SHSAT Sunset

Six SHSAT Questions

3 English and 3 math

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Question 1 of 6

2024 Test B English, question 27
Passage 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

How do paragraphs 11–13 affect the plot of the excerpt?

  • AThey establish a feeling of mystery by presenting different points of view about the arriving people.
  • BThey contribute to the rising action by developing the idea that the arriving people could present a problem.
  • CThey lead to a turning point by describing the moment that the crowd realizes why the strangers have come.
  • DThey explain the cause of the main conflict by providing background information about the strangers.
Show the answer

Correct answer: B

The question asks how paragraphs 11–13 affect the plot of the excerpt.

A. Incorrect. The paragraphs contain questions about the arriving people. They do not present differing points of view.

B. CORRECT. The paragraphs ask questions in which the characters wonder whether the arriving people are friends or enemies, or whether they might be bringing good news or bad news. By suggesting that the arriving people could present problems such as these, the paragraphs contribute to the rising action.

C. Incorrect. The paragraphs do not lead to a turning point. Instead, they lead to the climax—the moment the characters discover who the visitors are. This means the paragraphs contribute to the rising action.

D. Incorrect. Paragraphs 11–13 ask questions. They do not provide background information about the strangers; instead, they ask for it.

Question analysis

This plot question asks how paragraphs 11–13, a series of questions the villagers ask as riders approach their camp, affect the plot of the Nez Perce narrative. Students must know terms such as rising action and turning point and apply them precisely, since several choices describe these paragraphs in ways that sound right but place them wrongly in the story.

NYS standards
  • 7R3 plot elements and ideas
  • 7R5 structure and genre features
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 plot questions usually ask what one event causes. Here students apply terms such as rising action and turning point to three short paragraphs of questions, which matches the grade 8 focus on how parts of a story propel the action.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from several paragraphs or stanzas
Reading level
Passage: about grade 5, 885 words; question and choices: about grade 9, 77 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

Paragraphs 11–13 contain nothing but questions, so the answer depends on what questions do in a plot. Friends or enemies, good news or bad, they raise the chance of trouble and build toward the arrival. The trap is that they also create mystery, and the choice that says so fails only on one phrase about differing viewpoints. A second risk is jumping ahead: at this point no one even knows who the riders are, let alone why they came.

  1. A very close wrong answer major Choosing the answer

    The right choice and the one about mystery both see the questions building uncertainty about the riders; only the claim of “different points of view,” when the questions are “in every mind,” separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. How, not what moderate Working it out

    The stem asks what paragraphs 11–13 do for the plot, so students must place a run of unanswered questions in the story’s shape rather than retell them. (Anderson & Krathwohl, 2001)

  3. Something you must already know minor Working it out

    The choices rely on plot terms the excerpt never explains, rising action, turning point and main conflict, and judging them means knowing what each one names. (Anderson & Krathwohl, 2001)

Strategy: Note what the villagers know and do not yet know in paragraphs 11–13, then compare that with what happens later in the excerpt to place these paragraphs in the plot.

Traps

  • Half right: One half of a two-part choice is right and the other is not.
  • Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
  • 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.

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 asks how paragraphs 11–13 move the story along. You need some plot words. The plot is the order of events in a story. The rising action is the part where tension builds. Tension is the feeling that something big may happen. A turning point is the moment when things change.

  2. Read paragraphs 11 to 13

    These paragraphs are all questions. Paragraph 11 asks: “Were they friends or enemies? Did they bring good news or bad?” In paragraph 12, “Who can it be?” is one of “the questions in every mind.” Paragraph 13 asks if the riders are the hunters who left “two summers past.”

  3. Decide what the questions do

    The words “enemies” and “bad” news show the camp is worried. The riders might bring trouble. No one knows yet. So the questions build tension as the riders get closer. That is rising action. The answers come later. The crowd knows the hunters in paragraph 18. They learn who the woman is in paragraphs 22 and 23.

  4. Check each choice

    • B fits. The questions raise the chance of a problem and build tension.
    • A: there is mystery, but not “different points of view.” Everyone asks the same questions.
    • C: in these paragraphs, no one knows who the riders are yet, let alone why they came. So nothing here is the moment the crowd realizes anything.
    • D: the paragraphs ask about the strangers. They give no background facts about them.

  5. Pick the answer

    Watch out: A is very close. The questions do create mystery. But one wrong phrase, “different points of view,” rules it out. So the answer is B.

Why this answer and not the closest wrong one

A is the closest wrong answer, because the questions do make the scene mysterious. But the questions are “in every mind,” so there are no different points of view. B names what the questions do for the plot: they hint that the riders could bring trouble, and they build tension.

More questions like this

The original DOE question
Question 27 as printed on page 180 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 180, question 27. Open page 180 of the handbook (PDF)
The passage “Excerpt from Do Them No Harm!” as printed (pages 177–178)The passage “Excerpt from Do Them No Harm!” as printed on page 177 of the DOE handbook.The passage “Excerpt from Do Them No Harm!” as printed on page 178 of the DOE handbook.

Question 2 of 6

2026 Practice Test C English, question 45
Passage New York City and the Hudson Valley Brickmaking Industry

New York City and the Hudson Valley Brickmaking Industry

(1) Residents of New York City cannot help but notice that the buildings they are surrounded by are predominantly made of brick. (2) Indeed, one author described the bricks that make up some of New York City’s most recognizable landmarks, including the Empire State Building and the arches of the Brooklyn Bridge, as “an inescapable presence.” (3) Nearly 65 million people visit New York City every year to see these and other attractions. (4) Why was so much brick used in buildings constructed during the century before World War II? (5) The answer involves an explosion in New York City’s population that caused an increase in demand for bricks starting in the mid-1800s and lasting into the 1940s that affected the area more than 100 miles up the Hudson River Valley toward Albany. (6) When demand later declined, the industry would entirely disappear.

(7) New York City’s appetite for brick was fueled by surging population and demand for safer building materials. (8) The city’s population ballooned from about 313,000 people in 1840 to more than 7 million a century later. (9) Following large, destructive fires in the 1830s, the city had enacted laws requiring the use of fire-resistant materials. (10) Bricks turned out to be an inexpensive way for builders to quickly throw together a lot of structures while keeping building owners and safety inspectors happy.

(11) The brickmaking industry in the Hudson Valley skyrocketed. (12) At the industry’s peak, approximately 130 brick factories crammed themselves into a 120-mile stretch along the Hudson River. (13) These factories employed much of the region’s population. (14) Population growth in the Hudson River Valley after the mid-1800s was substantial, though percentage increases were smaller than those in New York City. (15) Even though the factories operated only during the summers, their production was immense. (16) Hudson Valley factories shipped over a billion bricks on barges down the river to New York City annually. (17) In fact, The New York Times declared, “It’s fairly safe to assume that any brick building constructed between 1800 and 1950 includes some form of sediment from the banks of the Hudson River.”

(18) History has proven, however, that boom times eventually end. (19) Shifting architectural trends halted not only the Hudson Valley’s brickmaking boom, but its entire industry. (20) Starting early in the twentieth century, architects and builders increasingly favored steel, glass, and concrete instead of brick. (21) By 1979, reduced demand for brick had driven every Hudson Valley brick factory out of business.

(22) The Hudson Valley’s brickmaking legacy lives on in the architecture of New York City. (23) Throughout the city, famous and ordinary buildings stand as evidence of the remarkable productivity achieved by the Valley’s brickmakers. (24) The brick buildings surrounding New Yorkers today serve as a reminder that, as one amateur brick historian put it, “The skyline of New York City used to be the landscape of the Hudson River Valley.”

Question

Which two sentences present information irrelevant to the ideas in the passage and should be deleted?

Move the two correct answers to the box.

Should Be Deleted

On the test you drag tiles into the box; here, choose them below. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

  • A“Nearly 65 million people visit New York City every year to see these and other attractions.” (sentence 3)
  • B“The city’s population ballooned from about 313,000 people in 1840 to more than 7 million a century later.” (sentence 8)
  • C“Population growth in the Hudson River Valley after the mid-1800s was substantial, though percentage increases were smaller than those in New York City.” (sentence 14)
  • D“History has proven, however, that boom times eventually end.” (sentence 18)
  • E“Throughout the city, famous and ordinary buildings stand as evidence of the remarkable productivity achieved by the Valley’s brickmakers.” (sentence 23)
Show the answer

Correct answer: A and C

The question asks which two sentences present information irrelevant to the ideas of the passage and should be deleted.

CORRECT. “Nearly 65 million people visit New York City every year to see these and other attractions” (sentence 3) presents irrelevant information and should be deleted. While the statistic is interesting, it focuses on tourism, which is not relevant to either the topic of the passage—the growth and decline of the brickmaking industry—or to the ideas in the first paragraph (sentences 1–6), which focus on the prominence of brick in New York City’s buildings.

Incorrect. “The city’s population ballooned from about 313,000 people in 1840 to more than 7 million a century later” (sentence 8) is relevant to the ideas of the passage and should not be deleted. The numerical details in sentence 8 about the rapid growth in New York City provide relevant evidence that supports a central idea of the passage: that the brickmaking industry of the Hudson River Valley expanded quickly to supply the demands of builders who were, in turn, meeting the demands of a booming population.

CORRECT. “Population growth in the Hudson River Valley after the mid-1800s was substantial, though percentage increases were smaller than those in New York City” (sentence 14) presents irrelevant details and should be deleted. The information in sentence 14 comparing population growth rates between the region and New York City is not needed to develop the passage’s central focus on the rise and decline of the Hudson River Valley’s brickmaking industry. The sentence also does not support ideas in the third paragraph (sentences 11–17). Although it mentions that brick factories employed much of the Hudson River Valley population, the productivity of the factories, rather than the Valley’s population, is the topic of the paragraph.

Incorrect. “History has proven, however, that boom times eventually end” (sentence 18) is relevant to the ideas of the passage and should not be deleted. Sentence 18 effectively introduces the central idea of the passage: the decline of the Hudson River Valley brickmaking industry. The sentence also serves as a clear transition from the third paragraph (sentences 11–17), which provides evidence of the industry’s boom, to the fourth paragraph (sentences 18–21), which provides details about how the industry came to an end.

Incorrect. “Throughout the city, famous and ordinary buildings stand as evidence of the remarkable productivity achieved by the Valley’s brickmakers” (sentence 23) is relevant to the ideas of the passage and should not be deleted. The mention of famous and ordinary buildings provides an effective link to introductory sentences 1 and 2, strengthening the passage’s conclusion, while also emphasizing the productivity of the Hudson Valley brickmaking industry detailed in the third paragraph (sentences 11–17).

Question analysis

This revising question asks which two of five sentences from the passage on the Hudson Valley brickmaking industry present irrelevant information and should be deleted. On the test students move two sentence tiles into a box labeled Should Be Deleted, and both must be right. The five sentences come one from each paragraph, so each must be judged against its own paragraph and the whole passage.

NYS standards
  • 7W2b developing a topic
Grade level
Grade 7 reading and task
Compared with grade 7
Above grade 7 Grade 7 revision usually asks for one off-topic sentence within one paragraph. Here students check five sentences spread over the whole passage, must find two, and get credit only if both are right, a broader demand 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: about grade 12, 465 words; question and choices: about grade 10, 114 words.
Corrected from the DOE handbook
  • A missing hyphen in a quotation in the explanation was restored.
  • A grammatical error 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

Sentence 3 announces itself: tourism has nothing to do with brick. The second deletion is the trap, because sentence 14 sits after a sentence about factories employing the region’s population and looks like a continuation. Students who reject population as a topic often delete sentence 8 instead, which measures a cause the passage names, and students who judge by sound cut the general “boom times eventually end,” which turns the passage toward the industry’s decline.

  1. A very close wrong answer moderate Choosing the answer

    Sentences 8 and 14 are both population statistics; what separates them is that sentence 8 measures the “surging population” sentence 7 names as a cause, while the Valley’s growth in sentence 14 explains nothing. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. How sentences fit together moderate Working it out

    Each kept sentence earns its place by its links: sentence 18 turns from boom to decline, sentence 23 echoes the landmarks of sentences 1–2, and sentence 14 merely follows sentence 13’s “region’s population.” (Halliday & Hasan, 1976)

  3. Much to keep track of minor Working it out

    Five tiles, one from each paragraph, are judged against both their paragraph and the whole passage, and both picks must be right, so two relevance tests run side by side. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Sum up the passage’s topic in one phrase, then read each of the five sentences in its paragraph and ask whether removing it would leave a gap in the ideas around it; delete only sentences that leave none.

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.
  • Off the point: Adds or keeps something that does not support the paragraph's or the passage's focus, or an introduction or conclusion that misses the main point.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s explanation quotes sentence 14 as “Population growth in the Hudson River Valley after the mid 1800s was substantial.” The passage and the tile say “mid-1800s,” so the site now quotes it that way.
  • Explanation: The handbook’s explanation says “The information in sentence 14 comparing population growth rates between the region and New York City are not needed.” The subject is “information,” so the site now says “is not needed.”

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    You need to find two sentences that don’t belong. A sentence is irrelevant when it has nothing to do with the main ideas. You can delete it without leaving a gap. On the test you drag two tiles into the Should Be Deleted box. Here, you choose two of the five. Both must be right.

  2. Name the topic of the passage

    The passage explains why New York City has so many brick buildings. It tells how the Hudson Valley brick industry grew to meet that demand, then died out. Each paragraph has its own job. Paragraph 1 asks why so much brick was used. Paragraph 2 gives the causes. Paragraph 3 shows how much the factories made. Paragraph 4 tells how the industry ended. Paragraph 5 sums up.

  3. Test sentences 3 and 14

    Sentence 3 says: “Nearly 65 million people visit New York City every year to see these and other attractions.” That is about tourists, not bricks. Sentence 14 says: “Population growth in the Hudson River Valley after the mid-1800s was substantial, though percentage increases were smaller than those in New York City.” Paragraph 3 is about how much the factories made, “over a billion bricks” a year. How fast the Valley grew explains nothing about bricks.

  4. Test the other three sentences

    • Sentence 8 stays. Sentence 7 says the hunger for brick was “fueled by surging population.” Sentence 8 gives the numbers that show it.
    • Sentence 18 stays. It turns the passage from the boom to the decline. A boom is a time of fast growth.
    • Sentence 23 stays. It links back to the famous buildings in sentences 1 and 2 and praises the brickmakers.

    Watch out: Sentence 14 comes right after one about factories that “employed much of the region’s population.” So it seems to fit. But it changes the topic from factories to population.

  5. Choose the answers

    Sentences 3 and 14 add facts that don’t help explain the brick industry. So the answers are A and C. Move these two tiles into the Should Be Deleted box. A is “Nearly 65 million people visit New York City every year to see these and other attractions.” (sentence 3). C is “Population growth in the Hudson River Valley after the mid-1800s was substantial, though percentage increases were smaller than those in New York City.” (sentence 14).

Why this answer and not the closest wrong one

B is the closest wrong answer, since sentences 8 and 14 are both about population. But sentence 8 measures the “surging population” that sentence 7 names as a cause of the demand for brick. Sentence 14 explains nothing about bricks. So delete 14 and keep 8.

More questions like this

The original DOE question
Question 45 as shown on the DOE’s digital test, page 66 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C English, page 66, question 45. This is the DOE’s version, before the corrections listed in the analysis.
The passage “New York City and the Hudson Valley Brickmaking Industry” as printed (page 64)The passage “New York City and the Hudson Valley Brickmaking Industry” as printed on page 64 of the DOE handbook.

Question 3 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.

More questions like this

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 4 of 6

2024 Test B Math, question 41

Question

A bag contains 2 green, 4 blue, and 3 yellow disks. Alana selects 4 disks from the bag at random, one at a time, without replacing them. What is the probability that she selects all 4 blue disks?

  • A82,187\frac{8}{2{,}187}
  • B1126\frac{1}{126}
  • C2566,561\frac{256}{6{,}561}
  • D169\frac{16}{9}
Show the answer

Correct answer: B

If Alana picks 4 blue disks in a row, the number of blue disks and the total number of disks will both decrease by 1 after each selection.

For the first selection, there are 4 blue disks out of 9 total disks, so the probability of selecting the first blue disk is 49\frac{4}{9}.

The probabilities of selecting blue for the next three selections are 38\frac{3}{8}, 27\frac{2}{7}, and 16\frac{1}{6}.

49×38×27×16=243,024=1126\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}=\frac{24}{3{,}024}=\frac{1}{126}

Question analysis

A bag holds green, blue and yellow disks, and Alana draws four of them one at a time without putting any back. Students find the probability that all four disks are blue. It tests the probability of dependent events, and the difficulty is keeping track of how the bag changes after each draw.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 compound probability is usually found with lists or tree diagrams for two stages. Four dependent draws, each changing the counts, are too many to list and need the multiplication rule, which grade 7 does not formally teach.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 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

Students who know the multiplication rule for successive draws still have two adjustments to make on every draw, and each wrong choice skips at least one of them. Leaving the bag at 9 disks, or putting each disk back, both produce fractions over 6,561 that look as exact as the right one. Writing the four fractions in a row before multiplying, and cancelling as you go, keeps the arithmetic small.

  1. Something you must already know moderate Working it out

    The rule for successive draws without replacement is assumed: each draw's chance is multiplied in, with both the blue count and the total lowered by one. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic moderate Working it out

    The product 4/9 × 3/8 × 2/7 × 1/6 = 24/3,024 has to be reduced to 1/126, and two wrong choices hide behind 9 to the fourth power, 6,561. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the chance of blue for each of the four draws in turn, lowering both the blue count and the total by one each time, and multiply.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of wanted outcomestotal number of outcomesP = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}
P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row
P(A then B)=P(A)×P(B after A)P(A \text{ then } B) = P(A) \times P(B \text{ after } A)
A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

  1. Sort out the question

    The bag holds 2 green, 4 blue and 3 yellow disks. That is 9 disks in all. Alana takes 4 disks, one at a time.

    “Without replacing them” means a disk she takes does not go back in the bag. You need the chance that all 4 disks are blue.

  2. Find the chance for each pick

    For all 4 to be blue, each pick must be blue. After each blue pick, one blue disk is gone. One disk is gone from the total, too. So the top and the bottom each go down by 1.

    PickBlue disks leftDisks leftChance of blue
    1st4949\frac{4}{9}
    2nd3838\frac{3}{8}
    3rd2727\frac{2}{7}
    4th1616\frac{1}{6}

  3. Multiply the four chances

    All four picks must go right, one after another. Use the rule for events in a row and multiply.

    49×38×27×16=4×3×2×19×8×7×6=243,024\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{4 \times 3 \times 2 \times 1}{9 \times 8 \times 7 \times 6} = \frac{24}{3{,}024}

  4. Simplify the fraction

    Divide the top and the bottom by 24. That keeps the value the same.

    24÷243,024÷24=1126\frac{24 \div 24}{3{,}024 \div 24} = \frac{1}{126}

    Watch out: Using 49\frac{4}{9} for every pick gives 2566,561\frac{256}{6{,}561}, choice C. That is right only if each disk goes back in. Keeping 9 on the bottom each time gives 82,187\frac{8}{2{,}187}, choice A. Choice D is more than 1, and no chance can be more than 1.

    So the answer is B, 1126\frac{1}{126}.

Check your answer

Cancel before you multiply. On top, 4×2=84 \times 2 = 8 cancels the 8 on the bottom. That leaves 39×7×6=3378\frac{3}{9 \times 7 \times 6} = \frac{3}{378}. Divide the top and bottom by 3 to get 1126\frac{1}{126} again.

More questions like this

The original DOE question
Question 98 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 98. Open page 211 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).

Question 5 of 6

2024 Test B Math, question 53

Question

Which graph represents the solution to |x+1|≥2|x+1|\geq 2?

  • ANumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1. The line is shaded from −3 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • BNumber line from −3 to 3 with a tick at every integer. Closed dots at −1 and 1. The line is shaded from −1 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • CNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1, with the segment between them shaded.
  • DNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 3. The line is shaded from −3 to the left, ending in an arrow, and from 3 to the right, ending in an arrow.
Show the answer

Correct answer: A

Since |x+1|≥2|x+1|\geq 2, either x+1≤−2x+1\leq -2 or x+1≥2x+1\geq 2.

Solving x+1≤−2x+1\leq -2 for x gives x≤−3x\leq -3.

Solving x+1≥2x+1\geq 2 for x gives x≥1x\geq 1.

A filled circle on a number line represents “or equal to.” So the correct number line representation has filled circles at −3 and 1, with the number line filled to the left of −3 for “less than” and to the right of 1 for “greater than.”

Question analysis

Students choose the number line graph that shows every solution of an inequality in which x + 1 sits inside absolute value bars. It tests absolute value as distance and graphing inequalities. The tricky part is turning one absolute value statement into two pieces and placing each boundary point and shaded ray correctly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 9, 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

Reading |x + 1| as the distance from x to −1 makes this quick: the points 2 units away are −3 and 1, and greater than or equal means outside them. Students who solve only one case and mirror it across 0 land on graphs centered on 0, and students who find both endpoints may still shade between them. The graphs differ in just one dot or one shading choice, so each must be read carefully.

  1. Something you must already know moderate Working it out

    Solving |x + 1| ≥ 2 needs two cases, x + 1 ≤ −2 or x + 1 ≥ 2, joined by or; absolute value inequalities go beyond grade 7, and the stem gives no hint. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The right graph and the segment from −3 to 1 share both filled endpoints, so only deciding between outside and between the boundaries separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Four number lines must be read for filled dots and arrow directions, and two of them differ from the right one only in where a single dot sits. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the absolute value as distance from −1 on the number line, find the points exactly 2 units away, and decide whether the solutions lie between or outside them.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value as distance
|x−a|=distance from x to a|x - a| = \text{distance from } x \text{ to } a
x and a are points on the number line. The bars mean absolute value. |x − a| is how far apart the two points are, which is never negative. Since x + 1 = x − (−1), |x + 1| is the distance from x to −1.

  1. Sort out the question

    You need the graph of every x that makes |x+1|≥2|x + 1| \geq 2 true. The sign ≥ means greater than or equal to. The bars mean absolute value.

    On a number line, a filled dot means that point counts. A thick line with an arrow means all the numbers that way count too.

  2. Read it as a distance

    Adding 1 is the same as subtracting −1, so x+1=x−(−1)x + 1 = x - (-1). By the formula, |x+1||x + 1| is the distance from x to −1.

    So the inequality says that x is at least 2 units away from −1.

  3. Find the two boundary points

    Start at −1 and go 2 units each way.

    −1−2=−3-1 - 2 = -3

    −1+2=1-1 + 2 = 1

    At −3 and at 1, the distance is exactly 2. The words or equal to let them count, so they get filled dots.

  4. Decide which side to shade

    Being at least 2 away means being farther out. Numbers between −3 and 1 are too close to −1. For example, 0 is only 1 unit away. So shade to the left of −3 and to the right of 1.

    In algebra, that is x+1≤−2x + 1 \leq -2 or x+1≥2x + 1 \geq 2. These give x≤−3x \leq -3 or x≥1x \geq 1.

  5. Match the graph

    Choice A has filled dots at −3 and 1, with arrows going outward. That matches.

    Watch out: Choice C shades between −3 and 1. Those are the points within 2 of −1, the answer to |x+1|≤2|x + 1| \leq 2. Choices B and D are centered on 0, but the center here is −1.

    So the answer is A, the graph shaded outward from −3 and 1.

Check your answer

Test a point from each part. For x = 2, |2+1|=3|2 + 1| = 3, and 3 is at least 2. For x = 0, |0+1|=1|0 + 1| = 1, which is less than 2. For x = −3, |−3+1|=2|-3 + 1| = 2, which counts. Graph A agrees with all three.

More questions like this

The original DOE question
Question 110 as printed on page 214 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 110. Open page 214 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).

Question 6 of 6

2018 Test B Math, question 44

Question

A number line with eight evenly spaced hash marks. Point M is on the second mark, labeled −1/4, and point T is on the seventh mark, labeled 5/8. Point R, with no label, is on the fifth mark, three marks to the right of M.

The hash marks on the number line above are evenly spaced. What is the coordinate of point RR?

  • A740\frac{7}{40}
  • B940\frac{9}{40}
  • C1140\frac{11}{40}
  • D2140\frac{21}{40}
Show the answer

Correct answer: C

There are 5 sections between M and T. To find the length of one of these sections, find the distance between M and T and divide by 5:

(58−(−14))÷51=\left(\frac{5}{8}-\left(-\frac{1}{4}\right)\right)\div\frac{5}{1}=

(58+28)÷51=\left(\frac{5}{8}+\frac{2}{8}\right)\div\frac{5}{1}=

78÷51=78×15=740\frac{7}{8}\div\frac{5}{1}=\frac{7}{8}\times\frac{1}{5}=\frac{7}{40}

R is 3 sections away from M, so add:

−14+3(740)=−1040+2140=1140-\frac{1}{4}+3(\frac{7}{40})=-\frac{10}{40}+\frac{21}{40}=\frac{11}{40}

R is located at 1140\frac{11}{40} .

Question analysis

Students find the fraction at point R on a number line with evenly spaced hash marks, where M is at −1/4 and T is at 5/8. It tests fraction arithmetic with negative numbers. Working out the size of one gap and counting gaps from M to R takes several careful steps.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
  • 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 place fractions and negatives on number lines, but here they must subtract a negative fraction, divide by five, and then add back. Four fraction steps with unlike denominators make it harder than typical classwork.
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 6, 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

Fractions, a negative starting point and a count of spaces all have to come out right together. Students lose points by counting tick marks instead of spaces, by subtracting −1/4 as if it were positive, or by measuring R from 0 instead of from M. Converting both labels to fortieths first, −10/40 and 25/40, makes each gap a whole number of fortieths.

  1. Error-prone arithmetic major Working it out

    The gap length needs 5/8 − (−1/4) = 7/8, then 7/8 ÷ 5 = 7/40, then −10/40 + 21/40: negatives and three different denominators meet in one problem. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The hash marks have to be counted as spaces: five from M to T, with R three spaces past M, while the unlabeled marks outside M and T play no part. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the distance from M to T by subtracting their coordinates, divide it by the number of equal gaps, then step from M toward T by the right number of gaps.

Traps

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right number−left number\text{distance} = \text{right number} - \text{left number}
The right number is the point farther to the right on the number line. The left number is the other point. Subtracting tells how many units apart they are. Taking the left from the right keeps the answer positive.
Dividing a fraction by a whole number
ab÷c=ab×c\frac{a}{b} \div c = \frac{a}{b \times c}
a over b is the fraction, and c is the whole number you divide by. Splitting a fraction into c equal parts makes each part c times smaller. So you multiply the bottom number, b, by c.

  1. Read the number line

    M is at −14-\frac{1}{4} and T is at 58\frac{5}{8}. The hash marks are evenly spaced, so every space is the same length. Count the spaces. From M to T there are 5 spaces. From M to R there are 3 spaces. You need the number at point R.

  2. Find the distance from M to T

    Use the distance formula. First write −14-\frac{1}{4} in eighths, as −28-\frac{2}{8}.

    58−(−28)=58+28=78\frac{5}{8} - \left(-\frac{2}{8}\right) = \frac{5}{8} + \frac{2}{8} = \frac{7}{8}

    Taking away a negative is the same as adding. M and T are 78\frac{7}{8} apart.

  3. Find the length of one space

    The 5 spaces are equal, so divide by 5.

    78÷5=78×5=740\frac{7}{8} \div 5 = \frac{7}{8 \times 5} = \frac{7}{40}

    Each space is 740\frac{7}{40} long.

    Watch out: this is the length of one space, not where R is. Stopping here gives choice A.

  4. Step from M to R

    R is 3 spaces to the right of M.

    3×740=21403 \times \frac{7}{40} = \frac{21}{40}

    Now add that to M. Write −14-\frac{1}{4} as −1040-\frac{10}{40}.

    −1040+2140=1140-\frac{10}{40} + \frac{21}{40} = \frac{11}{40}

    Watch out: 2140\frac{21}{40} is how far R is from M, not where R is. Choosing it gives choice D.

  5. Give the answer

    Point R is at 1140\frac{11}{40}, a little to the right of 0. So the answer is C, 1140\frac{11}{40}.

Check your answer

Go backward from T instead. T is 2 spaces past R, and 58=2540\frac{5}{8} = \frac{25}{40}. So R is at 2540−1440=1140\frac{25}{40} - \frac{14}{40} = \frac{11}{40}. Both ways give the same point.

More questions like this

The original DOE question
Question 101 as printed on page 161 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 101. Open page 161 of the handbook (PDF)

Sample #504776