicative, 1.5 cups of blue for every cup of red, and one multiplication finishes the question. The trap is additive thinking: red grows by 7 cups from the second row to the third, and adding the same 7 to the blue column gives a listed answer. Testing a rule on every row exposes the additive one, since blue is 0.5 more than red in the first row but 2 more in the second.
Reading a figure, table or graph minor Understanding the question
The table's first row, 1 cup of red to 1.5 cups of blue, hands over the constant of proportionality; the other two rows only confirm it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Strategy: Use the row with 1 cup of red paint to read the cups of blue per cup of red, multiply that rate by 11, and check the rate against the row with 4 cups.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- Stops one step early: A choice is a number from partway through the solution.
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Scales only the jump from 4 to 11 cups of red, 7 × 1.5 = 10.5, and forgets to add the 6 cups of blue already there.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Thinks additively: red goes up 7 cups from 4 to 11, so adds the same 7 to the 6 cups of blue, getting 13.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Multiplies the two numbers in the second row, 4 × 6 = 24, without finding how the two columns are related.
Worked solution, step by step
Formulas used
- Constant of proportionality
- x and y are two amounts that change together. k is the constant of proportionality: how much y goes with 1 of x. In a proportional relationship, y divided by x gives the same k every time.
- Proportional relationship
- x is the cups of red paint, and y is the cups of blue paint. k is the cups of blue for each cup of red. In a proportional relationship, y is always k times x. Use it to find y for any amount of red.
Sort out the table
The table shows cups of red and blue paint for the same purple. 1 cup of red goes with 1.5 cups of blue. 4 cups of red go with 6 cups of blue. You need y, the cups of blue for 11 cups of red. The same color means the paints mix at the same rate in every row.
Find the blue for each cup of red
Both rows give 1.5. So there are always 1.5 cups of blue for each cup of red. This number is the constant of proportionality.
Use the rate for 11 cups
Think of it as 11 cups plus half of 11 cups: . So y is 16.5 cups of blue.
Avoid the traps and answer
Watch out: red goes up 7 cups from 4 to 11. Adding the same 7 cups to the 6 cups of blue gives 13, choice B. Paint mixes by multiplying, not by adding. Watch out: , choice A, is only the extra blue for the 7 extra cups of red.So the answer is C, 16.5.
Check your answer
Divide to check: . That matches the other rows. Also, every row has more blue than red, so y must be more than 11. That rules out 10.5.
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The original DOE question

Question 71 of 120
2024 Test A Math, question 57Question
An ice cream shop sells 16 cups of ice cream for $48 and 4 cups for $12. There is a proportional relationship between the number of cups of ice cream and the cost. What is the constant of proportionality for this relationship?
- A3 cups per dollar
- B3 dollars per cup
- C4 cups per dollar
- D4 dollars per cup
Show the answer
Correct answer: B
There is a proportional relationship between the cost and the number of cups of ice cream.
Divide the cost by the number of cups of ice cream to find the constant of proportionality.
The constant is the cost per cup, 3 dollars.
Question analysis
An ice cream shop sells 16 cups for $48 and 4 cups for $12, and the cost is proportional to the number of cups. This proportional relationship question asks for the constant of proportionality, with its units. The choices pair two numbers with two kinds of units, so the units of the rate matter as much as the number.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 students find constants of proportionality in tables, graphs and equations. Here one division gives the number, as in grade 6 unit pricing, and the real demand is attaching the right units to it.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Something you must already know minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 4, 50 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One division gives the number, so the question is really about units. The cost depends on the number of cups, and dividing $48 by 16 cups gives the dollars for each cup; reversing the label describes a shop where one dollar buys three cups. The second offer, 4 cups for $12, is a built-in check that the rate stays the same.
A very close wrong answer moderate Choosing the answer
Two choices carry the number 3, "3 cups per dollar" and "3 dollars per cup", so finding 48 ÷ 16 is not enough; the units decide. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that dividing dollars by cups gives dollars per cup: a quotient's units read as the top quantity per the bottom one. (Anderson & Krathwohl, 2001)
Strategy: Divide one quantity by the other for one offer, check that the other offer gives the same quotient, and label it with the numerator's units per the denominator's units.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Gets 3 from 48 ÷ 16 but labels it cups per dollar, turning the rate upside down; at these prices a dollar buys only 1/3 of a cup.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Divides the two cup amounts, 16 ÷ 4 = 4, and attaches cups per dollar, a ratio of like quantities given made-up units.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Divides the two costs, $48 ÷ $12 = 4, or the two cup counts, and calls the result dollars per cup.
Worked solution, step by step
Formulas used
- Constant of proportionality
- In a proportional relationship, y is always the same number, k, times x. Here x is the number of cups and y is the cost in dollars. So k is y divided by x. Its units are the top’s units per the bottom’s units: dollars per cup.
Sort out the question
16 cups cost $48, and 4 cups cost $12. The cost is proportional to the cups. That means the cost is always the same number times the number of cups. That number is the constant of proportionality. You need it with the right units.
Divide cost by cups
For each offer, divide the cost in dollars by the number of cups.
Both offers give 3. So the constant is 3.
Name the units
Dollars are on top and cups are on the bottom. So the units are dollars per cup. 3 dollars per cup means each cup costs $3. Watch out: two choices use 3. Choice A, 3 cups per dollar, flips the rate. Then 16 cups would cost only about $5, not $48.
Avoid the traps and answer
Watch out: dividing two costs, , or two cup counts, , gives 4. That compares like amounts, so it is not a cost for each cup. It leads to choice C or D.So the answer is B, 3 dollars per cup.
Check your answer
Multiply back. At 3 dollars per cup, 16 cups cost , which is $48. And 4 cups cost , which is $12. Both match the shop’s prices.
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The original DOE question

Question 72 of 120
2024 Test B English, question 3Question
Read this sentence.
The Appalachian Trail is a really long trail that a lot of people do each year.
Which revision of the sentence uses the most precise language?
- AThe Appalachian Trail is an extremely long trail that millions of people do each year.
- BThe Appalachian Trail is a 2,200-mile trail that more than a million people hike each year.
- CThe Appalachian Trail is a 2,200-mile trail that two million people hike each year.
- DThe Appalachian Trail is a lengthy trail that a couple million people do each year.
Show the answer
Correct answer: C
The question asks for the revision of the sentence that uses the most precise language.
A. Incorrect. The word “extremely” does not specify the precise length of the trail (2,200 miles); “millions” is less precise than the specific number of people who hike the trail each year (two million); and the verb “do” imprecisely describes the purpose of visiting the trail (hiking).
B. Incorrect. While the revision specifies the precise length of the trail and the verb “hike” precisely describes the purpose of visiting the trail, “more than a million” is less precise than the specific number of people who hike the trail each year (two million).
C. CORRECT. The length of the trail (2,200 miles), the specific number of people who hike the trail each year (two million), and the action of the visitors (hiking) are clear and precise in this revision of the sentence.
D. Incorrect. The word “lengthy” is less precise than specifying the actual length of the trail (2,200 miles); “a couple million” is less precise than the specific number of people who hike the trail each year (two million); and the verb “do” imprecisely describes the purpose of visiting the trail (hiking).
Question analysis
This precise-language question gives a casual sentence saying the Appalachian Trail is “a really long trail that a lot of people do each year” and asks which revision is most precise. Several revisions swap in more specific words or numbers, so students must check every part of each sentence, not just the first improvement they notice.
- NYS standards
- 7L3a precise and concise language
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Choosing precise wording is grade 7 language work, but this item is one short sentence in everyday words. Typical grade 7 revision sets the same choice inside a paragraph, with more wording to weigh.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly evaluate
- What makes it hard
- A very close wrong answer moderate
- Every choice must be checked minor
- Evidence
- Answered from one sentence or line
- Reading level
- Question and choices: about grade 6, 89 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The question is quick once each revision is tested on three points: how long the trail is, how many people hike it, and what they do there. The near miss catches students who stop checking once they see the mileage and the verb “hike,” though its count of hikers is still a range. Words such as “extremely” and “lengthy” are the other snare, since they sound sharper without saying any more.
A very close wrong answer moderate Choosing the answer
The right revision and the one with “more than a million” share the 2,200-mile length and the verb “hike,” and differ only in the hiker count: an exact two million or an open-ended floor. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
Each revision repairs a different part of “a really long trail that a lot of people do,” so all four have to be checked for the length, the count and the verb. (Embretson & Wetzel, 1987)
Strategy: Check each revision on three points: how long the trail is, how many people visit, and what they do there. Keep only a version that is exact on all three.
Traps
- Wordy, vague or informal: A revision is grammatical but wordy, imprecise or too casual for the passage.
The wrong answers, classified
A Wordy, vague or informal (Connors & Lunsford, 1988)
Treats “extremely long” and “millions” as upgrades, though neither gives a length or a count, and the vague verb “do” is still there.
B Wordy, vague or informal (Connors & Lunsford, 1988)
Sees the exact length and the verb “hike” and stops checking, missing that “more than a million” is looser than a specific number of hikers.
D Wordy, vague or informal (Connors & Lunsford, 1988)
Takes the fancier “lengthy” and the casual “a couple million” for sharper wording, though neither is exact and “do” still does not say what visitors do.
Working it out, step by step
Figure out what the question asks
You need the revision with the most precise language. Precise means exact. A precise word tells the reader exactly what or how much. For example, a lot of homework is vague, but three pages of math homework is precise. The sentence has three vague parts: “a really long trail,” “a lot of people,” and the verb “do.”
Make a three-part test
The most precise revision must fix all three parts.
- How long is the trail? Look for an exact length.
- How many people go? Look for an exact number.
- What do they do there? Do is vague. People hike a trail.
A choice that fixes only one or two parts is not the most precise.
Check each choice
- C gives “a 2,200-mile trail,” “two million people,” and “hike.” It passes all three tests.
- B gives the length and “hike.” But “more than a million” is not exact. It could mean 1.1 million or 5 million.
- A says “extremely long” and “millions.” Neither is exact, and it keeps “do.”
- D says “lengthy” and “a couple million.” These sound fancier but say no more. It also keeps “do.”
Pick the answer
Watch out: B is very close. It has the length and the right verb, so many students stop there. But its count of people is still loose. Only C is exact on all three parts. So the answer is C.
Why this answer and not the closest wrong one
B is the closest wrong answer. It matches C on the length, 2,200 miles, and on the verb “hike.” The only difference is the number of hikers. “More than a million” sets a low limit but gives no exact count. “Two million” tells the reader exactly how many, so C is more precise.
More questions like this
The original DOE question

Question 73 of 120
2024 Test B English, question 5Passage Find Time to Volunteer
Find Time to Volunteer
(1) Many people believe that they gain a greater sense of purpose by giving their time to serve others. (2) These are the people who spend their free time volunteering at various places within the community. (3) Those who are served benefit from the work of these volunteers. (4) However, research shows that those who volunteer also receive benefits.
(5) For many students, this proposition may sound impossible. (6) Schoolwork can require hours of study each week. (7) Volunteering at a retirement center or homeless shelter requires both time and energy. (8) Extracurricular activities fill up after-school time. (9) Family obligations and part-time jobs often have to be worked into the schedule, too. (10) However, if students can find time to volunteer, even for a few hours a month, they may find that the benefits outweigh the scheduling difficulties.
(11) To begin with, students who engage in volunteer activities acquire valuable skills and experience, which can help them focus on potential fields of study and career options. (12) Furthermore, when students list volunteer work on college applications, admissions counselors see applicants who care about making their community and college campus a better place.
(13) Volunteerism benefits more than students applying to college. (14) This type of experience is equally important for students moving directly into the workforce after high school. (15) Volunteering demonstrates initiative, dedication, and a strong work ethic, qualities that companies value in a potential employee. (16) In a CareerBuilder study, 60 percent of managers regarded volunteerism as a significant asset when considering applicants.
(17) Students who volunteer undoubtedly make a positive investment in their future, but they also affect their present quality of life. (18) Studies have shown that volunteering is good for both the mind and the body. (19) Volunteering can boost mood levels and reduce anxiety. (20) According to the Mayo Clinic,1 the feeling of doing something meaningful and the appreciation received from that action can reduce stress.
(21) Even the busiest of students can find a way to donate some time to a worthy cause. (22) Though managing different commitments may seem overwhelming at times, students are sure to feel a sense of fulfillment in volunteerism.
1Mayo Clinic: a nonprofit medical organization dedicated to treatment and research
Question
Which revision of sentence 2 uses the most precise language?
- AThese are the people who spend their free time volunteering at animal shelters, helping with activities in community centers, or cleaning up parks.
- BThese are the people who spend their free time helping others in numerous ways at a variety of places, events, or organizations that need support.
- CThese are the people who spend their free time working at local establishments that help people, animals, or other groups in need of assistance.
- DThese are the people who spend a lot of time volunteering at places where they can help people in many ways.
Show the answer
Correct answer: A
The question asks which revision of sentence 2 uses the most precise language.
A. CORRECT. In sentence 2, the author states that people who volunteer in their free time do so “at various places.” The revision to sentence 2 provides the most precise language because it references three specific places (animal shelters, community centers, and parks) where people often volunteer.
B. Incorrect. Although this sentence mentions “a variety of places, events, or organizations” at which people volunteer, the revision does not use the most precise language, because it does not list specific places, events, or organizations.
C. Incorrect. Although this sentence mentions “local establishments that help people, animals, or other groups,” the revision does not use the most precise language, because it does not name the specific types of establishments at which people can volunteer.
D. Incorrect. Although this sentence mentions “places where [volunteers] can help people in many ways,” the revision does not use the most precise language, because it does not provide specific examples of where people can volunteer.
Question analysis
This revising question uses a persuasive essay urging high school students to find time to volunteer. It asks which rewrite of the second sentence, about the people who spend free time volunteering at various places, uses the most precise language. The trap is that longer, list-like wording can look more specific than it really is.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Grade 7 writers learn to replace vague wording with precise language. Here that skill is applied to one sentence of a plain persuasive essay, with no grammar to fix, which makes it lighter than typical grade 7 revision work.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly evaluate
- What makes it hard
- Evidence
- Answered from one sentence or line
- Reading level
- Passage: about grade 11, 354 words; question and choices: about grade 10, 102 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Sentence 2’s vague phrase is “various places,” and only one revision replaces it with places a reader can picture: animal shelters, community centers and parks. The wrong revisions grow longer or list the kinds of people helped, which can pass for precision on a quick read. Asking whether a revision names a real place settles it.
Every choice must be checked minor Choosing the answer
Every revision is longer than sentence 2 and sounds more detailed, so each has to be checked for whether it names actual places or only adds general words such as “organizations” and “establishments.” (Embretson & Wetzel, 1987)
Strategy: Find the vague phrase in sentence 2, then ask of each revision whether it tells the reader something more definite or simply uses more words to say the same thing.
Traps
- Wordy, vague or informal: A revision is grammatical but wordy, imprecise or too casual for the passage.
The wrong answers, classified
B Wordy, vague or informal (Connors & Lunsford, 1988)
Mistakes a longer string of general categories, “places, events, or organizations,” for precision, though none is named and the sentence only grows wordier.
C Wordy, vague or informal (Connors & Lunsford, 1988)
Is drawn by the three-part list of people, animals and other groups, but “local establishments” still leaves the actual places unnamed.
D Wordy, vague or informal (Connors & Lunsford, 1988)
Accepts “a lot of time” and “many ways,” which keep the original’s vagueness and even drop the point that volunteers give their free time.
Working it out, step by step
Figure out what the question asks
You need to rewrite sentence 2 with more precise language. Precise means exact. Precise words help the reader picture something real. Sentence 2 says: “These are the people who spend their free time volunteering at various places within the community.” Look for the part that is vague.
Find the vague phrase
The vague phrase is “at various places.” Various means different kinds. But the sentence never says which places. You can’t picture them. A better sentence would name real places. Example: I help out at different spots is vague. I help out at the library and the food bank is precise.
Check each choice
- A names three real places: “animal shelters,” “community centers,” and “parks.” You can picture each one.
- B says “a variety of places, events, or organizations.” That is a longer list of general words. It still names nothing real.
- C says “local establishments.” An establishment is a business or group. It still doesn’t say which ones.
- D says “a lot of time” and “many ways.” Those words are vague too. D even drops the idea of free time.
Pick the answer
Watch out: C has a list, “people, animals, or other groups,” so it looks detailed. But that list tells who gets help, not where volunteers go. Only one choice names the places. So the answer is A.
Why this answer and not the closest wrong one
C is the closest wrong answer. Its list of people, animals and other groups sounds specific. But “local establishments” could be any place at all. A replaces “various places” with three places a reader can picture: animal shelters, community centers and parks. That makes A the most precise.
More questions like this
The original DOE question

The passage “Find Time to Volunteer” as printed (page 163)

Question 74 of 120
2024 Test B English, question 23Passage Excerpt from Do Them No Harm!
Excerpt from Do Them No Harm!
by Zoa L. Swayne
This narrative is about the Nez Perce, an American Indian tribe, in what is now northern Idaho. The tribe is preparing for a gathering before the coming winter.
In the moon of Ta-Yum, the hottest days of summer, when salmon spawn in the little streams and huckleberries ripen in the high mountains, people from many villages of the Chopunnish Nation gathered in the Oyaip Prairie for the work and festivities of their annual camas1 harvest.
From far and near The People came. . . .
Should a stranger enter their homeland and ask, “Where are you from?” the reply was always, “We are Nee-mee-poo, The People who live here in this place.”
It was a time of Lawtiwa-mah-ton—a time of being friends together—when The People came for this last chance to enjoy being together before the Cold Moons kept them close to their fires. The visiting and trading, the foot racing and horse racing, and the gambling and stick games would be remembered and talked about long after they had forgotten the drudgery of digging and roasting camas, picking berries, or drying meat and fish. Lawtiwa-mah-ton! It was good to be friends together.
As was their custom since wahk-kee-ma, a time far back beyond the memory of man, they set up their camps in the same locations their parents and grandparents had occupied before them. Red Bear’s people, from Kamiah, made their camp near the trail that came out of the mountains. Their neighbors in Kamiah Valley, The People from Tee-e-lap-a-lo, had their camp close by. Across the wide meadow, by the great roasting pits, the camps of the Te-wap-poo and Ask-kah-poo were located. The tepees and ish-nash, brush shelters, of other groups nestled in their accustomed areas in and among the pines in such numbers that they encircled the entire meadow land.
Red Bear’s people had traveled all summer with neighboring bands, gathering and preparing roots, picking and drying berries, drying and smoking meat and fish for their winter food supply. Now they were at the Oyaip camp. The women worked hard to dig and cure as many bags of roots as they could during the warm, sunny days, for the sharp night air brought warnings that WARM was going and COLD was coming.
Everyone helped in some way. Most of the men fished or hunted for meat. While many of the women dug and roasted camas, other women and older children picked and dried berries.
And the younger children played. They played at hunting. They played with the babies. They played with their horses and puppies. They learned how to live through their play.
This sun, happiness, peace, and quiet blessed the Red Bear camp. All were busy with their daily tasks, until sudden cries came from the children playing by the trail.
“People coming! People coming! People coming on the trail from the high mountains!” they called as they ran to their elders, who looked sharply at the figures of approaching horsemen.
Were they friends or enemies? Did they bring good news or bad?
“Who can it be? What brings them here?” were the questions in every mind.
“Could they be the four hunters who had gone to Buffalo Country two summers past? Would they have news of the families who had gone long ago to Buffalo Country and never returned?”
It was customary for a hunting party to be gone for more than one season.
“Looks like hunters,” the older men agreed. “Looks like they had good hunting. Maybe our four hunters. [They have] been gone many moons.”
“Looks like five people—not four,” others observed.
Excitement grew as the riders came close enough to be recognized.
“A-a-a-a-a, they are our four hunters! But who is the fifth person?” they asked.
“Looks like a woman. Who is she?”
The hunters rode up to the welcoming crowd, proud to show off the loads of meat, hides, and other trophies of their hunt. They paraded around the encampment for all to see how strong their Hunting Power had been—what great hunters they, themselves, were.
Red Bear’s people rejoiced at their hunters’ success. Good hunters brought good to everybody. The meat meant plenty of food and the hides meant soft-tanned robes to give comfort through the Cold Moons. But it was the sight of the frail figure of the woman that aroused their curiosity. Who was she? Where had she come from? . . .
“Belongs to Red Bear people. Gone then come back,” the hunters said, as they dismounted and unloaded their packs. . . .
Now they could see! She was the daughter of the family gone so long ago! The girl-child who had left came back now—a grown woman.
“Wat-ku-ese!” the women cried. “Gone-from-Home-then-Come-Back. Wat-ku-ese!” And Wat-ku-ese was her name from that time on.
Gentle arms lifted Wat-ku-ese from her horse. The women brought her food and made a place for her to rest. For many suns they cared for her until she became stronger.
One evening Wat-ku-ese told her story for all to hear.
From DO THEM NO HARM!: Lewis and Clark Among the Nez Perce by Zoa L. Swayne. Published by Caxton Press. Copyright © 1990 by Zoa L. Swayne. Orofino, Idaho and Legacy House, Inc. Orofino, Idaho. All rights reserved.
1camas: a wild, edible plant
Question
Paragraph 1 contributes to the setting of the excerpt by establishing that
- Athe story’s location is a plentiful place that allows The People to enjoy a comfortable gathering.
- Bthe events in the story occurred in the past and are being remembered by The People.
- Cthe camp in the story is changing and that the changes are causing problems for The People.
- Dthe story’s plot begins as The People are observing their land and what it offers them.
Show the answer
Correct answer: A
The question asks how paragraph 1 contributes to the setting of the excerpt.
A. CORRECT. Paragraph 1 describes summer “when salmon spawn,” “huckleberries ripen,” and The People gather for “their annual camas harvest.” These descriptions show the abundance of food in the place where the excerpt occurs.
B. Incorrect. Although the setting refers to past events, the narration is not by The People. In addition, the voice narrating the excerpt is not an element of the setting.
C. Incorrect. Nothing in paragraph 1 suggests that the camp is changing in any way. The People gather annually in this place. In addition, the language used to describe the setting does not suggest any problems.
D. Incorrect. The People are not making observations about the setting in paragraph 1. They are described as performing actions, such as gathering “for the work and festivities of their annual . . . harvest.”
Question analysis
This setting question asks what the opening paragraph of Zoa L. Swayne’s narrative Do Them No Harm! establishes about the time and place of the Nez Perce camas harvest. The paragraph is one long sentence crowded with unfamiliar names, so students must look past them to the details and decide what kind of place and season they describe.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Seeing how an opening paragraph sets the scene matches the grade 6 standard, which names setting directly. The paragraph is one long sentence with unfamiliar Nez Perce names, but its details are concrete and the task is a single inference.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- Reading between the lines moderate
- Complex sentences minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 5, 885 words; question and choices: about grade 8, 76 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The paragraph is crowded with names, but its details are concrete: fish, berries, a harvest, festivities. Students who add them up see abundance and a happy gathering; those who look past them reach for ideas about who tells the story or about changes to come. The choice about The People observing their land is the subtle one, because the land’s gifts are there but no one in the paragraph is watching them.
Reading between the lines moderate Working it out
Paragraph 1 never calls the place plentiful; that has to be inferred from salmon that “spawn in the little streams,” huckleberries that “ripen,” and many villages gathering for the camas harvest. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Complex sentences minor Understanding the question
Paragraph 1 is a single sentence that opens with “In the moon of Ta-Yum,” a second time phrase and a “when” clause, so its subject, “people from many villages,” arrives only halfway through. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Strategy: Underline every detail in paragraph 1 about time, place and activity, then decide what overall impression of the setting they create before reading the choices.
Traps
- Half right: One half of a two-part choice is right and the other is not.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
The wrong answers, classified
B Half right (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Rightly notes that the story is set in the past, then adds that The People are remembering it, though an outside narrator tells the story.
C Reads too much in (Graesser, Singer & Trabasso, 1994)
Senses change in the coming cold of later paragraphs and reads trouble into the opening, though paragraph 1 describes a yearly gathering with no hint of problems.
D Twists what the text says (King et al., 2004)
Sees what the land offers, salmon, berries and camas, but has The People surveying it, when paragraph 1 shows them gathering to work and celebrate, not observing.
Working it out, step by step
Figure out what the question asks
The setting of a story is where and when it takes place. This asks what paragraph 1 tells you about the setting. The paragraph is one long sentence with many new names. Look past the names to the details.
Find the details
Paragraph 1 says it is “the hottest days of summer.” It is a time “when salmon spawn in the little streams and huckleberries ripen in the high mountains.” To spawn means to lay eggs. To ripen means to get ready to eat. People from many villages come to the Oyaip Prairie “for the work and festivities” of their yearly harvest. Festivities are celebrations.
Decide what the details show
Fish, berries and a harvest all mean food. So this is a place with plenty to eat. Many villages come together to work and to celebrate. So it is a happy gathering, too. The paragraph shows a rich, friendly place.
Check each choice
- A fits: a plentiful place where The People gather in comfort. Plentiful means having a lot.
- B: the story does happen in the past. But a narrator tells it. A narrator is the voice telling the story. The People are not remembering it.
- C: nothing in paragraph 1 is changing or causing problems. The gathering happens every year.
- D: The People are not studying the land. They have come to work and celebrate.
Watch out: D mentions what the land offers, which is in paragraph 1. But no one in the paragraph is watching the land.
Pick the answer
Paragraph 1 sets up a place full of food and good company. So the answer is A.
Why this answer and not the closest wrong one
D is a close wrong answer, because the land’s gifts, fish, berries and camas, are all in paragraph 1. But D says The People are observing the land. They are not. They have come “for the work and festivities” of the harvest. A sums up what the details show: plenty of food and a happy gathering.
More questions like this
The original DOE question

The passage “Excerpt from Do Them No Harm!” as printed (pages 177–178)


Question 75 of 120
2024 Test B English, question 28Uses the passage above: “Excerpt from Do Them No Harm!”.
Question
In paragraph 20, the phrases “trophies of their hunt” and “paraded around” affect the paragraph by
- Asuggesting that the hunters are rewarded for their leadership.
- Bemphasizing that The People are dependent on the hunters for food.
- Crevealing that the hunters are pleased with their success.
- Dindicating that The People have gathered specifically to welcome the hunters.
Show the answer
Correct answer: C
The question asks how the phrases “trophies of their hunt” and “paraded around” affect paragraph 20.
A. Incorrect. The phrase “trophies of their hunt” describes the animals that the hunters have successfully found. They have already received their rewards (the animals) and are showing them to The People.
B. Incorrect. The excerpt describes many ways that The People find and prepare their own food. While they appreciate the extra food, they are not dependent on the hunters.
C. CORRECT. The hunters “paraded around” because they were proud of the animals they had been able to get while hunting. Since a trophy is a recognition of success, they consider these animals their trophies.
D. Incorrect. The phrases show the pride the hunters feel. As paragraph 20 says, the hunters want “all to see how strong their Hunting Power had been.” The phrases are focused on the hunters rather than on those who have gathered around them.
Question analysis
This word-meaning question looks at the phrases “trophies of their hunt” and “paraded around” in the scene where four Nez Perce hunters ride back into camp, and asks what the phrases add to the paragraph. Students must work out what the words suggest in this context, not just their dictionary meanings, and weigh four plausible effects on the paragraph.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Grade 7 students analyze the connotations of words in context. These two phrases are familiar and sit in one short paragraph with helpful context nearby, so this is lighter than typical grade 7 word-choice work.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 5, 885 words; question and choices: about grade 8, 55 words.
- Corrected from the DOE handbook
- A quotation in the explanation was corrected to match the passage.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Paragraph 20 nearly states the answer: the hunters are “proud to show off” their loads and parade “for all to see how strong their Hunting Power had been.” Errors come from neighboring details instead: the welcoming crowd in the same paragraph, the food for everyone in the next, or the everyday sense of a trophy as an award. Reading the phrases inside their own sentences settles it.
Figurative or literary language minor Understanding the question
“Trophies of their hunt” turns meat and hides into prizes and “paraded around” suggests a show, and students must read both as signs of pride rather than of awards or ceremony. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
The right answer is reworded minor Choosing the answer
The right choice recasts “proud to show off” in new words, while one wrong choice echoes the “welcoming crowd” and another the “plenty of food” of paragraph 21. (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)
Strategy: Think about what each phrase usually suggests, then reread the whole of paragraph 20 to check that meaning against the context before looking at the choices.
Traps
- Dictionary meaning, not meaning here: A common meaning of the word that does not fit its use in this text.
- Too broad or too strong: A choice overstates, generalizes, or uses absolute words the text does not support.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
The wrong answers, classified
A Dictionary meaning, not meaning here (Perfetti, 2007)
Reads “trophies” as prizes handed out for an achievement and adds leadership, but the trophies are the meat and hides, and nothing mentions leading.
B Too broad or too strong (King et al., 2004)
Borrows paragraph 21’s point that good hunters bring food for all and inflates it into dependence, though The People gather much of their own food.
D Twists what the text says (King et al., 2004)
Is drawn by the “welcoming crowd” in the same paragraph, but The People came to Oyaip for the camas harvest and simply greet the hunters who arrive.
What was corrected from the DOE handbook, and why
- Explanation: The handbook’s explanation of Option C says the hunters “parade around.” Paragraph 20, and the question itself, say they “paraded around the encampment,” so the site now quotes “paraded around” and puts the rest of that sentence in the past tense to match.
Working it out, step by step
Figure out what the question asks
This asks what two phrases add to paragraph 20: “trophies of their hunt” and “paraded around.” Think about what each phrase suggests. Then check it against the whole paragraph.
Think about each phrase
A trophy is something you show off to prove you won, like a cup for a winning team. Here the trophies are the meat and hides from the hunt. To parade is to march around so people can watch, like in a parade. Both phrases suggest showing off something you are proud of.
Check the whole paragraph
Paragraph 20 says the hunters were “proud to show off the loads of meat, hides, and other trophies of their hunt.” Then “They paraded around the encampment for all to see how strong their Hunting Power had been.” An encampment is a camp. The paragraph ends “what great hunters they, themselves, were.” So the hunters are proud of how well they did.
Check each choice
- C fits. The hunters are pleased with their success.
- A: no one gives the hunters a prize. The trophies are what they caught, and nothing mentions leading.
- B: The People get much of their own food. They dig roots, pick berries and fish. They don’t depend on the hunters to eat.
- D: The People came for the camas harvest. They greet the hunters when they arrive. They did not gather just for them.
Watch out: D is tempting because the paragraph mentions a “welcoming crowd.” But the two phrases describe the hunters, not the crowd.
Pick the answer
Both phrases show the hunters proudly showing off the results of a good hunt. So the answer is C: the hunters are pleased with their success.
Why this answer and not the closest wrong one
D and B are the closest wrong answers, because they use nearby details: the “welcoming crowd” and the food the hunters bring. But the two phrases are about the hunters themselves. They are “proud to show off” what they caught. That pride is C.
More questions like this
The original DOE question

Question 76 of 120
2024 Test B English, question 45Passage Snowy Mountains
Snowy Mountains
by John Gould Fletcher
Higher and still more high,
Palaces made for cloud,
Above the dingy city-roofs
Blue-white like angels with broad wings,
Pillars of the sky at rest
The mountains from the great plateau
Uprise.
But the world heeds them not;
They have been here now for too long a time.
The world makes war on them,
Tunnels their granite cliffs,
Splits down their shining sides,
Plasters their cliffs with soap-advertisements,
Destroys the lonely fragments of their peace.
Vaster and still more vast,
Peak after peak, pile after pile,
Wilderness still untamed,
To which the future is as was the past,
Barrier spread by Gods,
Sunning their shining foreheads,
Barrier broken down by those who do not need
The joy of time-resisting storm-worn stone,
The mountains swing along
The south horizon of the sky;
Welcoming with wide floors of blue-green ice
The mists that dance and drive before the sun.
“Snowy Mountains” by John Gould Fletcher—Public Domain
Question
How does the poet develop the speaker’s point of view in the second stanza (lines 8–14)?
- Aby describing images of the mountains’ awe-inspiring size and strength
- Bby illustrating the differences among the various ways humans can affect the natural environment
- Cby criticizing society for taking careless, harmful courses of action against nature
- Dby demonstrating how the mountains and the people are able to benefit from each other
Show the answer
Correct answer: C
The question asks how the poet develops the speaker’s point of view in the second stanza.
A. Incorrect. The speaker’s descriptions of the mountains’ awe-inspiring size and strength mainly occur in the first and third stanzas, while in the second stanza, the speaker focuses on humanity and its negative effect on the mountains.
B. Incorrect. The wording of the second stanza—“makes war” (line 10), “Splits down their shining sides” (line 12), and “Destroys the lonely fragments” (line 14)—communicates the speaker’s position that people’s main impact on the mountains is destructive; the wording does not illustrate different ways people impact the natural environment.
C. CORRECT. In the second stanza, the speaker criticizes the destructive actions of “the world” (lines 8 and 10) against the mountains—actions such as tunneling “their granite cliffs” (line 11) and destroying “the lonely fragments of their peace” (line 14).
D. Incorrect. In the second stanza, the speaker does not describe a positive relationship between the mountains and the people.
Question analysis
This point-of-view question on the poem “Snowy Mountains” asks how the poet develops the speaker’s outlook in the second stanza, lines 8–14. Students must read the stanza on its own, decide what attitude its images and verbs express, and then find the choice that names both that attitude and the way it comes across.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Below typical grade 7 demand. Grade 7 point-of-view questions often require weighing subtle or shifting attitudes; this stanza is seven short lines with one consistent stance, carried by strong action words, so it suits grade 6.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- How, not what moderate
- Figurative or literary language minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: 161 words (a poem; reading level not measured); question and choices: about grade 10, 66 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The speaker’s anger in lines 8–14 is plain, and most students sense that the stanza blames people. Points are lost to choices that describe the poem’s admiration of the peaks, which belongs to the stanzas around this one, or that list the stanza’s actions without their hostility. Staying inside the seven lines and reading the attitude in verbs such as “makes war” and “Destroys” settles it quickly.
How, not what moderate Working it out
The stem asks how the viewpoint is built, and the answer lies in what the world does in lines 10–14: it “makes war,” “Tunnels,” “Splits,” “Plasters” and “Destroys.” (Anderson & Krathwohl, 2001)
Figurative or literary language minor Understanding the question
The stanza personifies the mountains as victims, ignored and robbed of “the lonely fragments of their peace,” so the criticism of people comes through sympathy for the peaks. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Strategy: Read lines 8–14 by themselves and put the speaker’s attitude into a few words of your own before looking at the choices.
Traps
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
The wrong answers, classified
A Wrong part of the text (Kirsch & Mosenthal, 1990)
Describes the first and third stanzas, which marvel at the peaks’ height and vastness; in lines 8–14 the “granite cliffs” and “shining sides” appear only as things being damaged.
B Twists what the text says (King et al., 2004)
Sees that the stanza names several human actions, tunneling, splitting and plastering, but they are all blows in one “war,” not different kinds of effect laid side by side.
D Wrong tone or attitude (Graesser, Singer & Trabasso, 1994)
Reads tunnels and advertisements as people putting the mountains to use, a kind of exchange, and misses the hostility of “The world makes war on them.”
Working it out, step by step
Figure out what the question asks
This asks about point of view. That is how the speaker, the voice talking in the poem, feels about the subject. It also asks how the poet shows that feeling. Look only at the second stanza, lines 8 to 14.
Read the second stanza closely
The stanza opens with “But the world heeds them not” (line 8). To heed something is to pay attention to it. Then “the world” starts to act. It “makes war on them,” “Tunnels their granite cliffs,” and “Splits down their shining sides.” It covers them with ads. Finally, in line 14, it “Destroys the lonely fragments of their peace.”
Decide what the speaker thinks
Here “the world” means people in general. People ignore the mountains, yet keep hurting them. Each action is an attack: making war, tunneling, splitting, plastering, destroying. The speaker is clearly on the mountains’ side. By listing these acts, the speaker blames people for being careless and harmful. Blaming someone for what they do wrong is called criticism.
Check each choice
- A does not fit. Size and strength fill the first and third stanzas, as in “Higher and still more high” and “Vaster and still more vast.” In lines 8 to 14, the mountains are only victims.
- B does not fit. The stanza lists several acts, but it does not compare them. They are all part of one “war.”
- C fits. The stanza blames society for ignoring and harming the mountains.
- D does not fit. The stanza shows no give-and-take. People only harm the mountains, and the mountains get nothing back.
Choose the answer
The speaker’s view comes through a list of careless, harmful acts that the poem plainly condemns. So the answer is C.
Why this answer and not the closest wrong one
B is the closest wrong choice, because the stanza really does name several ways people affect the mountains. But the speaker is not studying how they differ. Each one is another blow in the same attack, and the speaker is upset by all of them. C names that anger as criticism, which is what the stanza does.
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The original DOE question

The passage “Snowy Mountains” as printed (page 142)

Question 77 of 120
2024 Test B Math, question 10Question
- A
- B
- C
- D
Show the answer
Correct answer: B
Question analysis
A fractions question asks for 4 2/3 divided by 2 1/2, with mixed numbers on both sides of the division sign. Students convert to improper fractions, multiply by the reciprocal and simplify. Mixed numbers invite shortcuts, so every step has to follow the rule for dividing fractions.
- NYS standards
- 6.NS.1 dividing fractions by fractions
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Dividing fractions is a grade 6 standard, and this is one division with small mixed numbers. Grade 7 work usually adds negative numbers or a word problem around it, so this sits below grade 7.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- An answer in an unexpected form minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 16+, 5 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Dividing mixed numbers has a fixed procedure, and every wrong choice is a shortcut that skips part of it. Estimating first helps: 4 2/3 is a little less than twice 2 1/2, so the quotient is a little under 2, which already rules out 15/28 and 3 1/3. The remaining near miss, 2 1/3, comes from dividing by the whole number 2 alone.
Error-prone arithmetic moderate Working it out
Both mixed numbers must become improper fractions, 14/3 and 5/2, before the second is flipped, and 14/3 × 2/5 = 28/15 leaves an improper fraction to deal with. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The choices are written as mixed numbers and one proper fraction, so 28/15 has to be converted to a mixed number before it can be matched. (Enright, Morley & Sheehan, 2002)
Strategy: Rewrite both mixed numbers as improper fractions, multiply the first by the reciprocal of the second, then simplify and turn the improper fraction back into a mixed number.
Traps
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
The wrong answers, classified
A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Flips the first fraction instead of the second, computing 3/14 × 5/2 = 15/28, which is 2 1/2 divided by 4 2/3.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Divides 4 2/3 by the whole number 2 and drops the 1/2 of the divisor, so 14/3 ÷ 2 = 7/3, or 2 1/3.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Divides the whole parts and the fraction parts separately, 4 ÷ 2 = 2 and 2/3 ÷ 1/2 = 4/3, then adds them to get 3 1/3.
Worked solution, step by step
Formulas used
- Mixed number to fraction
- a is the whole number, and b over c is the fraction part. Multiply the whole number by c to turn the wholes into pieces. Then add the b extra pieces. The bottom number stays c.
- Dividing fractions
- a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.
Sort out the problem
You need to divide by . Both are mixed numbers, a whole number and a fraction together. Mixed numbers are hard to divide as they are. So first change each one into an improper fraction. That is a fraction whose top is bigger than its bottom.
Change the mixed numbers
Watch out: do not divide the whole parts and the fraction parts on their own. Doing and , then adding, gives , choice D.Flip the second fraction and multiply
Multiply the tops, , and the bottoms, . Watch out: flip only the second fraction. Flipping the first one gives , choice A.
Change back to a mixed number
The choices are mixed numbers, so change 28/15 back. 15 goes into 28 one time, with 13 left over.
Watch out: dividing by just 2 and dropping the one half gives , choice C.So the answer is B, .
Check your answer
Estimate. is about 4.7, and is 2.5. Two groups of 2.5 make 5, so the answer is a bit less than 2. is about 1.9, which fits. Or multiply back: .
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The original DOE question

Question 78 of 120
2024 Test B Math, question 11Question
In a sports league, each team has 36 players and 3 coaches. There are also a number of team assistants. The ratio of team assistants to players is . What is the ratio of coaches to assistants?
- A
- B
- C
- D
Show the answer
Correct answer: B
Let x be the number of assistants.
Since the ratio of assistants to players is , then:
There are a total of 6 assistants.
So the ratio of coaches to assistants is .
Question analysis
A ratio word problem about a sports team with 36 players, 3 coaches and an unknown number of assistants, where the ratio of assistants to players is 1 to 6. Students find how many assistants there are, then form a new ratio of coaches to assistants. Keeping the order of each ratio straight matters.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a missing amount from a ratio and then writing a second ratio is grade 6 ratio reasoning with whole numbers. Typical grade 7 work adds fractions, rates or percents, which this two-step problem does not.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Turning words into math minor
- Much to keep track of minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 39 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Two short steps, finding 6 assistants and then writing 3 to 6, make this one of the easier ratio questions. The care needed is in keeping each ratio's order and its two groups straight, since the given ratio compares assistants with players and the asked one compares coaches with assistants. None of the wrong choices matches a common slip closely, so students who find the 6 assistants seldom go wrong.
Turning words into math minor Understanding the question
“The ratio of team assistants to players is 1/6” must become 6 assistants for 36 players, and the question then asks for a ratio in a new order, coaches first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Three groups are named, 36 players, 3 coaches and an unknown number of assistants, and each ratio pairs a different two of them. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Strategy: Use the 1 to 6 ratio to find the number of assistants from the 36 players, then write coaches over assistants, in that order, and simplify.
Traps
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Compares the 9 coaches and assistants together with the 36 players, 9/36 = 1/4, a ratio of all the staff to players that the question never asks for.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 2/3; it equals the 6 assistants over all 9 staff, which puts the wrong group on top and a total underneath.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reads the ratio 1/6 as the assistants' share of everyone and gives the other 5/6, as if that remaining share described the coaches.
Worked solution, step by step
Formulas used
- Ratio as a fraction
- A ratio compares two amounts. The group named first goes on top, and the group named second goes on the bottom. Order matters. Assistants to players is not the same as players to assistants.
- Equal ratios
- a and b are the two numbers in a ratio, written as a fraction. n is any number except zero. Multiplying both numbers by the same n gives an equal ratio. This is how you scale a small ratio up to real counts.
Sort out what you know
Each team has 36 players and 3 coaches. It also has some assistants, but you are not told how many. The ratio of assistants to players is 1/6. That means 1 assistant for every 6 players. You need the ratio of coaches to assistants.
Find the number of assistants
Write the ratio as a fraction, assistants on top and players on the bottom. To get from 6 players to 36, multiply by 6. Do the same to the top.
So there are 6 assistants for the 36 players.
Write the new ratio
Now the order changes. Coaches come first, so coaches go on top.
Divide the top and the bottom by 3 to simplify.
Avoid the traps and answer
Watch out: use only coaches and assistants. Putting all 9 staff over the 36 players gives 1/4, choice A. Putting the 6 assistants over all 9 staff gives 2/3, choice C. And 1/6 is not a share of everyone, so 5/6, choice D, has no meaning here.So the answer is B, .
Check your answer
A ratio of 1/2 means 1 coach for every 2 assistants. With 3 coaches, that makes 6 assistants. Then 6 assistants to 36 players is , the ratio the question gave.
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The original DOE question

Question 79 of 120
2024 Test B Math, question 12Question
Lena is building a fence. She will need to dig holes to help support the posts that hold up the fence. The holes need to have a depth of feet below the ground. Each post is 10 feet long. What is the height of the part of the post that is above the ground?
- A feet
- B feet
- C10 feet
- D feet
Show the answer
Correct answer: A
The posts are each 10 feet long, and feet of each post is below the ground, so the height of the part of the posts that is above the ground is the difference between the two amounts:
Question analysis
A fence-building word problem: each post is 10 feet long, and 3 1/3 feet of it goes into a hole in the ground. Students find how much of the post stands above ground by subtracting a mixed number from a whole number. Regrouping the whole number into thirds is the step that trips students up.
- NYS standards
- 7.NS.3 four operations with rational numbers
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Subtracting a mixed number from a whole number is fraction work from before grade 7, set here in a one-step story. It asks much less than grade 7 problems, which add negative numbers or several steps.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic minor
- Solution
- 1 step
- Reading level
- Question and choices: about grade 2, 61 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The situation is plain, a post partly buried in a hole, and the subtraction 10 − 3 1/3 is the whole task. The only real trap is the regrouping, since there are no thirds to take the 1/3 from until one foot is split into three. Checking that the two parts of the post add back to 10 feet catches the slip.
Error-prone arithmetic minor Working it out
Taking 3 1/3 from the whole number 10 requires regrouping 10 as 9 3/3, the step where students subtract the whole numbers and simply carry the 1/3 along. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Subtract the buried part from the whole post, rewriting 10 as 9 and 3/3, or as 30/3, so that the thirds can be taken away.
Traps
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Subtracts the whole numbers, 10 − 3 = 7, then carries the 1/3 along instead of taking it away too, getting 7 1/3.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Gives the post's full 10-foot length, the whole post instead of the part above ground, overlooking the 3 1/3 feet sunk in the hole.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Adds the depth of the hole to the length of the post, 10 + 3 1/3, as if digging the hole made the post taller.
Worked solution, step by step
Formulas used
- Parts of a whole
- The whole is the full length of the post. Below is the part in the hole, and above is the part you can see. The two parts add up to the whole, so subtract to find one part.
- One whole in thirds
- One whole can be cut into 3 equal thirds. So 1 is the same as 3 over 3. Use this when you need to take thirds away from a whole number.
Sort out the question
Each post is 10 feet long. Part of it goes into a hole feet deep. The rest sticks up above the ground. You need the height of the part above the ground.
Decide to subtract
The part below and the part above make up the whole post. So take the part below away from the whole.
Watch out: do not add. , choice D, makes the post longer than it is.Turn one whole into thirds
The 10 has no thirds to take 1/3 from. So take 1 whole from the 10 and cut it into 3 thirds.
Now there are thirds to subtract from.
Subtract wholes and thirds
Subtract the whole numbers, then the thirds.
That is and . Watch out: the one third must be taken away too. Doing 10 − 3 = 7 and then keeping the one third gives , choice B.
So the answer is A, feet.
Check your answer
Add the two parts back together: . That is the whole post. Also, the answer must be less than 10 feet, since part of the post is under the ground.
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The original DOE question

Question 80 of 120
2024 Test B Math, question 21Question
Which number shown above is the greatest?
- A
- B
- C
- D
Show the answer
Correct answer: A
All the numbers are negative, so they are all less than zero. Consider the absolute value of each number and where it falls on a number line. The negative number closest to 0 will be the greatest.
The negative number with the greatest absolute value is −49, so it has the least value.
The negative number with the next greatest absolute value is −1, so it would fall next in order from least to greatest.
Compare the absolute values of and .
Therefore, the numbers listed from least to greatest are:
So the number is the greatest.
Question analysis
Four negative numbers are listed: negative 1/49, negative 49, negative 1/7 and negative 1. Students decide which of them is the greatest. Comparing negative fractions with negative whole numbers tests whether students order numbers by their position on the number line or by their size without the sign.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Ordering positive and negative rational numbers on a number line is grade 6 content in New York. This is a one-step comparison of four numbers, lighter than typical grade 7 work with signed numbers.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- A very close wrong answer moderate
- Something you must already know minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 10, 12 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
With no arithmetic to do, the question tests how firmly students picture negative numbers on a line. Comparing sizes without the sign picks the number farthest from 0, and comparing the two fractions by habit picks the wrong one of the pair. Placing all four as points left of 0, with −1 between −49 and the fractions, makes the order plain.
A very close wrong answer moderate Choosing the answer
−1/49 and −1/7 differ only in the denominator, and for negative numbers the larger denominator gives the greater value, the reverse of what their sizes suggest. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must bring the number-line fact that a negative number's value falls as its distance from 0 grows, so −49 is the least of the four, not the greatest. (Anderson & Krathwohl, 2001)
Strategy: Picture each number's place on the number line: among negative numbers, the one closest to 0 is the greatest, so compare their distances from 0.
Traps
- Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
B Negatives and direction (Brown & Burton, 1978)
Picks the number with the largest digits, 49, judging its size without the minus sign that puts it farthest below 0.
C Negatives and direction (Brown & Burton, 1978)
Knows 1/7 is greater than 1/49 and keeps that order for the negatives, missing that the order reverses on the left of 0.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Places −1/7 and −1/49 below −1, as if dividing a negative number made it more negative, so −1 looks like the greatest.
Worked solution, step by step
Formulas used
- Comparing negative numbers
- a and b are positive numbers. Putting a minus sign on both flips their order. The one with the smaller size is closer to 0, so it is greater. For example, 2 is less than 5, but −2 is greater than −5.
Sort out the question
All four numbers are negative, which means less than 0. You need the greatest one, the one with the highest value. On a number line, greater numbers are farther to the right.
Think about distance from 0
For negative numbers, the one closest to 0 is the greatest. The size of a number without its sign is its absolute value. It is the number’s distance from 0. So look for the smallest size.
Compare the sizes
The sizes are 1/49, 49, 1/7 and 1. Both 1/49 and 1/7 have 1 on top. Cutting 1 into 49 pieces gives much smaller pieces than cutting it into 7. So 1/49 is less than 1/7.
So 1/49 is the smallest size. That puts closest to 0.
Put the numbers in order
Flip the order of the sizes to order the negative numbers, from least to greatest.
Watch out: −49 has the biggest digits, but it is the least, so choice B is wrong. Watch out: 1/7 is more than 1/49, but −1/7 is less than −1/49. So choice C is wrong.So the answer is A, .
Check your answer
Change the fractions to decimals: and . Think of money you owe. Owing 2 cents is better than owing 14 cents, $1 or $49. So −1/49 is the greatest.
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The original DOE question

Question 81 of 120
2024 Test B Math, question 23Question
Each loaf of bread a baker makes requires ounce of salt. How many loaves can be made if 40 ounces of salt are used?
- A8
- B20
- C45
- D200
Show the answer
Correct answer: D
Each loaf of bread requires ounce of salt. Therefore, the baker can make 5 loaves of bread with 1 ounce of salt. Multiply to find the number of loaves that can be made with 40 ounces of salt:
Question analysis
A baking word problem: each loaf of bread needs 1/5 ounce of salt, and the baker uses 40 ounces. Students find how many loaves that much salt makes, working with a whole number and a unit fraction. Deciding whether to multiply or divide is the whole question.
- NYS standards
- 6.NS.1 dividing fractions by fractions
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Dividing a whole number by a fraction in a word problem is a grade 6 standard. This takes one step with a unit fraction, simpler than typical grade 7 rational number problems.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Turning words into math minor
- Solution
- 1 step
- Reading level
- Question and choices: about grade 3, 24 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The only decision is whether to multiply or divide, and thinking about one ounce first settles it: one ounce holds five fifths, so it makes five loaves. Students who multiply 40 by 1/5 get a small number, but a tiny amount of salt per loaf should give many loaves, not few. That reasonableness check leaves a single choice.
Turning words into math minor Understanding the question
“How many loaves” from 40 ounces at 1/5 ounce each asks how many fifths fit in 40, a division that many students turn into multiplying by 1/5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Work out how many loaves one ounce of salt makes, then scale up to 40 ounces; dividing 40 by 1/5 gives the same count.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Multiplies 40 by 1/5 instead of dividing, finding a fifth of the salt rather than how many fifths of an ounce it holds.
B An arithmetic slip (Junker & Sijtsma, 2001)
Knows to multiply 40 by 5 but drops a zero, writing the product as 20 instead of 200.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Adds the 5 from 1/5 to the 40 ounces, combining the numbers in the problem without a model of the situation.
Worked solution, step by step
Formulas used
- Number of equal groups
- Total is the whole amount you have. Size of one group is how much each one uses. Dividing tells how many groups fit into the total. Here one group is the salt for one loaf.
- Dividing by a unit fraction
- a is any number. 1 over n is a unit fraction, a fraction with 1 on top. Dividing by it asks how many nths fit into a. Each whole holds n of them, so you multiply by n.
Sort out what you know
Each loaf uses ounce of salt. The baker uses 40 ounces of salt. You need how many loaves that makes. That means how many fifths of an ounce fit into 40 ounces.
Decide to divide
You are splitting the salt into equal groups of 1/5 ounce. Each group makes one loaf. So divide the total salt by the salt for one loaf.
Watch out: do not multiply by 1/5. , choice A, is a fifth of the salt, not a number of loaves.Divide by the fraction
Each ounce holds 5 fifths, so 1 ounce makes 5 loaves. Dividing by 1/5 is the same as multiplying by 5.
The salt makes 200 loaves. Watch out: keep both zeros. 20, choice B, drops one.
So the answer is D, 200.
Check your answer
Multiply back. 200 loaves at ounce each use ounces. That is all the salt. Also, each loaf uses less than 1 ounce, so there should be more loaves than ounces. 200 is more than 40.
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The original DOE question

Question 82 of 120
2024 Test B Math, question 38Question
Michael received a check for $213 for working 20 hours at his part-time job. The amount on the check was his total pay after $32 in taxes was deducted. What was Michael’s hourly pay rate before taxes were deducted?
- A$9.05
- B$10.65
- C$12.25
- D$21.35
Show the answer
Correct answer: C
Add the taxes to the amount of the check to find Michael’s total pay before taxes were deducted:
Divide the total pay by the number of hours to find the hourly pay rate:
Question analysis
Michael's paycheck shows what he earned for 20 hours after taxes were taken out, and the tax amount is given. Students find his hourly pay rate before taxes. It tests unit rates in a money word problem, where students must decide what to do with the tax that was deducted.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Unit rate problems are grade 6 content, and this one needs one addition and one division with a decimal result. It is easier than typical grade 7 work, apart from deciding which amount is the pay before taxes.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Turning words into math minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 36 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Two operations in the right order, adding the tax back and dividing by the hours, are all this takes. The danger is the word deducted, which pulls students toward subtracting the $32 a second time, and the phrase before taxes, which rules out the take-home rate from dividing the check alone. Checking that $12.25 × 20 − $32 gives back $213 confirms the answer.
Turning words into math minor Understanding the question
"The amount on the check was his total pay after $32 in taxes was deducted" makes $213 what remained, so the $32 is added back, not taken away. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Rebuild Michael's pay before taxes from the check and the tax that was taken out, then divide that total by the 20 hours worked.
Traps
- Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
A Follows a key word (Hegarty, Mayer & Monk, 1995)
Follows the word deducted and subtracts the $32 from the $213 check again, (213 − 32) ÷ 20 = 9.05, instead of adding it back.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Divides the $213 check by 20 hours, which gives Michael's hourly pay after taxes rather than before.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No standard working gives $21.35, which equals 32 − 10.65; dividing $213 by 10 hours instead of 20, $21.30, comes closest, so it mostly draws guesses.
Worked solution, step by step
Formulas used
- Hourly pay rate
- rate is the pay for one hour. total pay is the money earned for all the hours. hours is the time worked. Dividing shares the total pay equally among the hours. Use it when the pay for each hour is the same.
Sort out the question
Michael’s check was for $213. That is what was left after $32 in taxes was deducted, which means taken away. He worked 20 hours.
You need his hourly pay before taxes. So first find his full pay before the $32 was taken out.
Add the taxes back
The $32 was taken away from his full pay to leave $213. To undo taking away, add it back.
His pay before taxes was $245. Watch out: The word “deducted” tempts you to subtract again. That gives , choice A.
Divide by the hours
He earned $12.25 for each hour before taxes. Watch out: Dividing the $213 check by 20 gives $10.65, choice B. That is his pay for each hour after taxes.
So the answer is C, $12.25.
Check your answer
Work forward. , so 20 hours pay $245. Take away $32 in taxes: . That is the amount on the check.
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The original DOE question

Question 83 of 120
2024 Test B Math, question 51Question
Rita used 125 small solid cubes to build a large solid cube. The measure of each edge of the small cubes equals 1 inch. What is the measure of each edge of the large cube, in inches?
- A5
- B25
- C41.33
- D125
Show the answer
Correct answer: A
The volume of a cube with a side length of s is
Since 125 small cubes were used to build the large cube, .
Each side of the large cube is made up of 5 small cubes.
Since 5 small cubes make up each edge of the large cube, and the edge of each small cube measures 1 inch, the measure of each edge of the large cube is .
Question analysis
Rita stacks 125 one-inch cubes into a single large solid cube. Students find the length of each edge of the large cube in inches. It tests volume and perfect cubes with a building-block model, and the care needed is in telling apart the cubes along an edge, across a face and in the whole solid.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Building a large cube from unit cubes is the model grade 6 uses to introduce perfect cubes. It is a single step with small numbers, easier than typical grade 7 geometry, which works with composite solids and formulas.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 3, 35 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is one of the easiest questions on the test, since 125 is a familiar cube and the small cubes make the large one easy to picture as 5 rows of 5 in 5 layers. The wrong choices name a layer instead of an edge, a division by 3 for three dimensions, or the total count itself, and multiplying 5 × 5 × 5 as a check rules all of them out.
Something you must already know minor Working it out
Students must see that 125 one-inch cubes make a volume of 125 cubic inches, and know or find that 5 × 5 × 5 = 125. (Anderson & Krathwohl, 2001)
Strategy: Think of the large cube as equal rows, columns and layers of small cubes, and find the number that, used three times as a factor, gives 125.
Traps
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
The wrong answers, classified
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Gives the number of small cubes in one layer or on one face, 5 × 5 = 25, instead of the number along one edge.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Divides 125 by 3 for the cube's three dimensions instead of taking a cube root; that gives 41.67, and 41.33 is the only choice near it.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Gives back the 125 from the stem, the count of small cubes and the volume in cubic inches, rather than an edge length.
Worked solution, step by step
Formulas used
- Volume of a cube
- V is the volume, the space inside, in cubic units. s is the length of one edge. s³ means s × s × s. A cube’s length, width and height are all s, so length × width × height becomes s × s × s.
Sort out the question
Rita stacks 125 small cubes into one big solid cube. Each small cube has edges of 1 inch. You need the length of one edge of the big cube, in inches.
Find the big cube’s volume
Each small cube is cubic inch. The big cube is solid, with no gaps. So its volume is 125 cubic inches.
Find the edge length
Find the number that makes 125 when you use it three times. Try 5.
So 5 small cubes sit along each edge. Each one is 1 inch long, so the edge is 5 inches.
Rule out the other choices
Watch out: 25, choice B, is the number of small cubes in one layer, 5 rows of 5. 125, choice D, is the number of cubes, or the volume. Dividing 125 by 3 does not give an edge either. That is about 41.67, near choice C.
So the answer is A, 5.
Check your answer
Picture building it. A row of 5 cubes is 5 inches long. Five rows make a layer of 25 cubes. Five layers make cubes. That matches.
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The original DOE question

Question 84 of 120
2023 Test A English, question 1Question
Which edit should be made in the sentence?
On May 8, 2018, Boston University student Zach Prescott ran a mile in 4 minutes 43.2 seconds while juggling three lacrosse balls, breaking the 32-year-old world record in joggling a sport that combines jogging and juggling.
- ADelete the comma after 2018.
- BInsert a comma after student.
- CDelete the comma after lacrosse balls.
- DInsert a comma after joggling.
Show the answer
Correct answer: D
The question asks which edit should be made to correct the sentence.
A. Incorrect. This edit incorrectly deletes the comma that is necessary to set off the year in a date from the remainder of the sentence.
B. Incorrect. This edit introduces an unnecessary comma into the sentence by needlessly separating the adjective phrase “Boston University student” from the noun it modifies, “Zach Prescott.”
C. Incorrect. This edit would delete the comma needed to separate the participle phrase “breaking the 32-year-old world record in joggling” and its companion appositive “a sport that combines jogging and juggling” from the rest of the sentence. This complete element is nonessential, which means it requires separation from the rest of the sentence with a comma because its removal would not fundamentally change the meaning of the sentence.
D. CORRECT. The appositive “a sport that combines jogging and juggling” further describes the word “joggling.” An appositive requires separation with a comma because it is nonessential, meaning that its removal would not change the overall meaning of the sentence.
Question analysis
This punctuation question gives one long sentence about a Boston University student who ran a mile while juggling and broke a world record, and asks which comma edit it needs. Students weigh four marked spots, at a date, between a title and a name, and at two places near the end, deciding at each whether a comma belongs.
- NYS standards
- 7L2 capitalization, punctuation, and spelling
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Commas with dates and with extra information added to a sentence are grade 6 punctuation rules. This item tests them in a single 36-word sentence with one edit to choose, lighter than grade 7 work on phrases, clauses and modifiers.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- A rule that doesn't sound wrong moderate
- Hard words minor
- Complex sentences minor
- Evidence
- Answered from one sentence or line
- Reading level
- Question and choices: about grade 7, 60 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The one missing comma sits at the end of a long sentence, right after an unfamiliar word, and the sentence still sounds acceptable without it. The strongest pull applies the appositive rule one spot too early, between a descriptive title and the name it introduces, while the other two wrong edits remove commas that the date and the added phrase about the record both need. Bracketing each piece of added information, from where it starts to where it ends, exposes the gap after the sport’s name.
A rule that doesn't sound wrong moderate Working it out
Two quiet rules decide the question: the closing appositive “a sport that combines jogging and juggling” must be set off from “joggling,” while a title placed straight before a name, as in “Boston University student Zach Prescott,” takes no comma. (Connors & Lunsford, 1988)
Hard words minor Understanding the question
“Joggling” is a word few students have met, and without the comma “in joggling a sport” can read like a verb and its object rather than the name of a sport followed by its definition. (Perfetti, 2007; Freedle & Kostin, 1993)
Complex sentences minor Understanding the question
The 36-word sentence strings together a date, a title and name, a “while” phrase, the phrase “breaking the 32-year-old world record” and a closing appositive, so every comma has to be judged against where these parts begin and end. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Strategy: Find the main subject and verb, then take the four underlined spots in order and ask what the words on each side are doing and whether a comma belongs between them.
Traps
- Punctuation: A comma, semicolon, colon, dash or apostrophe used where it does not belong, or missing where it does.
The wrong answers, classified
A Punctuation (Connors & Lunsford, 1988)
Treats the comma after “2018” as extra, not knowing that a full date takes a comma after the year as well as between the day and the year.
B Punctuation (Connors & Lunsford, 1988)
Applies the appositive rule to “student Zach Prescott,” but a title placed directly before a name is not set off, and a lone comma there would cut the title off from the name.
C Punctuation (Connors & Lunsford, 1988)
Deletes the comma before “breaking the 32-year-old world record” as if the phrase belonged to the juggling, though without it the lacrosse balls seem to break the record instead of Prescott.
Working it out, step by step
Figure out what the question asks
This is a punctuation question. Punctuation means marks like commas and periods. The sentence has four underlined spots. Each choice adds or takes away one comma. Only one edit fixes the sentence. To find it, ask what each group of words is doing.
Learn the rule for renaming words
Sometimes a sentence names a thing and then tells you what it is. The words that rename it are called an appositive. An appositive gets a comma to set it off from the rest of the sentence.
Example: My uncle plays the tuba, a big brass horn. The words a big brass horn tell you what a tuba is, so a comma comes before them.
Find where a comma is missing
Look at the end of the sentence: “breaking the 32-year-old world record in joggling a sport that combines jogging and juggling.” Joggling is the name of a sport. The words “a sport that combines jogging and juggling” tell you what joggling is. So they are an appositive. But nothing sets them off from “joggling.” A comma belongs right after “joggling.”
Check each choice
- A is wrong. When more of the sentence follows a full date, the year needs a comma after it too. Example: On June 2, 2020, we moved.
- B is wrong. A title that comes right before a name takes no comma. You write Coach Rivera with no comma. In the same way, “Boston University student” is a title for Zach Prescott.
- C is wrong. The comma before “breaking” sets off the part about the record. Without it, the lacrosse balls seem to break the record, not Zach.
- D is right. It adds the comma that sets off the appositive.
Watch out: B is tempting, because “student” does describe Zach. But a title placed right before a name stays joined to it.
Choose the answer
With the new comma, the end reads: breaking the 32-year-old world record in joggling, a sport that combines jogging and juggling. So the answer is D: insert a comma after joggling.
Why this answer and not the closest wrong one
B is the closest wrong answer. It puts one comma between a title and a name. That comma would cut “Boston University student” off from “Zach Prescott.” D fixes a real gap. It puts a comma before “a sport that combines jogging and juggling,” the words that tell what joggling is.
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The original DOE question

Question 85 of 120
2023 Test A Math, question 6Question
- A40
- B370
- C2,482
- D10,000
Show the answer
Correct answer: C
Question analysis
This exponents question asks for the sum of 3 to the fourth power and 7 to the fourth power. Students evaluate each power separately and then add. The traps are treating an exponent as a multiplier and adding the bases before raising them to a power, and the larger power takes careful multiplication.
- NYS standards
- 6.EE.1 whole-number exponent expressions
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Evaluating whole-number powers is grade 6 work, and this question asks only for two of them and a sum. The one demand beyond routine practice is computing a fourth power of 7 accurately without a calculator.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- Something you must already know minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 16+, 1 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The sum is short, but it tests exactly what an exponent means and when to apply it. The wrong choices come from three misreadings of the notation: treating the power as a multiplier, using one factor too few, and adding the bases before raising them. A size check settles it: 7⁴ alone is 49 × 49, a little under 2,500, so 40 and 370 are far too small and 10,000 far too large.
Error-prone arithmetic moderate Working it out
7⁴ means 7 × 7 × 7 × 7 = 2,401, a four-digit product built by hand, most easily as 49 × 49; 3⁴ = 81 is quick by comparison. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
An exponent counts repeated factors rather than acting as a multiplier, so 3⁴ is 3 × 3 × 3 × 3 = 81, not 3 × 4 = 12; the item assumes that meaning. (Anderson & Krathwohl, 2001)
Strategy: Compute each power on its own, squaring and then squaring again, and add the two results only at the end.
Traps
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
The wrong answers, classified
A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Multiplies each base by its exponent, 3 × 4 + 7 × 4 = 12 + 28 = 40, instead of using each base as a factor four times.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Uses each base as a factor three times instead of four, 27 + 343 = 370, so both powers stop one factor short.
D Order of operations (Brown & Burton, 1978)
Adds the bases before applying the exponent, treating 3⁴ + 7⁴ as (3 + 7)⁴ = 10,000, though powers come before addition.
Worked solution, step by step
Formulas used
- A power
- a is the base, the number being multiplied. The small 4 is the exponent. It tells how many times the base is used as a factor, a number being multiplied. So the exponent counts factors. It does not mean multiply by 4.
- Powers come before adding
- a and b are any two positive numbers. The sign ≠ means is not equal to. Powers come before adding. So raise each number to its power first, then add. Adding the bases first gives a much bigger number.
Sort out the question
You need the value of . Each number has a small 4 above it. This 4 is an exponent. Work out each power on its own. Then add the two results.
Find 3 to the fourth power
3⁴ means four 3s multiplied together. Pair them up to make it easier.
Watch out: 3⁴ is not 3 × 4 = 12. Doing that to both powers gives 12 + 28 = 40, choice A.
Find 7 to the fourth power
7⁴ means four 7s multiplied together. Pair them up the same way.
To find 49 × 49, do 49 × 50 = 2,450. Then take away one 49. That gives 2,401. Watch out: count four 7s, not three. Three 7s make 343. Using three factors in both powers gives 370, choice B.
Add the two powers
Watch out: don’t add 3 and 7 first. That gives , choice D. Powers come before adding. So the answer is C, 2,482.
Check your answer
Estimate. 7⁴ is 49 × 49, a little less than 50 × 50 = 2,500. Adding 81 keeps it near 2,500. Only 2,482 is close. 40 and 370 are far too small, and 10,000 is far too big.
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The original DOE question

Question 86 of 120
2023 Test A Math, question 7Question
In one week, inches of rain fell on Monday, inches fell on Tuesday, and inch fell on Wednesday. How many inches of rain fell during those three days?
- A
- B
- C
- D
Show the answer
Correct answer: A
To solve, add the three given fractions:
The common denominator is 24:
Question analysis
A rainfall word problem gives the inches of rain on Monday, Tuesday and Wednesday as two mixed numbers and a fraction, and asks for the total over the three days. Students add fractions with unlike denominators. Finding a common denominator for fourths, thirds and eighths, and regrouping the fractional part into wholes, is where errors come in.
- NYS standards
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Adding fractions and mixed numbers with unlike denominators is taught before grade 7, and this is a direct word problem. It takes more care than a typical exercise only because three different denominators must be rewritten over a common one.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- Many steps minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 35 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The setup is plain addition, so the difficulty is all in the fractions. Students who know to rename over a common denominator rarely miss it, while those who add the fraction parts straight across, or put the numerators over 24 without scaling them, reach choices with a whole part of 3. Since 1 3/4 + 2 2/3 alone is more than 4, any total below 4 can be ruled out at a glance.
Error-prone arithmetic moderate Working it out
The fractional parts have denominators 4, 3 and 8, so they must be renamed as 18, 16 and 21 twenty-fourths, and their sum, 55/24, regrouped as 2 7/24 before the wholes are added. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Renaming, adding the numerators, regrouping the improper fraction and adding the whole numbers are four separate moves, and losing the 2 carried from 55/24 drops the total by two inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Strategy: Rewrite the three fractional parts over a common denominator of 24, add the whole numbers and the fractions separately, and regroup any extra twenty-fourths as a whole.
Traps
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
The wrong answers, classified
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No single renaming mistake gives 49 twenty-fourths; this choice sits exactly 1/4 below the right total, so it mostly catches guesses near it.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Adds the whole numbers to get 3, then adds the fractions straight across, (3 + 2 + 7)/(4 + 3 + 8) = 12/15 = 4/5.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Finds the common denominator 24 but adds the numerators without renaming them, (3 + 2 + 7)/24 = 1/2, and adds that to the wholes, 3.
Worked solution, step by step
Formulas used
- Equivalent fractions
- a over b is a fraction, and k is any whole number except 0. Multiplying the top and the bottom by the same number gives an equal fraction. You cut each piece into k smaller pieces. So you have k times as many pieces, each k times smaller.
- Adding fractions with the same bottom
- d is the denominator, the bottom number. It tells the size of the pieces. a, b and c are the numerators, the top numbers. They count the pieces. When the pieces are the same size, add the counts. The bottom stays the same.
- Fraction to mixed number
- a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.
Sort out the question
Rain fell on three days: inches, inches and inch. You need the total rain, in inches. So add the three amounts. Add the whole numbers and the fractions separately. The whole numbers are 1 and 2, so they add to 3 inches.
Find a common denominator
The fractions have denominators, or bottom numbers, 4, 3 and 8. You can only add pieces of the same size. So find the smallest number that 4, 3 and 8 all divide into. Count by 8s: 8, 16, 24. 24 works, since 24 = 4 × 6 and 24 = 3 × 8. So use 24ths.
Rename each fraction in 24ths
Multiply the top and the bottom by the same number.
Watch out: change the tops too. Keeping 3, 2 and 7 over 24 gives and a total of , choice D.
Add the fractions
Now the pieces are the same size, so add the tops.
That is more than 1 whole. 24 goes into 55 twice, with 7 left over.
Watch out: never add the bottoms. Adding straight across gives and a total of , choice C.
Put the parts together
Add the 3 whole inches from before.
So the answer is A, inches.
Check your answer
Use decimals. is 1.75, is about 2.67, and is 0.875. They add to about 5.29. And is about 0.29. So fits, and is too small.
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The original DOE question

Question 87 of 120
2023 Test A Math, question 9Question
A revolving sign makes 1 complete revolution every 90 minutes. If the sign starts moving at 2:30 p.m., at what time will the sign complete 8 revolutions?
- A1:00 a.m.
- B2:30 a.m.
- C4:00 p.m.
- D12:00 midnight
Show the answer
Correct answer: B
First, use a proportion to find out how many minutes it will take the sign to complete 8 revolutions. Let x equal the total number of minutes:
Next, convert 720 minutes into hours and minutes:
Finally, 12 hours after 2:30 p.m. is 2:30 a.m.
Question analysis
A revolving sign turns once every 90 minutes and starts at 2:30 in the afternoon. Students find the clock time when it finishes 8 full revolutions. The question combines a rate with converting minutes to hours, and then adding the elapsed time to the starting time on a 12-hour clock.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 A constant rate, a multiplication and a conversion from minutes to hours are grade 6 skills. The only step beyond routine work is adding the elapsed time to a starting clock time.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Many steps minor
- Something you must already know minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 4, 31 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Each step is routine, so most points are lost in keeping elapsed time and clock time apart. The 12 hours is a duration, and the clock reading that ends it is 12 hours after 2:30 in the afternoon, not 12:00. A sense check helps: 12 hours after any time shows the same digits on the clock with a.m. and p.m. swapped.
Many steps minor Working it out
Three moves: 8 × 90 = 720 minutes, 720 ÷ 60 = 12 hours, then 12 hours counted on from the 2:30 start, which crosses midnight into the morning. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Counting hours past midnight means switching from p.m. to a.m. on a 12-hour clock, which the stem leaves to the student; the choices mix both halves of the day. (Anderson & Krathwohl, 2001)
Strategy: Multiply 90 minutes by 8 to get the total time, change it to hours, and then count that many hours forward from 2:30 p.m.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Counting the ends (Brown & Burton, 1978)
Counts only 7 revolutions, 630 minutes or 10½ hours, as if the first turn were already done at the start, and lands at 1:00 a.m.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Adds a single revolution's 90 minutes to the 2:30 start, giving 4:00 in the afternoon, without multiplying by the 8 revolutions.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds that 8 revolutions take 12 hours and picks 12:00 midnight, reading the elapsed time as a clock time instead of counting on from 2:30.
Worked solution, step by step
Formulas used
- Total time
- Time for one is how long one revolution takes. How many is the number of revolutions. Each one takes the same time, so multiply to get the total time.
- Minutes to hours
- There are 60 minutes in 1 hour. So to change minutes to hours, divide by 60. Each group of 60 minutes makes 1 hour.
Sort out the question
The sign turns all the way around once every 90 minutes. One full turn is called a revolution. The sign starts at 2:30 p.m. You need the clock time when it finishes 8 revolutions.
Find the total time in minutes
Each revolution takes 90 minutes, and there are 8 of them.
The 8 revolutions take 720 minutes. Watch out: use all 8 revolutions. Counting only 7 gives 630 minutes and 1:00 a.m., choice A. Adding just one 90 minutes gives 4:00 p.m., choice C.
Change the minutes to hours
Clock times are easier to count in hours. There are 60 minutes in an hour.
The sign turns for 12 hours.
Count forward from 2:30 p.m.
Start at 2:30 p.m. and add 12 hours. When you pass midnight, p.m. changes to a.m. So 12 hours after 2:30 p.m. is 2:30 a.m. the next morning. Watch out: 12 hours is an amount of time, not a clock time. Reading it as 12:00 midnight gives choice D. So the answer is B, 2:30 a.m.
Check your answer
From 2:30 p.m. to 2:30 a.m. is one full trip around the clock, 12 hours. 12 hours is minutes. And , the number of revolutions in the question.
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The original DOE question

Question 88 of 120
2023 Test A Math, question 10Question
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Points Y and Z are not shown on the number line above. If X is the midpoint of , and Y is the midpoint of , where on the number line would point Z be located?
- A2
- B4
- C6
- D8
Show the answer
Correct answer: D
First, find where Y is located on the number line. Point X is the midpoint of . Since there are 2 units between W and X, there must also be 2 units between X and Y. Thus, Y is located at 4 on the number line.
Point Y is the midpoint of . Use the same reasoning as above. Since there are 4 units between W and Y, there must be 4 units between Y and Z. So, Z is located at 8 on the number line.
Question analysis
A number line shows point W at 0 and point X at 2, and two more points, Y and Z, are described but not drawn. Students use two midpoint facts, one after the other, to find where Z lies. Each midpoint fact is about a different segment, so keeping track of which point is the middle one matters.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Midpoints on a number line rest on distance, which grade 7 students find as a difference. Here every point lies at a whole number of zero or more, so the work is two short steps with no negative numbers.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Many steps minor
- Reading a figure, table or graph minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 7, 36 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Each midpoint statement doubles a distance from W, so the problem is two short doublings, 2 to 4 and then 4 to 8. The risk is stopping halfway, since Y's position is found first and appears among the choices, or reusing X as an endpoint in the second statement. Rereading which point the question asks about, and checking that Y is halfway between W and Z, catches both.
Many steps minor Working it out
The two midpoint statements must be used in order: the first places Y at 4, and only then can the second, about segment WZ, place Z. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
Only W at 0 and X at 2 are marked on the 0-to-10 line, and the overbars name WY and WZ as segments, so Y and Z have to be placed from the statements alone. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Strategy: Use the first midpoint to place Y as far beyond X as X is from W, then use the second midpoint the same way to place Z beyond Y.
Traps
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Gives 2, the position of point X, the only labeled point besides W, without using either midpoint statement.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Uses the first statement to place Y at 4 and answers with Y's position instead of going on to locate Z, the point the question asks about.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Treats Y as the midpoint of XZ instead of WZ, so Z lands 2 units past Y, at 6, continuing the even steps from W to X to Y.
Worked solution, step by step
Formulas used
- Midpoint of a segment
- A and B are the ends of a segment, a straight piece of a line. M is its midpoint, the point exactly halfway. AM is the distance from A to M, and MB is the distance from M to B. A midpoint splits a segment into two equal parts, so these are equal.
Sort out the question
The number line shows W at 0 and X at 2. Points Y and Z are not shown. X is the midpoint of segment WY. Y is the midpoint of segment WZ. A midpoint is the point exactly halfway along a segment. You need the number where Z sits.
Place point Y
X is halfway from W to Y. So X to Y is as long as W to X.
So Y is 2 units past X.
Point Y is at 4. Watch out: this is Y, not Z. Stopping here gives 4, choice B.
Place point Z
Now Y is halfway from W to Z. W to Y is 4 units, so Y to Z is also 4 units.
Watch out: the second midpoint is for segment WZ, not XZ. Taking just 2 more units from Y puts Z at 6, choice C. So the answer is D, 8.
Check your answer
Check both midpoints. Halfway from 0 to 4 is , which is X. Halfway from 0 to 8 is , which is Y. Both facts in the question are true.
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The original DOE question

Question 89 of 120
2023 Test A Math, question 11Question
What value of n makes the equation above true?
- A1
- B
- C5
- D
Show the answer
Correct answer: B
| Cross multiply. | |
| Apply the multiplicative inverse property; divide both sides of the equation by 81. | |
| Then simplify the fraction. |
Question analysis
This question gives the proportion 81/10 = 9/n and asks for the value of n that makes it true. Students solve a proportion by cross-multiplying and dividing, then simplify the result. Reducing the resulting fraction and matching its form to the choices is where students slip.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Solving a proportion is familiar from grades 6 and 7, and this one needs only cross-multiplying and one division. Its one extra demand is simplifying the result and writing it in the form the choices use.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- An answer in an unexpected form moderate
- A very close wrong answer minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 7, 12 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The proportion solves in one cross-multiplication, but the result is an improper fraction that must be simplified and rewritten to match. Students who notice that 81 is 9 × 9 can shortcut it: the numerator shrinks by a factor of 9 from left to right, so the denominator must shrink too, giving 10 ÷ 9. The mixed-number form is where answers drift, by dropping the leftover ninth or by writing 10 × 1/9 as 10 1/9.
An answer in an unexpected form moderate Choosing the answer
Cross-multiplying gives n = 90/81, which has to be reduced to 10/9 and rewritten as the mixed number 1 1/9 before it matches any choice. (Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The choices 1 and 1 1/9 differ by one ninth, and estimating 90 ÷ 81 as about 1 cannot separate them; only keeping the remainder does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Strategy: Cross-multiply to get 81n = 90, divide both sides by 81, and simplify the fraction before comparing it with the choices.
Traps
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Answer in another form: The right answer is written differently from the way you worked it out.
The wrong answers, classified
A Rounds too early (King et al., 2004)
Finds n by dividing 10 by 9 but keeps only the whole-number part of the quotient, 1, dropping the remainder that the exact answer needs.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No cross-multiplication or scaling of 81/10 gives 5, which would make 9/n equal 1.8 instead of 8.1, so it mostly catches guesses.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reaches n = 10 × 9/81 = 10 × 1/9 but writes the product as the mixed number 10 1/9, as if a whole number beside a fraction meant multiplication.
Worked solution, step by step
Formulas used
- Cross-multiplying
- a over b and c over d are two equal fractions, and b and d are not 0. Multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.
- Simplifying a fraction
- a over b is a fraction, and k is a number that divides both the top and the bottom. Dividing the top and the bottom by the same number gives an equal fraction with smaller numbers.
- Fraction to mixed number
- a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.
Sort out the question
The equation says two fractions are equal: and . n is an unknown bottom number. You need the value of n that makes them equal. Some choices are mixed numbers, so n may not be a whole number.
Cross-multiply the two fractions
Multiply each top by the other fraction’s bottom. This gets n out of the bottom of the fraction.
Divide both sides by 81
81n means 81 times n. To undo multiplying by 81, divide both sides by 81.
Now simplify. 90 and 81 are both in the 9 times table, so divide the top and bottom by 9.
Write it as a mixed number
10 ÷ 9 is 1 with 1 left over. The leftover stays over 9.
Watch out: keep the leftover. Dropping it gives 1, choice A. Also, ten ninths is not 10 and one ninth. That mixed number, , is choice D, which is far too big. So the answer is B, .
Check your answer
Estimate. is about 8. For to be about 8, n must be a little more than 1. Check exactly: . It matches.
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The original DOE question

Question 90 of 120
2023 Test A Math, question 13Question
The product of two positive integers is 65. Which number could be the sum of the two integers?
- A5
- B18
- C24
- D52
Show the answer
Correct answer: B
First, find the pairs of positive integers that multiply to 65: and .
The sums of the pairs are 66 and 18. 18 is the option that is provided in the answer choices.
Question analysis
Two positive integers multiply to 65, and this question asks which number in the choices could be their sum. Students list the factor pairs of 65 and add each pair. The words could be signal that more than one pair exists, and only one of the sums appears among the choices.
- NYS standards
- 6.NS.4 common factors and multiples
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding factor pairs of a two-digit number is review by grade 7. The question takes two quick steps, listing the pairs and adding each, with nothing to compute beyond small products and sums.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 17 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
With only two factor pairs, 1 × 65 and 5 × 13, the work is a short list and two additions. The word could matters: the sum depends on which pair is chosen, and only one of the two sums appears among the choices. The main slip is answering with a factor rather than with the sum the question asks for.
Something you must already know minor Working it out
The pairs come from factoring 65 as 5 × 13, a fact the stem leaves to the student; the pair 1 and 65 also counts, but its sum, 66, is not offered. (Anderson & Krathwohl, 2001)
Strategy: List every pair of positive integers whose product is 65, starting from 1, and add the two numbers in each pair.
Traps
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Finds the factor pair 5 and 13 and answers with the factor 5, one of the two integers, instead of their sum.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No pair of positive integers with product 65 adds to 24; reaching it needs a wrong pair such as 3 and 21, whose product is 63, so it mostly catches guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No factor pair of 65 adds to 52, which is 65 − 13 or 4 × 13, and no common working combines them that way, so it mostly draws guesses.
Worked solution, step by step
Formulas used
- Factor pair
- a and b are positive whole numbers that multiply to make 65. Such a pair is called a factor pair of 65. To find them all, try 1, 2, 3 and so on. Stop when the factors start to repeat.
- Dividing by 5
- A whole number splits into groups of 5, with none left over, when its last digit is 0 or 5. So 65 divides by 5, but 64 does not.
Sort out the question
Two positive integers multiply to make 65. Positive integers are the counting numbers 1, 2, 3 and so on. The product is the answer to a multiplication. You need a number that could be their sum, the answer when you add them.
List the factor pairs of 65
Try each small number. See if it divides 65 with nothing left over.
- 1 works: .
- 2 and 4 don’t work, because 65 is odd. 3 doesn’t work either: , with 2 left over.
- 5 works, since 65 ends in 5: .
- 6 and 8 are even, so they don’t work. 7 doesn’t either: .
Since , any factor bigger than 8 pairs with one smaller than 8. You already tried those. So the pairs are 1 and 65, and 5 and 13.
Add each pair
The possible sums are 66 and 18. Only 18 is a choice. Watch out: the question asks for the sum, not one of the numbers. Picking the factor 5 is choice A. So the answer is B, 18.
Check your answer
Work back from 18. Look for two numbers that add to 18 and multiply to 65. and . Both match. For 24 or 52, no pair of whole numbers that multiplies to 65 adds up to them.
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The original DOE question

Question 91 of 120
2023 Test A Math, question 14Question
If n is an odd integer that is less than −3.25, what is the greatest possible value of n?
- A−1
- B−2
- C−3
- D−5
Show the answer
Correct answer: D
The two greatest integers less than −3.25 are −4 and −5. Since −4 is even, the answer is −5.
Question analysis
This question asks for the greatest odd integer that is less than −3.25. Students compare negative numbers on a number line and apply two conditions at once, odd and less than. With negatives, the greater number is the one closer to zero, which makes the two conditions easy to misapply.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Ordering negative numbers is grade 6 content, and the question adds only the condition that the integer be odd. It is lighter than typical grade 7 work, though the order of negatives still trips up students who compare their sizes.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic minor
- Much to keep track of minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 7, 18 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The question is short, but it asks students to reason on the negative side of 0, where greater means closer to 0 and the size of the digits misleads. The integers just below −3.25 are −4 and then −5, and the first of those is even. Sketching a number line, or reading less than as to the left of, keeps the direction straight.
Error-prone arithmetic minor Working it out
On the negative side, less means farther from 0, so the integers below −3.25 run −4, −5, −6; −3, though its digit is closer to 3.25, lies to the right. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
Three conditions apply at once, odd, less than −3.25 and the greatest such value, and checking only two of them leads to a listed choice. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Strategy: Picture −3.25 on a number line, list the integers to its left starting with the closest, and take the first one that is odd.
Traps
- Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
The wrong answers, classified
A Misreads the question (Hegarty, Mayer & Monk, 1995)
Picks the greatest odd number offered, −1, without checking the condition that n be less than −3.25; −1 lies to its right on the number line.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
The choice −2 is even, so it breaks the condition that n be odd, and it is also greater than −3.25; it mostly catches guesses.
C Negatives and direction (Brown & Burton, 1978)
Looks for the odd integer just past −3.25 but steps toward 0, taking −3, though −3 is greater than −3.25, not less.
Worked solution, step by step
Formulas used
- Comparing negative numbers
- a and b are positive numbers, and a is bigger. On a number line, numbers get smaller as you move left. A negative number farther from 0 is farther left. So it is smaller, even though its digits look bigger.
Sort out the question
n has to pass three tests. It is an integer, a whole number with no decimal part. It is odd, so it is not a multiple of 2. And it is less than −3.25. Of all the numbers that pass, you need the greatest one. That is the one farthest to the right on a number line.
Picture −3.25 on a number line
−3.25 sits between −4 and −3. Less than −3.25 means to the left of it.
So −4 is less than −3.25, but −3 is greater. Watch out: −3 is not less than −3.25. It is to the right, closer to 0. For negative numbers, the one closer to 0 is greater. Picking −3 is choice C.
Find the odd integers to the left
The integers less than −3.25, starting with the closest, are −4, −5, −6 and so on. The greatest one is −4. But −4 is even, since 4 = 2 × 2. So skip it. The next one, −5, is odd. Watch out: −1 is odd, but it is greater than −3.25, so choice A breaks the rule. So the answer is D, −5.
Check your answer
Check all three tests for −5. It is an integer. It is odd. It is to the left of −3.25, so it is less. The only integer between −5 and −3.25 is −4, and it is even. So no greater number works.
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The original DOE question

Question 92 of 120
2023 Test A Math, question 16Question
Between which two consecutive integers is the fraction ?
- A2 and 3
- B3 and 4
- C4 and 5
- D5 and 6
Show the answer
Correct answer: C
First, convert to a mixed number: . The two consecutive integers on either side of are 4 and 5.
Question analysis
This question asks between which two consecutive whole numbers the fraction 29/7 lies. Students divide to turn the improper fraction into a mixed number, or compare it with nearby multiples of 7. It is a quick check of number sense with fractions, where a remainder must be read correctly.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Changing an improper fraction to a mixed number and placing it between whole numbers is review well before grade 7. This takes one division with a small divisor and nothing more.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 13 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Dividing 29 by 7 gives 4 with a remainder of 1, so the fraction is 4 1/7 and lies just past 4. The one decision is which side of 4 the remainder puts it on, and a positive remainder always moves the value up. Writing the mixed number before looking at the choices makes the pair obvious.
A very close wrong answer minor Choosing the answer
The pairs 3 and 4 and 4 and 5 both contain 4, so knowing that 29/7 is about 4 is not enough; only the remainder, 29 − 28 = 1, puts it above 4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Strategy: Find the largest multiple of 7 that is not more than 29, write the fraction as a mixed number, and read off the whole numbers on each side.
Traps
- Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
The wrong answers, classified
A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No division of 29 by 7 lands below 3, since 7 × 3 = 21 is far short of 29; only misreading 29/7 as 2.9 gets there, so it mostly catches guesses.
B Counting the ends (Brown & Burton, 1978)
Finds the whole-number part, 4, but takes it as the upper end of the pair, as if 29/7 fell just short of 4 instead of just past it.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Nothing puts 29/7 above 5, since 7 × 5 = 35 is more than 29; even rounding 4 1/7 up gives 5 as the upper end, so it mostly catches guesses.
Worked solution, step by step
Formulas used
- Fraction to mixed number
- a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.
Sort out the question
Consecutive integers are whole numbers that come one right after the other, like 4 and 5. You need the two that falls between. So find the whole number just below it and the one just above it.
Divide 29 by 7
A fraction bar means divide. How many 7s fit in 29?
28 is the biggest multiple of 7 that is not more than 29. A multiple of 7 is 7 times a whole number. Then . So 29 ÷ 7 is 4 with 1 left over.
Write it as a mixed number
is a little more than 4 and less than 5. Watch out: it is just past 4, not just short of it. So 4 is the lower number of the pair. Picking 3 and 4 is choice B. So the answer is C, 4 and 5.
Check your answer
Multiply each end by 7. and . The top of the fraction, 29, is between 28 and 35. So is between 4 and 5. As a decimal it is about 4.14.
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The original DOE question

Question 93 of 120
2023 Test A Math, question 19Question
For what value of x is the proportion shown above true?
- A27
- B25
- C16
- D13
Show the answer
Correct answer: B
Write the ratios in fraction form and solve for x:
| Cross multiply. | |
| Apply the multiplicative inverse property; divide both sides by 28. | |
Question analysis
This question gives the proportion x:35 = 20:28 and asks for the value of x that makes it true. Students use equivalent ratios or cross-multiplication. The trap is reasoning by adding or subtracting the difference between terms rather than multiplying, which gives numbers that look reasonable but break the ratio.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a missing term in a proportion is grade 6 ratio reasoning, and here the given ratio simplifies to small numbers. It is easier than typical grade 7 proportional reasoning, which usually involves fractions or a real context.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic minor
- A very close wrong answer minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 12 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is a single proportion with whole numbers, and the only real trap is reasoning by differences instead of by a common factor. Since 35 is 7 more than 28, adding or subtracting 7 to 20 is tempting, but equal ratios scale; they do not shift. Reducing 20:28 to 5:7 shows at once that the scale factor to 35 is 5.
Error-prone arithmetic minor Working it out
Cross-multiplying gives 28x = 700, and 700 ÷ 28 is not a table fact; simplifying 20:28 to 5:7 first turns it into 5 × 5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
27 is only 2 more than 25, so a rough sense that x is a little more than 20 fits both; only the exact scale factor, 35/28 = 5/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Strategy: Simplify 20:28 to lowest terms first, then scale it up so its second term becomes 35, or cross-multiply and divide by 28.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
The wrong answers, classified
A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reasons with differences: since 35 is 7 more than 28, adds 7 to 20, keeping the gap between the terms instead of their ratio.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Scales 20 by 28/35 instead of 35/28, so 20 × 28 ÷ 35 = 16, which turns the ratio of the second terms upside down.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reasons with differences and also goes the wrong way, subtracting 7 from 20, though x must exceed 20 because 35 exceeds 28.
Worked solution, step by step
Formulas used
- Ratio as a fraction
- a:b is a ratio, read a to b. It compares two amounts by division. You can write it as the fraction a over b. Two equal ratios make a proportion.
- Equivalent fractions
- a over b is a fraction, and k is any number except 0. Multiplying or dividing the top and the bottom by the same number gives an equal fraction.
Sort out the question
is a proportion. A proportion says two ratios are equal. A ratio compares two numbers, like 20 to 28. You need the value of x that makes the two ratios equal.
Write the ratios as fractions
Write each ratio as a fraction. The first number of the ratio goes on top. The second goes on the bottom.
Simplify the known fraction
20 and 28 both divide by 4. Smaller numbers are easier to work with.
So now .
Scale up so the bottom is 35
7 times 5 is 35. To keep the fraction equal, multiply the top by 5 too.
So x = 25. Watch out: a ratio is about multiplying, not adding. 35 is 7 more than 28, but adding 7 to 20 gives 27, choice A. Also, x must be more than 20, because 35 is more than 28. So 16 and 13 are too small. So the answer is B, 25.
Check your answer
Cross-multiply, as the test’s own explanation does. and . The two products are equal, so the ratios 25:35 and 20:28 are equal.
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The original DOE question

Question 94 of 120
2023 Test A Math, question 23Question
If a person chooses a number at random from the set above, what is the probability that the number is less than 0.005?
- A
- B
- C
- D
Show the answer
Correct answer: C
Three of the five numbers listed (0.001, 0.0001, and 0.00001) are less than 0.005. So, the probability of a person selecting a number less than 0.005 is .
Question analysis
A set holds five decimals, from one tenth down to one hundred-thousandth, and one number is chosen at random. Students find the probability that the number is less than 0.005. Comparing decimals with many zeros is the real task here, along with putting the count over the right total.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Probability with equally likely outcomes is grade 6 content in New York, and this needs one comparison per number and one fraction. The only real demand is comparing decimals with many zeros correctly.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- Something you must already know minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 6, 27 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Each number in the set is a tenth of the one before, so only the three smallest fall below 0.005, which lies between 0.01 and 0.001. Students lose the point either by counting the numbers above 0.005 instead of below it or by writing the count over the wrong total. Lining the decimals up with the same number of places makes every comparison easy.
Error-prone arithmetic moderate Working it out
0.005 has to be compared with decimals from 0.1 down to 0.00001; written to five places, 0.00500 against 0.01000 and 0.00100, the order becomes a comparison of whole numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The probability is the count of numbers below 0.005 over all 5 numbers in the set; the stem leaves that part-to-whole rule to the student. (Anderson & Krathwohl, 2001)
Strategy: Write each number with five decimal places, compare it with 0.00500, count those that are smaller, and divide by the five numbers in the set.
Traps
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Gives 1/5, the chance of picking any one particular number from the set, without counting how many of the five are below 0.005.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Counts the 2 numbers greater than 0.005, 0.1 and 0.01, instead of those less than it, finding the chance of the opposite event.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Writes a part-to-part ratio, the 2 numbers above 0.005 over the 3 below it, instead of dividing by all 5 numbers in the set.
Worked solution, step by step
Formulas used
- Probability
- P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.
- Adding zeros to a decimal
- Zeros added to the right end of a decimal don’t change its value. One tenth is the same as ten hundredths. Giving every number the same number of decimal places lets you compare them like whole numbers.
Sort out the question
The set has five numbers. A person picks one at random, so each number has the same chance. You need the probability, or chance, that the number picked is less than 0.005.
Give every number five decimal places
Decimals are easier to compare with the same number of places. Add zeros to the end of each.
Number With five places 0.1 0.10000 0.01 0.01000 0.001 0.00100 0.0001 0.00010 0.00001 0.00001 0.005 0.00500 Count the numbers below 0.005
Compare each number with 0.00500 like whole numbers: 10000, 1000, 100, 10 and 1 against 500. Two numbers are bigger: 0.1 and 0.01. Three are smaller: 0.001, 0.0001 and 0.00001. Watch out: 0.01 has fewer digits than 0.005, but it is bigger. Counting the 2 bigger numbers gives , choice B.
Write the probability
Watch out: divide by all 5 numbers in the set. Comparing the 2 bigger numbers with the 3 smaller ones gives , choice D. So the answer is C, .
Check your answer
The numbers below 0.005 and the numbers above it make up the whole set: 3 + 2 = 5. So the two chances, and , add up to 1, as they should.
More questions like this
The original DOE question

Question 95 of 120
2023 Test A Math, question 28Question
It took Lars 2 hours to ride his bicycle 48 kilometers. What was his average speed in miles per hour? (Use the approximation 1 mile = 1.6 kilometers.)
- A1.5
- B15.0
- C30.0
- D38.4
Show the answer
Correct answer: B
First, calculate Lars’s average speed in kilometers per hour:
Then, convert kilometers to miles:
| Cross multiply. | |
| Apply the multiplicative inverse property; divide both sides of the equation by 1.6. | |
Question analysis
Lars rides 48 kilometers in 2 hours, and this rate question asks for his average speed in miles per hour, using 1 mile = 1.6 kilometers. Students find a unit rate and convert it to other units. Deciding whether to multiply or divide by 1.6, and remembering the 2 hours, are where students go wrong.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a speed and converting units with a given rate are grade 6 ratio skills. The question takes two short steps with friendly numbers, lighter than typical grade 7 rate problems with fractional quantities.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Combines ideas minor
- Error-prone arithmetic minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 7, 23 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The question tests the direction of a conversion more than the arithmetic. A mile is longer than a kilometer, so a distance has fewer miles than kilometers, and multiplying by 1.6 runs the conversion backward. The other trap is converting the 48 kilometers to 30 miles and forgetting that the ride took 2 hours, which gives a distance instead of a speed.
Combines ideas minor Working it out
A rate and a unit conversion are combined: 48 ÷ 2 = 24 kilometers per hour, then 24 ÷ 1.6 = 15 miles per hour, and the two steps can come in either order. (Embretson, 1998; Daniel & Embretson, 2010)
Error-prone arithmetic minor Working it out
Dividing by 1.6 means dividing 240 by 16; a decimal point in the wrong place makes the speed ten times too small, and that value is offered. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Find the speed in kilometers per hour first, then divide by 1.6, since a mile is longer than a kilometer and the number of miles must be smaller.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Stops one step early: A choice is a number from partway through the solution.
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
The wrong answers, classified
A An arithmetic slip (Junker & Sijtsma, 2001)
Divides 24 by 16 instead of 1.6, a decimal point in the wrong place that makes the speed ten times too small, 1.5.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Converts the 48 kilometers to 30 miles correctly and stops, forgetting to divide by the 2 hours, so it gives a distance rather than a speed.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Multiplies 24 kilometers per hour by 1.6 instead of dividing, converting in the wrong direction and getting more miles than kilometers.
Worked solution, step by step
Formulas used
- Average speed
- Distance is how far you go. Time is how long it takes. Speed is how far you go in each unit of time, such as each hour. Average speed shares the whole trip out evenly over the time.
- Kilometers to miles
- A kilometer is a unit of length used in most countries. 1 mile is about 1.6 kilometers, so a mile is longer. To find how many miles, find how many groups of 1.6 kilometers fit. So divide by 1.6.
Sort out the question
Lars rode 48 kilometers in 2 hours. You need his average speed in miles per hour. That means miles, not kilometers. Use the fact that 1 mile is about 1.6 kilometers.
Find the speed in kilometers per hour
Divide the distance by the time.
Lars went 24 kilometers each hour. This speed is still in kilometers, so you are not done yet.
Change kilometers to miles
Divide by 1.6. To make it easier, multiply both numbers by 10 first. Then you divide by the whole number 16.
Watch out: divide by 1.6, don’t multiply. A mile is longer, so there are fewer miles than kilometers. is choice D. Dividing 24 by 16 gives 1.5, choice A, with the decimal point in the wrong place.
Name the answer
Lars rode 15 miles each hour. Watch out: changing 48 kilometers to 30 miles is only part of the work. 30 is a distance, choice C. You still have to divide by the 2 hours. So the answer is B, 15.0 miles per hour.
Check your answer
Work backward. At 15 miles per hour for 2 hours, Lars rides 30 miles. And kilometers, the distance in the question. Also, 15 is smaller than 24, as it should be, since miles are longer than kilometers.
More questions like this
The original DOE question

Question 96 of 120
2023 Test A Math, question 30Question
Lamel has a jar containing 6 red chips, 10 blue chips, and 4 yellow chips. If he removes one chip at random, what is the probability that it will not be red?
- A
- B
- C
- D
Show the answer
Correct answer: B
There are currently 20 chips in the jar (). Of those chips, 14 are not red. So the probability of choosing a chip that is not red is .
Question analysis
In this probability word problem, Lamel's jar holds red, blue and yellow chips, and students find the chance that one chip drawn at random is not red. It tests counting the favorable outcomes against the total and writing the probability as a fraction. Missing the bold word not, or counting the wrong colors, leads to tempting wrong fractions.
- NYS standards
- 6.SP.8a uniform probability models
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Probability of a single event with equally likely outcomes is grade 6 content in New York. Students add three counts and form one fraction, and the only extra care is the bold word not, so it sits below grade 7.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 33 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is a one-event probability with small whole numbers, so the bold not carries the question. A student who skims past it computes the chance of red, a tidy fraction that is waiting among the choices. The other wrong fractions come from removing the wrong color or keeping only one of the two colors that are not red. Adding blue and yellow first, then dividing by all 20 chips, avoids each of them.
Turning words into math minor Understanding the question
The bold not turns the question around: the chips that count are the 10 blue and 4 yellow ones, 14 in all, or equivalently 1 minus the red share of 6/20. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The count of 14 out of 20 matches no choice until it is reduced to 7/10, so a correct fraction left unsimplified can look missing from the list. (Enright, Morley & Sheehan, 2002)
Strategy: Add all the chips for the total, then count the chips that are blue or yellow, or subtract the chance of red from 1, and simplify the fraction.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Leaves out the 4 yellow chips instead of the 6 red ones, finding the chance of not yellow, 16/20 = 4/5.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Overlooks the bold not and gives the chance that the chip is red, 6 out of 20, which reduces to 3/10.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Counts only the 4 yellow chips as not red and forgets the 10 blue ones, so it finds 4/20 = 1/5, the chance of yellow alone.
Worked solution, step by step
Formulas used
- Probability
- P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. It works when each outcome is equally likely, like pulling one chip without looking.
Sort out the question
Lamel’s jar holds 6 red chips, 10 blue chips and 4 yellow chips. He takes one chip without looking. You need the probability that it is not red. That means the chip is blue or yellow.
Count all the chips
There are 20 chips in all. That is the bottom of the fraction, because any of the 20 could come out.
Count the chips that are not red
The blue and yellow chips are the ones that are not red.
So 14 chips are not red. Watch out: the question asks for not red. Counting the 6 red chips gives , choice C. Counting only the 4 yellow chips gives , choice D.
Find and simplify the probability
Dividing the top and bottom by 2 puts the fraction in lowest terms. So the answer is B, .
Check your answer
A chip is either red or not red, so the two chances must add up to 1. The chance of red is . Then . It works.
More questions like this
The original DOE question

Question 97 of 120
2023 Test A Math, question 41Question
Yesterday Sarah read 15% of her entire book. Today she read another 17% of the entire book. In lowest terms, what fraction of the book is left for her to read?
- A
- B
- C
- D
Show the answer
Correct answer: C
So far, Sarah has read of the entire book. That means she has left to read.
Question analysis
This percent word problem says Sarah read 15% of her book yesterday and 17% today, and asks what fraction of the book she still has to read, in lowest terms. Students combine percents of the same whole, find the part remaining, and convert a percent to a simplified fraction. Rounding, or giving the part already read, leads to wrong choices.
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 percent work often takes a percent of a percent or finds a percent change. Here both percents are of the same whole book, so they combine directly; the rest is a subtraction and simplifying a fraction, lighter than typical.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 5, 33 words.
- Corrected from the DOE handbook
- Missing percent signs in the explanation were restored.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The percent work is a sum and a subtraction from 100, so the question mostly tests the last step, writing 68% as a fraction in lowest terms. Students who round to 30% read and 70% left to get a friendlier fraction land on a listed choice, as do those who give the part read. Dividing 68 and 100 by their common factor 4 keeps the answer exact.
Turning words into math minor Understanding the question
Both percents are “of the entire book,” so they simply add to 32%, and “left for her to read” asks for the rest, 100% − 32%, not the part already read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The result has to be written as a fraction in lowest terms, so 68/100 must be divided by 4 to reach a form that appears among the choices. (Enright, Morley & Sheehan, 2002)
Strategy: Add the two percents, since both are of the whole book, subtract the total from 100%, then write the result over 100 and simplify it fully.
Traps
- Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No sum or difference of 15% and 17% gives 28%, so 7/25 mostly draws students who match the right fraction's 25 without checking the numerator.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Gives the part already read rather than the part left, and rounds that 32% to 30% to write it as 3/10.
D Rounds too early (King et al., 2004)
Finds that 68% is left but rounds it to 70% to make an easier fraction, giving 7/10 instead of simplifying 68/100 exactly.
What was corrected from the DOE handbook, and why
- Explanation: The handbook writes “So far, Sarah has read 15 + 17 = 32 of the entire book. That means she has 100 − 32 = 68 left to read.” (p. 143, question 98), with no percent signs, so 32 and 68 read as amounts of the book (pages, say) that the question never gives. The question gives 15% and 17%, and the handbook’s next line begins “68% =”, so the site now reads 15% + 17% = 32% and 100% − 32% = 68%. The answer, 17/25, is unchanged.
Worked solution, step by step
Formulas used
- Percent as a fraction
- p is the percent. Percent means “out of 100,” so p percent is p hundredths. For example, 50% is 50 out of 100, which is one half.
- Simplifying a fraction
- a is the top number, b is the bottom, and n is a number that divides both evenly. Dividing the top and bottom by the same number keeps the value the same. A fraction is in lowest terms when no number bigger than 1 divides both.
Sort out the question
Sarah read 15% of her book yesterday and 17% today. Both are percents of the whole book. You need the part still left to read, as a fraction in lowest terms. That means a fraction that can’t be simplified any more.
Find how much she has read
Both percents are of the same whole book, so you can add them. She has read 32% of it.
Find how much is left
The whole book is 100%.
Watch out: 32% is the part she has read, not the part left. Rounding 32% to 30% gives , choice B.
Write 68% as a fraction
This fraction is not in lowest terms yet, because 68 and 100 are both even.
Write it in lowest terms
The number 4 divides both 68 and 100 evenly.
Watch out: don’t round 68% to 70%. That gives , choice D. So the answer is C, .
Check your answer
Turn the answer back into a percent: , which is 68%. Then 68% + 32% = 100%, the whole book. Also, 17 and 25 share no factor except 1, so the fraction is in lowest terms.
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The original DOE question

Question 98 of 120
2023 Test A Math, question 43Question
Kim jogs 8 kilometers in 1 hour 40 minutes. At that rate, how many meters does she jog per minute?
- A0.08
- B80
- C800
- D8,000
Show the answer
Correct answer: B
First, calculate Kim’s jogging time in minutes:
1 hour 40 minutes = 100 minutes
Then, convert kilometers to meters:
8 kilometers = 8,000 meters
Now simplify the ratio to find the answer:
Question analysis
This rate question says Kim jogs 8 kilometers in 1 hour 40 minutes and asks how many meters she jogs per minute. Students convert a mixed time to minutes and kilometers to meters before finding a unit rate. Skipping a conversion, or stopping before dividing, gives answers that are off by factors of ten or more.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Unit rates and metric conversions are grade 6 topics. This needs two conversions, hours and minutes to minutes and kilometers to meters, then one division with whole numbers, simpler than grade 7 rate problems built on fractions.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Something you must already know minor
- Many steps minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 4, 17 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The rate itself is one division; what makes the question work is that the distance and the time both arrive in the wrong units. Converting only the time gives a speed in kilometers per minute, and converting only the distance and stopping gives the whole trip in meters. Writing the units beside every number, meters over minutes, shows when the answer is ready.
Something you must already know minor Working it out
The conversions are not given: 1 kilometer is 1,000 meters and 1 hour is 60 minutes, so 8 kilometers becomes 8,000 meters and 1 hour 40 minutes becomes 100 minutes. (Anderson & Krathwohl, 2001)
Many steps minor Working it out
Two conversions, one for distance and one for time, come before the single division, and skipping the distance conversion leaves an answer a thousand times too small. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Strategy: Rewrite the time entirely in minutes and the distance entirely in meters, then divide distance by time; check that the answer's units are meters per minute.
Traps
- Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
- Stops one step early: A choice is a number from partway through the solution.
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
The wrong answers, classified
A Units or scale (King et al., 2004)
Divides the 8 kilometers by the 100 minutes but never changes kilometers to meters, giving a speed of 0.08 kilometer per minute.
C An arithmetic slip (Junker & Sijtsma, 2001)
Converts both units correctly but cancels only one zero when dividing 8,000 by 100, getting a rate ten times too large.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Converts 8 kilometers to 8,000 meters and stops, giving the whole distance before dividing by the 100 minutes it took.
Worked solution, step by step
Formulas used
- Kilometers to meters
- Each kilometer is 1,000 meters, so multiply the number of kilometers by 1,000 to get meters. The word part kilo means a thousand.
- Rate
- The rate, or speed, is how far you go in one unit of time. Distance is how far, and time is how long. Dividing shares the distance equally among the minutes, so you get the distance for one minute.
Sort out the question
Kim jogs 8 kilometers in 1 hour 40 minutes. You need her speed in meters per minute: how many meters she jogs in one minute. The question gives kilometers and hours, so change both units first.
Change the time to minutes
One hour is 60 minutes. Add the extra 40 minutes.
Kim jogs for 100 minutes.
Change the distance to meters
Kim jogs 8,000 meters. Watch out: 8,000 is the whole distance, not meters per minute. Stopping here gives choice D.
Divide the distance by the time
She jogs 80 meters each minute. Watch out: if you skip the change to meters, you get 8 ÷ 100 = 0.08 kilometer per minute, choice A. So the answer is B, 80 meters per minute.
Check your answer
Work backward. At 80 meters a minute, 100 minutes gives meters. That is 8 kilometers, just as the question says.
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The original DOE question

Question 99 of 120
2023 Test A Math, question 47Question
Which percentage is closest in value to 0.0099?
- A0%
- B0.1%
- C1%
- D100%
Show the answer
Correct answer: C
Since 0.0099 is equal to 0.99%, the answer option closest to that value is 1%.
Question analysis
This percent question asks which of four percentages is closest in value to the decimal 0.0099. Students convert a small decimal to a percent by multiplying by 100, then compare it with choices that differ by factors of ten. Moving the decimal point the wrong number of places, or reading the digits 99 as a percent, leads to wrong answers.
- NYS standards
- 6.RP.3c percent as a rate per 100
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Converting a decimal to a percent is grade 6 work. This decimal has two zeros before its first digit, so the point must move carefully, but it is a single conversion followed by picking the nearest choice, below typical grade 7 demand.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 6, 7 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One conversion answers the question, and the choices are spaced by powers of ten, so the work is all in the decimal places. A decimal this small looks like nothing, and students who forget to multiply by 100 see it as nearly 0%. Others move the point the wrong number of places or read the digits 99 as a percent. Seeing that 0.01 is exactly 1% puts 0.0099 just below it.
Error-prone arithmetic minor Working it out
The decimal 0.0099 has two zeros before its digits, so moving the point two places gives 0.99, and miscounting the places lands on 0.1% or 99%. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must know that a percent is a number of hundredths, so 0.0099 is multiplied by 100 to become 0.99%, a rule the question takes for granted. (Anderson & Krathwohl, 2001)
Strategy: Multiply the decimal by 100 to write it as a percent, moving the point two places to the right, then choose the percent nearest to that value.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reads 0.0099 as if it were already a percent, 0.0099%, which is nearest 0%, never multiplying by 100 to change the decimal into a percent.
B An arithmetic slip (Junker & Sijtsma, 2001)
Moves the decimal point one place instead of two, turning 0.0099 into about 0.1, and writes that as 0.1%, ten times too small.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Sees the digits 99, thinks of 99 percent, and picks the nearest choice, ignoring the zeros that make 0.0099 less than one hundredth.
Worked solution, step by step
Formulas used
- Decimal to percent
- A percent is a number of hundredths. To change a decimal into a percent, multiply by 100. That moves the decimal point two places to the right. For example, 0.25 × 100 = 25, so 0.25 is 25%.
Sort out the question
The number 0.0099 is a decimal, but the choices are percents. To compare them, change 0.0099 into a percent first. Then find the choice nearest to it.
Change the decimal to a percent
So 0.0099 is 0.99%. The decimal point moved two places to the right. Watch out: 0.0099 is not already a percent. Reading it as 0.0099% points to 0%, choice A.
Find the nearest choice
Find how far 0.99% is from each choice.
Choice Distance from 0.99% A, 0% 0.99 B, 0.1% 0.89 C, 1% 0.01 D, 100% 99.01 Watch out: moving the point only one place gives about 0.1, choice B. So the answer is C, 1%.
Check your answer
Work the other way. 1% is 1 out of 100, or 0.01. And 0.0099 is just a tiny bit less than 0.0100. So 0.0099 is very close to 1%, just a little less.
More questions like this
The original DOE question

Question 100 of 120
2023 Test A Math, question 50Question
A rectangular concrete driveway is 30 feet long, 8 feet wide, and 6 inches thick. What is the volume of the concrete?
- A44 cu ft
- B48 cu ft
- C120 cu ft
- D240 cu ft
Show the answer
Correct answer: C
First convert 6 inches to 0.5 foot. Then, calculate the volume, in cubic feet:
Question analysis
This volume word problem describes a concrete driveway 30 feet long, 8 feet wide and 6 inches thick, and asks for the volume of the concrete in cubic feet. Students convert inches to feet and multiply three dimensions. Mixing units, or stopping at the area of the driveway's surface, is the main way to go wrong.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Volume of a rectangular prism with a fractional edge is grade 6 content. The only extra step is changing inches to feet before multiplying, so it is lighter than grade 7 volume work with composite or triangular prisms.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 2, 27 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Volume of a box is familiar, so the question is decided by the mix of units: two dimensions in feet and one in inches. Multiplying the three numbers as printed gives 1,440, which is not a choice, and that should send students back to convert. The listed wrong answers instead come from leaving out the thickness or combining the numbers without the volume rule.
Something you must already know minor Working it out
The stem gives no formula and no conversion, so students must bring both the volume rule, length times width times height, and the 12 inches in a foot. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The thickness becomes the fraction 1/2 foot, so the product is 30 × 8 × 1/2, or, if the inches are kept, 1,440 divided by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Change 6 inches to a fraction of a foot first so that all three measurements share a unit, then multiply length, width and thickness.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Adds the three numbers in the stem, 30 + 8 + 6 = 44, instead of multiplying the dimensions.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Multiplies only the width and the thickness, 8 × 6 = 48, combining two of the stem's numbers and leaving out the 30-foot length.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the area of the driveway's surface, 30 × 8 = 240, and stops before multiplying by the half-foot thickness.
Worked solution, step by step
Formulas used
- Inches to feet
- There are 12 inches in 1 foot. So to change inches into feet, divide the number of inches by 12.
- Volume of a rectangular box
- V is the volume, the space inside a box shape. l is the length, w is the width and h is the height, here the thickness. All three must use the same unit. The answer is in cubic units, such as cubic feet.
Sort out the question
The driveway is a flat box of concrete. It is 30 feet long, 8 feet wide and 6 inches thick. You need its volume, the space the concrete fills, in cubic feet. The thickness is in inches, so change it to feet first.
Change the thickness to feet
So 6 inches is half a foot. Now all three measures are in feet.
Multiply the three measures
The volume is 120 cubic feet. Watch out: 30 × 8 = 240 is only the area of the top surface, in square feet. Stopping there gives choice D. Adding 30 + 8 + 6 = 44 gives choice A.
So the answer is C, 120 cu ft.
Check your answer
Think of it another way. A slab 30 feet by 8 feet and a full foot thick would hold 240 cubic feet. This one is only half a foot thick, so it holds half as much, 120 cubic feet. Using 6 instead of one half would give 1,440, far too big.
More questions like this
The original DOE question

Question 101 of 120
2023 Test B English, question 12Passage Excerpt from “The Food Business Incubator”
Excerpt from “The Food Business Incubator”
by Mandalit Del Barco
“La Cocina” means “the kitchen” in Spanish. It’s also the name of a business incubator1 based in San Francisco’s Mission District. Since it began in 2005, it’s been helping local food entrepreneurs, many of whom are low-income immigrant women, develop their small businesses.
Over the years, many of its alumni have found success: more than 50 chefs in its program have become self-sufficient business owners, and many of them have opened their own brick-and-mortar restaurants. Two alumnae of its culinary program . . . were even recognized as semi-finalists for prestigious James Beard awards.2
A new cookbook, We Are La Cocina: Recipes in Pursuit of the American Dream, tells some of their stories.
Executive director Caleb Zigas says the nonprofit La Cocina grew out of two grassroots economic development organizations3 who found many people cooking at their homes and selling food on the streets. The vendors needed an affordable commercial kitchen space and technical assistance in order for their businesses to be legally viable. La Cocina provided just such a space, in addition to helping them develop business plans, pull city permits and more.
Zigas says as many as eight businesses can work in the kitchen space at La Cocina at any one time. Some can prep for a farmers’ market sale, corporate catering gigs or weddings, while others might be making and packaging their food products.
“It’s just an incredible and exciting range of techniques, flavors, perspectives, age, language. And that’s a really beautiful thing,” he says. . . .
Twice a year, La Cocina hosts “F&B: Voices from the Kitchen,” a storytelling project where their chefs can tell their own stories, as they do in the new cookbook. Some of the chefs will be on tour to promote the book, whose proceeds will go to support La Cocina entrepreneurs.
NPR4 caught up with . . . La Cocina chefs, who shared their stories:
Mariko Grady, Aedan Fermented Foods
At La Cocina, you can often hear Mariko Grady singing or humming as she prepares miso, koji, and amasake. Her fermented products come in four different flavors, including mushroom and chicken, to be used in soups and sauces. She originally brought the fermenting rice and barley koji seeds from Japan, where she had a 30-year career as a singer and dancer with the prestigious modern theatrical dance group she founded, Pappa Tarahumara. They performed around the world, and 16 years ago, had a one-night show in San Francisco. The man who would become her husband was in the audience. She soon joined him in San Francisco, often returning to Tokyo to rehearse. But after the earthquake and tsunami hit Japan in 2011, her company disbanded. . . . Grady focused on nourishing her family and creating a line of fermented products that she sells online, at local Bay Area stores and at the Ferry Plaza Farmers Market. She began at La Cocina in 2012 and named her business after her son, Aedan. (Written in Kanji characters, the name means “wisdom” and “handed down from generation to generation.”) . . . Her fermented products are “also full of wisdom about how to relate to nature and how to create a healthy life,” she says. . . .
Shani Jones, Peaches Patties
Shani Jones is a native San Franciscan. Her father was born in New Orleans, her mother, in Jamaica. She says their home was always filled with a variety of spices and dishes like jerk chicken and Jamaican patties—savory pastries filled with beef or chicken. Jones says she learned to cook from her mother, whose nickname was Peaches. After returning from college in Atlanta, Jones worked on her doctorate in organizational leadership and management while driving a Lyft5 car. She often told passengers about her idea of opening a catering company with her mom’s recipes. They steered her to La Cocina, where she developed her own business, named after her mother. Five years later, she caters and runs a kiosk6 at a small food cooperative in Bernal Heights, where some of her handmade patties have an Ethiopian twist, “because my husband is Ethiopian.” Jones has big aspirations for Peaches Patties: “The ultimate goal,” she says, “is to be the patty kingpin7 of the West Coast.”
From “The Food Business Incubator That Helps Immigrant Women Pursue The American Dream” by Mandalit Del Barco from WEEKEND EDITION SUNDAY, June 19, 2019. Copyright © 2019 NPR.
QUALIFICATIONS FOR LA COCINA APPLICANTS
| Level of Income | La Cocina considers assets and access to opportunity when screening applicants. La Cocina’s mission is to support entrepreneurs who face barriers to becoming successful in the food industry. Applicants must be classified as low- or very-low-income earners according to the Housing and Urban Development (HUD) Guidelines for San Francisco. |
|---|---|
| Business Plan | Applicants must have a business plan that is ready to be put into action. La Cocina will help applicants refine an existing business plan. Applicants without a business plan should contact one of La Cocina’s partner organizations for guidance in developing one. |
| Viable Product | The applicant’s food must not only be delicious but also stand out. For a business to make a profit, there also has to be a market for the food. |
| Entrepreneurial Work Ethic | La Cocina is excited to work with entrepreneurs who are motivated and passionate about growing their business. Succeeding in the food industry requires a lot of heart and hard work. |
| Community Spirit | The 30+ entrepreneurs who share La Cocina’s kitchen are from many different backgrounds. La Cocina values the community it has created and expects each entrepreneur to be active in bringing the spirit of La Cocina to the wider world. In order to succeed in the La Cocina kitchen, accepted applicants must be flexible and willing to work closely with others. |
Source: La Cocina
1business incubator: company that helps new and startup companies develop by providing services such as management training and office space
2James Beard awards: an award named after American cook and author James Beard, considered one of the highest honors in the culinary industry
3economic development organizations: organizations whose mission is to promote economic improvement within a specific geographical area
4NPR: National Public Radio, a nonprofit media organization that produces and distributes news and cultural programming
5Lyft: ride-sharing company where people use their own cars to provide rides to customers
6kiosk: cart or small structure with one or more open sides that is used to vend merchandise or services
7kingpin: the leader in a group or undertaking
Question
The quotation from Caleb Zigas in paragraph 6 conveys a central idea of the excerpt by
- Ashowing appreciation for the vibrant diversity of La Cocina.
- Bcommunicating La Cocina’s goal of helping people of varying abilities.
- Cdescribing a typical scenario at La Cocina.
- Demphasizing La Cocina’s focus on creating products.
Show the answer
Correct answer: A
The question asks how the quotation from Caleb Zigas in paragraph 6 conveys a central idea of the excerpt.
A. CORRECT. A central idea of the excerpt is that La Cocina’s entrepreneurs are a group of people cooking a diverse range of cuisines. This quotation from the executive director of La Cocina conveys this idea by emphasizing the diversity of both the cooking methods and the chefs themselves.
B. Incorrect. While La Cocina’s goal is “helping local food entrepreneurs, many of whom are low-income immigrant women” (paragraph 1), the organization does not necessarily have a goal of helping people without solid cooking skills. Per the table that describes the “Qualifications for La Cocina Applicants,” the applicant’s food must be delicious, which speaks to the skill required for all participants.
C. Incorrect. The quotation describes the diversity of people and cooking techniques found at La Cocina, but it does not reference “a typical scenario.” Paragraphs 9 and 10 more closely describe examples of typical La Cocina chef entrepreneurs.
D. Incorrect. Applicants to La Cocina must have a food that is a “viable product,” but this quotation emphasizes the diversity of chefs and cuisines, not the requirements chefs must meet to enter the program.
Question analysis
This question looks at paragraph 6 of the La Cocina article, a two-sentence quotation in which executive director Caleb Zigas sums up what he sees at the nonprofit. Students must decide how the quotation conveys a central idea of the excerpt, which means restating Zigas’s words in their own and connecting them to the article as a whole.
- Grade level
- Grade 7 reading and task
- Compared with grade 7
- Below grade 7 Showing how a quotation supports a central idea is grade 6 and 7 work. This quotation is two sentences in everyday words, and its meaning sits close to the surface, so the demand is lighter than typical grade 7 reading.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly analyze
- What makes it hard
- Reading between the lines minor
- How, not what minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Passage: about grade 10, 1057 words; question and choices: about grade 12, 48 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The quotation is short and warm, and most students hear its enthusiasm for variety at once. The wrong choices come from reading it with paragraph 5 still in mind, a kitchen busy with market and catering prep and with food products being packaged, or from turning a range of cooking styles into a range of skill. Students who restate the quotation in their own words before reading the choices avoid all three.
Reading between the lines minor Working it out
Zigas never says “diversity”; students must see that his list, “techniques, flavors, perspectives, age, language,” describes variety among the cooks and their food, and that “a really beautiful thing” expresses appreciation. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
How, not what minor Working it out
The stem asks how the quotation conveys a central idea, so students must name what Zigas’s words do for the excerpt, not just what they say, and link them to the mix of chefs profiled later. (Anderson & Krathwohl, 2001)
Strategy: Reread the quotation and say in your own words what Zigas finds exciting; then choose the option that matches your words and also fits the rest of the article.
Traps
- Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
- Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
The wrong answers, classified
B Twists what the text says (King et al., 2004)
Turns Zigas’s “range of techniques” and “age” into people of differing ability, though he celebrates variety in cooking styles and backgrounds, and the table requires every applicant’s food to be delicious.
C Wrong part of the text (Kirsch & Mosenthal, 1990)
Carries paragraph 5’s scene, eight businesses prepping for markets, catering and weddings, into the quotation, which describes no typical scenario but sums up the variety Zigas sees.
D Twists what the text says (King et al., 2004)
Links “flavors” and “techniques” to the packaged food products of paragraph 5 and makes products the point, though Zigas is praising the mix of people and cooking styles.
Working it out, step by step
Know what a central idea is
A central idea is an important idea that runs through a whole text. This question asks how one quotation from Caleb Zigas, the head of La Cocina, shows a central idea. Start by putting his words into your own.
Read the quotation in paragraph 6
Zigas says, “It’s just an incredible and exciting range of techniques, flavors, perspectives, age, language.” A range is a wide mix. Techniques are ways of cooking. Perspectives are ways of seeing the world. Then he adds, “And that’s a really beautiful thing.”
Say it in your own words
In plain words, Zigas loves how different everyone at La Cocina is. The cooks are young and old. They speak different languages. They cook in different ways and make food with different flavors. He calls this mix “beautiful,” so he is praising it. A word for this kind of mix is diversity. The chef stories show it too. Mariko Grady brought her koji seeds from Japan. Shani Jones gives some patties an Ethiopian twist.
Check each choice
- A is right. Vibrant means lively and full of energy. Zigas shows appreciation, or praise, for the lively mix of people and foods.
- B is wrong. A range of cooking styles is not a range of skill. The table says every applicant’s food “must not only be delicious but also stand out.”
- C is wrong. The busy kitchen scene is in paragraph 5. The quotation sums up a feeling. It doesn’t describe a usual day.
- D is wrong. Zigas is praising people and styles, not products for sale.
Choose the answer
The quotation praises the wide mix of people and cooking at La Cocina, an idea the chef stories also show. So the answer is A.
Why this answer and not the closest wrong one
B is the closest wrong answer, because Zigas does mention “age” and “techniques.” But he is talking about variety, not about how good the cooks are. A matches both halves of his quotation: the mix of people and food, and the word “beautiful.”
More questions like this
The original DOE question

The passage “Excerpt from “The Food Business Incubator”” as printed (pages 158–160)



Question 102 of 120
2023 Test B Math, question 1Question
LAST YEAR’S TOTAL SALES
| Vehicles | Number |
|---|---|
| Four-door cars | 5,850 |
| Two-door cars | 2,250 |
| Trucks | 900 |
| Total | 9,000 |
The table above shows the number of each type of vehicle sold at a dealership last year. If 1,000 vehicles will be sold next month, what is the best estimate (based on last year’s sales) of the number of two-door cars that will be sold?
Grid-in question: work out the answer and enter it as a number.
Show the answer
Correct answer: 250
Let x represent the number of two-door cars that will be sold next month. Use the information in the table to set up a proportion:
Question analysis
A table shows last year's sales at a car dealership: four-door cars, two-door cars and trucks, 9,000 vehicles in all. This grid-in asks for the best estimate of how many two-door cars will be sold next month if 1,000 vehicles are sold. Students set up a proportion from the table; most of its rows are not needed.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 proportion problems often take several steps or awkward numbers. This one scales a single share from a table to a new total, with numbers that divide cleanly, which is closer to grade 6 ratio work.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 4, 55 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Once the share is seen as 2,250 out of 9,000, a quarter, the arithmetic is light, which makes this one of the easier grid-ins. Points are lost by entering the share itself as a percent, by reading the wrong row of the table, or by building a monthly average when the question supplies next month’s total. With no choices to compare, checking that the entry is a quarter of 1,000 is the only safeguard.
Turning words into math minor Understanding the question
The phrase “best estimate (based on last year’s sales)” has to become a proportion: two-door cars keep the same share of next month’s 1,000 vehicles as of last year’s 9,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table lists three vehicle types and a total; only the two-door row, 2,250, and the 9,000 total are needed, so the four-door and truck rows are distractions. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Strategy: Find what fraction of last year's 9,000 vehicles were two-door cars, simplify it, then take that fraction of the 1,000 vehicles expected next month.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Information you don't need: The question gives numbers or facts that play no part in the answer.
The wrong answers, classified
25 Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds that 2,250 of 9,000 sales is 25 percent and enters that percent, stopping before taking a quarter of next month’s 1,000 vehicles.
650 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reads the four-door row instead of the two-door row and scales it correctly: 5,850 of 9,000 is 650 of 1,000.
187.5 Misreads the question (Hegarty, Mayer & Monk, 1995)
Divides last year’s 2,250 two-door sales by 12 months, a monthly average that ignores the stated 1,000 vehicles for next month.
Worked solution, step by step
Formulas used
- Proportion
- A proportion says two fractions are equal. Each fraction here is a part over its whole: two-door cars over all vehicles sold. Use it when a share stays the same while the total changes. Then you can find the missing part.
Sort out the question
The table shows last year’s sales. In all, 9,000 vehicles were sold, and 2,250 of them were two-door cars. Next month, 1,000 vehicles will be sold. You need about how many of them will be two-door cars.
“Based on last year’s sales” means two-door cars keep the same share of all sales. A share is a part out of a whole. The four-door and truck rows don’t matter here.
Find last year’s share of two-door cars
Last year, 2,250 out of 9,000 vehicles were two-door cars. Write that as a fraction and simplify it.
This works because . So 1 out of every 4 vehicles sold was a two-door car.
Set up the proportion and solve
Let x be the number of two-door cars next month. Next month’s share must match last year’s share.
To get x alone, multiply both sides by 1,000.
So about 250 two-door cars will be sold.
Give the answer
Watch out: one fourth is the same as 25%. Gridding 25 gives the percent, not the number of cars. Also be sure to use the two-door row. The four-door row would give 650. So you would grid 250.
Check your answer
Try it another way. Next month’s 1,000 vehicles are of last year’s 9,000. So take of the 2,250 two-door cars: . Both ways give 250 cars.
More questions like this
The original DOE question

Question 103 of 120
2023 Test B Math, question 3Question
A kindergarten teacher has 72 crayons, 60 pencils, and 84 sheets of paper to distribute to the children in her class. If each child receives an equal number of each item and there are no items remaining, what is the greatest possible number of children in the class?
Grid-in question: work out the answer and enter it as a number.
Show the answer
Correct answer: 12
The greatest possible number of children in the class is the greatest common factor of 72, 60, and 84. Write the prime factorization of each number:
Then, find the factors that all three have in common, which is .
Question analysis
A kindergarten teacher shares 72 crayons, 60 pencils and 84 sheets of paper so that every child gets the same number of each item, with nothing left over. This grid-in asks for the greatest possible number of children in the class. Students must recognize a greatest common factor problem and not confuse it with a least common multiple.
- NYS standards
- 6.NS.4 common factors and multiples
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Turning words into math moderate
- Something you must already know minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 11, 45 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The story has to be recognized as a sharing problem before any arithmetic starts, and the word greatest points to the largest divisor shared by all three counts. Students who spot a small shared factor can stop too soon, while those who think of a common multiple head the wrong way entirely. Checking that 72, 60 and 84 all divide by the entry, and that no larger number does, settles it.
Turning words into math moderate Understanding the question
Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)
Strategy: Write 72, 60 and 84 as products of primes, multiply the primes all three share, and check that each amount divides evenly by the result.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- Stops one step early: A choice is a number from partway through the solution.
The wrong answers, classified
6 Stops one step early (King et al., 2004; Gierl et al., 2017)
Spots that 6 divides 72, 60 and 84 and stops there, a common factor not yet multiplied by the extra 2 that all three share.
2520 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Finds the least common multiple of 72, 60 and 84, 2,520, confusing a number all three counts divide into with one that divides all three.
4 Stops one step early (King et al., 2004; Gierl et al., 2017)
Takes only the shared 2 × 2 from the prime factorizations and drops the factor 3 that 72, 60 and 84 also share.
Worked solution, step by step
Formulas used
- Greatest common factor
- A factor of a number divides it with nothing left over. The GCF is the greatest number that is a factor of every number in the group. A prime number has only 1 and itself as factors, like 2, 3, 5 and 7. Multiply the primes that all the numbers share, as many times as each number has them.
Sort out the question
The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.
So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.
Break each number into primes
Write each number as a product of primes. A product is the answer to a multiplication.
You can check each one by multiplying it back out.
Pick out the shared primes
Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.
Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.
Give the answer
Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.
Check your answer
Share the items among 12 children. Each child gets crayons, pencils and sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.
More questions like this
The original DOE question

Question 104 of 120
2023 Test B Math, question 9Question
Bryan completed a 100-meter race in 11.74 seconds. Luis completed the same race in 11.69 seconds. What was the difference between their times, expressed as a fraction of a second?
- A
- B
- C
- D
Show the answer
Correct answer: A
First, subtract Luis’s time from Bryan’s time: .
Then, convert to a fraction:
.
Question analysis
Bryan ran a 100-meter race in 11.74 seconds and Luis ran the same race in 11.69 seconds. Students find the difference between their times and express it as a fraction of a second. It tests subtracting decimals and rewriting a decimal as a fraction in lowest terms.
- NYS standards
- 6.NS.3 multi-digit decimal operations
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Subtracting decimals and writing a decimal as a fraction are skills from before grade 7. The two times are close, so the subtraction is quick, and the only care needed is with place value and simplifying.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 32 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Two quick skills settle this question: lining up the decimals to subtract, and naming the result as a fraction in lowest terms. The two times differ only in their last two digits, so the subtraction is short, but it needs a borrow that some students skip. The fraction step is where a decimal’s digit gets mistaken for a denominator.
Error-prone arithmetic minor Working it out
Subtracting 11.69 from 11.74 needs a borrow in the hundredths, 14 − 9, after which the tenths cancel to 0, the step where place value goes wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The difference, 0.05, must be read as five hundredths and reduced as a fraction; the question gives no help with converting a decimal. (Anderson & Krathwohl, 2001)
Strategy: Line up the decimal points, subtract, write the difference as a number of hundredths over 100, and reduce the fraction.
Traps
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
The wrong answers, classified
B An arithmetic slip (Junker & Sijtsma, 2001)
Looks only at the tenths, 11.7 − 11.6 = 0.1, dropping the hundredths, a place-value slip that doubles the true gap.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Subtracts the smaller digit from the larger in each place, 9 − 4 = 5 and 7 − 6 = 1, getting 0.15, which is 3/20.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Gets 0.05 but reads its digit 5 as a denominator, writing 1/5, instead of five hundredths, 5/100.
Worked solution, step by step
Formulas used
- Decimals as hundredths
- The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths, and you can write its digits over 100. For example, 0.37 is 37/100. Then simplify by dividing the top and bottom by the same number.
Sort out the question
Bryan ran the race in 11.74 seconds. Luis ran it in 11.69 seconds. The difference is how much longer one time is than the other. You need it as a fraction of a second.
Subtract the two times
Subtract the smaller time from the larger one. Line up the decimal points.
In the hundredths place, 4 is less than 9. So regroup: trade 1 tenth for 10 hundredths. Then 14 − 9 = 5 hundredths. The tenths are now 6 − 6 = 0.
Watch out: don’t just take the smaller digit from the larger one in each place. That gives 0.15, which is , choice C.
Write the difference as a fraction
The 5 is in the hundredths place, so 0.05 is 5 hundredths.
Now simplify. Divide the top and bottom by 5.
Give the answer
Watch out: 0.05 is not . One fifth is 0.2, which is four times as big. So the answer is A, .
Check your answer
Turn back into a decimal: . Then add it to Luis’s time: . That is Bryan’s time, so the difference is right.
More questions like this
The original DOE question

Question 105 of 120
2023 Test B Math, question 11Question
What is the value of n expressed as a fraction?
- A
- B
- C
- D
Show the answer
Correct answer: B
Question analysis
The number n equals 0.13 plus 207/100, and students must write n as a single fraction. This question mixes a decimal with a fraction in hundredths. Converting both to the same form, adding, and reducing to lowest terms are the steps, and the four fractions offered look quite different at first glance.
- NYS standards
- 6.NS.3 multi-digit decimal operations
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Converting decimals to fractions and simplifying are skills taught before grade 7. Both numbers here are in hundredths, so the addition is immediate and only the final reduction takes thought.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 7, 15 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Converting 0.13 to hundredths makes this a single addition, 13 + 207 = 220 hundredths, so the question tests little beyond reducing a fraction. Two of the wrong choices sit close to 2.2 but come from no common working, and the third only echoes the digits of 0.13. A student who checks each choice as a decimal finds only one equal to 2.2.
An answer in an unexpected form minor Choosing the answer
The sum comes out as 220/100 or 2.2, but the choices are in lowest terms, so it must be reduced, or each choice converted, before anything matches. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
One term is a decimal and the other a fraction; writing 0.13 as 13/100 puts both in hundredths, and a misplaced digit there changes the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Write 0.13 as 13/100 so both terms are in hundredths, add the numerators, then divide the top and bottom by their greatest common factor.
Traps
- Answer in another form: The right answer is written differently from the way you worked it out.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Echoes the digits of 0.13, reading it as the fraction 1/3, and ignores the 207/100 altogether.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 9/4, which is 2.25; adding 13 and 207 hundredths gives 2.2, so this mostly draws guesses near the sum.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Nothing leads to 23/10 except 13 + 207 coming out as 230, a slip with no reason behind it, so it mostly catches guesses.
Worked solution, step by step
Formulas used
- Decimals as hundredths
- The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
- Adding fractions with the same bottom
- a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.
Sort out the question
n equals 0.13 plus . One part is a decimal and one is a fraction. You need n as a fraction. The choices are in lowest terms, which means they can’t be simplified any more.
Write the decimal as a fraction
The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.
Now both numbers are in hundredths. Watch out: 0.13 is not . That only copies the digits, choice A.
Add the hundredths
Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.
Simplify the fraction
is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.
So the answer is B, .
Check your answer
Change both to decimals. . And . They match. Choices C and D are close, but and .
More questions like this
The original DOE question

Question 106 of 120
2023 Test B Math, question 16Question

This wheel has 15 spokes, equally spaced. What is the measure of the angle between adjacent spokes?
- A30°
- B24°
- C15°
- D12°
Show the answer
Correct answer: B
The total number of degrees in a circle is 360. The measure of the angle between adjacent spokes is .
Question analysis
A wheel has 15 spokes, equally spaced around its hub, as the figure shows. Students find the measure of the angle between two neighboring spokes. The question tests knowing how many degrees make a full turn around a point and sharing them equally among the spokes.
- NYS standards
- 7.G.5 angle relationships and equations
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding an angle by dividing a full turn into equal parts is early angle measurement. It takes one division, and the only fact needed is the number of degrees in a circle, so it sits below grade 7.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 5, 16 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
One fact and one division answer this question, so it is among the easiest on the test. The wrong choices catch students who use a straight angle’s 180° for the full turn, who take the number of spokes as the angle, or who guess a familiar angle. Multiplying the answer by 15 to get back to 360° is a quick check.
Something you must already know minor Working it out
The question never says that a full turn is 360°; knowing it, and that the 15 spokes split it into 15 equal angles, is the whole method. (Anderson & Krathwohl, 2001)
Strategy: The spokes split the full turn around the hub into 15 equal angles, so divide the number of degrees in a circle by 15.
Traps
- Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No working with 15 spokes gives 30°, which is 360 ÷ 12, the angle between hours on a clock, so it mostly draws guesses at a familiar angle.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Takes the number of spokes, 15, as the angle in degrees, matching a number in the question without any calculation.
D Wrong formula (King et al., 2004)
Divides 180 by 15, using a straight angle’s 180° as the full turn around the hub instead of 360°.
Worked solution, step by step
Formulas used
- Equal angles around a point
- One full turn around a point is 360 degrees. If lines from the point split that turn into equal angles, each angle gets the same share. So divide 360 by how many angles there are.
Sort out the question
The wheel has 15 spokes. They all start at the hub, the center of the wheel, and they are equally spaced. Adjacent spokes are spokes right next to each other. You need the angle between two adjacent spokes.
Count the angles around the hub
Go once around the hub. Between each spoke and the next is one angle. So 15 spokes make 15 angles, all the same size. Together they make one full turn, which is 360 degrees.
Watch out: a full turn is 360 degrees, not 180. Using 180 gives 12 degrees, choice D.
Share the full turn
Use the formula for equal angles.
Each angle is 24 degrees. Watch out: 15 is the number of spokes, not the size of the angle, so choice C is wrong. So the answer is B, 24°.
Check your answer
Multiply back: 15 angles of 24 degrees make degrees, a full turn. Also, a clock’s 12 hour marks are 30 degrees apart. This wheel has more spokes than that, so each angle must be smaller than 30 degrees.
More questions like this
The original DOE question

Question 107 of 120
2023 Test B Math, question 17Question
For the equation above, what is the sum of the values of m when , when , and when ?
- A12
- B72
- C132
- D135
Show the answer
Correct answer: C
First, find the value of m for each value of t:
For :
For :
For :
Then, add those three values together:
Question analysis
The equation m • t = 72 links two variables, and the raised dot means multiplication. Students find m when t is 1, 2 and 3, and then add those three values of m. Each value is a one-step division, and the trap is combining the t values before dividing.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Solving equations like 3m = 72 is grade 6 work. This item repeats that step three times and adds the results, more bookkeeping than difficulty, so it is below typical grade 7 demand.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Turning words into math minor
- Many steps minor
- Solution
- 4 steps
- Reading level
- Question and choices: about grade 7, 21 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Each value of m comes from one division, so the work is bookkeeping rather than algebra. The trap is structural: adding 1, 2 and 3 first and dividing once feels efficient but answers a different question, since m does not change in step with t. Writing the three values of m in a row before adding them keeps the steps separate.
Turning words into math minor Understanding the question
“The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Strategy: Divide 72 by each value of t in turn to get three separate values of m, then add the three quotients.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Adds the three t values to get 6 and divides 72 by 6, as if the sum of the m values were m at the sum of the t values.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Gives back 72, the constant in the equation and the value of m only when t = 1, without the values for t = 2 and t = 3.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 135, which is 3 more than the right sum; it would need a slip such as 72 ÷ 3 = 27, so it mostly draws guesses.
Worked solution, step by step
Formulas used
- Division undoes multiplication
- a, b and c are numbers, and b is not 0. If you know a product and one of the numbers that made it, divide to find the other one. For example, 4 × 5 = 20, so 4 = 20 ÷ 5.
Sort out the question
The equation is . The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.
Get m by itself
Since m times t is 72, m is 72 divided by t.
Now put in each value of t, one at a time.
Find m for each value of t
Each time, divide 72 by the value of t. A table keeps the three answers in order.
Value of t Value of m 1 2 3 Add the three values
Watch out: don’t add the t values first. is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.
Check your answer
Check each pair by multiplying: , and . All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.
More questions like this
The original DOE question

Question 108 of 120
2023 Test B Math, question 23Question
FRUIT SOLD
AT STORE XYZ
| Fruit | Number Sold |
|---|---|
| Apples | 8 |
| Bananas | 6 |
| Cherries | 12 |
| Oranges | 6 |
| Peaches | ? |
| Total | 50 |
Peaches made up what percentage of the total number of pieces of fruit sold at Store XYZ, according to the table above?
- A18%
- B28%
- C36%
- D64%
Show the answer
Correct answer: C
First, find the number of peaches sold:
Then, find the percentage:
Question analysis
A table lists the fruit sold at Store XYZ, apples, bananas, cherries and oranges, with the number of peaches shown as a question mark and a total of 50. Students find what percentage of the fruit sold were peaches. They first work out the missing count, then turn it into a percent of the total.
- NYS standards
- 6.RP.3c percent as a rate per 100
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Finding a missing value and writing a part as a percent of a whole are grade 6 skills. With a total of 50, the percent step is quick, so the task is lighter than typical grade 7 percent problems.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Reading a figure, table or graph minor
- Many steps minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 4, 34 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The total of 50 makes the percent step easy, since each piece of fruit is 2 percent, but the question first hides the count it needs. Students who find the 18 peaches and stop, or who give the share of the four listed fruits, reach listed choices. Checking that the peaches’ percent and the other fruits’ percent add to 100 confirms the answer.
Reading a figure, table or graph minor Understanding the question
The table gives four counts, a question mark for peaches and a total of 50, so the missing count must be found by subtraction before any percent. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Adding 8, 6, 12 and 6, subtracting from 50, and turning 18 out of 50 into a percent are three steps, and the count of 18 is itself a listed percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Strategy: Add the four known counts, subtract from 50 to get the peaches, then double that count, since each piece out of 50 is 2 percent.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the 18 peaches but gives that count as the percent, forgetting to compare it with the total of 50 pieces.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 28%, which would need 14 peaches; it takes a slip such as 36 for 8 + 6 + 12 + 6, so it mostly draws guesses.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Finds the share of the other four fruits, 32 of 50 or 64 percent, instead of the share of the peaches.
Worked solution, step by step
Formulas used
- Percent of a total
- The part is the amount you are asking about, here the peaches. The whole is the total, here all 50 pieces of fruit. Dividing gives the fraction of the whole. Multiplying by 100 turns it into a percent, which means out of 100.
Sort out the question
The table shows how many of each fruit Store XYZ sold. The number of peaches is missing, shown by a question mark. The total is 50 pieces of fruit. You need the percent of the total that were peaches.
Add the fruits you know
Apples, bananas, cherries and oranges add up to 32 pieces of fruit. That is every fruit except the peaches.
Find the number of peaches
The peaches are whatever is left of the 50.
So 18 peaches were sold. Watch out: 18 is a count, not a percent. Choosing 18% is choice A.
Turn the count into a percent
Peaches were 36% of the fruit sold. Watch out: 32 out of 50 is 64%, choice D. That is the other four fruits, not the peaches. So the answer is C, 36%.
Check your answer
Out of 50 pieces, each piece is 2% of the total, because . So 18 peaches are percent. Also, 36% for peaches plus 64% for the rest makes 100%.
More questions like this
The original DOE question

Question 109 of 120
2023 Test B Math, question 24Question
A prism is made up of 9 congruent cubes. Each of the cubes has side lengths measuring 4 units. What is the volume, in cubic units, of the prism?
- A108
- B144
- C480
- D576
Show the answer
Correct answer: D
First, determine the volume of each congruent cube. All the side lengths of the cubes are equal, so the length, width, and height of each cube is 4 units. To determine the volume of each cube, multiply :
So, the volume of each cube is 64 cubic units.

The prism is comprised of 9 of these cubes, so multiply the volume of each cube (64) by the total number of cubes (9) to determine the volume, in cubic units, of the prism:
Question analysis
A prism is built from 9 identical cubes, and each cube has edges 4 units long. Students find the volume of the prism in cubic units. It tests the volume of a cube and how volumes add when solids are put together, and it rewards keeping area, surface area and volume apart.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Volume of cubes and rectangular prisms is grade 5 and 6 content. This item needs one cube's volume and a multiplication by 9, two short steps, so it is below typical grade 7 work.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 3, 27 words.
- Corrected from the DOE handbook
- A letter x printed in place of a multiplication sign in the explanation was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The question needs one formula and one multiplication, and the stem does not even say how the cubes are arranged, because the volume does not depend on it. The wrong choices come from mishandling the cube’s volume: the exponent multiplied instead of applied, the area of one face, or a surface area. Writing the volume out as edge times edge times edge, in cubic units, rules all three out.
Something you must already know minor Working it out
A cube’s volume, edge × edge × edge, is not given; with it, each 4-unit cube holds 64 cubic units, and the arrangement of the 9 cubes does not matter. (Anderson & Krathwohl, 2001)
Strategy: Find the volume of one cube by multiplying its three edges, then multiply by the number of cubes; how the cubes are arranged does not matter.
Traps
- Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
- A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
The wrong answers, classified
A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Takes 4³ as 4 × 3 = 12, multiplying the edge by the exponent, so each cube seems to hold 12 cubic units and 9 × 12 = 108.
B Wrong formula (King et al., 2004)
Uses the area of one face, 4 × 4 = 16 square units, as each cube’s volume, so 9 × 16 = 144.
C Wrong formula (King et al., 2004)
Pictures the 9 cubes as a 3-by-3 slab, 12 by 12 by 4 units, and finds its surface area, 480 square units, instead of its volume.
What was corrected from the DOE handbook, and why
- Explanation: The handbook gives the volume of a cube as “length x width x height”, with the letter x where it means “times”. In a math explanation the letter x reads as a variable, so the site writes length × width × height, the sign the handbook itself uses on the next line (4 × 4 × 4 = 64). Nothing else changes: each cube is 64 cubic units, and the 9 cubes make 64 × 9 = 576.
Worked solution, step by step
Formulas used
- Volume of a cube
- V is the volume, the space inside a solid, in cubic units. s is the length of one edge. All the edges of a cube are equal, so its length, width and height are all s. Multiplying them counts how many unit cubes fit inside.
Sort out the question
A prism, a solid with flat faces, is made of 9 congruent cubes. Congruent means exactly the same size and shape. Each cube has sides of 4 units. You need the volume of the whole prism, in cubic units.
The question doesn’t say how the cubes are stacked. You don’t need to know. Volume is the space inside, and 9 cubes fill the same space however they are arranged.
Find the volume of one cube
Each cube has a volume of 64 cubic units. Watch out: means , not . That slip leads to 108, choice A. Also, is only the area of one face. It leads to 144, choice B.
Multiply by the number of cubes
The prism is 9 of these cubes, so its volume is 9 times as big.
The prism’s volume is 576 cubic units.
Give the answer
Watch out: volume counts the space inside, not the outside faces. Suppose the cubes formed one flat 3-by-3 layer. Adding up the areas of its outside faces gives 480, choice C. That is surface area, not volume. So the answer is D, 576.
Check your answer
Estimate: , and one cube fewer is . You can also test one arrangement. A flat 3-by-3 layer is 12 by 12 by 4 units, and . Same answer.
More questions like this
The original DOE question

Question 110 of 120
2023 Test B Math, question 34Question
There are 3 different highways from City W to City X, 4 different highways from City X to City Y, and 3 different highways from City Y to City Z. How many different routes are there for a trip from City W to City X to City Y to City Z?
- A10
- B12
- C24
- D36
Show the answer
Correct answer: D
Multiply the numbers of highways between each pair of cities:
Question analysis
A counting question about highways: there are 3 highways from City W to City X, 4 from X to Y and 3 from Y to Z. Students count the different routes from W to Z through X and Y. It tests the counting principle, and deciding whether to add or multiply the numbers of highways is the key.
- NYS standards
- 7.SP.8b sample spaces for compound events
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Grade 7 students build tree diagrams and lists for compound events. This item needs only the counting principle across three stages, with no restriction on the choices, which makes it shorter than typical grade 7 work.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 1 step
- Reading level
- Question and choices: about grade 11, 48 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Once a route is seen as one highway on each of three stretches, the count is a single product, 3 × 4 × 3. The wrong choices come from adding the counts, from stopping after two stretches, and from adding two-stretch products. A small tree for the first two stretches shows why the numbers multiply: each of the 3 highways to City X branches into 4.
Something you must already know minor Working it out
The counting principle, multiplying the 3, 4 and 3 highway choices because each route uses one highway on every stretch, is assumed; the stem never says to multiply. (Anderson & Krathwohl, 2001)
Strategy: A route uses one highway on each of the three stretches, so count the choices for each stretch and multiply; a small tree for two stretches shows why.
Traps
- Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
- Stops one step early: A choice is a number from partway through the solution.
The wrong answers, classified
A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Adds the highways, 3 + 4 + 3 = 10, counting the roads available rather than the ways to choose one road on each stretch.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Multiplies the first two stretches, 3 × 4 = 12 routes from City W to City Y, and stops before the 3 highways on to City Z.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Counts routes two stretches at a time, 3 × 4 = 12 for W to Y and 4 × 3 = 12 for X to Z, and adds them instead of multiplying all three.
Worked solution, step by step
Formulas used
- Counting principle
- a, b and c are the numbers of choices for the first, second and third parts of a task. Every choice for one part can go with every choice for the others. So you multiply. Use it when you make one choice at each stage and want to count all the ways.
Sort out the trip
A route goes from City W to X, then to Y, then to Z. It uses exactly one highway on each of three stretches:
- W to X: 3 highways
- X to Y: 4 highways
- Y to Z: 3 highways
You need to count all the different routes.
Count routes from W to Y
Start small. Pick any of the 3 highways to City X. From there, you can take any of the 4 highways to City Y. So each first highway leads to 4 routes.
There are 12 routes from City W to City Y.
Add the last stretch
Each of those 12 routes can end with any of the 3 highways to City Z. So multiply again. This is the counting principle with all three stretches.
There are 36 routes in all.
Avoid the common traps
Watch out: Adding 3 + 4 + 3 = 10 counts highways, not routes. That gives choice A. Stopping at 12 forgets the last stretch, choice B. Adding 12 + 12 = 24 mixes two partial counts, choice C. A route is built by choosing one highway after another, so you multiply.
So the answer is D, 36.
Check your answer
Picture one highway from W to X. After it, there are 4 × 3 = 12 ways to finish the trip. Each of the 3 highways from W to X gives 12 ways. That makes 12 + 12 + 12 = 36 routes.
More questions like this
The original DOE question

Question 111 of 120
2023 Test B Math, question 36Question
The side length of square S is shorter than the side length of square T by 1 centimeter. The area of square S is 9 square centimeters. What is the difference, in square centimeters, between the area of square T and the area of square S?
- A3
- B7
- C8
- D10
Show the answer
Correct answer: B
Let x represent the side length of square T, in centimeters. Then the side length of square S is .
The area of square S is:
So the side length of square T is 4 cm, and the area of square T is 16 sq cm.
The difference between the areas, in square centimeters, is .
Question analysis
Two squares differ in side length by 1 centimeter, and the smaller one has an area of 9 square centimeters. Students find how much greater the larger square's area is. It tests the link between a square's side and its area, and the extra area is easy to miscount when picturing the border that the longer sides add.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 The area of a square and recognizing 9 as a perfect square are grade 6 ideas. This takes three short steps with small whole numbers, lighter than grade 7 work on composite figures, though the first step runs backward from area to side.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Many steps minor
- Turning words into math minor
- Solution
- 3 steps
- Reading level
- Question and choices: about grade 6, 44 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Every number is small, so the work is keeping track of which quantity the question wants: a difference of areas, not a side or a single area. The added area forms an L-shaped strip around square S, and students who picture it as two strips 4 centimeters long count the corner square twice. Finding both areas, 9 and 16, and subtracting avoids reasoning about the strip at all.
Many steps minor Working it out
The work runs from area to side and back: the square root of 9 gives S’s side, adding 1 gives T’s, squaring gives 16, and subtracting gives the difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The phrase “shorter than the side length of square T by 1 centimeter” describes S from T’s point of view, so T’s side is S’s side plus 1, not minus 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Strategy: Find the side of square S from its area, add 1 centimeter to get the side of square T, find that square's area, and subtract the smaller area from the larger.
Traps
- Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
- Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Finds the side of square S, 3 centimeters, from its area of 9 and gives that length instead of a difference between two areas.
C Overlapping groups (King et al., 2004)
Pictures the added area as two 4-by-1 strips along two sides of square S, 8 in all, counting the 1-by-1 corner where they overlap twice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Adds the stem’s two numbers, 9 + 1 = 10, as if a side 1 centimeter longer made the area 1 square centimeter larger, and gives that area.
Worked solution, step by step
Formulas used
- Area of a square
- A is the area, the space inside the square, in square units. s is the length of one side. All four sides of a square are equal, so the area is the side times itself. To go backward from an area, find the number that times itself gives the area.
Sort out the two squares
Square S has an area of 9 square centimeters. Its side is 1 centimeter shorter than the side of square T. You need the area of square T minus the area of square S, in square centimeters.
Find the side of square S
Use the area formula backward. Which number times itself makes 9?
So each side of square S is 3 centimeters. This number is called the square root of 9.
Find the side of square T
Square T’s side is 1 centimeter longer than square S’s side.
Each side of square T is 4 centimeters.
Find the area of square T
Use the area formula with a side of 4 centimeters.
The area of square T is 16 square centimeters.
Subtract the areas
Take the smaller area away from the larger one.
The difference is 7 square centimeters.
Watch out: Picturing the new area as two strips, each 4 by 1, gives 8, choice C. But those strips share a 1-by-1 corner, so it gets counted twice. Also, 3 is a side length, not a difference of areas. That is choice A.
So the answer is B, 7 square centimeters.
Check your answer
Draw square T and mark square S in one corner. The extra space is an L shape. It is a 3-by-1 strip, another 3-by-1 strip and a 1-by-1 corner. That is 3 + 3 + 1 = 7 square centimeters. It matches.
More questions like this
The original DOE question

Question 112 of 120
2023 Test B Math, question 47Question
Four slips of paper numbered 1, 2, 3, and 4 were placed in a bag:
- One slip was selected from the bag, and the number recorded.
- Without replacement, a second slip was chosen.
- This process continued until no more slips remained in the bag.
- One part of a tree diagram representing the sample space is shown.

Which number should be placed in the box to complete this part of the tree diagram?
- A1
- B2
- C3
- D4
Show the answer
Correct answer: C
The “without replacement” statement means that the bag contained four slips for the first draw, three slips for the second draw, two slips for the third draw, and one slip for the last draw. Following the branch from the top and navigating down to the box, the 1 was chosen first, the 2 was chosen second, and the 4 was chosen third. The only remaining number is 3. The element would be 1243.
Question analysis
Four slips numbered 1 to 4 are drawn from a bag one at a time without replacement, and part of a tree diagram of the outcomes is shown. Students decide which number belongs in the empty box at the end of one branch. It tests reading a tree diagram and understanding what drawing without replacement means.
- NYS standards
- 7.SP.8b sample spaces for compound events
- Grade level
- Grade 7 content, by NYS standard
- Compared with grade 7
- Below grade 7 Tree diagrams for compound events are a grade 7 topic, but here the tree is already drawn and only one box is empty. Following a single branch to that box is shorter than typical grade 7 work with sample spaces.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly understand
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 68 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Only one branch matters, and once it is read as the slips drawn so far, 1, 2 and 4, the last slip left in the bag is the only one missing. Errors come from not counting the top 1 as a draw, from forgetting that a drawn slip stays out, or from copying a nearby number in the diagram instead of following the path.
Reading a figure, table or graph minor Understanding the question
The box hangs below the branch 1, then 2, then 4, and students must read that path from the top as the order of the first three draws. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Strategy: Follow the branch from the top down to the box, list the slips already drawn on that path, and name the one still left in the bag.
Traps
- Without replacement: In a second pick there is one fewer item, which changes the probability.
- Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
The wrong answers, classified
A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Treats the 1 at the top as the start of the diagram rather than the first slip drawn, so 1 seems still to be in the bag.
B Without replacement (Fischbein & Schnarch, 1997)
Lets slip 2, already drawn second on this branch, come up again, missing that a slip drawn without replacement stays out of the bag.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Copies 4, the number directly above the box, reading the diagram as a pattern of numbers rather than as the order of the draws.
Worked solution, step by step
Sort out the question
The bag holds four slips: 1, 2, 3 and 4. Slips are drawn one at a time. “Without replacement” means a drawn slip does not go back in the bag. So each number can be drawn only once.
Read the tree diagram
A tree diagram shows the orders the draws can happen in. Each row going down is one draw. The 1 at the top is the first slip drawn. The numbers below it are the choices for the second draw, and so on.
Follow the path to the box
Start at the top and go down the lines to the box.
- First draw: 1
- Second draw: 2
- Third draw: 4
- Fourth draw: the box
Slips 1, 2 and 4 are already out of the bag.
Find the slip that is left
The bag started with 1, 2, 3 and 4. Only slip 3 has not been drawn. So it must be the fourth draw.
Watch out: The 1 at the top counts as a draw, so 1 is not left. That trap is choice A. Slip 2 was already drawn, so choice B can’t be right. And 4 is just the number above the box, choice D.
So the answer is C, 3.
Check your answer
Read the whole path: 1, 2, 4, 3. Each number from 1 to 4 shows up exactly once. That is just what should happen when no slip goes back in the bag.
More questions like this
The original DOE question

Question 113 of 120
2023 Test B Math, question 49Question
A rectangular swimming pool of uniform depth is 20 yards long and 14 yards wide. If the water is 6 feet deep, what is the volume of water in the pool, in cubic yards?
- A40
- B286
- C560
- D1,680
Show the answer
Correct answer: C
First, change 6 feet to yards:
Then, multiply the dimensions of the pool to find the volume, in cubic yards:
Question analysis
A rectangular swimming pool is 20 yards long and 14 yards wide, and the water is 6 feet deep. Students find the volume of water in cubic yards. It tests the volume of a rectangular prism, and the depth is in a different unit from the other measurements, so it has to be converted first.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 The volume of a rectangular prism is grade 5 and 6 content, and the conversion here gives a whole number. Grade 7 volume work is usually longer, with composite or triangular prisms; here the only extra step is one unit conversion.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Combines ideas minor
- Something you must already know minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 6, 31 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The only real trap is the mixed units: 20 and 14 are yards, 6 is feet, and the answer is wanted in cubic yards. Multiplying the three numbers as printed gives a listed choice three times too large. Underlining the unit of each measurement before multiplying, and converting the odd one out, takes seconds.
Combines ideas minor Working it out
The length and width are in yards but the depth is in feet, so a conversion, 6 feet = 2 yards, must come before the volume formula is used. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Neither 3 feet = 1 yard nor volume = length × width × depth is given; both are assumed in a stem that asks for cubic yards. (Anderson & Krathwohl, 2001)
Strategy: Change the depth from feet to yards first, so that all three measurements are in yards, then multiply the length, width and depth.
Traps
- Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
- Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Adds the three measurements, 20 + 14 + 6 = 40, with no conversion and no volume formula, combining the stem’s numbers.
B Wrong formula (King et al., 2004)
Finds the area of the water’s surface, 20 × 14 = 280, then adds the depth of 6 instead of multiplying by it.
D Units or scale (King et al., 2004)
Multiplies 20 × 14 × 6 without changing 6 feet to 2 yards, mixing yards with feet and getting a volume three times too large.
Worked solution, step by step
Formulas used
- Volume of a rectangular box
- V is the volume, the space inside, in cubic units. l is the length, w is the width, and h is the height or depth. All three must be in the same unit. Multiplying them counts the unit cubes that fill the box.
- Feet to yards
- There are 3 feet in 1 yard. So to change feet to yards, divide the number of feet by 3. A yard is longer than a foot, so you get a smaller number.
Sort out the measurements
The pool is shaped like a box. Uniform depth means the water is equally deep everywhere. The pool is 20 yards long and 14 yards wide. The water is 6 feet deep. You need the volume, the space the water fills, in cubic yards.
Change the depth to yards
Two sides are in yards, but the depth is in feet. All three must use the same unit before you multiply. The answer is in cubic yards, so change feet to yards.
The water is 2 yards deep.
Find the volume
Use the volume formula with all three measurements in yards.
The pool holds 560 cubic yards of water.
Give the answer
Watch out: Multiplying 20 × 14 × 6 mixes yards with feet. It gives 1,680, choice D, which is three times too big.
So the answer is C, 560 cubic yards.
Check your answer
Check the units. Yards times yards times yards gives cubic yards, just what the question asks for. Also, 1,680 ÷ 3 = 560. So using 6 feet unchanged makes the answer exactly three times too big.
More questions like this
The original DOE question

Question 114 of 120
2023 Test B Math, question 51Question
A number, n, is divided by 25. If the quotient is 10 with a remainder of 5, what is the quotient expressed as a decimal?
- A10.20
- B10.25
- C10.50
- D10.75
Show the answer
Correct answer: A
A quotient of 10 with a remainder of 5 means the quotient is
Question analysis
A number is divided by 25, giving a quotient of 10 with a remainder of 5, and students write the quotient as a decimal. It tests what a remainder means and how it becomes part of a decimal quotient. Connecting the remainder to the decimal part of the answer is the key idea.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Something you must already know moderate
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 22 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The question checks one idea: a remainder is a fraction of the divisor, not a string of digits to place after the point. Students who write the remainder 5 as tenths, or set the divisor 25 after the point, land on listed choices. Rebuilding n = 25 × 10 + 5 = 255 and dividing by 25 confirms the decimal.
Something you must already know moderate Working it out
A remainder of 5 means 5 out of the 25 needed for another whole, so it becomes 5/25 = 0.2, a link between remainders and decimals the question assumes. (Anderson & Krathwohl, 2001)
Strategy: Rebuild n from the quotient and the remainder, then divide it by 25 to a decimal; or write the remainder as a fraction of the divisor.
Traps
- Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Writes the divisor, 25, after the decimal point, setting the stem’s numbers 10 and 25 side by side instead of turning the remainder into a fraction.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Writes the remainder 5 as the tenths digit, 10.5, believing a remainder becomes the digits after the decimal point.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 10.75; it treats the remainder as three quarters of the divisor, which nothing in the stem suggests, so it mostly draws guesses.
Worked solution, step by step
Formulas used
- Quotient with a remainder
- d is the divisor, the number you divide by. q is the whole-number part of the quotient, the answer to the division. r is the remainder, what is left over. The leftover still has to be divided by d, so it becomes the fraction r over d.
Sort out the words
n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.
Turn the remainder into a fraction
The 5 left over still has to be divided by 25. So it becomes the fraction . Use the formula.
Change the fraction to a decimal
Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.
So the quotient is 10 + 0.20 = 10.20.
Give the answer
Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.
So the answer is A, 10.20.
Check your answer
Rebuild n. Ten 25s plus the 5 left over is . Now divide: . That matches 10.20.
More questions like this
The original DOE question

Question 115 of 120
2022 Test A English, question 5Question
Which revision corrects the error in sentence structure in the paragraph?
In 1967 Kathrine Switzer signed up for the Boston Marathon using her first and middle initials instead of her full name, at that time, only men were permitted to officially register and receive a number for the legendary race. Once officials realized a woman was attempting to run in the race, they made efforts to remove her from the competition. Switzer prevailed and finished in just over four hours, paving the way for the official rule change that allowed for the inclusion of women. In 2017, to mark the fiftieth anniversary of this pioneering event, a seventy-year-old Switzer repeated her run, wearing the number 261, the same number she had worn in that first run in 1967.
- Aname. At
- Brace. They
- Chours. Paving
- D261. The
Show the answer
Correct answer: A
The question asks for the revision that corrects the error in sentence structure in the paragraph.
A. CORRECT. As written, the first sentence (“In 1967 Kathrine Switzer . . . legendary race.”) is a run-on sentence because a comma is used to combine two independent clauses. Independent clauses are complete sentences with a subject and a verb. The clauses—“In 1967 Kathrine Switzer signed up for the Boston Marathon using her first and middle initials instead of her full name” and “at that time, only men were permitted to officially register and receive a number for the legendary race”—should be separated with a period instead. This revision correctly places a period after the word “name” and capitalizes the word “At” to separate the two independent clauses into two sentences.
B. Incorrect. The revision introduces an error into the paragraph and does not revise the run-on in the first sentence. The clause “Once officials realized a woman was attempting to run in the race” is a dependent clause functioning as an adverb of the independent clause that follows it. Changing the comma after the word “race” to a period would incorrectly separate the adverbial dependent clause from the independent clause.
C. Incorrect. The revision introduces an error into the paragraph and does not revise the run-on in the first sentence. The sentence “Switzer prevailed and finished in just over four hours, paving the way for the official rule change that allowed for the inclusion of women” combines an independent clause (“Switzer prevailed and finished in just over four hours”) and a participial phrase (“paving the way for the official rule change that allowed for the inclusion of women”). The phrase “paving the way for the official rule change that allowed for the inclusion of women” is not a complete sentence and must stay connected to the first part of the sentence. Changing the comma after “hours” to a period would incorrectly separate the participial phrase from the independent clause.
D. Incorrect. The revision introduces an error into the paragraph and does not revise the run-on in the first sentence. Changing the comma after “261” to a period would incorrectly separate the appositive phrase “the same number she had worn in that first run in 1967” from the words it describes (“number 261”). Additionally, the phrase “the same number she had worn in that first run in 1967” is not a complete sentence.
Question analysis
This sentence-structure question gives a four-sentence paragraph about Kathrine Switzer, who entered the 1967 Boston Marathon under her initials when only men could register and ran it again at seventy. Students pick the revision that corrects its error in sentence structure, and because every choice turns an underlined comma into a period, each spot needs its own check.
- Grade level
- Grade 6 reading and task
- Compared with grade 7
- Below grade 7 Run-ons and fragments are taught before grade 7, and this item checks that skill at four underlined spots in one short paragraph. The clauses and phrases around the correct commas take some grammar sense, but the core skill is basic.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- A rule that doesn't sound wrong moderate
- Every choice must be checked minor
- Complex sentences minor
- Evidence
- Answered from one paragraph or stanza
- Reading level
- Question and choices: about grade 7, 131 words.
- Corrected from the DOE handbook
- A misspelled name in the paragraph and its explanation was corrected.
- Grammar terms in the explanations of two wrong options were corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Every choice makes the same kind of change, so the question comes down to telling one real comma splice from three commas doing legitimate work. The most dangerous spot is after “race,” where a subject and verb sit on both sides of the comma, just as at the real error, and only the word “Once” marks the first part as dependent. Students who test each side for a complete thought, not just for a subject and a verb, find the splice in the opening sentence quickly.
A rule that doesn't sound wrong moderate Working it out
The run-on hides behind the short phrase “at that time,” which makes the comma after “name” sound like an ordinary pause; only checking that each side can stand alone exposes the comma splice. (Connors & Lunsford, 1988)
Every choice must be checked minor Choosing the answer
All four revisions swap a comma for a period, so every underlined spot needs the same test: could the words before it and the words after it each stand as a sentence? (Embretson & Wetzel, 1987)
Complex sentences minor Understanding the question
The three correct commas guard three different structures, an opening “Once” clause, a participial phrase beginning with “paving” and an appositive after “261,” and each has to be recognized as unable to stand alone. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Strategy: At each underlined comma, read the words before it and the words after it as separate pieces and ask whether each could stand alone as a sentence; that shows where a period belongs.
Traps
- Fragment or run-on: Creates or keeps a sentence fragment, a run-on sentence or a comma splice.
The wrong answers, classified
B Fragment or run-on (Connors & Lunsford, 1988)
Finds a subject and verb on both sides of the comma after “race” and calls it a splice, missing that “Once” makes the first part a dependent clause that cannot stand alone.
C Fragment or run-on (Connors & Lunsford, 1988)
Treats the long third sentence as a run-on and ends it after “hours,” leaving “Paving the way for the official rule change” as a fragment with no subject or main verb.
D Fragment or run-on (Connors & Lunsford, 1988)
Spots the subject and verb “she had worn” after the comma following “261,” but they sit inside a phrase that renames the number, so the new sentence would be a fragment.
What was corrected from the DOE handbook, and why
- Stem and explanation: The handbook spells the runner’s name “Katherine Switzer” in the paragraph and twice in the explanation. She is Kathrine Switzer, who entered the 1967 Boston Marathon as “K. V. Switzer,” so the site now spells her name that way, in the question’s title too. No answer depends on the spelling.
- Explanation: The DOE’s explanation of Option C calls “paving the way for the official rule change that allowed for the inclusion of women” a dependent clause, and on that basis calls the sentence a complex sentence. The words have no subject or finite verb of their own: they are a participial phrase, which the site now calls them, and the site no longer calls the sentence complex for that reason. Its explanation of Option D calls “the same number she had worn in that first run in 1967” a modifying clause; it is an appositive phrase, a noun phrase that renames “number 261,” and the site now says so. Neither phrase can stand alone as a sentence, so Options C and D are still wrong, and the answer, Option A, is unchanged.
Working it out, step by step
Figure out what the question asks
This is a sentence structure question. In one spot, two full sentences are joined by only a comma. That mistake is called a run-on. Every choice turns one underlined comma into a period. Only one of those changes fixes a run-on. The other three would break off a piece that can’t stand alone.
Learn the rule for run-ons
An independent clause is a group of words that makes sense alone. It has a subject and a verb. The subject is who or what the sentence is about. The verb tells what the subject does or is. A comma alone can’t join two independent clauses. Use a period instead.
Example: It rained, we stayed inside. Fix: It rained. We stayed inside.
Test the comma after name
Read the words before that comma: “In 1967 Kathrine Switzer signed up for the Boston Marathon using her first and middle initials instead of her full name.” That makes sense alone. Now read the words after it: “at that time, only men were permitted to officially register.” That makes sense alone too. Its subject is “only men,” and its verb is “were permitted.” So two full sentences are joined by a comma. That is the run-on.
Test the other three commas
- B (after “race”): The first part is “Once officials realized a woman was attempting to run in the race.” The word “Once” makes it a dependent clause. That is a group of words with a subject and verb that can’t stand alone.
- C (after “hours”): The words starting “paving the way for the official rule change” are a participial phrase. The phrase starts with an -ing verb form and adds information. It has no subject of its own.
- D (after “261”): “the same number she had worn in that first run in 1967” is an appositive phrase. It renames the noun right before it, the number 261. It can’t stand alone either.
Choose the answer
Watch out: B is the trickiest. It has a subject and a verb on each side of its comma, “officials realized” and “they made.” But “Once” keeps the first part from standing alone. Only the comma after “name” joins two full sentences. A period goes there, and “At” gets a capital letter. So the answer is A: name. At.
Why this answer and not the closest wrong one
B is the closest wrong answer. Like the comma after “name,” its comma has a subject and a verb on each side. But “Once” means the first part needs the words after the comma to finish its thought. At the comma after “name,” both sides already stand alone. So that comma is the real run-on.
More questions like this
The original DOE question

Question 116 of 120
2022 Test A Math, question 24Question
- A1.62
- B16.2
- C162
- D1,620
Show the answer
Correct answer: C
Since the greatest number of decimal places is 2 (hundredths), multiply both terms by 100, and then divide:
Question analysis
This question asks students to divide 12.96 by 0.08, a decimal division with no story around it. It tests dividing multi-digit decimals with the standard algorithm. The choices differ only in where the decimal point falls, so moving the point the same number of places in both numbers is the whole challenge.
- NYS standards
- 6.NS.3 multi-digit decimal operations
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Dividing multi-digit decimals is grade 6 fluency that grade 7 work builds on rather than teaches. The division comes out exactly, so the real demand is placing the decimal point correctly, a single routine skill.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 1, recall and reproduction · mainly apply
- What makes it hard
- Error-prone arithmetic moderate
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 16+, 1 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Every choice uses the digits 1, 6 and 2, so the question tests only where the decimal point goes. Dividing by a number less than one gives a result larger than 12.96, which feels wrong to students who expect division to shrink a number. Estimating first, 12.96 ÷ 0.08 is about 13 × 12.5, puts the result in the low hundreds and rules out the other three at once.
Error-prone arithmetic moderate Working it out
The divisor 0.08 has two decimal places, so both numbers must be multiplied by 100, turning the problem into 1,296 ÷ 8, and the quotient comes out larger than 12.96. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Multiply the dividend and the divisor by the same power of 10 so that the divisor becomes a whole number, then carry out the long division.
Traps
- An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
The wrong answers, classified
A An arithmetic slip (Junker & Sijtsma, 2001)
Divides 12.96 by 8, leaving out the decimal point in the divisor 0.08, so the quotient, 1.62, is 100 times too small.
B An arithmetic slip (Junker & Sijtsma, 2001)
Moves the decimal point one place instead of two, in effect dividing 12.96 by 0.8, so the quotient, 16.2, is ten times too small.
D An arithmetic slip (Junker & Sijtsma, 2001)
Shifts the decimal point three places in 12.96 but only two in 0.08, dividing 12,960 by 8, so the quotient is ten times too big.
Worked solution, step by step
Formulas used
- Dividing by a decimal
- a is the number being divided. b is the divisor, the number you divide by. Multiplying both by the same number, like 100, does not change the answer. Use 100 when the divisor has two decimal places, so it becomes a whole number.
Sort out the question
You need . The divisor, the number you divide by, is 0.08. It has two decimal places. Dividing by a decimal is hard, so make the divisor a whole number first. All four choices use the digits 1, 6 and 2. So the real question is where the decimal point goes.
Make the divisor a whole number
Multiply both numbers by 100. This moves each decimal point two places to the right.
The new problem has the same answer. Watch out: move both points the same number of places. Moving the point three places in 12.96 but two in 0.08 gives 1,620, choice D.
Divide by 8
Split 1,296 into parts that 8 divides easily.
Watch out: ignoring the point in 0.08 means dividing 12.96 by 8. That gives 1.62, choice A. Moving the point only one place gives 16.2, choice B. So the answer is C, 162.
Check your answer
Multiply back: . That matches. Also, 0.08 is less than 1. Dividing by a number less than 1 makes the answer bigger. And , which is far too big.
More questions like this
The original DOE question

Question 117 of 120
2022 Test A Math, question 29Question
Dakota has four yardsticks and two 12-inch rulers. If she lays all four of the yardsticks and both rulers end-to-end, what is the maximum number of feet she can measure accurately at once?
- A12 ft
- B14 ft
- C16 ft
- D18 ft
Show the answer
Correct answer: B
A yardstick is 1 yard in length.
A 12-inch ruler is 1 foot in length.
Two 12-inch rulers are 2 feet in length.
Question analysis
Dakota lays four yardsticks and two 12-inch rulers end to end and wants to know how many feet she can measure at once. Students convert yards and inches to feet and add the lengths. Mixing up the units, or leaving out one kind of measuring stick, is the main risk in this unit conversion question.
- NYS standards
- 6.RP.3d converting measurement units
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Converting yards and inches to feet is grade 6 measurement work. With friendly numbers and only two conversions to add, this is easier than typical grade 7 tasks, though the wording about measuring at once may slow some readers.
- Estimated difficulty
- Routine: one familiar step, little to misread.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 3, 37 words.
- Corrected from the DOE handbook
- A grammatical slip in the explanation was corrected.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
This is a unit conversion dressed as a measuring story: four yardsticks and two rulers are laid end to end, so the total is simply a sum of lengths. Points are lost by forgetting that the rulers count too, or by giving the rulers the yardsticks' length. Writing every piece in feet before adding, 3 + 3 + 3 + 3 + 1 + 1, shows the total at a glance.
Something you must already know minor Working it out
Neither conversion is given: a yardstick is 1 yard, or 3 feet, and each 12-inch ruler is 1 foot, so 4 yards and 24 inches become 12 + 2 feet. (Anderson & Krathwohl, 2001)
Strategy: Change every length to feet before adding: a yard is 3 feet and 12 inches is 1 foot. Then total the six pieces laid end to end.
Traps
- Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
- Stops one step early: A choice is a number from partway through the solution.
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Converts the four yardsticks to 12 feet and stops, forgetting to add the 2 feet of the two 12-inch rulers laid at the end.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 16 feet; it would need each 12-inch ruler to be 2 feet long, so it mostly draws guesses between nearby totals.
D Units or scale (King et al., 2004)
Converts the rulers with the yard factor too, counting all six sticks as 3 feet long, 6 × 3 = 18, though a 12-inch ruler is 1 foot.
What was corrected from the DOE handbook, and why
- Explanation: The handbook says “Two 12-inch rulers is 2 feet in length.” (p. 142, question 86), repeating the verb of the line before it, “A 12-inch ruler is 1 foot in length.” The site now reads “Two 12-inch rulers are 2 feet in length.” The working and the answer, 14 ft, are unchanged.
Worked solution, step by step
Formulas used
- Yards to feet
- y is a length in yards. f is the same length in feet. One yard is 3 feet, so each yard adds 3 feet. A yardstick is exactly 1 yard long.
- Inches to feet
- i is a length in inches. f is the same length in feet. One foot is 12 inches, so divide the inches by 12. A 12-inch ruler is 1 foot long.
Sort out the question
Dakota lays 4 yardsticks and 2 rulers end to end, in one long line. End to end means each piece starts where the last one stops. So the total length is the sum of all six pieces. You need it in feet. A yardstick is 1 yard long. Each ruler is 12 inches long.
Change the yardsticks to feet
The four yardsticks make 4 yards.
The yardsticks make 12 feet. Watch out: don’t stop here. 12 feet is choice A, but the rulers count too.
Change the rulers to feet
Two rulers of 12 inches make 24 inches.
The rulers make 2 feet. Watch out: a ruler is not as long as a yardstick. Counting all six pieces as 3 feet gives feet, choice D.
Add the lengths
Dakota can measure 14 feet at once. So the answer is B, 14 ft.
Check your answer
List each piece in feet: . That matches. Or use inches. A yard is 36 inches, so 4 yards is inches. Add the rulers: inches. Then feet.
More questions like this
The original DOE question

Question 118 of 120
2022 Test A Math, question 36Question
When the three-digit whole number 27 ? is divided by 3 or by 4, the remainder is zero. What is the missing digit?
- A0
- B2
- C3
- D6
Show the answer
Correct answer: D
Each of the options 0, 2, 3, and 6 must be considered:
Option A: In the case of 0, the three-digit whole number is 270. Divide 270 by both 3 and 4:
Dividing 270 by 3 results in a remainder of 0, but dividing 270 by 4 results in a remainder of 2, so 0 cannot be the missing digit.
Option B: In the case of 2, the three-digit whole number is 272. Divide 272 by both 3 and 4:
Dividing 272 by 4 results in a remainder of 0, but dividing 272 by 3 results in a remainder of 2, so 2 cannot be the missing digit.
Option C: In the case of 3, the three-digit whole number is 273. Divide 273 by both 3 and 4:
Dividing 273 by 3 results in a remainder of 0, but dividing 273 by 4 results in a remainder of 1, so 3 cannot be the missing digit.
Option D: In the case of 6, the three-digit whole number is 276. Divide 276 by both 3 and 4:
Dividing 276 by 3 results in a remainder of 0, and dividing 276 by 4 results in a remainder of 0, so 6 is the missing digit.
Question analysis
A three-digit number begins with 27 and is missing its last digit, and it divides evenly by both 3 and 4. Students find the missing digit. It tests factors, multiples and divisibility rules; a digit that works for one divisor may fail the other, so both tests must be applied.
- NYS standards
- 6.NS.4 common factors and multiples
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Divisibility by 3 and by 4 is grade 6 number sense. Testing four possible digits against two conditions is quick, much like a grade 6 exercise, as long as both divisors are actually checked.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 19 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Every candidate passes at least one of the two tests, which makes the question a matter of finishing the checks rather than finding a method. Students who test only by 3, or only by 4, find a fitting number and stop. Seeing that a number divisible by both 3 and 4 is divisible by 12, and that 276 = 12 × 23, settles it in one step.
Turning words into math minor Understanding the question
The words “divided by 3 or by 4, the remainder is zero” mean that both divisions come out even, though or can be read as either one being enough. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
Each candidate, 270, 272, 273 and 276, has to be tested against two divisors, and three of them pass one test, so stopping after one check is costly. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
The divisibility tests are not given: a number is divisible by 3 when its digit sum is, and by 4 when the number formed by its last two digits is. (Anderson & Krathwohl, 2001)
Strategy: Use the divisibility tests: a number is divisible by 4 when its last two digits are, and by 3 when its digit sum is. Keep only a digit that passes both.
Traps
- Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
- Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
The wrong answers, classified
A Misreads the question (Hegarty, Mayer & Monk, 1995)
Checks only division by 3, which 270 passes with a digit sum of 9, and never tests 4, though 270 ÷ 4 leaves a remainder of 2.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Checks only division by 4, which 272 passes because 72 is a multiple of 4, but its digit sum is 11, so 272 ÷ 3 leaves 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Matches the missing digit to the divisor 3 in the question; 273 is a multiple of 3 but leaves a remainder of 1 when divided by 4.
Worked solution, step by step
Formulas used
- Test for dividing by 3
- The digit sum is what you get when you add up a number’s digits. Divides by 3 means it divides evenly, with a remainder of 0. If the digit sum divides evenly by 3, so does the whole number. Use it to test a big number quickly.
- Test for dividing by 4
- Look only at the number made by the last two digits. If it divides evenly by 4, so does the whole number. This works because every 100 divides evenly by 4, so only the last two digits matter.
Sort out the question
The number is 27 followed by a missing digit. Remainder zero means the division comes out even, with nothing left over. The remainder is zero when the number is “divided by 3 or by 4.” That means both divisions come out even. So the number must pass both tests. The choices give four digits to try: 0, 2, 3 and 6.
Test each number for 3
Add the digits of each number.
Number Digit sum Divides by 3? 270 yes 272 no 273 yes 276 yes So 270, 273 and 276 pass. 272 fails.
Test each number for 4
Look at the last two digits: 70, 72, 73 and 76.
So 272 and 276 pass. 270 and 273 fail.
Keep the digit that passes both
Only 276 passes both tests. Check by dividing:
Watch out: 0 and 3 pass only the test for 3, so A and C fail. 2 passes only the test for 4, so B fails. And 3 is not right just because the question says 3. So the answer is D, 6.
Check your answer
A number that divides evenly by both 3 and 4 also divides evenly by 12, since . And . So 276 is a multiple of 12. None of 270, 272 or 273 is.
More questions like this
The original DOE question

Question 119 of 120
2022 Test A Math, question 37Question

Based on the circle graph, how many pounds of milk and eggs does the average American consume each year?
- A300
- B368
- C460
- D552
Show the answer
Correct answer: D
First, determine the missing percentage for “milk and eggs” consumed, based on the circle graph.
Then, determine 30% of 1,840 pounds. Multiply:
Question analysis
A circle graph shows how the 1,840 pounds of food the average American eats in a year divide among six groups, with the percent for milk and eggs left as x. Students find how many pounds of milk and eggs that is. They first work out the missing percent, then take that percent of the total.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Reading a circle graph and finding a percent of a total are grade 6 skills. Finding the missing sector from 100% adds one easy step, so this is shorter and simpler than typical grade 7 multistep percent problems.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Reading a figure, table or graph moderate
- Error-prone arithmetic minor
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 9, 19 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
Two short steps decide the question: find the missing percent from the other five, then take that percent of 1,840 pounds. The drawing does not help, since the milk-and-eggs sector is drawn larger than the 38% sector beside it, so the share must come from the labels. Students who find the 30 and then pick the choice whose digits match it skip the second step.
Reading a figure, table or graph moderate Understanding the question
Milk and eggs is labeled only x%, so its share comes from the five labeled sectors, 100 − 70 = 30, and the 1,840-pound total sits in a caption below the circle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Taking 30% of 1,840 by hand, 0.3 × 1,840 = 552, means a decimal times a four-digit number, best done as 184 × 3. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Strategy: Subtract the five labeled percents from 100% to find x, then multiply the 1,840-pound total by that percent written as a decimal.
Traps
- Stops one step early: A choice is a number from partway through the solution.
The wrong answers, classified
A Stops one step early (King et al., 2004; Gierl et al., 2017)
Finds the missing share, 30%, and stops; 300 is the choice whose digits match it, though 30% of 1,840 pounds is still to be taken.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 368, which is 20% of 1,840; it would need a slip such as 80 for the labeled percents' sum, so it mostly draws guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 460, a quarter of 1,840; the sector is drawn larger than a quarter, so even judging by eye does not lead here.
Worked solution, step by step
Formulas used
- Parts of a circle graph
- A circle graph shows one whole split into parts. Each part, or section, shows its percent. The whole is 100%, so all the percents add up to 100. Use it to find a missing section.
- Percent of a number
- p is the percent. The whole is the full amount. The part is the piece that matches the percent. Percent means out of 100, so p% is p hundredths of the whole.
Sort out the question
The circle graph splits all the food an average American eats in a year into six sections. The total is 1,840 pounds per person. Five sections show a percent. Milk and eggs shows only x%. You need the pounds of milk and eggs. First find x. Then find x% of 1,840 pounds.
Trust the labels, not the drawing
This graph is not drawn to scale, though it does not say so. The milk and eggs section looks bigger than the fruits and vegetables section, which is labeled 38%. But you will see that milk and eggs is smaller. So don’t judge the size of any section by eye here. Work only from the printed numbers.
Find the missing percent
All six sections add to 100%. Put in the five labeled percents and x.
Milk and eggs is 30% of the total. That is less than 38%, though it is drawn bigger.
Find 30% of 1,840 pounds
Another way: 10% of 1,840 is 184. 30% is 3 times that, . Watch out: don’t stop at 30. Choice A, 300, only looks like 30. So the answer is D, 552 pounds.
Check your answer
Estimate. 30% is a bit less than one third. One third of 1,840 is about 613. So the answer should be a bit under 613, and 552 is. Also, , which is 30%. Choice C, 460, is only one quarter, 25%.
More questions like this
The original DOE question

Question 120 of 120
2022 Test A Math, question 41Question
LAST YEAR’S TOTAL ATTENDANCE
| Attendees | Number |
|---|---|
| Children | 1,750 |
| Adults | 2,250 |
| Seniors | 1,000 |
| Total | 5,000 |
If 100 individuals will attend the museum next week, what is the best estimate (based on last year’s attendance) of the number of children who will be in that group?
- A30
- B34
- C35
- D50
Show the answer
Correct answer: C
First, determine what percentage of the total attendees last year were children.
The best estimate based on last year’s attendance is 35% of 100.
Question analysis
A table gives last year's attendance by children, adults and seniors, 5,000 people in all. If 100 people attend the museum next week, students estimate how many of them will be children. It tests using a ratio or percent from data to make a prediction; only the children's row and the total matter.
- Grade level
- Grade 6 content, by NYS standard
- Compared with grade 7
- Below grade 7 Using last year's share of visitors to predict a smaller group is grade 6 ratio and probability work. A group of exactly 100 makes the scaling easy, so the arithmetic is lighter than in typical grade 7 proportion problems.
- Estimated difficulty
- Straightforward: two steps or one careful reading.
- Thinking it demands
- Depth of knowledge 2, skills and concepts · mainly apply
- What makes it hard
- Solution
- 2 steps
- Reading level
- Question and choices: about grade 5, 38 words.
- Students, last 30 days
- Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)
Why it is hard
The share 1,750 out of 5,000 converts neatly to 35 out of 100, and next week's group happens to be exactly 100, so the arithmetic is nearly done once the proportion is written. Points slip away when a student works out the scale factor between the two totals and gives it, or estimates the share instead of computing it. Dividing 1,750 by 50 checks the result.
Turning words into math minor Understanding the question
The phrase “best estimate (based on last year’s attendance)” has to become a proportion: children keep the share of next week's 100 that they had of last year's 5,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
Only the children row, 1,750, and the 5,000 total are needed; the adults and seniors rows add numbers that play no part in the estimate. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Strategy: Find what fraction of last year's 5,000 visitors were children, rewrite it as an equivalent fraction out of 100, and apply it to next week's group.
Traps
- Stops one step early: A choice is a number from partway through the solution.
- Information you don't need: The question gives numbers or facts that play no part in the answer.
The wrong answers, classified
A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 30, which would need 1,500 children last year; it mostly draws students who round the share down or guess a round number.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
No common working gives 34; splitting 100 evenly among three groups gives 33 1/3, the closest route, so it mostly draws guesses.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Divides 5,000 by 100 to get the scale factor, 50, and gives it, stopping before dividing the 1,750 children by that factor.
Worked solution, step by step
Formulas used
- Percent of a total
- The part is the number in one group, here children. The whole is the total of all groups. Dividing gives the fraction of the whole that the group makes up. Times 100 turns it into a percent, a number out of 100.
- Percent of a number
- p is the percent. The whole is the full amount. The part is the piece that matches the percent. Percent means out of 100, so p% is p hundredths of the whole.
Sort out the question
Last year 5,000 people came in all. 1,750 of them were children. Next week 100 people will come. You need the best estimate of how many will be children. Best estimate means children make up the same share as last year. You don’t need the adults and seniors rows.
Find last year’s percent of children
So 35% of last year’s visitors were children. That means 35 out of every 100.
Use that percent for 100 people
About 35 of the 100 people should be children. Watch out: tells how many times bigger last year’s crowd was. It is not a number of children. Don’t pick 50, choice D. So the answer is C, 35.
Check your answer
Shrink both numbers the same way. Since , divide the children by 50 too: . That matches. Also, 1,750 is a bit more than one third of 5,000, and 35 is a bit more than one third of 100.
More questions like this
The original DOE question
