SHSAT Sunset

Six SHSAT Questions

3 English and 3 math

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Question 1 of 6

2024 Test A English, question 28
Passage Champion of the Channel

Champion of the Channel

In 1926 an editor at the London Daily News predicted that Gertrude Ederle, an American swimmer with eighteen world records and three Olympic medals, would fail in her attempt to swim across the English Channel. He claimed that “even the most uncompromising champion of the rights and capacities of women must admit that in contests of physical skill, speed and endurance they must remain forever the weaker sex.” Yet, at only nineteen years old, Ederle not only became the first woman to accomplish this feat, she also broke the men’s record by two hours. Gertrude Ederle’s triumphant swim across the English Channel was a testimony to her determination, innovative spirit, and passion for swimming.

Crossing the English Channel is a daunting task for any swimmer. At its narrowest point, the channel measures twenty-one miles across. Its icy waters hover around sixty degrees Fahrenheit, and its unruly tides and currents toss swimmers about like bobbing corks. Stinging jellyfish, seaweed, and floating debris from shipwrecks and lost cargoes present added hazards.

For decades the channel’s perils have defeated countless swimmers. Ederle, too, failed in her first attempt to cross the channel in 1925. Just six miles short of finishing, she became ill, and her coach had to haul her out of the water. Undeterred, Ederle decided to try again. Ederle knew that if she did not complete the challenge this time, she might never get the opportunity to set this record, because a rival female swimmer was preparing to make her second attempt at the crossing as well.

To prepare for the marathon swim, Ederle found ways to improve her equipment. She and her sister Meg discovered that melted candle wax perfectly sealed goggle edges, effectively waterproofing Ederle’s goggles against hammering waves. The sisters also designed a two-piece silk swimsuit for Ederle. During her first channel-crossing attempt she had worn a standard one-piece swimsuit that, after the lengthy hours of swimming across the channel, had stretched out, filling with water and creating drag, making an already challenging task almost insurmountable. Unlike the cumbersome typical bathing suit, this silk invention weighed little and allowed for easy movement.

On August 6, 1926, Ederle waded into the channel near Cape Gris-Nez, France. At first she shivered in the bone-chilling water even though she had covered her body in eight layers of grease for insulation. Her limbs felt stiff. Her strokes were irregular. Driving forward, she fought to clear her mind and find what she called her “sphere,” a place where the sea became her only companion and the shrieks of gulls and the humming of boat engines faded away. Using a new overhand stroke called the American crawl, Ederle eventually settled into a steady pace, briskly breaking through waves.

Throughout Ederle’s swim, two tugboats accompanied her. One carried newspaper reporters who wired dispatches of her progress to shore. The other, displaying a sign that read “This way, ole kid!” with an arrow pointing forward, transported her coach, family, and friends. Her coach played songs, such as “Yes, We Have No Bananas,” on a phonograph so that Ederle could time her strokes to the rhythm. Using a net, her coach also passed her baby bottles of broth for nourishment.

For hours Ederle swam, dodging debris with an amused smile. However, as she neared the English shore, a sudden fierce storm erupted. The tides and waves forced Ederle backward, and she fought the stubborn swells for several hours. The salty water caused her tongue to swell and inflamed her ears. Yet Ederle felt indescribably happy as she churned through the sea. Finally, as she neared the English shore, the storm abated, and the tide turned. No longer fighting against her, the sea pushed her toward the shore and victory.

After fourteen hours and thirty-one minutes, Ederle, on wobbly legs, stepped onto the English shore. The waiting crowd roared, honked their automobile horns, blasted their tugboat whistles, and set off flares that flashed in the sky. Ederle had swum into history.

When Ederle returned to New York, she received a parade, where thousands of people shouted “Trudy!” Not only were everyday American citizens proud of Ederle, but she also inspired them to be more active. Over the next few years, more than 60,000 people credited her with motivating them to earn their American Red Cross swimming certificates. Gertrude Ederle’s accomplishment proved to the world that with determination and passion, it was possible for a person to achieve his or her goals.

SOLO SWIMS ACROSS THE ENGLISH CHANNEL

Earliest Speed Records
Year Swimmer Time
1875 Matthew Webb (M) 21 hours, 45 minutes
1923 Enrico Tiraboschi (M) 16 hours, 33 minutes
1926 Gertrude Ederle (F) 14 hours, 31 minutes
1926 Arnst Vierkotter (M) 12 hours, 38 minutes
Current Speed Records
Year Swimmer Time
2012 Trent Grimsey (M) 6 hours, 55 minutes
2006 Yvetta Hlavacova (F) 7 hours, 25 minutes

Question

Which statement describes how the author’s use of a problem-and-solution structure in paragraph 5 contributes to the development of ideas in the passage?

  • ADetailing the challenges presented by the cold channel waters highlights the effectiveness of Ederle’s training.
  • BDescribing Ederle’s physical difficulties during her swim provides evidence of the team effort required to ensure her safety.
  • CExplaining the difficulties that arose early in the effort helps predict the additional obstacles that occurred during Ederle’s attempt.
  • DIllustrating Ederle’s process of blocking out her discomfort shows that swimming the channel was both a mental and a physical challenge.
Show the answer

Correct answer: D

The question asks how a problem-and-solution structure in paragraph 5 contributes to the ideas presented in the passage.

A. Incorrect. The details in paragraph 5 do not provide a connection between the cold temperature of the water and the effectiveness of Ederle’s training.

B. Incorrect. The difficulties Ederle encountered were only partially relieved by her team, and her team’s efforts to ensure Ederle’s safety are not explained in the passage.

C. Incorrect. The paragraph does not focus on the relationship between the problems Ederle encountered at the start of her swim and the problems she encountered during or near the end of her swim.

D. CORRECT. The paragraph describes how the water’s temperature made Ederle uncomfortable and made it difficult for her to regulate her stroke. The paragraph continues with the explanation of how she overcame these issues by focusing her mind on the sea and tuning out the distractions in her surroundings (“the sea became her only companion and the shrieks of gulls and the humming of boat engines faded away” [paragraph 5]).

Question analysis

Gertrude Ederle's cold, stiff first hours in the English Channel are told as problem and solution, and this text-structure question asks what that pattern adds to the passage's ideas. Students must spot the troubles she met in the cold water and how she handled them, then pick the abstract statement that accounts for both halves of the pattern.

NYS standards
  • 7R5 structure and genre features
  • 7R3 plot elements and ideas
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 A problem-and-solution question about one paragraph is familiar grade 7 fare, yet these four answers are general claims about the swim, each tied to a real hardship, so students need a clear map of the paragraph before judging them.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from one paragraph or stanza
Reading level
Passage: about grade 9, 790 words; question and choices: about grade 14, 95 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The opening halves of the choices are all accurate, which moves the whole decision onto their second halves. Readers who notice only the physical side of the paragraph, the cold and the stiffness, reach for training, teamwork or foreshadowing to explain it. The paragraph's turning point is a mental one, clearing her mind until the noise faded, and only the statement that names both a mental and a physical challenge accounts for the problem and the solution together.

  1. How, not what major Working it out

    The question asks what paragraph 5's pairing of problems (bone-chilling water, stiff limbs, irregular strokes) with solutions (finding her “sphere”, the American crawl) does for the passage, so retelling the hardships earns nothing. (Anderson & Krathwohl, 2001)

  2. Every choice must be checked moderate Choosing the answer

    Each choice opens with a fair description of paragraph 5, the cold water, her physical difficulties, the early struggle, so only what each claims that description shows can be tested. (Embretson & Wetzel, 1987)

  3. Figurative or literary language minor Understanding the question

    The mental half of the solution is told in images, a place where “the sea became her only companion” and gulls and engines “faded away”, which must be read as blocking out discomfort. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

Strategy: Sort paragraph 5 into two lists, what went wrong for Ederle early in the swim and what she did about it, then ask what that pairing shows about the swim as a whole.

Traps

  • Wrong purpose or structure: Names a purpose, structure or effect that the part does not serve, or describes what a part says when the question asks what it does.
  • Outside knowledge: A choice is plausible in the real world but not supported by this text.
  • Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Learn what the question asks

    A problem-and-solution structure tells about a difficulty and then how someone deals with it. The question asks what this pattern in paragraph 5 adds to the ideas of the passage. So first sort the paragraph into problems and solutions.

  2. List the problems in paragraph 5

    When Ederle waded in, “she shivered in the bone-chilling water.” Next, “Her limbs felt stiff. Her strokes were irregular.” Limbs are arms and legs. Irregular means uneven, not steady. All of these are troubles with her body.

  3. List her solutions

    Ederle “fought to clear her mind.” She looked for what she called her “sphere.” There, “the sea became her only companion and the shrieks of gulls and the humming of boat engines faded away.” In other words, she shut out the noise and her discomfort. Then, using the American crawl, she “settled into a steady pace.”

  4. Connect problems and solutions

    The problems were physical, which means of the body. A big part of the fix was mental, which means of the mind. She calmed her thoughts so her body could keep working. So the paragraph shows that the swim tested her mind as well as her body.

  5. Check each choice

    • D fits: blocking out her discomfort shows the swim was “both a mental and a physical challenge.”
    • A is wrong because paragraph 5 never mentions her training.
    • B is wrong because her team on the tugboats appears in paragraph 6, and her safety is never discussed.
    • C is wrong because the early troubles are solved inside this paragraph instead of setting up later ones.

    Watch out: A storm does hit in paragraph 7, so C may look right. But paragraph 5 does not point ahead to it.

  6. Choose the answer

    Only D explains both the problem and the way Ederle solved it. So the answer is D.

Why this answer and not the closest wrong one

C is tempting, since a storm really does strike in paragraph 7. But paragraph 5 never hints at it. The paragraph ends with its problem solved, as Ederle is “briskly breaking through waves.” D explains both halves of the pattern: troubles of the body, eased by the mind.

More questions like this

The original DOE question
Question 28 as printed on page 54 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 54, question 28. Open page 54 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 “Champion of the Channel” as printed (pages 56–57)The passage “Champion of the Channel” as printed on page 56 of the DOE handbook.The passage “Champion of the Channel” as printed on page 57 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.

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 3 of 6

2026 Practice Test C English, question 24
Passage Bird Talk

Bird Talk

by Carl Sandburg

And now when the branches were beginning to be heavy,
It was the time when they once had said, “This is the
  beginning of summer.”
The shrilling of the frogs was not so shrill as in the
  first weeks after the broken winter;
The birds took their hops and zigzags a little more
  anxious; a home is a home; worms are worms.
The yellow spreads of the dandelions and buttercups
  reached across the green pastures.
Tee whee and tee whee came on the breezes, and the grackles
  chuzzled their syllables.
And it was the leaves with a strong soft wind over them
  that talked most of all and said more than any others
  though speaking the fewest words.
It was the green leaves trickling out the gaunt nowhere
  of winter, out on the gray hungry branches—
It was the leaves on the branches, beginning to be heavy,
  who said as they said one time before, “This is the be-
  ginning of summer.”

We shall never blame the birds who come
  where the river and the road make the Grand Crossing
  and talk there, sitting in circles talking bird talk.
If they ask in their circles as to who is here
  and as to who is not here and who used to be here,
Or if instead of counting up last year as against
  this year, they count up this year as against next
  year, and have their bird chatter about who is here
  this year who won’t be here next year,
We shall never blame the birds.

If I have put your face among leaf faces, child,
Or if I have put your voice among bird voices,
Blame me no more than the bluejays.

“Bird Talk” by Carl Sandburg—Public domain.

Question

Throughout the poem, the poet conveys the speaker’s point of view by

  • Agradually moving the speaker’s attention from one element of nature to another.
  • Bproviding the speaker’s motivation for describing the landscape in great detail.
  • Chighlighting the speaker’s emotional response as each season passes.
  • Dshifting the speaker’s focus from external observations to personal reflection.
Show the answer

Correct answer: D

This question asks for an explanation of how the poet conveys the speaker’s point of view on the natural world and the change of seasons as presented in the poem.

A. Incorrect. Though in the first stanza the speaker describes various elements of nature such as branches, frogs, birds, flowers, and leaves, these elemental shifts are not precisely gradual and occur only in the first stanza. Further, the first stanza, though appreciative of the natural world, is primarily descriptive and does not develop the speaker’s perspective to the degree that other parts of the poem do.

B. Incorrect. While it may be inferred that the speaker describes the landscape to such a degree because it has revealed a seasonal shift, that this is indeed the speaker’s motivation is never explicitly conveyed by the poet.

C. Incorrect. Multiple seasons do not pass in the poem; only the shift to the “ ‘beginning of summer’ ” (line 3) occurs in an endless cycle. Further, the speaker does not reveal an emotional response; it is the responses of nature that the speaker describes.

D. CORRECT. The first stanza is the speaker’s observations of the natural world embracing a new season. In the second stanza, the speaker begins to reflect on how the natural world, as embodied by the birds, calmly accepts seasonal change in its perpetual sameness. The reflection continues in the third stanza, where it is revealed that the speaker has been addressing these observations to a child.

Question analysis

This point-of-view question asks how the poet of “Bird Talk” conveys the speaker’s point of view throughout the poem. Students must keep in mind the long opening description of early summer and the two shorter stanzas after it, and decide which description of the poet’s method holds from the first line to the last.

NYS standards
  • 7R6 point of view and author position
  • 7R5 structure and genre features
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 point-of-view questions often ask what a speaker thinks. This one asks how the poet conveys that view across all three stanzas of a long free-verse poem, and the choices name methods in general terms, closer to grade 8 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: 296 words (a poem; reading level not measured); question and choices: about grade 10, 54 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 first stanza takes up 19 of the poem’s 32 lines and moves steadily from branches to frogs, birds, flowers and leaves, so a choice that describes it alone feels right. The point of view shows only when the poem turns to “We” and then to “I” addressing a child, and that third stanza is three hard lines. Students must keep the whole poem’s course in mind and choose the method that holds from first line to last.

  1. Evidence in several places moderate Working it out

    The answer depends on all three stanzas at once: the long catalog of the season in lines 1–19, the shift to “We shall never blame the birds,” and the closing “If I have put your face … child.” (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)

  2. A very close wrong answer moderate Choosing the answer

    The choice about moving from one element of nature to another and the key both describe a shift in the speaker’s attention; they differ in what shifts, nature to nature or the outside world to the speaker’s own thoughts. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. How, not what moderate Working it out

    The stem asks how the poet conveys a point of view, a method, so a choice must describe what the speaker does across the poem, not what the speaker sees. (Anderson & Krathwohl, 2001)

  4. Figurative or literary language minor Understanding the question

    The last stanza is cryptic, “Blame me no more than the bluejays,” and students who miss that the speaker is now talking about putting a child into the poem lose the move to reflection. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

Strategy: Write one phrase for what the speaker attends to in each stanza, then keep only a choice that describes the whole sequence, not one stanza.

Traps

  • Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
  • Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.

The wrong answers, classified

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 about point of view: how the speaker, the voice talking in the poem, sees things. It also asks how the poet shows that view across the whole poem. So look at what the speaker pays attention to in each of the three stanzas. A stanza is a group of lines set apart by a blank line.

  2. Follow the speaker through the stanzas

    • Stanza 1, lines 1–19: the speaker describes the world outside. There are branches, frogs, birds, flowers and leaves at the start of summer.
    • Stanza 2, lines 20–29: the speaker starts to judge and think. “We shall never blame the birds.”
    • Stanza 3, lines 30–32: the speaker says “I” and talks to a child: “If I have put your face among leaf faces, child.”

  3. Name the change in focus

    The poem starts with external observations, or things the speaker sees and hears outside. It ends with personal reflection, or the speaker’s own thoughts. In the last stanza, the speaker admits to putting a child’s face and voice into the poem. The speaker asks to be blamed “no more than the bluejays.” Since “We shall never blame the birds,” that means not at all. This move from looking outward to looking inward is how the speaker’s view comes through.

  4. Check each choice

    • A does not fit. Moving from branches to frogs to birds happens only in stanza 1. It misses the turn to “We” and “I.”
    • B does not fit. The speaker never says why the landscape is described.
    • C does not fit. Only one change of season happens, the start of summer. The anxious feelings belong to the birds, not the speaker.
    • D fits. The speaker moves from describing nature to thinking about it.

    Watch out: A describes stanza 1 very well. But the question says “Throughout the poem,” so the answer must cover all three stanzas.

  5. Choose the answer

    The speaker starts by watching nature and ends by thinking aloud to a child. So the answer is D, “shifting the speaker’s focus from external observations to personal reflection.”

Why this answer and not the closest wrong one

A is the closest wrong choice. Stanza 1 really does move from one part of nature to another. But that is only 19 of the 32 lines. In stanzas 2 and 3, the speaker turns to “We” and then “I,” and shares personal thoughts. Only D describes the speaker’s path from the first line to the last.

More questions like this

The original DOE question
Question 24 as shown on the DOE’s digital test, page 37 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C English, page 37, question 24.
The passage “Bird Talk” as printed (page 32)The passage “Bird Talk” as printed on page 32 of the DOE handbook.

Question 4 of 6

2024 Test B Math, question 17

Question

A bowl contains 12 green candies, 4 yellow candies, and some red candies. The probability of choosing a green candy at random is twice as great as the probability of choosing a red candy at random. What is the probability of choosing a yellow candy at random?

  • A211\frac{2}{11}
  • B29\frac{2}{9}
  • C14\frac{1}{4}
  • D311\frac{3}{11}
Show the answer

Correct answer: A

The probability of choosing a green candy is twice as great as the probability of choosing a red candy. There are 12 green candies. Let r = the number of red candies.

12=2r12=2r

r=6r=6

So there are 6 red candies. 12 green candies + 4 yellow candies + 6 red candies = 22 total candies.

There are 4 yellow candies out of a total of 22 candies. So the probability of choosing a yellow candy is

422=211\frac{4}{22}=\frac{2}{11}.

Question analysis

A probability question about a bowl of 12 green candies, 4 yellow candies and an unknown number of red ones. A green candy is twice as likely to be picked as a red one, and students find the chance of picking a yellow candy. The missing red count has to be worked out before any probability can be found.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Simple probability is grade 6 content in New York, and grade 7 problems usually give every count. Working backward from a comparison of two probabilities to a missing count, then forward to a third probability, is more than typical grade 7 work.
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 7, 49 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 withholds the red count, and everything depends on recovering it from the comparison of probabilities. Students who cannot make that step tend to leave red out altogether, while those who find 6 red candies can still slip at the end by using the wrong total or the wrong color. Writing all three counts and the total of 22 before forming any fraction prevents both endings.

  1. Turning words into math moderate Understanding the question

    “Twice as great as the probability of choosing a red candy” must become 12 = 2 × the red count, which works only because both probabilities share one total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Combines ideas moderate Working it out

    An unknown count has to be found from a ratio before any probability can be written, joining proportional reasoning with probability as a part over the total. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Turn twice as likely into a statement about numbers of candies to find the red count, add all three colors for the total, then put the yellow count over that total and simplify.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of the kind you wanttotal numberP = \frac{\text{number of the kind you want}}{\text{total number}}
P is the probability, the chance that something happens. The top is how many candies are the kind you want. The bottom is how many candies there are in all. Each candy is just as likely to be picked, so this fraction gives the chance.

  1. Sort out what you know

    The bowl has 12 green candies, 4 yellow candies and some red candies. You are not told how many are red. Let r be the number of red candies.

    The chance of picking green is twice the chance of picking red. You need the probability (the chance) of picking yellow.

  2. Compare green and red

    Write each chance with the probability formula.

    P=number of the kind you wanttotal numberP = \frac{\text{number of the kind you want}}{\text{total number}}

    Green and red have the same bottom number, the total number of candies. So if the chance of green is twice the chance of red, there are twice as many green candies as red ones.

    12=2×r12 = 2 \times r

  3. Find the red candies and the total

    r=12÷2=6r = 12 \div 2 = 6

    There are 6 red candies. Now add up all the candies.

    12+4+6=2212 + 4 + 6 = 22

    There are 22 candies in all. Watch out: do not leave out the red candies just because their number was not given. Using 16 as the total gives 1/4, choice C.

  4. Find the chance of yellow

    There are 4 yellow candies out of 22.

    P=422=211P = \frac{4}{22} = \frac{2}{11}

    Divide the top and the bottom by 2 to simplify. Watch out: the question asks for yellow, not red. 6/22 = 3/11 is the chance of red, choice D. The bottom must be all 22 candies. Using 18 gives 2/9, choice B.

    So the answer is A, 211\frac{2}{11}.

Check your answer

Check the clue. The chance of green is 1222\frac{12}{22}, and the chance of red is 622\frac{6}{22}. Twelve is twice 6, so green is twice as likely. Also, the three chances add up to 1: 1222+422+622=2222=1\frac{12}{22} + \frac{4}{22} + \frac{6}{22} = \frac{22}{22} = 1.

More questions like this

The original DOE question
Question 74 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 74. Open page 203 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

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.

More questions like this

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

2024 Test B Math, question 57

Question

If x and y are both negative integers with x<yx<y, which expression has the least value?

  • Ax2yx^{2}y
  • Bxy2xy^{2}
  • C−x2y-x^{2}y
  • D−xy2-xy^{2}
Show the answer

Correct answer: A

Since x and y are both negative integers, both x2x^{2} and y2y^{2} are positive.

Therefore, both x2yx^{2}y and xy2xy^{2} are negative, and both −x2y-x^{2}y and −xy2-xy^{2} are positive.

Because x<yx<y and both x and y are negative, |y|<|x||y|<|x| and y2<x2y^{2}<x^{2}. So |y2|<|x2||y^{2}|<|x^{2}|.

Since |y|<|x||y|<|x| and x and y are integers, |xy2|<|x2y||xy^{2}|<|x^{2}y|.

So x2y<xy2x^{2}y<xy^{2}.

−xy2=|xy2| and −x2y=|x2y|-xy^{2}=|xy^{2}|\text{ and }-x^{2}y=|x^{2}y|

Therefore, x2y<xy2<−xy2<−x2yx^{2}y<xy^{2}<-xy^{2}<-x^{2}y.

Question analysis

Two negative integers x and y, with x less than y, appear in four expressions built from squares and minus signs. Students find the expression with the least value. It tests multiplying negative numbers and comparing their sizes, and the difficulty is tracking how each minus sign and each square changes both the sign and the size of the result.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.NS.7d absolute value versus order
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students multiply negative numbers, but this asks them to compare four variable expressions for any pair of negative integers. Deciding how the squares compare when x is less than y goes beyond typical grade 7 sign practice.
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 8, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two facts settle the question: squares of negatives are positive, and the more negative number has the larger square. Students who track the minus signs carelessly think a leading minus makes an expression negative, and students who assume the larger y has the larger square choose the wrong negative expression. Substituting x = −3 and y = −2 gives −18, −12, 18 and 12, which orders all four at once.

  1. Variables instead of numbers moderate Understanding the question

    No values are given, only that x and y are negative integers with x < y, so students must reason about signs or invent a pair such as −3 and −2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    x²y and xy² use the same letters with the exponent moved; both are negative, and only comparing x² with y² separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Every choice must be checked minor Choosing the answer

    All four expressions have to be placed in order, because the minus signs in front of the last two turn those expressions positive. (Embretson & Wetzel, 1987)

Strategy: Pick two negative integers with x less than y, such as −3 and −2, evaluate all four expressions, and confirm the order with a second pair.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.
Squaring a number
x2=x×xx^2 = x \times x
x is any number. x² (x squared) means x times itself. The square of a number other than 0 is always positive, because two negatives or two positives multiply to a positive.

  1. Sort out the question

    x and y are negative integers, such as −3 and −2. Since x < y, x is to the left of y on the number line. So x is more negative, farther from 0.

    You need the expression with the least value. That is the most negative one.

  2. Find the sign of each expression

    Squares are positive, so x2x^2 and y2y^2 are positive.

    • x2yx^2 y is a positive times a negative, so it is negative.
    • xy2x y^2 is a negative times a positive, so it is negative.
    • −x2y-x^2 y and −xy2-x y^2 are the opposites of these, so they are positive.

    So the least value must be A or B. Watch out: The minus sign in front of C and D does not make them negative. It turns a negative into a positive.

  3. Compare A and B with numbers

    Try x = −3 and y = −2, since −3 < −2.

    x2y=(−3)2×(−2)=9×(−2)=−18x^2 y = (-3)^2 \times (-2) = 9 \times (-2) = -18

    xy2=(−3)×(−2)2=(−3)×4=−12x y^2 = (-3) \times (-2)^2 = (-3) \times 4 = -12

    Since −18 < −12, A is smaller.

  4. See why it always works

    Both expressions contain x × y, which is positive. x2y=(xy)×xx^2 y = (xy) \times x and xy2=(xy)×yx y^2 = (xy) \times y. Multiplying the same positive number by the more negative x gives the more negative result.

    Watch out: y is bigger than x, but y2y^2 is smaller than x2x^2: 4 is less than 9. Thinking y² is bigger leads to choice B.

    So the answer is A, x2yx^2 y.

Check your answer

Try another pair, x = −5 and y = −1. Then A is 25×(−1)=−2525 \times (-1) = -25, B is (−5)×1=−5(-5) \times 1 = -5, C is 25 and D is 5. A is still the least.

More questions like this

The original DOE question
Question 114 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 114. Open page 215 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).

Sample #366693