Six SHSAT Questions

3 English and 3 math

Here are 6 random questions to try. Takers earning below 75% are unlikely to receive an offer. You have 1 minute 48 seconds per question on the real exam.

Question 1 of 6

2026 Practice Test B English, question 41
Passage Letter from Brooklyn

Letter from Brooklyn

by Jacob Scheier

I can already see how this will end.
How I will grow tired of the bridge’s
steep incline, and the absent-minded tourists
wandering into the bicycle path.
The weather will turn cold.
But that all happens later.
For now it is the early edge of fall,
leaves green still while the air narrows,
is slightly crisp, almost grazing
the hair of my arm like a passing stranger,
as though the air has been forced into intimacy
by the brevity of daylight.
But when it starts darkening at 4,
this closeness, I know, will be a felt distance,
like someone drawing your attention
to their lack of intimacy.
These days I am still walking at a cathedral pace
beneath the branches bending across avenues,
brownstones like rows of lived-in chapels,
like a pop-up picture book I could have had as a child,
but didn’t. How Brooklyn makes me nostalgic
for the moment I am walking inside of.
These late afternoons filled
with a loneliness that makes me feel
distinctly myself, and an awareness
of how rare that is.

“Letter from Brooklyn” from LETTER FROM BROOKLYN: POEMS by Jacob Scheier, published by ECW Press. Copyright © 2013 by Jacob Scheier All rights reserved.

brevity: briefness
cathedral: meditative
chapels: private places of prayer or worship
nostalgic: yearning for the past

Question

Read these lines from the poem.

I can already see how this will end. (line 1)

How Brooklyn makes me nostalgic
for the moment I am walking inside of.
(lines 21–22)

The lines develop a central idea of the poem by

  • Asuggesting the speaker’s desire to find a way to break free from the repetitive pattern of daily life.
  • Brevealing the speaker’s feelings of disappointment over the predictable change in season.
  • Cemphasizing the speaker’s awareness of the future significance of the present moment in the setting.
  • Dshowing the speaker’s anticipation of specific emotions caused by an intimate knowledge of the setting.
Show the answer

Correct answer: C

The question asks how line 1 and lines 21–22 develop a central idea of the poem.

A. Incorrect. Although these lines suggest that life is governed by repetitive patterns, such as the change of seasons, they do not suggest that the speaker wishes to break free from these natural cycles of change, and a desire for freedom is not a central idea of the poem.

B. Incorrect. Although the speaker acknowledges that colder weather will inevitably bring disappointment, the speaker does not suggest in these lines that the predictability of changing weather is itself disappointing. Rather, the speaker recognizes the value of enjoying the present moment.

C. CORRECT. These lines connect the present warmer weather with the future colder weather and develop the central idea that the present “moment I am walking inside of” must be embraced because the speaker will be “nostalgic” for it in the future.

D. Incorrect. In these lines, the speaker is aware that specific emotions will occur in the future, during cold weather, but this awareness is only part of the more complex central idea of the poem, which is that current emotions and feelings must be recognized, experienced, and appreciated.

Question analysis

This central-idea question pairs the first line of “Letter from Brooklyn” with lines 21–22, where Brooklyn makes the speaker nostalgic “for the moment I am walking inside of,” and asks how they develop a central idea of the poem. Students must link excerpts from opposite ends of the poem, untangle a paradoxical image and choose among four abstract statements.

NYS standards
  • 7R2 theme or central idea
  • 7R5 structure and genre features
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 students analyze how lines develop a central idea. Here two excerpts twenty lines apart must be read together, one of them built on a paradox, and the four choices are long and abstract, well beyond typical grade 7 work.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Evidence
Answered from several paragraphs or stanzas
Reading level
Passage: 217 words (a poem; reading level not measured); question and choices: about grade 7, 99 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 among the hardest questions in the set because the two excerpts must be joined through a paradox and then matched to four long, abstract statements. Line 1 foresees an ending; lines 21–22 miss a moment while it is still happening, so both look at now from the future. Students who see only the foresight pick the choice about anticipated emotions, while those who hear gloom in “how this will end” settle on disappointment or a wish to escape.

  1. Figurative or literary language major Understanding the question

    Lines 21–22 are a paradox: “nostalgic,” glossed as yearning for the past, is aimed at “the moment I am walking inside of,” so the speaker already feels the present as a future memory. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

  2. A very close wrong answer moderate Choosing the answer

    The choice about anticipating “specific emotions” shares the key’s link between future and present; it differs by tracing those feelings to knowing the setting well, which neither excerpt says. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Evidence in several places moderate Working it out

    The excerpts are the poem’s first line and lines 21–22, twenty lines apart, and what joins them, the speaker looking ahead at the present, shows only when they are read side by side. (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)

Strategy: Paraphrase each excerpt on its own, using the glossary's meaning of “nostalgic,” then decide what the first line and the last two share and match that to a choice.

Traps

  • Half right: One half of a two-part choice is right and the other is not.
  • Wrong tone or attitude: Misjudges the tone, attitude or point of view, such as reading admiration as criticism, or taking a character's or subject's feelings for the author's.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This question gives two pieces of the poem: line 1 and lines 21–22. You need the central idea they build together. A central idea is the main point the poem keeps working toward. Read each piece alone first, then join them.

  2. Read line 1

    Line 1 says “I can already see how this will end.” The speaker looks ahead. Lines 2–5 then list what will come: growing tired of the bridge, and the cold. So the speaker already pictures the future while living in the present.

  3. Read lines 21–22

    The lines say “How Brooklyn makes me nostalgic / for the moment I am walking inside of.” The glossary says nostalgic means “yearning for the past.” Yearning means longing. But the speaker longs for now, a moment still happening. That is a paradox, an idea that seems impossible but makes sense. The speaker knows this walk will be a precious memory.

  4. Join the two pieces

    Both pieces look at the present from the future. In line 1, the speaker sees the end coming. In lines 21–22, the speaker already misses the walk. So the speaker knows this present moment will matter later. Significance means importance.

  5. Check each choice

    • A is wrong. The speaker never wants to break free from daily routine.
    • B is wrong. Line 1 can sound gloomy. But the speaker isn’t upset with the seasons. The speaker treasures the walk.
    • C is right. The speaker sees that this moment in Brooklyn will be precious later.
    • D is wrong. The speaker does foresee feelings. But neither piece says they come from knowing the place well.

    Watch out: D gets the looking-ahead part of line 1 right. But lines 21–22 long for the moment itself, not for feelings caused by knowing the city.

  6. Choose the answer

    So the answer is C: “emphasizing the speaker’s awareness of the future significance of the present moment in the setting.”

Why this answer and not the closest wrong one

D is the closest wrong answer, because line 1 does show the speaker foreseeing feelings. But D says they come from an “intimate knowledge of the setting.” Neither piece says that. Lines 21–22 miss the moment while it is still happening. C captures that: the present will matter in the future.

The original DOE question
Question 41 as shown on the DOE’s digital test, page 56 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B English, page 56, question 41.
The passage “Letter from Brooklyn” as printed (page 50)The passage “Letter from Brooklyn” as printed on page 50 of the DOE handbook.

Question 2 of 6

2024 Test A English, question 24
Passage At Dusk

At Dusk

by Natasha Trethewey

At first I think she is calling a child,
my neighbor, leaning through her doorway
at dusk, street lamps just starting to hum
the backdrop of evening. Then I hear
the high-pitched wheedling1 we send out
to animals who know only sound, not
the meanings of our words—here here—
nor how they sometimes fall short.
In another yard, beyond my neighbor’s
sight, the cat lifts her ears, turns first
toward the voice, then back
to the constellation of fireflies flickering
near her head. It’s as if she can’t decide
whether to leap over the low hedge,
the neat row of flowers, and bound
onto the porch, into the steady circle
of light, or stay where she is: luminous
possibility—all that would keep her
away from home—flitting before her.
I listen as my neighbor’s voice trails off.
She’s given up calling for now, left me
to imagine her inside the house waiting,
perhaps in a chair in front of the TV,
or walking around, doing small tasks;
left me to wonder that I too might lift
my voice, sure of someone out there,
send it over the lines stitching here
to there, certain the sounds I make
are enough to call someone home.

“At Dusk” from NATIVE GUARD by Natasha Trethewey, published by Houghton Mifflin Company. Copyright © 2006 by Natasha Trethewey. All rights reserved.

1wheedling: speaking in a flattering manner

Question

How does the setting affect the events of the poem?

  • AIt enables the speaker to regularly overhear the neighbor calling for her cat at dusk.
  • BIt allows the speaker to observe the neighbor quietly without interacting with her.
  • CIt shows how living in close proximity helps the speaker feel connected to the neighbor.
  • DIt shows how feeling isolated causes the speaker to imagine interactions with the neighbor.
Show the answer

Correct answer: B

The question asks how the setting affects the events discussed in the poem.

A. Incorrect. In line 2, the speaker refers to the woman calling for her cat as “my neighbor,” which clearly suggests that the speaker lives next door to the woman. However, although the two live next door to each other, line 1 suggests that perhaps the speaker does not regularly overhear the neighbor because the speaker initially thinks that the neighbor is “calling a child” and only later realizes the neighbor is calling to a cat. Although the setting places the speaker and the neighbor in close proximity, the poem does not indicate the frequency or regularity with which the speaker overhears the neighbor.

B. CORRECT. In the first four lines, the speaker observes the actions of the neighbor as the neighbor leans out to call for the cat at dusk. Despite the speaker’s observation of the neighbor, there is no indication that the neighbor notices the speaker. The reader can conclude that because of the time of day and the speaker being in the house next door, the speaker can observe the neighbor without being observed by her or having to interact with her.

C. Incorrect. The setting of the poem is a neighborhood at nightfall, as indicated by “at dusk” in line 3, the reference to “another yard” in line 9, and the flickering fireflies in line 12. However, while the speaker is close enough to the neighbor to overhear her calling to the cat, there is no indication that the speaker feels connected to the neighbor. Instead, the speaker expresses curiosity about the neighbor after the neighbor goes inside, and the speaker is “left . . . / to imagine her inside the house waiting” (lines 21–22). The fact the speaker initially thought that the neighbor was calling a child indicates that the speaker is not familiar with the neighbor and her habits, and there is no indication that the speaker feels connected to the neighbor.

D. Incorrect. In lines 22–24, the speaker imagines what the neighbor may be doing while the neighbor is waiting inside for the cat to return, but the speaker is imagining the neighbor’s solitary experience after the neighbor has momentarily “given up” on calling to the cat (line 21). The speaker does not imagine personally interacting with the neighbor but rather briefly imagines what the neighbor is doing at home while the cat is away.

Question analysis

This reading question asks how the setting of At Dusk, a neighborhood at nightfall where a woman leans from her doorway to call her cat, affects the events of the poem. Because so little happens, students must infer the setting's role from small details, and several choices sound reasonable without being supported by the poem.

NYS standards
  • 7R3 plot elements and ideas
  • 7R1 textual evidence and inferences
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 students analyze how setting shapes events, usually in a story with a clear plot. This poem has almost no action, so the student must reason from a handful of details about place and time, which is harder than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from the whole text
Reading level
Passage: 236 words (a poem; reading level not measured); question and choices: about grade 8, 67 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

Little happens in this poem, so the setting's role must be built from small details: a neighbor at her doorway at dusk, a cat in another yard, a speaker who hears and watches but never speaks. The strongest wrong answer is right about overhearing and wrong only in claiming it happens regularly, which the speaker's first mistaken guess argues against. The other choices supply a feeling of closeness or of isolation that the poem never states.

  1. A very close wrong answer major Choosing the answer

    The overhearing choice describes what the setting allows almost as well as the right one and fails on a single word, “regularly,” since the poem records one evening and the speaker first mistakes the call. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Reading between the lines moderate Working it out

    The setting's effect is never stated; it has to be inferred from the speaker hearing and watching from nearby while the neighbor, “leaning through her doorway,” never seems to notice anyone. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

Strategy: Picture the scene from the speaker's position, then check each choice word by word against the poem and reject any that claims something the poem never states.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    The setting is where and when a story or poem takes place. You need to see how the setting shapes what happens in the poem.

  2. Picture the setting

    It is “dusk” (line 3), the dim time just after sunset. The neighbor is “leaning through her doorway” (line 2). The cat is “In another yard” (line 9). The speaker is close enough to hear and watch it all. But the speaker never speaks to the neighbor. Nothing shows the neighbor notices the speaker.

  3. Work out the effect

    Living next door puts the speaker close by. In the fading light, the speaker can watch quietly. So the speaker sees the neighbor and the cat without joining in. The whole poem is what the speaker sees, hears and imagines from there.

  4. Check each choice

    • A is wrong. The word “regularly” is the problem. The poem shows one evening.
    • B is right. The speaker watches from nearby and never talks with the neighbor.
    • C is wrong. The speaker never says they feel close to the neighbor.
    • D is wrong. The poem never says the speaker feels lonely. The speaker pictures the neighbor alone inside, not the two of them talking.

    Watch out: A is almost right, since living nearby does let the speaker overhear. One word makes it wrong.

  5. Choose the answer

    The setting lets the speaker watch quietly from close by. So the answer is B.

Why this answer and not the closest wrong one

A is the closest wrong answer. Living next door does let the speaker hear the call. But nothing says it happens often. The poem tells about one evening. At first the speaker even thinks a child is being called. B fits what the poem shows: the speaker watches without ever talking to the neighbor.

The original DOE question
Question 24 as printed on page 50 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 50, question 24. Open page 50 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).
The passage “At Dusk” as printed (page 48)The passage “At Dusk” as printed on page 48 of the DOE handbook.

Question 3 of 6

2026 Practice Test B English, question 45
Passage The End of an Era

The End of an Era

(1) In 2004, the National Aeronautics and Space Administration (NASA) landed a rover, or robotic vehicle, named Opportunity on the surface of Mars. (2) The rover’s mission was to search for evidence of water and life on the planet. (3) The rover was designed to gather data on Mars for about 90 days, but Opportunity’s assignment did not come to an end until nearly 15 years later in 2019.

(4) During the rover’s mission, a high-resolution camera on the rover’s robotic arm took magnified photographs of the surface of Mars. (5) The photographs revealed small round rocks scattered across the surface of the planet. (6) The scientists nicknamed these rocks “blueberries” based on the rocks’ resemblance to the fruit. (7) These rocks were important because their spherical shape suggested that liquid water may have flowed over them for a substantial amount of time.

(8) Opportunity continued to provide valuable data about craters and other surface features. (9) Throughout the mission, the rover traveled a total of 28.06 miles across Mars’ surface. (10) When Opportunity had traveled nearly the length of a marathon, scientists celebrated the accomplishment by naming the valley that the rover was in Marathon Valley. (11) However, in 2018, a dust storm spread all over the planet, putting the rover’s progress on hold.

(12) Prior to this planet-wide dust storm, minor dust storms had periodically deposited a layer of dust onto the solar panels of the rover. (13) These prevented it from charging its batteries. (14) Scientists then had to rely on the passing winds and dust devils to clean off the solar panels. (15) After the massive 2018 dust storm subsided, scientists expected to be able to reestablish communication with Opportunity. (16) They made repeated attempts but to no avail. (17) In 2019, NASA finally declared Opportunity’s mission complete.

(18) Opportunity was an important asset in scientists’ study of Mars; it returned valuable data about our neighbor planet and survived powerful dust storms. (19) NASA described the mission of Opportunity as “one of the most successful and enduring feats of interplanetary exploration.” (20) Even though this rover’s journey ended, other rovers still roam the surface of Mars.

Question

Which revision of sentence 11 best maintains the formal style established in the passage?

  • AHowever, in 2018, a dust storm covered the planet, bringing the rover’s progress to a standstill.
  • BHowever, in 2018, a dust storm shrouded the planet, impermanently halting the rover’s progress.
  • CHowever, in 2018, a dust storm enveloped the whole of the planet, arresting the rover’s progress for a time.
  • DHowever, in 2018, a dust storm covered up the whole planet, which quickly brought a stop to the progress of the rover.
Show the answer

Correct answer: A

The question asks for the revision of sentence 11 that best maintains the formal style established in the passage.

A. CORRECT. This sentence uses clear and formal writing throughout. The phrase “covered the planet” is accurate and formal without being overly poetic, and “standstill” is a precise word for putting something “on hold,” as expressed by the original sentence, while maintaining the formal style of the passage.

B. Incorrect. Although the sentence is formal in style, the word “shrouded” is overly poetic, and the phrase “impermanently halting” is too stilted and formal for the style and tone of the passage.

C. Incorrect. Although the sentence is formal in style, the words “enveloped” and “arresting” are overly poetic, especially when used in the same sentence, and the phrases “the whole of the planet” and “for a time” are needlessly wordy and imprecise, compared with the formal scientific language of the passage.

D. Incorrect. The phrases “covered up the whole planet” and “quickly brought a stop to the progress of the rover” are less formal and unnecessarily wordy, compared with the formal style established by the rest of the passage.

Question analysis

This style question asks which revision of sentence 11, about the 2018 dust storm that covered Mars and stopped the Opportunity rover, best maintains the formal style of the passage. All four versions say the same thing, so students must compare word choice and wordiness and decide which one reads like the rest of the passage.

NYS standards
  • 7W2f informative writing style
  • 7L3a precise and concise language
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Keeping a formal style is grade 7 writing work, usually tested by spotting slang or casual wording. Here all four revisions keep the meaning and differ in register by degrees, so students must judge the passage's own style closely, more than typical.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from one sentence or line
Reading level
Passage: about grade 11, 337 words; question and choices: about grade 9, 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 question turns on what “formal” means. Many students equate it with impressive words and choose the revision with “enveloped” and “arresting,” or the one with “shrouded,” missing that both are more poetic than a passage that reports dates and distances. The casual revision is easy to reject, so the real test is preferring the plainest formal sentence to the grandest one.

  1. Something you must already know moderate Working it out

    Students must know that a formal informational style is plain and precise, so words such as “shrouded,” “enveloped” and “arresting” are too ornate for a passage that reports miles and dates. (Anderson & Krathwohl, 2001)

  2. Every choice must be checked moderate Choosing the answer

    All four revisions keep the meaning of sentence 11 and differ only in register and length, so each must be heard against the passage’s voice rather than checked for errors. (Embretson & Wetzel, 1987)

  3. Hard words minor Understanding the question

    “Impermanently,” “shrouded” and “arresting,” in its sense of stopping, are uncommon words, and students must judge how they sound in the passage, not only what they mean. (Perfetti, 2007; Freedle & Kostin, 1993)

Strategy: Reread a few sentences around sentence 11 to hear the passage's voice, then check each revision for casual phrasing, needless words and words that would sound out of place in that voice.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This question asks which version of sentence 11 keeps the passage’s formal style. Formal writing is serious and clear, like a textbook. It avoids slang and casual words. But formal doesn’t mean fancy. Formal writing about science uses plain, exact words. For example, The game got called off is casual. The contest was halted indefinitely is too fancy. The game was canceled is formal and plain.

  2. Hear the passage’s voice

    Sentence 9 says “the rover traveled a total of 28.06 miles across Mars’ surface.” Sentence 12 says minor storms “deposited a layer of dust onto the solar panels.” The words are exact and plain. The passage gives dates and numbers. It doesn’t paint poetic pictures.

  3. Check each revision

    • A is right. “Covered the planet” is plain and exact. A standstill is a complete stop, a clear way to say “on hold.”
    • B is wrong. “Shrouded” is poetic. A shroud is a cloth wrapped around a body. “Impermanently halting” sounds stiff and odd.
    • C is wrong. “Enveloped” and “arresting” are fancy words. Here arresting means stopping. “The whole of the planet” and “for a time” add extra words.
    • D is wrong. “Covered up” is casual. “Brought a stop to the progress of the rover” is wordy, or longer than it needs to be.

    Watch out: Many students think formal means big words, so they choose C. But a passage of dates and miles calls for plain words.

  4. Choose the answer

    Only A is both formal and plain, like the rest of the passage. So the answer is A: “However, in 2018, a dust storm covered the planet, bringing the rover’s progress to a standstill.”

Why this answer and not the closest wrong one

C is the closest wrong answer, because it has no slang and sounds serious. But “enveloped” and “arresting” are poetic, and “the whole of the planet” is wordy. The passage reports facts in plain words. A says the same thing in the plainest formal way.

The original DOE question
Question 45 as shown on the DOE’s digital test, page 63 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B English, page 63, question 45.
The passage “The End of an Era” as printed (page 60)The passage “The End of an Era” as printed on page 60 of the DOE handbook.

Question 4 of 6

2024 Test B Math, question 56

Question

A rectangular field had a length of 80 meters and a width of 50 meters. The field was recently enlarged, with the length increased by 15% and the width increased by 10 meters. By what percentage was the area of the field increased?

  • A26.5%
  • B35%
  • C38%
  • D42.5%
Show the answer

Correct answer: C

The original area of the field was 4,000 square meters.

80×50=4,00080\times 50=4{,}000

When the field was enlarged, the length increased by 15%, so the new length is 92 meters.

80×15%80\times 15\%

80×0.15=1280\times 0.15=12

80+12=9280+12=92

The width increased by 10 meters, so the new width is 60 meters.

50+10=6050+10=60

The new area of the field is 5,520 square meters.

92×60=5,52092\times 60=5{,}520

The percent increase is found by finding the difference between the two areas and dividing the increase by the original area.

5,520−4,000=1,5205{,}520-4{,}000=1{,}520

1,520÷4,000=0.38=38%1{,}520\div 4{,}000=0.38=38\%

Question analysis

A rectangular field 80 meters long and 50 meters wide has its length increased by 15 percent and its width by 10 meters. Students find the percent by which the field's area grew. It tests percent increase with rectangle area, and the tricky part is that one side changes by a percent and the other by a fixed length.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.G.1 area of polygons
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually apply one percent change to one quantity. This changes a length by a percent and a width by a fixed amount, then asks for the percent change in area, a five-step chain with two kinds of change.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
5 steps
Reading level
Question and choices: about grade 6, 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

Each step is ordinary, but the two changes are stated in different units, one a percent and one in meters, which tempts students to force them into the same form. Adding the two percents is the most natural shortcut and lands close to the right value, because it misses only the small extra area where both changes overlap. Computing the old and new areas directly is slower but leaves nothing to chance.

  1. Combines ideas moderate Working it out

    Percent change has to be combined with area: the length grows by a percent and the width by meters, and the question asks for the percent change of their product. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Many steps moderate Working it out

    Five computations are chained: the new length 92, the new width 60, the areas 4,000 and 5,520, and the increase 1,520 as a share of 4,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The 35% from adding 15% and 20% sits only 3 points below 38%, so a rough estimate of the new area cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compute the new length and the new width, find the old and new areas, and divide the increase in area by the original area.

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.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, the flat space inside. l is the length and w is the width. Multiplying counts how many unit squares fit, row by row.
Percent increase
percent increase=new−oldold×100\text{percent increase} = \frac{\text{new} - \text{old}}{\text{old}} \times 100
old is the starting amount, and new is the amount after. new − old is how much it grew. Dividing by old compares the growth with the start. Multiplying by 100 turns that into a percent.
  1. Sort out the question

    The field was 80 meters long and 50 meters wide. The length grew by 15%, a percent. The width grew by 10 meters, a number of meters. These are two different kinds of increase.

    You need the percent by which the area grew.

  2. Find the old area

    You need the old area to compare with the new one.

    A=80×50=4,000A = 80 \times 50 = 4{,}000

    The old field covered 4,000 square meters.

  3. Find the new length and width

    15% of 80 is 0.15×80=120.15 \times 80 = 12. So the new length is 80+12=9280 + 12 = 92 meters.

    The new width is 50+10=6050 + 10 = 60 meters.

    Watch out: 10 meters is not 10%. Treating it as 10% leads to 26.5%, choice A. Treating 15% as 15 meters leads to 42.5%, choice D.

  4. Find the new area

    Multiply the new length by the new width.

    A=92×60=5,520A = 92 \times 60 = 5{,}520

    The new field covers 5,520 square meters. That is 5,520−4,000=1,5205{,}520 - 4{,}000 = 1{,}520 square meters more.

  5. Find the percent increase

    percent increase=5,520−4,0004,000×100=1,5204,000×100=38\text{percent increase} = \frac{5{,}520 - 4{,}000}{4{,}000} \times 100 = \frac{1{,}520}{4{,}000} \times 100 = 38

    The area grew by 38%. Watch out: The width grew by 20%, since 10 is 20% of 50. Adding 15% + 20% = 35% gives choice B. That misses the extra corner piece where both new strips meet.

    So the answer is C, 38%.

Check your answer

Use growth factors. The length is multiplied by 1.15. The width goes from 50 to 60, so it is multiplied by 1.2. The area is multiplied by 1.15×1.2=1.381.15 \times 1.2 = 1.38, which is 38% more.

The original DOE question
Question 113 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 113. Open page 215 of the handbook (PDF)

Question 5 of 6

2018 Test B Math, question 50

Question

Which number line below shows the solution set for 2x−2≤y≤4x+102x-2\leq y\leq 4x+10 when y=1y=1?

  • AA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to −1 1/2; both ends are closed dots.
  • BA number line from −3 to 3 marked in quarter units, with a solid segment from −1 1/2 to 2 1/4; both ends are closed dots.
  • CA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to 1 1/2; both ends are closed dots.
  • DA number line from −3 to 3 marked in quarter units, with a closed dot at 1 1/2 and a solid ray from it running left past −2 1/2 to an arrowhead.
Show the answer

Correct answer: C

Separate the compound inequality into two pieces:

2x−2≤y2x-2\leq y and y≤4x+10y\leq 4x+10

Substitute y=1y=1 into each inequality and solve for xx:

2x−2≤12x-2\leq1

2x≤32x\leq 3

x≤32x\leq\frac{3}{2}

1≤4x+101\leq 4x+10

−9≤4x-9\leq4x

−94≤x-\frac{9}{4}\leq x

The solution is the number line that shows that xx is greater than or equal to −214-2\frac{1}{4} and less than or equal to 1121\frac{1}{2} .

Question analysis

Students substitute y = 1 into the compound inequality 2x − 2 ≤ y ≤ 4x + 10 and pick the number line that shows the values of x that work. It tests solving two-step inequalities and graphing the solution. The four graphs differ in their endpoints and in whether the shading is closed off on both sides, so each endpoint has to be found exactly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph one two-step inequality. This compound inequality in two variables requires substituting, solving two inequalities and combining them into one interval, a step toward algebra 1.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 9, 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

Substituting y = 1 is easy; the work is solving two inequalities and matching a fractional interval to a graph. Students who solve only the left inequality find a matching ray, and sign slips on either endpoint produce intervals that shift or flip across zero. The graphs differ by quarter-unit ticks, so reading them carefully matters as much as the algebra.

  1. Many steps moderate Working it out

    The compound inequality must be split into 2x − 2 ≤ 1 and 1 ≤ 4x + 10, each solved for x, and the two results joined into one interval. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Both endpoints are fractions, 3/2 and −9/4, and the second comes from dividing −9 by 4, where a lost sign moves the endpoint across zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Reading a figure, table or graph moderate Understanding the question

    The graphs are marked in quarter units, so −2 1/4 and 1 1/2 have to be found by counting small ticks between the labeled integers. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. A very close wrong answer minor Choosing the answer

    The ray ending at 1 1/2 looks almost like the right segment, sharing its right end; only the left end, an arrow instead of a closed dot at −2 1/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Split the compound inequality into two separate inequalities, solve each for x after putting 1 in for y, then combine the two results into a single interval and match it to a graph.

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.)
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality
if a≤b, then a+c≤b+c\text{if } a \leq b, \text{ then } a + c \leq b + c
a, b and c are numbers. The sign ≤ means is less than or equal to. Adding or subtracting the same number c on both sides keeps it true. Dividing both sides by the same positive number also keeps it true.
  1. Sort out the question

    The sign ≤\leq means is less than or equal to. An inequality compares two amounts that may not be equal. The statement 2x−2≤y≤4x+102x-2 \leq y \leq 4x+10 is two inequalities joined together. You put 1 in place of yy. Then you need every xx that makes both parts true.

  2. Split it into two parts

    With y=1y = 1, the statement becomes two inequalities.

    2x−2≤12x - 2 \leq 1

    1≤4x+101 \leq 4x + 10

    The answer must make both of them true at the same time.

  3. Solve the first part

    Add 2 to both sides. Then divide both sides by 2.

    2x≤32x \leq 3

    x≤32x \leq \frac{3}{2}

    So xx is at most 1121\frac{1}{2}. This is the right end of the answer.

    Watch out: the right end is a positive number. Choice A ends its bar at −112-1\frac{1}{2} instead.

  4. Solve the second part

    Subtract 10 from both sides. Then divide both sides by 4.

    −9≤4x-9 \leq 4x

    −94≤x-\frac{9}{4} \leq x

    So xx is at least −214-2\frac{1}{4}. This is the left end.

    Watch out: keep the minus sign. Choice B uses the opposite signs and ends at 2142\frac{1}{4}.

  5. Put the two parts together

    The answer is every number from −214-2\frac{1}{4} to 1121\frac{1}{2}, with both ends included. On the graph, that is a solid bar with a dot at each end. The small ticks are quarters. So count 1 tick left of −2 and 2 ticks right of 1.

    Watch out: choice D uses only the first part. Its arrow goes left forever.

  6. Give the answer

    Choice C shows a bar from −214-2\frac{1}{4} to 1121\frac{1}{2}, with dots at both ends. So the answer is C.

Check your answer

Test a number on the bar, like x=0x = 0. Then 2(0)−2=−22(0)-2 = -2 and 4(0)+10=104(0)+10 = 10, and 1 is between them. Now test x=−3x = -3, which is off the bar. Then 4(−3)+10=−24(-3)+10 = -2, and 1 is not less than −2. So the bar must stop on the left, which rules out D.

The original DOE question
Question 107 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 107. Open page 163 of the handbook (PDF)

Question 6 of 6

2026 Practice Test D Math, question 19

Question

A store manager purchases a piece of furniture for $456.00. To determine the selling price, the manager increases the purchase cost by 125%. A customer buys the furniture and pays an additional 8% sales tax. How much does the customer pay for the furniture?

  • A$615.60
  • B$627.48
  • C$1,108.08
  • D$1,846.80
Show the answer

Correct answer: C

Given a purchase price of $456.00, the 125% increase in price is calculated as:

456×1.25=570456 \times 1.25 = 570

The selling price equals the purchase price plus the increase, or:

456+570=1,026456 + 570 = 1{,}026

The sales tax on the selling price is:

1,026(0.08)=82.081{,}026(0.08) = 82.08

The customer pays the selling price plus the sales tax:

1,026+82.08=1,108.081{,}026 + 82.08 = 1{,}108.08

Question analysis

A store manager buys a piece of furniture for $456.00 and raises the price by 125% to set the selling price, and the customer also pays 8% sales tax. Students find the total the customer pays. It tests percent increase and sales tax in sequence, and reading an increase of more than 100% correctly is the main challenge.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students find markups and sales tax, usually with increases under 100%. An increase of more than 100% is unusual and easy to misread, and two percent changes must be applied in order, which goes beyond routine grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 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

A percent above 100 is unusual in grade 7 work, and it is the center of the question: the markup more than doubles the price. Students who treat “by 125%” as “125% of” still find a listed answer after adding tax, and so do students who add $125. A rough check helps: the selling price must be more than twice $456, about $1,000, and the tax adds less than a tenth of that.

  1. Turning words into math moderate Understanding the question

    “Increases the purchase cost by 125%” adds 125% of $456 to the $456 itself, a multiplier of 2.25; reading it as 125% of the cost leaves the original price out. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps moderate Working it out

    Four steps in a row, the increase, the selling price, the tax and the total, each depend on the one before, and two of them are easy to merge or skip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    Taking 8% of $1,026 means multiplying by 0.08, and a decimal point one place off, 0.8, makes the tax ten times too large. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn each percent change into a multiplier, remembering that an increase by a percent is added to the original amount, apply the markup and then the tax, and estimate to check the size.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
p% of n=p100×np\% \text{ of } n = \frac{p}{100} \times n
n is the starting amount, and p is the percent. Percent means out of 100, so 125% is 125/100 = 1.25 and 8% is 8/100 = 0.08. Multiply n by that decimal to find the part.
Price after an increase
new price=old price+increase\text{new price} = \text{old price} + \text{increase}
The old price is what you start with. The increase is the amount added on. To increase by a percent, find that percent of the old price, then add it to the old price. Sales tax is added the same way.
  1. Sort out the question

    The manager pays $456.00 for the furniture. The selling price is that cost increased by 125%. The customer then pays 8% sales tax on top. Sales tax is an extra charge, a percent of the price. You need the total the customer pays.

  2. Find the increase

    Find 125% of $456. As a decimal, 125% is 1.25, which is 1 and a quarter. A quarter of 456 is 114:

    456×1.25=456+114=570456 \times 1.25 = 456 + 114 = 570

    The increase is $570.

  3. Find the selling price

    Increases by means the increase is added to the cost:

    456+570=1,026456 + 570 = 1{,}026

    The selling price is $1,026.

    Watch out: $570 is only the increase. Using it as the price gives $615.60, choice A. And 125% is a percent, not $125. Adding $125 leads to $627.48, choice B.

  4. Find the sales tax

    As a decimal, 8% is 0.08. Find 8% of the selling price:

    1,026×0.08=82.081{,}026 \times 0.08 = 82.08

    The tax is $82.08.

    Watch out: 8% is 0.08, not 0.8. Using 0.8 makes the tax ten times too big and gives $1,846.80, choice D.

  5. Add the tax to the price

    The customer pays the selling price plus the tax:

    1,026+82.08=1,108.081{,}026 + 82.08 = 1{,}108.08

    So the answer is C, $1,108.08.

Check your answer

Estimate. The price more than doubles: $456 plus $570 is about $1,000. Tax of 8% adds about $80. So the total should be a bit over $1,080, and only $1,108.08 is close. Or multiply in one go: 456 × 2.25 × 1.08 = 1,108.08.

The original DOE question
Question 19 as shown on the DOE’s digital test, page 92 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 92, question 19.

Sample #872734