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Six SHSAT Questions

3 English and 3 math

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Question 1 of 6

2026 Practice Test C English, question 31
Passage Serpent Mound

Serpent Mound

The Great Serpent Mound, located in Adams County, Ohio, is a human-made mound of earth that researchers believe was created between 300 B.C. and A.D. 1100 by an indigenous culture.

Ohio, 1846

Brush Creek stood low when the museum men came
with their measuring tapes and sketchbooks.
 It was winter. Fringed with ice,
 the creek doubled back on itself
as if it had forgotten something.
 Pa was in Cincinnati, or else on his way home,
  so Ma told me to lead the men
  into the marshy low grounds. It being winter,
    there was little underbrush to speak of—
    in the summer there would have been
  briars, poison ivy, biting flies. I listened
for the swish of a beaver’s heavy tail,
 the chitter of a chickadee, or the cry of a hawk,
 but the winter silence of the creek pressed
 down on all of us like a weight.
The humps in the ground were all but
invisible until you were right up on them. The figure
was even less obvious: the sinuous body,
 the tail coiled three times around,
 and at the other end, the mouth wide open.
  In the summer the creek bottom was crowded
  with so much life that you could trip over
    the ridges of earth before you saw
    anything at all. In winter you could climb
    a tree and get some idea of the whole thing:
  the serpent’s body undulating, slithering
silently across the ancient
 earth. At the mouth end, there was an oval mound
 as if the snake were about to swallow an egg—
as snakes sometimes did in our rickety
henhouse—my Pa always said, or
as if swallowing the sun, one of the museum
 men suggested, taking notes with his quill pen,
 an old-style inkhorn slung at his side.
 I liked that: swallowing the sun,
  just the sort of thing a snake might do,
  might want to do. When, later, I told my sister Ruth,
    she disagreed. It is singing to the sun,
    she insisted. That is why its mouth
  is wide open. She said, “Sometimes I think
I hear it on summer nights. Not swallowing, singing.”

Question

What impact do the words “swish,” “chitter,” and “cry” in lines 12–13 have in the poem?

  • AThey illustrate how lively the mound’s surroundings are during the summer.
  • BThey show how the mound disrupts natural life in the marsh.
  • CThey indicate the variety of wildlife found in the area around the mound.
  • DThey suggest that the speaker prefers the mound’s appearance in winter.
Show the answer

Correct answer: A

The question asks what impact the words “swish,” “chitter,” and “cry” in lines 12–13 have in the poem.

A. CORRECT. The words highlight the sounds of the animals that visit the mound in the summer: the “swish of a beaver’s heavy tail” (line 12), the “chitter of a chickadee” (line 13), and the “cry of a hawk” (line 13). These words help contrast the quiet stillness of the setting around the mound in the winter.

B. Incorrect. While the words in lines 12–13 convey the sounds certain animals make in the summer, the sounds of the animals that live near the mound are a natural part of life in the area, not a disruption.

C. Incorrect. The speaker is contrasting the sounds of the animals around the mound during the rest of the year with the silence of the winter months, not merely indicating the variety of wildlife.

D. Incorrect. The speaker expresses the idea that the mound is more visible in the winter because of the lack of brush (“It being winter, / there was little underbrush to speak of” [lines 8–9]). Then lines 12–13 describe sounds, not the appearance of the mound in winter.

Question analysis

This question about word choice asks what impact the sound words “swish,” “chitter” and “cry” in lines 12–13 of “Serpent Mound” have in the poem. They belong to a beaver, a chickadee and a hawk near the creek, and more than one choice sounds like a fair account of them, so students must read the surrounding lines closely.

NYS standards
  • 7R4 word meaning and choice
  • 7L5 figurative language and word relationships
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 students often explain what sensory words add to a description. These three words sit inside a long sentence that runs across several lines, so their effect has to be worked out from the lines around them, beyond typical grade 7 work.
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: 354 words (a poem; reading level not measured); question and choices: about grade 6, 61 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Three vivid animal sounds invite the easy reading that the poem is showing off the creek’s wildlife, and that choice is true as far as it goes. The words matter because they are missing: the speaker listens for them, and “the winter silence” answers. Students must also accept that the key’s “during the summer” comes from the poem’s later picture of a crowded summer creek, not from lines 12–13 themselves.

  1. A very close wrong answer major Choosing the answer

    The variety choice is literally true, since a beaver, a chickadee and a hawk are three kinds of animal; it differs from the key in missing why the sounds are named, as a contrast with winter. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Reading between the lines moderate Working it out

    The speaker listens for these sounds and does not hear them, so students must infer that they stand for the creek’s life in warmer months, which lines 21–22 call “crowded / with so much life.” (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

  3. Complex sentences minor Understanding the question

    The three words sit in one sentence that runs from “I listened” in line 11 to “like a weight” in line 15 and turns on “but,” so their point arrives only at its end. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Read the full sentence that holds these words, from “I listened” in line 11 to the end of line 15, then decide what the three sound words add to the scene.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This asks what three sound words do in the poem. They are “swish,” “chitter” and “cry,” in lines 12–13. Read the whole sentence they sit in. It starts at “I listened” in line 11 and ends with “like a weight” in line 15.

  2. Read the full sentence

    Lines 11–15 say “I listened / for the swish of a beaver’s heavy tail, / the chitter of a chickadee, or the cry of a hawk, / but the winter silence of the creek pressed / down on all of us like a weight.” A chickadee is a small bird, and a chitter is its quick, chirping sound. The speaker listens for these sounds. But the word “but” tells you they never come. Only silence answers.

  3. Work out why the sounds are named

    A beaver, a chickadee and a hawk are signs of a creek full of life. The speaker hears none of them in winter. Later lines tell you when the creek is lively. Lines 21–22 say “In the summer the creek bottom was crowded / with so much life.” So the missing sounds remind you how lively the place is in summer. They also make the winter silence feel even emptier.

  4. Check each choice

    • A fits. The sounds stand for the busy life around the mound in the warm months.
    • B does not fit. The quiet comes from the season: “the winter silence of the creek.” The mound does not cause it.
    • C does not fit. There are three kinds of animals, but the point is that their sounds are missing.
    • D does not fit. These lines are about sounds. The speaker says nothing about liking how the mound looks in winter.

    Watch out: C is true as far as it goes. But it misses why the speaker names the sounds at all: to set them against the silence.

  5. Choose the answer

    The sounds the speaker listens for, and doesn’t hear, show how lively the creek is in summer. So the answer is A, “They illustrate how lively the mound’s surroundings are during the summer.”

Why this answer and not the closest wrong one

C is the closest wrong choice, because a beaver, a chickadee and a hawk really are three kinds of wildlife. But the speaker listens for them and hears nothing. The word “but” turns the sentence to “the winter silence.” The sounds matter because they are missing. They show the creek in its lively season, which line 21 names as summer.

More questions like this

The original DOE question
Question 31 as shown on the DOE’s digital test, page 46 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C English, page 46, question 31.
The passage “Serpent Mound” as printed (page 40)The passage “Serpent Mound” as printed on page 40 of the DOE handbook.

Question 2 of 6

2026 Practice Test B English, question 37
Passage Letter from Brooklyn

Letter from Brooklyn

by Jacob Scheier

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

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

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

Question

Which lines support the idea that a change in the weather will lessen the speaker’s appreciation for the city?

  • A“as though the air has been forced into intimacy / by the brevity of daylight.” (lines 11–12)
  • B“But when it starts darkening at 4, / this closeness, I know, will be a felt distance,” (lines 13–14)
  • C“like someone drawing your attention / to their lack of intimacy.” (lines 15–16)
  • D“These days I am still walking . . . / beneath the branches bending across avenues,” (lines 17–18)
Show the answer

Correct answer: B

The question asks which lines from the poem support the idea that a change in the weather will lessen the speaker’s appreciation for the city.

A. Incorrect. These lines focus on the positive way the air feels to the speaker during “the early edge of fall” (line 7). They suggest that the speaker appreciates this experience currently, while being aware that the feeling might not last.

B. CORRECT. In these lines, the speaker describes the shift in the “closeness” of the air as the weather changes and “it starts darkening at 4.” The speaker clearly indicates that the positive emotions surrounding this closeness, with “leaves green still” and the air “slightly crisp” (lines 8–9), will soon be “a felt distance,” suggesting that the turn in the weather will make the city less appealing.

C. Incorrect. Although these lines compare the darkening, changing weather to a person revealing a “lack of intimacy,” that specific comparison does not focus on a diminishing appreciation for the city.

D. Incorrect. These lines describe the “cathedral pace” (line 17) at which the speaker walks, emphasizing a meditative or thoughtful experience. Such a description suggests the speaker still appreciates the city, not that the speaker’s appreciation will lessen.

Question analysis

This evidence question asks which lines from “Letter from Brooklyn” support the idea that a change in the weather will lessen the speaker's appreciation for the city. All four choices come from the middle of the poem, where the speaker describes the feel of the early-fall air through figurative comparisons, so students must read each pair of lines closely for both parts of the idea.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R4 word meaning and choice
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Choosing lines that support an idea is standard grade 7 evidence work. Here the idea is carried by figurative comparisons about the air, and the four choices sit within a few lines of one another, so separating them takes closer reading than is typical.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from one sentence or line
Reading level
Passage: 217 words (a poem; reading level not measured); question and choices: about grade 6, 74 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

All four choices come from lines 11–18, and three speak of closeness, distance or intimacy, so the vocabulary of the claim is everywhere. The idea has two parts, a change in the weather and a weaker feeling for the city, and only one pair of lines holds both. The strongest pull is the second half of the same sentence, which puts the loss of intimacy most vividly but leaves the darkening behind in the lines before.

  1. A very close wrong answer moderate Choosing the answer

    Lines 13–14 and 15–16 are one sentence split in two: the first half names the darkening at 4 and the coming distance, the second only describes what that distance feels like. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Figurative or literary language moderate Understanding the question

    The weaker feeling for the city is never stated; it comes through images, the air’s “closeness” turning into “a felt distance,” like a person pointing out their “lack of intimacy.” (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

  3. Reading between the lines minor Working it out

    Students must accept “when it starts darkening at 4” as the change in the weather the stem names, since the lines speak of daylight, not of cold or rain. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

Strategy: Split the idea into its two parts, a change in the weather and a weaker feeling for the city, and choose the lines that hold both on their own, without help from the lines around them.

Traps

  • Supports only part of the claim: In a question about evidence: a line that backs one half of the claim but not the other, or a related point rather than the claim itself.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This is an evidence question. The idea has two parts. First, the weather will change. Second, the speaker will enjoy the city less. You need the lines that hold both parts by themselves.

  2. Read the lines around the choices

    Lines 9–12 describe the early-fall air now. It is “almost grazing / the hair of my arm like a passing stranger.” Grazing here means lightly touching. The speaker says the air feels close, like “intimacy.” Intimacy means a warm, close feeling between people. Lines 13–16 then look ahead to shorter days.

  3. Find both parts in lines 13–14

    Lines 13–14 say: “But when it starts darkening at 4, / this closeness, I know, will be a felt distance”. Darkening at 4 means the sun setting at 4 o’clock, as days get shorter. That is the change in the weather. “This closeness” is the warm feeling the speaker has now. It will turn into distance. So the good feeling will fade.

  4. Check each choice

    • A is wrong. Lines 11–12 describe the closeness the speaker enjoys now. Nothing has faded yet.
    • B is right. It names the change, the early dark, and the loss, closeness turning into “a felt distance.”
    • C is wrong. Lines 15–16 finish the comparison: the distance feels like a person pointing out their “lack of intimacy.” But these lines never mention the weather.
    • D is wrong. Lines 17–18 show the speaker enjoying a slow walk now. They say nothing about the weather turning.

    Watch out: C is tempting because “lack of intimacy” sounds like a lost feeling. But the change in weather is back in lines 13–14.

  5. Choose the answer

    Only one pair of lines holds both the change and the fading feeling. So the answer is B, lines 13–14, which begin “But when it starts darkening at 4”.

Why this answer and not the closest wrong one

C is the closest wrong answer, because it shows the good feeling slipping away. But lines 15–16 only describe what that distance feels like. They don’t say what causes it. Lines 13–14 give both parts: the early dark and the fading closeness.

More questions like this

The original DOE question
Question 37 as shown on the DOE’s digital test, page 52 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B English, page 52, question 37.
The passage “Letter from Brooklyn” as printed (page 50)The passage “Letter from Brooklyn” as printed on page 50 of the DOE handbook.

Question 3 of 6

2026 Practice Test B English, question 25
Passage Looking for the Smallest Spark of Everything

Looking for the Smallest Spark of Everything

Located underground near Geneva, Switzerland, the Large Hadron Collider (LHC) helps scientists study and understand how the smallest particles of matter interact with one another. The LHC propels atomic particle beams along a 17-mile-long ring.

There are the things the world is made of,
things we can see and feel, and then
there are the things even smaller, things that seem
to exist only when we are looking right at them.

So beneath the level of molecules are atoms,
and beneath the level of atoms, we find
protons, electrons, neutrons; and beneath that—
what, exactly? Because we are talking about
everything here: your fingernail,
the candy at the back of your mouth,
the coffee your teacher drank this morning,
your little sister, and the stuffed dog
she used to carry around with her everywhere.
And everywhere.

Everything and everywhere
are made up of the same stuff, whatever
it is. How do we find it? We can listen for it
in the wavelengths from deep space,
talking back to us from unimaginable
distances. Or we can build long,
deep circular tunnels beneath the surface
of this Earth and race particles
(like racing cars, only very tiny cars
flashing along at close to the speed of light)
until everything we think we know
bangs against everything else we know.

And there, in the explosion, in the darkness,
briefly incandescent, they appear:
the quarks, the leptons, and the bosons;
the baryons and the mesons. Their names
sound like dinosaurs or maybe bands
playing terrible music in someone’s garage.

The one thing we know for sure
is that they spin. This is how gravity
enters our world, how our world is held
both together and apart, what keeps
together the pencil in your hand right now
as well as separate from, say,
Jupiter. They spin, and it is only down there
in the darkness—in the vast garage
where physicists jot down
what they can, whatever seems most real—
that they let us perceive their wild dancing,
combusting to the music they make.

Question

Read lines 27–28 from the poem.

And there, in the explosion, in the darkness,
briefly incandescent, they appear:

The imagery in the lines reveals that the speaker

  • Aenvies the physicists who research colliding particles.
  • Bhas personally observed colliding particles.
  • Cadmires the powerful reactions that occur when particles collide.
  • Dhas a simplistic understanding of how particles collide.
Show the answer

Correct answer: C

The question asks what the imagery in lines 27–28 reveals about the speaker.

A. Incorrect. The physicists mentioned in the fifth stanza (lines 33–44) are described as mere observers: they “jot down / what they can” (lines 41–42). In lines 27–28, the speaker is focused on the reaction between the colliding particles, not on the physicists who study them.

B. Incorrect. The speaker describes the particle collision in detail, but the text does not imply that the speaker has personally witnessed a collision. The speaker implies that his or her understanding of the collision is based on the experiences of the physicists observing the phenomenon (“They spin, and it is only down there / in the darkness—in the vast garage / where physicists jot down / what they can, whatever seems most real— / that they let us perceive their wild dancing” [lines 39–43]).

C. CORRECT. The word “explosion” and the phrase “briefly incandescent” create a picture of rapid changes and short bursts of bright light, revealing how powerful the new particles are. The language the speaker uses is positive, revealing that the speaker admires the power of the reaction and stands in awe of the new particles.

D. Incorrect. The poem reveals the speaker’s deep knowledge about subatomic particles. For example, the speaker is able to name the new particles discovered—“the quarks, the leptons, and the bosons; / the baryons and the mesons” (lines 29–30)—and is able to use imagery to create a powerful picture that reveals the complexity of a scientific process that humans have never been able to see.

Question analysis

This imagery question asks what lines 27–28 of “Looking for the Smallest Spark of Everything” reveal about the poem’s speaker, as particles appear “briefly incandescent” in an explosion inside a particle collider. Students must infer what a vivid image says about the person describing it. The word “incandescent” may be unfamiliar, and the choices are about very different things.

NYS standards
  • 7R4 word meaning and choice
  • 7R6 point of view and author position
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 students infer a speaker’s attitude from imagery, but these two lines turn on the uncommon word “incandescent,” state nothing about the speaker outright, and the choices range over the speaker’s knowledge, experience and feelings, 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 one paragraph or stanza
Reading level
Passage: 346 words (a poem; reading level not measured); question and choices: about grade 9, 55 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Nothing in the two lines names a feeling, so students must judge the speaker’s attitude from the image alone, and its key word is uncommon. The vividness is itself a trap: a scene described this sharply can seem to be one the speaker watched, though the poem says the particles are perceived only through the physicists’ work. The choices range over envy, experience, admiration and understanding, so each must be tested against the lines rather than against an overall impression.

  1. Reading between the lines major Working it out

    The lines describe particles, not the speaker, and contain no word of feeling; admiration has to be inferred from the drama of “the explosion” and the glow of “briefly incandescent.” (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

  2. Hard words moderate Understanding the question

    “Incandescent,” glowing with heat and light, is the word that makes the image dazzling, and a student who does not know it sees only an explosion in the dark. (Perfetti, 2007; Freedle & Kostin, 1993)

  3. Figurative or literary language minor Understanding the question

    The two lines are pure imagery, a burst of light in darkness as the particles “appear,” and the question asks what that picture reveals about the person painting it. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

Strategy: Picture the scene in lines 27–28, including what “incandescent” means, decide what the description suggests about the speaker, and test each choice against these two lines.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This asks what the imagery in lines 27–28 shows about the speaker. Imagery is language that helps you see, hear or feel a scene. The lines describe particles, not the speaker. So you must infer, or figure out from clues, how the speaker feels.

  2. Read lines 27–28

    The lines say “And there, in the explosion, in the darkness, / briefly incandescent, they appear:” Briefly means for a short time. Incandescent means glowing with heat and light. So picture a burst of bright light flashing in the dark. The particles have just crashed into each other, or collided. In that flash, the tiny particles appear.

  3. Decide how the speaker feels

    The words are bold and bright: “explosion,” “incandescent,” “appear.” The speaker makes the moment sound dramatic and amazing. People describe something this way when they are in awe of it. So the speaker admires, or looks up to, the power of the crash.

  4. Check each choice

    • A does not fit. Physicists come up only later, in line 41. Nothing shows envy, or wanting what they have.
    • B does not fit. Nothing says the speaker was there. Lines 39–42 place the sight “down there,” where “physicists jot down / what they can.”
    • C fits. The glowing burst in the dark shows the speaker’s wonder.
    • D does not fit. Right after, the speaker names quarks, leptons, bosons, baryons and mesons. That is not a simple understanding.

    Watch out: B is tempting because the scene is so vivid. But a writer can picture something without seeing it in person.

  5. Choose the answer

    The bright, explosive picture shows a speaker who is amazed by the crash. So the answer is C, “admires the powerful reactions that occur when particles collide.”

Why this answer and not the closest wrong one

B is the closest wrong choice, because the scene feels like an eyewitness report. But the lines never say the speaker was in the tunnel. What they do show is a feeling. Words like “explosion” and “briefly incandescent” make the moment thrilling. That is admiration, so C fits.

More questions like this

The original DOE question
Question 25 as shown on the DOE’s digital test, page 36 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B English, page 36, question 25.
The passage “Looking for the Smallest Spark of Everything” as printed (pages 32–33)The passage “Looking for the Smallest Spark of Everything” as printed on page 32 of the DOE handbook.The passage “Looking for the Smallest Spark of Everything” as printed on page 33 of the DOE handbook.

Question 4 of 6

2018 Test B Math, question 42

Question

There are 1,000 cubic centimeters in 1 liter, and 1,000 cubic millimeters in 1 milliliter. How many cubic millimeters are there in 1,000 cubic centimeters?

  • A1,000
  • B10,000
  • C100,000
  • D1,000,000
Show the answer

Correct answer: D

If 1 L = 1,000 cu cm, then 1 L = 1,000 mL. Set up a proportion, letting xx = the amount of cubic millimeters in 1,000 cubic centimeters.

1,000 cu mm1 mL=x cu mm1,000 mL\frac{1{,}000\text{ cu mm}}{1\text{ mL}}=\frac{x\text{ cu mm}}{1{,}000\text{ mL}}

Solve for xx: 1,000,000 cubic millimeters are in 1,000 cubic centimeters.

Question analysis

The question gives two metric facts, cubic centimeters per liter and cubic millimeters per milliliter, and asks how many cubic millimeters are in 1,000 cubic centimeters. It tests chaining unit conversions with volume units. The hard part is that the two facts do not connect without a link the question leaves out.

NYS standards
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 6 and 7 students convert units with one given factor. This needs three linked conversions, one using a fact the question does not state, and involves cubic units, which most grade 7 students find unfamiliar.
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 13, 20 words.
Corrected from the DOE handbook
  • The units in the explanation’s proportion were corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives two conversion facts that do not connect until the student supplies the link between liters and milliliters. Students who instead reach for 1 centimeter = 10 millimeters often scale a volume like a length or an area and land short by a factor of 10 or 100. Writing the unit at every step, so the units cancel, keeps both factors of 1,000 in view.

  1. Something you must already know moderate Working it out

    The chain needs 1 liter = 1,000 milliliters, which the stem never states, yet it is the only link between its two conversion facts. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Three conversions in a row, cubic centimeters to liters, liters to milliliters and milliliters to cubic millimeters, multiply out to 1,000 × 1,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Build a chain from cubic centimeters to liters, liters to milliliters, and milliliters to cubic millimeters, writing the units at every step so they cancel.

Traps

  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s proportion is 1,000 cu mm / 1 L = x cu mm / 1,000 cu mm. But 1,000 cubic millimeters make 1 milliliter, not 1 liter, and the second denominator should be the 1,000 mL that equal 1,000 cubic centimeters. The site uses 1,000 cu mm / 1 mL = x cu mm / 1,000 mL, which gives x = 1,000,000 cubic millimeters.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube
V=s×s×sV = s \times s \times s
V is the volume, the space inside the cube. s is the length of one side. A cube has the same length, width and height, so you multiply the side by itself three times.

  1. Sort out what you know

    You are told two facts. 1 liter (L) is 1,000 cubic centimeters (cu cm). 1 milliliter (mL) is 1,000 cubic millimeters (cu mm). You need the number of cu mm in 1,000 cu cm.

    One fact is missing. 1 liter is 1,000 milliliters. The word milli means one thousandth.

  2. Change cubic centimeters to liters

    The first fact says 1,000 cu cm is exactly 1 liter. So the amount in the question is 1 liter of space.

  3. Change liters to milliliters

    One liter is 1,000 milliliters. So 1,000 cu cm is the same as 1,000 mL.

    Watch out: you are not done yet. The question asks for cubic millimeters. Stopping at 1,000 gives choice A.

  4. Change milliliters to cubic millimeters

    Each milliliter is 1,000 cu mm. There are 1,000 milliliters, so multiply.

    1,000×1,000=1,000,0001{,}000 \times 1{,}000 = 1{,}000{,}000

    So 1,000 cu cm is 1,000,000 cu mm. The answer is D, 1,000,000.

Check your answer

Use the cube formula. A cube 1 cm on each side is 10 mm on each side. Its volume is V=10×10×10=1,000V = 10 \times 10 \times 10 = 1{,}000 cu mm. So 1,000 cu cm is 1,000×1,000=1,000,0001{,}000 \times 1{,}000 = 1{,}000{,}000 cu mm. Using only one 10 gives choice B. Using two 10s gives choice C.

More questions like this

The original DOE question
Question 99 as printed on page 160 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 99. Open page 160 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 5 of 6

2026 Practice Test C Math, question 10

Question

Two parallel lines, n and p, rise from lower left to upper right, and both cross a horizontal line m. Where line n crosses line m, the angle above line m, between the left part of m and the upper part of n, is marked r°. Farther right, a vertical segment runs from line p down to line m, meeting m at a right angle (marked with a small square); the angle at the top, between line p and the vertical segment, is marked 73°.

In the figure above, n∥pn \parallel p. Straight line m intersects both line n and line p. What is the value of r?

  • A17
  • B117
  • C163
  • D173
Show the answer

Correct answer: C

The triangle formed by line p, line m, and the line segment connecting the two contains a 73° angle, a right angle (90°), and a missing angle measuring x°. The sum of the interior angles of a triangle is 180°, so begin by finding the value of x:

180∘=73∘+90∘+x180^{\circ} = 73^{\circ} + 90^{\circ} + x

180∘=163∘+x180^{\circ} = 163^{\circ} + x

17∘=x17^{\circ} = x

Line m is a straight line, so it has a measure of 180°. The 17° angle from the triangle and the adjacent angle outside the triangle are supplementary, so they have a sum of 180°. Let s represent the measure of the adjacent angle outside the triangle. The value of s can be determined by subtracting:

s=180∘−17∘s = 180^{\circ} - 17^{\circ}

s=163∘s = 163^{\circ}

Given that n and p are parallel lines and that they have been cut by a transversal, line m, then corresponding angles are congruent. Angles r and s are corresponding angles, so the value of r is congruent to the value of s, which is 163°.

Question analysis

Two parallel lines, n and p, cross a straight line m, and a segment drawn at a right angle to m forms a small triangle with lines m and p, where a 73° angle is marked. Students find the angle r at line n. It combines a triangle's angle sum, a straight angle and parallel lines, and the angle wanted sits away from the marked ones.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle problems use supplementary, vertical and adjacent angles. This one also needs the angle sum of a triangle and the angles formed by parallel lines and a transversal, grade 8 facts, chained across a busy figure.
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 3, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each fact needs only a subtraction, but the student has to plan a route from the 73° angle at line p over to r at line n. The first number the route gives, the triangle's 17° angle, is listed, and it is the acute partner of r rather than r itself. The figure helps here: r is drawn plainly obtuse, and 163 is also 90 + 73, the triangle's exterior angle, which confirms it.

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

    The 73° sits between line p and the vertical segment, not between p and m, and r is the obtuse angle above m on the left of line n, far from every marked angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Three facts chain together: the triangle's angles sum to 180°, the angles along straight line m sum to 180°, and corresponding angles at the parallel lines n and p are equal. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Something you must already know minor Working it out

    The angle pairs formed by a line crossing two parallel lines are a grade 8 topic, and the question uses them without naming them. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of the right triangle, use the straight line m to get the angle beside it, then carry that angle over to line n using the parallel lines.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles of a triangle
a+b+c=180∘a + b + c = 180^{\circ}
a, b and c are the three angles inside any triangle. They always add up to 180°. If you know two of them, subtract their sum from 180° to find the third.
Angles on a straight line
a+b=180∘a + b = 180^{\circ}
a and b are two angles side by side on a straight line. Together they make a half turn, which is 180°. Use it to find the angle next to one you already know.
Corresponding angles
n∥p⇒corresponding angles are equaln \parallel p \Rightarrow \text{corresponding angles are equal}
The symbol ∥ means is parallel to. When line m crosses two parallel lines, it makes the same angles at each crossing. Angles in the same position at the two crossings are called corresponding angles. They are equal.

  1. Sort out the question

    Lines n and p are parallel, and line m crosses both. Near line p, a straight-up segment meets line m. The small square shows a right angle, 90°. Line p, line m and that segment make a small triangle. Its top angle is 73°.

    You need r, the wide angle where line n crosses line m.

  2. Find the triangle’s third angle

    The third angle is where line p crosses line m. Call it x. Use the triangle rule:

    73+90+x=18073 + 90 + x = 180

    x=180−163=17x = 180 - 163 = 17

    So that angle is 17°.

    Watch out: this is not r. Stopping here gives choice A, 17.

  3. Use the straight line m

    The 17° angle sits above line m, on the right of line p. Beside it, on the left of line p, is a wide angle. Call it s. The two angles fill the straight line m, so they add to 180°:

    s=180−17=163s = 180 - 17 = 163

    So s is 163°.

    Watch out: borrow carefully. 180 − 17 is 163, not 173, choice D.

  4. Carry the angle to line n

    Angle s is above line m and to the left of line p. Angle r is above line m and to the left of line n. They are in the same position, so they are corresponding angles. Since n and p are parallel:

    r=s=163r = s = 163

    So the answer is C, 163.

Check your answer

Look at the figure. Angle r is wider than a right angle, so it must be more than 90°. That rules out 17. Also, the narrow angle beside r, on the right of line n, matches the 17° angle at line p. And 163+17=180163 + 17 = 180, as it should on a straight line.

More questions like this

The original DOE question
Question 10 as shown on the DOE’s digital test, page 83 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 83, question 10.

Question 6 of 6

2026 Practice Test C Math, question 35

Question

If 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a, what is the value of a in terms of b?

Enter your answer in the space provided. Enter only your answer.

On the test you build the answer on an on-screen keypad; here, type it. Use ^ for a power, sqrt( ) or √ for a root, pi for π, and <= or >= for ≤ and ≥.

Show the answer

Correct answer: a = 10b

To find the value of a in terms of b, solve 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a for a. Begin by distributing 4 on the left side of the equation:

(4×a)+(4×2b)=28b+2a(4 \times a) + (4 \times 2b) = 28b + 2a

4a+8b=28b+2a4a + 8b = 28b + 2a

Then, get the a terms on one side and the b terms on the other:

4a−2a=28b−8b4a - 2a = 28b - 8b

Solve for the value of a in terms of b by combining like terms and then dividing each side by 2:

2a=20b2a = 20b

a=10ba = 10b

Question analysis

An algebra question gives an equation with a and b on both sides, one side written as 4 times a sum in parentheses, and students type a in terms of b after a printed "a =". It tests the distributive property and solving linear equations. The answer is typed, so a slip in distributing or moving terms is not caught by any choices.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve equations like 4(x + 2) = 28 with numbers. This one has letters on both sides, needs the distributive property and like terms, and asks for a in terms of b, which is grade 8 algebra.
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 4, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, yet there are four of them, each with a classic slip, and the typed answer catches none. Multiplying the 4 into only the a, or carrying 8b across without changing its sign, gives another whole multiple of b that looks finished. Testing b = 1, where 4(a + 2) and 28 + 2a must come out equal, confirms an entry quickly.

  1. Variables instead of numbers moderate Understanding the question

    Both letters appear on both sides, and the answer must stay a multiple of b, so there is no number to check an entry against unless students invent one. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Many steps moderate Working it out

    Four moves run in a chain, distributing the 4, gathering the a terms, gathering the b terms and dividing by 2, and an error at any one carries into the typed result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. An answer in an unexpected form minor Choosing the answer

    The answer is typed after a printed “a =”, with no choices, so 20b, the line before dividing by 2, can be entered without anything on screen questioning it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute the 4 to both terms in the parentheses, gather the a terms on one side and the b terms on the other, then divide by the coefficient of a.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
k(x+y)=kx+kyk(x + y) = kx + ky
k is the number outside the parentheses. x and y are the terms inside. The number outside multiplies every term inside, not just the first one. It works because 4 groups of (x + y) hold 4 x’s and 4 y’s.
Doing the same thing to both sides
if A=B, then A−C=B−C and A2=B2\text{if } A = B, \text{ then } A - C = B - C \text{ and } \frac{A}{2} = \frac{B}{2}
A and B are the two equal sides of an equation. C is any amount. Taking the same amount from both sides, or dividing both by the same number, keeps them equal, like a balanced scale.

  1. Sort out the question

    The equation is 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a. You need a in terms of b. That means a alone on one side, with only b and numbers on the other.

    The answer box already shows a =, so you type only what comes after it.

  2. Multiply out the parentheses

    Use the distributive property. The 4 multiplies both terms inside:

    4×a+4×2b=28b+2a4 \times a + 4 \times 2b = 28b + 2a

    4a+8b=28b+2a4a + 8b = 28b + 2a

    Watch out: multiplying only the a by 4 gives 4a + 2b. That leads to the wrong answer 13b.

  3. Gather the a terms

    Like terms are terms with the same letter, such as 4a and 2a. Put the a terms on the left. Subtract 2a from both sides:

    4a−2a+8b=28b4a - 2a + 8b = 28b

    2a+8b=28b2a + 8b = 28b

  4. Gather the b terms

    Now subtract 8b from both sides:

    2a=28b−8b2a = 28b - 8b

    2a=20b2a = 20b

    Watch out: moving 8b without changing its sign gives 28b+8b=36b28b + 8b = 36b. That leads to 18b, which is wrong.

  5. Divide both sides by 2

    2a means 2 times a. Divide both sides by 2 to get a alone:

    a=20b2=10ba = \frac{20b}{2} = 10b

    Watch out: 20b is equal to 2a, not a. Do not stop there.

  6. Enter it after the printed a =

    On the test, the box already shows a =. Press 1, then 0, then the b key. You do not need a times sign. Do not type a = again. On this site, type 10b in the box.

    So you would enter 10b, which makes a = 10b.

Check your answer

Pick b = 1, so a = 10. The left side is 4(10+2)=4×12=484(10 + 2) = 4 \times 12 = 48. The right side is 28+2×10=4828 + 2 \times 10 = 48. The sides match, so 10b works.

More questions like this

The original DOE question
Question 35 as shown on the DOE’s digital test, page 108 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 108, question 35.

Sample #525687