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Six SHSAT Questions

3 English and 3 math

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Question 1 of 6

2018 Test B English, question 48
Passage Invention of the Telegraph

Samuel Morse, an American inventor, is credited with creating the electronic telegraph, a communication device that allows users to send messages using a system of short and long pulses that represent letters, numbers, and punctuation. In 1843 the United States Congress passed the Telegraph Bill, which provided Morse with the funds to build an electric telegraph system.

Invention of the Telegraph

Earlier Signal Systems

Long before Samuel F. B. Morse electrically transmitted his famous message “What hath God wrought?” from Washington to Baltimore on May 24, 1844, there were signaling systems that enabled people to communicate over distances. Most were visual or “semaphore” systems using flags or lights. In the eighteenth century, such systems used an observer who would decipher a signal from a high tower on a distant hill and then send it on to the next station. The young American republic wanted just such a system along its entire Atlantic coast and offered a prize of $30,000 for a workable proposal. The framers of this legislation1^1 had no way of knowing that when they used the word “telegraph” to refer to this visual semaphore system, they would be offered an entirely new and revolutionary means of communication—electricity.

The Growth of an Idea

The idea of using electricity to communicate over distance is said to have occurred to Morse during a conversation aboard ship when he was returning from Europe in 1832. Michael Faraday’s recently invented electromagnet was much discussed by the ship’s passengers, and when Morse came to understand how it worked, he speculated that it might be possible to send a coded message over a wire. While a student at Yale College years before, he had written his parents a letter about how interesting he found the lectures on electricity. Despite what he had learned at Yale, Morse found when he began to develop his idea that he had little real understanding of the nature of electricity, and after sporadic attempts to work with batteries, magnets, and wires, he finally turned for help to a colleague at the University of the City of New York, Leonard D. Gale.

Gale was a professor of chemistry and familiar with the electrical work of Princeton’s Joseph Henry, a true pioneer in the new field. Well before Morse had his shipboard idea about a telegraph, Henry rang a bell at a distance by opening and closing an electric circuit. In 1831, he had published an article, of which Morse was unaware, that contained details suggesting the idea of an electric telegraph. Gale’s help and his knowledge of this article proved crucial to Morse’s telegraph system because Gale not only pointed out flaws in the system but showed Morse how he could regularly boost the strength of a signal and overcome the distance problems he had encountered by using a relay system Henry had invented. Henry’s experiments, Gale’s assistance, and, soon after, hiring the young technician Alfred Vail were keys to Morse’s success.

Obstacles and Opportunities

By December 1837, Morse had enough confidence in his new system to apply for the federal government’s appropriation, and during the next year he conducted demonstrations of his telegraph both in New York and Washington.

However, when the economic disaster known as the Panic of 1837 took hold of the nation and caused a long depression, Morse was forced to wait for better times. It was during this period that Morse visited Europe again and tried not only to secure patent protection overseas but to examine competing telegraph systems in England. . . .

By 1843, the country was beginning to recover economically, and Morse again asked Congress for the $30,000 that would allow him to build a telegraph line from Washington to Baltimore, forty miles away. The House of Representatives eventually passed the bill containing the Morse appropriation, and the Senate approved it in the final hours of that Congress’s last session. With President Tyler’s signature, Morse received the cash he needed and began to carry out plans for an underground telegraph line.

Realizing a Great Invention

Morse had hired the ingenious construction engineer Ezra Cornell to lay the pipe carrying the wire, and although Cornell did his job superbly, one of Morse’s partners, Congressman F. O. J. Smith, had purchased wire with defective insulation. Too much time had been wasted laying bad wire, and with the project on a rigid deadline, something had to be done quickly. Cornell suggested that the fastest and cheapest way of connecting Washington and Baltimore was to string wires overhead on trees and poles. The desperate Morse gave the go-ahead, and the line was completed in time for the dramatic and spectacularly successful link between the Supreme Court chamber of the Capitol building and the railroad station in Baltimore.

Soon, as overhead wires connected cities up and down the Atlantic coast, the dots-and-dashes method2^2 that recorded messages on a long moving strip of paper was replaced by the operator’s ability to interpret the code in real time. . . . Telegraph lines soon extended westward, and within Morse’s own lifetime they connected the continents of Europe and America.

“Invention of the Telegraph”—Public Domain/Library of Congress

1^1legislation: Telegraph Bill

2^2dots-and-dashes method: the short and long pulses of Morse code that are sent and received by telegraph operators

World map titled “British All-Red Telegraph Line, 1902,” with a compass rose (N, S, E, W) and a key showing a thick black line as an electric telegraph line. The line runs from London across the Atlantic Ocean and Canada to Vancouver, then west into the Pacific Ocean off the map’s left edge; from London and from Canada’s east coast to Bermuda, then to Barbados, across the Atlantic to Cape Town, and across the Indian Ocean to Perth, which a second line links to India; and from Perth across Australia to Brisbane and Norfolk, with branches to New Zealand and east into the Pacific off the map’s right edge. Washington, D.C., in the United States, is marked with a dot but no line reaches it.

The first transatlantic electric telegraph message was sent in 1858, and by 1902 the British All-Red Line connected most of the world.

Bar graph titled “Telegraph Messages Transmitted, 1870–1920,” with the number of messages transmitted, in millions, on the vertical axis (0 to 180 in steps of 20) and the years 1870, 1880, 1890, 1900, 1910 and 1920 on the horizontal axis. The bars show about 10 million messages in 1870, 30 million in 1880, 55 million in 1890, 65 million in 1900, 75 million in 1910 and 155 million in 1920.

Source: U.S. Bureau of the Census. Historical Statistics of the United States, Colonial Times to 1970, Bicentennial Edition, Part 2. Washington, D.C., 1975.

Question

Which statement describes how the author’s use of problem-solution in paragraph 2 contributes to the development of ideas in the passage?

  • AMorse’s discussion on a ship about Faraday’s electromagnet reminded him of the Yale College lectures on electricity, which he had enjoyed but had not fully understood, inspiring him to learn more about electricity from his colleague Gale.
  • BMorse’s discouragement over his lack of knowledge of electricity prompted him to experiment with batteries, magnets, and wires, which led to the development of a new long-distance communication system.
  • CMorse’s difficulty in understanding how Faraday’s electromagnet worked was frustrating, and it pushed him to create a system for sending signals over wires.
  • DMorse’s longtime fascination was not enough to make up for his lack of knowledge about electricity, so he eventually sought help from Gale.
Show the answer

Correct answer: D

The question asks for the best explanation of how the use of problem-solution in paragraph 2 helps develop the ideas in the passage. Option D is correct because it explains that Morse realized his problem was that he was not knowledgeable enough to continue his experimentation without help, so the solution was to seek help from a colleague.

Option A is incorrect because, while the discussion on the ship fascinated Morse and reminded him of the lectures he had attended at Yale College, it was his failed attempts with electricity that caused him to reach out to Gale for help.

Option B is incorrect because, while “Morse found when he began to develop his idea that he had little real understanding of the nature of electricity” (paragraph 2), the paragraph does not state that he was discouraged; it was his failed experimentation that resulted in him first realizing that he needed help.

Option C is incorrect because the paragraph states that “Morse came to understand how [Faraday’s electromagnet] worked,” and that understanding is what caused him to think about sending messages over a wire. Also, Morse’s difficulty was in understanding the properties of electricity in general once he began his experimentation.

Question analysis

In this text structure question about “Invention of the Telegraph,” students explain how the problem-and-solution pattern in paragraph 2 develops ideas about Samuel Morse. All four choices are long sentences built from real details about the ship, Yale and Gale, so students must check each one’s account of what led to what.

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 Identifying problem and solution is familiar from earlier grades, but these choices are long, detail-packed chains in which a single misplaced cause makes an option wrong, which demands closer checking than typical grade 7 items.
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 11, 910 words; question and choices: about grade 16, 132 words.
Corrected from the DOE handbook
  • A wrong paragraph number in the explanation was corrected.
  • A misspelled name in the explanation was corrected.
  • A punctuation slip in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every choice is assembled from things the paragraph really mentions, so recognition is useless and only the order of causes decides. The most tempting wrong choice ends with Gale, the right solution, but invents a reminder of the Yale lectures to get there; others turn the failed experiments into the solution or reverse what Morse understood. The question rewards reading each long choice clause by clause, which costs time.

  1. Every choice must be checked major Choosing the answer

    Each choice is a chain of two or three causes built from real details, the ship, Faraday, Yale, the batteries and wires, Gale, so every link has to be checked against paragraph 2’s order of events. (Embretson & Wetzel, 1987)

  2. A very close wrong answer moderate Choosing the answer

    The Yale-reminder choice and the key both join Morse’s enjoyment of electricity, his weak grasp of it and his turn to Gale; they part on one detail, the claim that the shipboard talk “reminded him” of Yale, which the paragraph never makes. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. The right answer is reworded moderate Choosing the answer

    The key turns the Yale letter and the shipboard speculation into “longtime fascination,” a phrase the passage never uses, while the wrong choices repeat its names and phrases and so look better supported. (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)

  4. A lot to read moderate Understanding the question

    The four choices run to about 110 words, the longest 37, and paragraph 2 itself packs a voyage, a college letter, failed experiments and a colleague into four sentences. (Kintsch, 1998; Gorin & Embretson, 2006; Freedle & Kostin, 1993)

  5. How, not what minor Working it out

    Students must name the paragraph’s problem, too little understanding of electricity, and its solution, Gale, before judging which account of the pattern is accurate. (Anderson & Krathwohl, 2001)

Strategy: Name the problem and the solution in paragraph 2 in a few words each, then choose the option whose sequence of events matches the paragraph exactly.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook places the sentence “Morse found when he began to develop his idea that he had little real understanding of the nature of electricity” in paragraph 3. It is in paragraph 2, the paragraph the question is about, so the site now says “(paragraph 2).”
  • Explanation: The handbook spells Leonard Gale’s name “Gail.” The passage spells it “Gale,” and so does the site now.
  • Explanation: The handbook capitalizes “Also” after a semicolon (“over a wire; Also, Morse’s difficulty…”). The site starts a new sentence instead: “over a wire. Also, Morse’s difficulty…”

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 problem-solution in paragraph 2. Problem-solution is a way to organize writing. First you learn about a problem, then how someone solved it. Find the problem and the solution. Then pick the choice that tells them in the right order.

  2. Find the problem in paragraph 2

    Paragraph 2 starts with “The idea of using electricity to communicate over distance.” It says Morse had enjoyed the electricity lectures at Yale. But then comes the problem: “Morse found when he began to develop his idea that he had little real understanding of the nature of electricity.”

  3. Find the solution

    The same sentence goes on: “after sporadic attempts to work with batteries, magnets, and wires, he finally turned for help to a colleague.” “Sporadic” means now and then, not steady. A colleague is a coworker. So Morse was interested in the subject, tried on his own, and did not succeed. His interest was not enough. His solution was to ask Gale for help.

  4. Check each choice

    • A is wrong. It ends with Gale, but the paragraph never says the ship talk “reminded him” of Yale. It also skips the failed tries.
    • B is wrong. It has the order backward. The experiments showed the problem. They did not build the system.
    • C is wrong. The paragraph says Morse “came to understand how it worked,” meaning the electromagnet.
    • D fits. Long interest was not enough, so he got help from Gale.

  5. Choose the answer

    The problem was Morse’s weak grasp of electricity, and the solution was Gale. So the answer is D, “Morse’s longtime fascination was not enough to make up for his lack of knowledge about electricity, so he eventually sought help from Gale.”

Why this answer and not the closest wrong one

A is the closest trap. Like D, it mentions Yale, Morse’s weak grasp of electricity and Gale. But it adds a link the paragraph never makes, that the ship talk reminded him of Yale. With long choices, check every part. One made-up link makes the whole choice wrong.

More questions like this

The original DOE question
Question 48 as printed on page 149 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 149, question 48. Open page 149 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). This is the DOE’s version, before the corrections listed in the analysis.
The passage “Invention of the Telegraph” as printed (pages 146–148)The passage “Invention of the Telegraph” as printed on page 146 of the DOE handbook.The passage “Invention of the Telegraph” as printed on page 147 of the DOE handbook.The passage “Invention of the Telegraph” as printed on page 148 of the DOE handbook.

Question 2 of 6

2018 Test B English, question 40
Passage Snowy Mountains

Snowy Mountains

by John Gould Fletcher

Higher and still more high,
Palaces made for cloud,
Above the dingy city-roofs
Blue-white like angels with broad wings,
Pillars of the sky at rest
The mountains from the great plateau
Uprise.

But the world heeds them not;
They have been here now for too long a time.
The world makes war on them,
Tunnels their granite cliffs,
Splits down their shining sides,
Plasters their cliffs with soap-advertisements,
Destroys the lonely fragments of their peace.

Vaster and still more vast,
Peak after peak, pile after pile,
Wilderness still untamed,
To which the future is as was the past,
Barrier spread by Gods,
Sunning their shining foreheads,
Barrier broken down by those who do not need
The joy of time-resisting storm-worn stone,
The mountains swing along
The south horizon of the sky;
Welcoming with wide floors of blue-green ice
The mists that dance and drive before the sun.

“Snowy Mountains” by John Gould Fletcher—Public Domain

Question

Read line 5 from the poem.

Pillars of the sky at rest

The line helps develop the theme of the poem by suggesting that the mountains

  • Aserve a noble and supportive purpose in the world.
  • Battract the clouds with their strength and permanence.
  • Cremain untamed through the ages.
  • Dprovide protection for the people.
Show the answer

Correct answer: A

The question asks how the line from the poem helps develop the theme of the poem. Line 5 helps develop the theme that the mountains are more important than people realize by suggesting that the mountains serve a noble, foundational purpose in the world, as explained in Option A.

Option B is incorrect because, while the poem does describe the strength and permanence of the mountains, the line does not suggest that these traits attract the clouds.

Option C is incorrect because, while the line states that the highest peaks are “at rest,” that phrase describes stillness, not wildness; the poem calls the mountains “still untamed” only later (line 17), and line 5 does not suggest it.

Option D is incorrect because, while pillars are strong and may indicate shelter, there is no suggestion in the line that the mountains are protecting the people. Additionally, this concept is not a theme of the poem.

Question analysis

This figurative language question asks how the metaphor “Pillars of the sky at rest” in “Snowy Mountains” helps develop the poem’s theme. Students must work out what calling mountains pillars of the sky suggests about them, and it is tricky because every choice sounds like something this poem could say.

NYS standards
  • 7R4 word meaning and choice
  • 7R2 theme or central idea
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 students interpret metaphors, but here they must also connect a short image to the poem’s theme and choose among abstract options, a step closer to grade 8 work on figurative meaning and theme.
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: 161 words (a poem; reading level not measured); question and choices: about grade 3, 52 words.
Corrected from the DOE handbook
  • A claim in the explanation that the poem contradicts was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is hard because every choice says something the poem could plausibly mean, and two of them borrow real phrases from other lines. Students who test choices against the poem as a whole find support for the wrong ones; the question rewards testing them against the single image of pillars holding up the sky. Seeing that pillars both support and ennoble a structure leads directly to the key.

  1. Figurative or literary language major Understanding the question

    The answer depends on unpacking one metaphor: pillars hold up a roof, so “Pillars of the sky” casts the mountains as supports of the heavens, and “at rest” adds a calm, dignified strength. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

  2. The right answer is reworded moderate Choosing the answer

    The key uses words the poem never does, “noble” and “supportive,” while wrong choices echo “cloud” from line 2 and “untamed” from line 17 and so look anchored in the text. (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)

  3. Combines ideas minor Working it out

    Students must interpret the image and then link it to the poem’s theme, that the mountains matter more than the world below recognizes. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Ask what a pillar is and why a poet might call mountains “pillars of the sky,” then test each choice against line 5 alone before thinking about the rest of the poem.

Traps

  • Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
  • Half right: One half of a two-part choice is right and the other is not.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook rejects Option C by saying the mountains “do not remain untamed or untouched,” but line 17 calls them “Wilderness still untamed,” and the handbook’s own explanation of the third-stanza question relies on that line. The site now says that “at rest” describes stillness, not wildness, and that the poem calls the mountains “still untamed” only later, in line 17, so line 5 does not suggest it.

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 one image, line 5, adds to the poem’s theme. A theme is the big message of a poem. Here the theme is that the mountains are great and important, even though people ignore them. You need to decide what “Pillars of the sky” says about the mountains.

  2. Look closely at line 5

    Line 5 reads “Pillars of the sky at rest.” It is part of the first stanza’s list of grand pictures. The list also has “Palaces made for cloud” in line 2. Line 5 is a metaphor. A metaphor calls one thing by the name of another to compare them, without using “like” or “as.”

  3. Explain what pillars do

    A pillar is a tall, strong column that holds up a roof. Big, important buildings, like temples and courthouses, have pillars. So calling the mountains “Pillars of the sky” makes them seem to hold up the sky. That is a supporting job. It is also a grand, noble one. “At rest” adds calm. The pillars stand still and steady.

  4. Check each choice

    • A fits. Holding up the sky is a noble, supportive job.
    • B is wrong. Clouds come from line 2. Line 5 says nothing about pulling clouds in.
    • C is wrong. “Wilderness still untamed” is line 17, not line 5. Pillars “at rest” suggest calm, not wildness.
    • D is wrong. Pillars hold up roofs, and roofs can shelter people. But the line never shows the mountains protecting anyone.

  5. Choose the answer

    Pillars hold things up and make a building grand. So the answer is A, “serve a noble and supportive purpose in the world.”

Why this answer and not the closest wrong one

C is the strongest trap, because the poem really does call the mountains untamed. But that is line 17. Test each choice against line 5 alone. A pillar is a support, and a pillar of the sky is a grand one. Only A comes straight from that picture.

More questions like this

The original DOE question
Question 40 as printed on page 143 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 143, question 40. Open page 143 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). This is the DOE’s version, before the corrections listed in the analysis.
The passage “Snowy Mountains” as printed (page 142)The passage “Snowy Mountains” as printed on page 142 of the DOE handbook.

Question 3 of 6

2018 Test A English, question 22
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

Paragraph 7 contributes to the development of the central idea of the passage by

  • Aillustrating that Ederle’s physical strength and mental fortitude allowed her to stay focused on her goal.
  • Bconveying that Ederle pushed herself to the edge of her physical capabilities in order to complete the swim.
  • Chighlighting the impact the severe weather had on Ederle’s emotions during her swim.
  • Demphasizing the surge of emotions Ederle felt as she came closer to achieving a personal goal.
Show the answer

Correct answer: D

The answer is (D)

The question asks for an explanation of how paragraph 7 contributes to the development of a central idea of the passage. Paragraph 7 is mainly about Ederle’s physical and emotional state as she pushed through a difficult portion of her swim and realized her goal was in sight. A central idea of the passage is that Ederle’s passion and determination allowed her to accomplish her goal of being the first woman to cross the English Channel.

Option D is correct because it emphasizes her passion by referring to the overwhelming emotions Ederle felt as she progressed during her swim, despite the unfavorable conditions (paragraph 7: “For hours Ederle swam, dodging debris with an amused smile” and “Yet Ederle felt indescribably happy as she churned through the sea.”).

Option A is incorrect because, although Ederle needed physical strength and mental fortitude to complete the swim, the paragraph contributes to a central idea because it highlights her emotion as she swam, not her physical strength or mental fortitude.

Option B is incorrect because the paragraph focuses on Ederle’s emotional response to the challenging circumstances and not the idea that Ederle was at the “edge of her physical capabilities.”

Option C is incorrect because Ederle’s feelings of happiness and amusement were caused by her anticipation and excitement as she approached achieving her goal, not by the severe weather.

Question analysis

This reading question asks how the paragraph about the storm near the end of Gertrude Ederle's English Channel swim contributes to the passage's central idea. Students must decide what the paragraph mainly does and connect it to the larger point, while every choice sounds like a fair description of a hard swim.

NYS standards
  • 7R5 structure and genre features
  • 7R2 theme or central idea
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Explaining how one paragraph develops a central idea is a grade 7 standard, but here several choices sound true of a hard swim through a storm, and the student must see what the paragraph is mainly about, beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from several paragraphs or stanzas
Reading level
Passage: about grade 9, 790 words; question and choices: about grade 11, 76 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Paragraph 7 is crowded with hardship, a storm, a swollen tongue and inflamed ears, so the obvious reading is that it shows toughness. The author's pointed “Yet Ederle felt indescribably happy” turns it toward emotion, which ties it to the passion named in the thesis. Several choices are fair about the swim in general, and this is one of the closer calls on the test, since strength is also on display.

  1. A very close wrong answer major Choosing the answer

    The right choice and the strength-and-fortitude one both describe Ederle's inner state as she nears her goal, and only whether paragraph 7 stresses her feelings or her toughness separates them, since she does fight “the stubborn swells for several hours”. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Evidence in several places moderate Working it out

    The central idea, her “determination, innovative spirit, and passion for swimming”, is stated in paragraph 1, six paragraphs before the storm it must be connected to. (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)

  3. How, not what moderate Working it out

    The question asks how paragraph 7 develops the central idea, so the student must weigh what its details stress, the “amused smile” and “indescribably happy”, not just what happens. (Anderson & Krathwohl, 2001)

Strategy: Mark the words in paragraph 7 that describe how Ederle felt and what she endured, decide which the author stresses, then connect that to the passage's stated central idea.

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

    You need to explain how paragraph 7 helps build the central idea, the main point of the passage. So first find the central idea. Then see what paragraph 7 stresses.

  2. Find the central idea

    The last sentence of paragraph 1 says: “Gertrude Ederle’s triumphant swim across the English Channel was a testimony to her determination, innovative spirit, and passion for swimming.” A testimony here means proof. Passion means a strong love for something.

  3. See what happens in paragraph 7

    A fierce storm pushes Ederle backward. She fights “the stubborn swells for several hours.” Swells are big, rolling waves. Salt water makes her tongue swell and her ears sore. Then the storm ends, and the sea pushes her “toward the shore and victory.”

  4. Notice what the author stresses

    Look at the feeling words. She swims “dodging debris with an amused smile.” In the middle of the pain, the author writes: “Yet Ederle felt indescribably happy as she churned through the sea.” Indescribably means too much to put into words. The word “Yet” shows a surprise. She felt joy in spite of the pain. That shows her passion for swimming.

  5. Check each choice

    • D: a “surge of emotions” as she came closer to her goal. A surge is a sudden rush. This matches the smile and the happiness.
    • A: strength and “mental fortitude,” or toughness. She was tough, but the paragraph stresses her joy.
    • B: the paragraph never says she reached “the edge of her physical capabilities.”
    • C says the storm caused her feelings. She was happy in spite of the storm, not because of it.

    Watch out: A is tempting because she fought for hours. But the “Yet” sentence turns to how she felt.

  6. Pick the answer

    Paragraph 7 builds the passage’s idea of her passion by showing her feelings. So the answer is D.

Why this answer and not the closest wrong one

A is the closest wrong answer, because she does fight “the stubborn swells for several hours.” But the paragraph’s key words are about feeling: “amused smile” and “indescribably happy.” Those link to her “passion for swimming” in paragraph 1. D names those emotions.

More questions like this

The original DOE question
Question 22 as printed on page 59 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 59, question 22. Open page 59 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 4 of 6

2024 Test B Math, question 49

Question

Suppose that w, x, y, and z are integers, none of which are zero. Under which set of conditions would the product x2yz3|w|x^{2}yz^{3}|w| always be positive?

  • Ax is positive and w is positive.
  • Bx is negative and w is negative.
  • Cy is negative and z is negative.
  • Dx and y are negative, and z and w are positive.
Show the answer

Correct answer: C

First, analyze the role of each variable in the expression x2yz3|w|x^{2}yz^{3}|w|.

x2x^{2} will always be a positive number.

y can be a positive or a negative number.

z3z^{3} can be a positive or a negative number.

|w||w| will always be a positive number.

The expressions x2x^{2} and |w||w| will always be positive. Therefore, making the values of y and z negative would result in the expression yz3yz^{3} being positive, since a negative multiplied by a negative is positive. Thus the given product would always be positive.

Question analysis

Four nonzero integers appear in a product that includes a square, a cube and an absolute value. Students choose the set of conditions that makes the product positive every time. It tests the sign rules for multiplication together with exponents and absolute value, and the trap is judging the sign by counting negative variables without looking at what is done to each.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.1 whole-number exponent expressions
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students learn sign rules for products of signed numbers. Here they must reason about four unknown integers under an even power, an odd power and an absolute value, and decide what is always true, which is more abstract than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 6, 58 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One insight decides the question: x² and |w| are positive whatever x and w are, so only the signs of y and z³ matter. Students who count negative letters instead of negative factors pick conditions that look balanced but leave y as the only negative factor. Students who test a single set of values can be fooled by a choice that leaves y and z unspecified.

  1. Variables instead of numbers moderate Understanding the question

    There are four variables and no numbers, and always positive must hold for every allowed value, so one test case can rule a choice out but never confirm it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Each set of conditions has to be checked against all four factors; the first two fix only x and w and leave y and z free to make the product negative. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    Students must know that a nonzero square and an absolute value are always positive, while an odd power such as z³ keeps the sign of z. (Anderson & Krathwohl, 2001)

Strategy: Work out, factor by factor, which signs each part of the product can take, then test each set of conditions against every factor rather than counting negatives.

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.

  1. Sort out the question

    w, x, y and z are integers, meaning whole numbers or their negatives. None of them is 0. The product x2yz3|w|x^2 y z^3 |w| means x2×y×z3×|w|x^2 \times y \times z^3 \times |w|.

    You need the conditions that make the product positive every time, for every number that fits them.

  2. Look at x squared and |w|

    x2x^2 means x × x. Two positives or two negatives multiply to a positive. So x2x^2 is always positive.

    |w||w| is the absolute value of w, its distance from 0. It is always positive, since w is not 0.

    So x and w never change the sign of the product. Their signs do not matter.

  3. Look at y and z cubed

    y can be positive or negative. z3z^3 means z × z × z, and it has the same sign as z. For example, (−2)3=−8(-2)^3 = -8.

    So the sign of the whole product is the sign of y×z3y \times z^3. It depends only on y and z.

  4. Test each choice

    • A and B say nothing about y and z. If y = −1 and z = 1, the product is negative. So they do not always work.
    • C: y is negative, and z3z^3 is negative. A negative times a negative is positive. This always works.
    • D: y is negative and z is positive. So the product is always negative.

    Watch out: D has two negative letters, x and y. But x2x^2 is positive anyway, so y is the only negative factor.

    So the answer is C, y is negative and z is negative.

Check your answer

Try numbers for C. Let x = −2, y = −3, z = −1 and w = −5. Then x2×y×z3×|w|=4×(−3)×(−1)×5=60x^2 \times y \times z^3 \times |w| = 4 \times (-3) \times (-1) \times 5 = 60, which is positive. Even with x and w negative, the product stays positive.

More questions like this

The original DOE question
Question 106 as printed on page 213 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 106. Open page 213 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 A Math, question 44

Question

Number line with points W, X, Y and Z marked from left to right. W is at −10 and Z is at 8; a tick mark at 0 lies between X and Y. X and Y have no values given.

On the number line above, points WW, XX, YY, and ZZ are integers, and WX:XY:YZ=4:2:3WX:XY:YZ = 4:2:3. What is the value of WY\overline{WY}?

  • A8
  • B11
  • C12
  • D18
Show the answer

Correct answer: C

Convert the ratios into fractions of WZWZ. Use the sum of the ratios for the denominator.

WX:XY:YZ=4:2:3WX:XY:YZ=4:2:3

WX=44+2+3WZ=49WZWX=\frac{4}{4+2+3}WZ=\frac{4}{9}WZ

XY=24+2+3WZ=29WZXY=\frac{2}{4+2+3}WZ=\frac{2}{9}WZ

The part of WZWZ that is WYWY is the sum of those fractions:

WY=49WZ+29WZ=69WZ=23WZWY=\frac{4}{9}WZ+\frac{2}{9}WZ=\frac{6}{9}WZ=\frac{2}{3}WZ

Find the length of WZWZ:

WZ=8−(−10)=18WZ=8-(-10)=18

The value of WYWY is 23(18)=12\frac{2}{3}(18)=12.

Question analysis

This number line question places integer points W, X, Y and Z on a line and gives the ratio of the segments WX, XY and YZ as 4 to 2 to 3. Students find the whole length from W to Z, split it by the ratio, and add the parts that make up WY. Mixing up a part with the whole is the main risk.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students find distances on a number line and use part-to-whole ratios, usually separately. This item joins them: a negative endpoint, a three-term ratio and a choice of which segments to combine.
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 10, 21 words.
Corrected from the DOE handbook
  • Equations in the explanation that mixed up a fraction of a length with the length itself 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

Each step is routine, but the question keeps asking for a different piece of the line: the whole WZ, the part WX, and finally WY, which is two parts added together. Two of the wrong choices are exactly those other pieces. Students who mark the actual positions, X at −2 and Y at 2, can check the ratio before answering.

  1. Combines ideas moderate Working it out

    A three-part ratio, 4:2:3, must be applied to a length found from a negative endpoint, 8 − (−10) = 18, joining ratio and number-line distance. (Embretson, 1998; Daniel & Embretson, 2010)

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

    Only W at −10 and Z at 8 are labeled, and the drawing is not to scale: Y is drawn nearer 5 than 2, so reading WY from the picture fails. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Find WZ, split it into 9 equal parts of 2, then add the 4 parts of WX and the 2 parts of XY. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length WZ from the endpoints on the number line, divide it by the total number of ratio parts, and multiply by the parts from W to Y.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes WX = 4/9, XY = 2/9 and WY = 2/3, and then says “The value of WY is 2/3(18) = 12”. Those fractions are parts of WZ, not the lengths themselves (WX is 8 and WY is 12), so WY cannot be both 2/3 and 12. The lines now read WX = 4/9 WZ, XY = 2/9 WZ and WY = 2/3 WZ. The method and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
d=b−ad = b - a
d is the distance between two points. b is the point on the right and a is the point on the left. Subtracting left from right always gives a positive distance. It works even when a is negative.
Size of one part of a ratio
u=Lpu = \frac{L}{p}
u is the size of one part. L is the whole length being shared. p is the total number of parts, found by adding the numbers in the ratio. Every part is the same size, so you share the length equally.

  1. Sort out what you know

    W is at −10 and Z is at 8. X and Y are between them. The ratio 4:2:3 means WX is 4 parts, XY is 2 parts and YZ is 3 parts. All the parts are the same size.

    You need the length WY, from W to Y. The drawing is not exact, so use the numbers, not the picture.

  2. Find the whole length WZ

    d=8−(−10)=8+10=18d = 8 - (-10) = 8 + 10 = 18

    WZ is 18 units long. Subtracting a negative number is the same as adding.

  3. Find the size of one part

    Add the ratio numbers to get the number of parts:

    4+2+3=94 + 2 + 3 = 9

    Share the 18 units equally among 9 parts:

    u=189=2u = \frac{18}{9} = 2

    Each part is 2 units long.

  4. Find WY

    WY is WX and XY together, so it has 4 + 2 = 6 parts:

    6×2=126 \times 2 = 12

    Watch out: WX alone is 4 × 2 = 8, which is choice A. And 18 is the whole line, choice D. So the answer is C, 12.

Check your answer

Place the points. W is at −10. X is 8 units to the right, at −2. Y is 4 more, at 2. Z is 6 more, at 8, which matches its label. And the distance from −10 to 2 is 2 − (−10) = 12.

More questions like this

The original DOE question
Question 101 as printed on page 89 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 101. Open page 89 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 6 of 6

2026 Practice Test C Math, question 18

Question

I. s−ts - t
II. st2st^{2}
III. sts^{t}

If s is a positive integer and t is a negative integer, which of the above expressions must be a positive number?

  • AII only
  • BI and II only
  • CII and III only
  • DI, II, and III
Show the answer

Correct answer: D

Each of the three expressions must be considered:

Expression I: s−ts - t, where s is a positive integer and t is a negative integer:

Subtracting a negative integer is the equivalent of adding a positive integer, so s−ts - t is equivalent to finding the value of s+(−t)s + (-t), which must result in a positive number.

Expression II: st2st^{2}, where s is a positive integer and t is a negative integer:

Squaring a negative integer results in a positive integer, so s(−t)2s(-t)^{2} is equivalent to finding the value of st2st^{2}, which must result in a positive number.

Expression III: sts^{t}, where s is a positive integer and t is a negative integer:

A negative exponent results in the reciprocal of the same exponential expression, so sts^{t} is equivalent to finding the value of 1s(−t)\frac{1}{s^{(-t)}}, which must result in a positive number.

Therefore, Expressions I, II, and III must all result in a positive number.

Question analysis

Three expressions, s − t, s times t squared, and s to the power t, are listed, and students decide which must be positive when s is a positive integer and t is a negative integer. It tests sign rules for subtraction and multiplication and the meaning of a negative exponent. Each expression has to be judged for every possible pair, not just one example.

NYS standards
  • 8.EE.1 integer exponent properties
  • 7.NS.1c rational number subtraction
  • 7.NS.2a rational number multiplication
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students learn sign rules for subtracting and multiplying signed numbers. Here they reason about letters rather than numbers, decide what must always hold, and handle a negative exponent, a grade 8 idea.
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 5, 42 words.
Corrected from the DOE handbook
  • Two steps of the explanation that treated t as a positive number were corrected.
  • A punctuation error in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The three expressions test three different sign facts: subtracting a negative, squaring a negative, and raising to a negative power. The last is the trap, since a negative exponent seems to promise a negative result when it really gives a fraction, and that belief alone leads to a listed choice. Trying s = 2 and t = −3 gives 5, 18 and 1/8, all positive, and the rules show why no other pair can change that.

  1. Variables instead of numbers moderate Understanding the question

    No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know moderate Working it out

    Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

Strategy: Try a pair such as s = 2 and t = −3 in each expression, then explain why its sign could or could not change for any other positive s and negative t.

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The question makes t the negative integer, but two steps of the DOE’s explanation write the negative integer as −t, as if t were positive. For Expression I it says “s − (−t) is equivalent to finding the value of s + t, which must result in a positive number”: s − (−t) is not Expression I, and with the question’s t, s + t need not be positive (s = 1 and t = −2 give −1). For Expression III it says “s^(−t) is equivalent to finding the value of 1/s^t”, about s^(−t) rather than Expression III. The site now says that s − t is equivalent to s + (−t), and that s^t is equivalent to 1/s^(−t). Since t is negative, −t is a positive integer, so s + (−t) is the sum of two positive integers and 1/s^(−t) is the reciprocal of a positive integer: both positive, and the answer, I, II and III, is unchanged. The step for Expression II, s(−t)² = st², is true for any t and is unchanged.
  • Explanation: The DOE’s explanation introduces Expression III as “s^t. where s is a positive integer and t is a negative integer”, with a period where Expressions I and II (and the 2022–23 guide, which first printed this explanation) have a comma. The site has the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a number
s−t=s+(−t)s - t = s + (-t)
s and t are integers, and −t is the opposite of t. Subtracting a number is the same as adding its opposite. When t is negative, −t is positive. So subtracting t adds a positive amount.
A negative exponent
st=1s−ts^{t} = \frac{1}{s^{-t}}
s is the base and t is the exponent. When t is negative, −t is positive. So a negative exponent means 1 over the base to a positive exponent. For example, 2 to the power −3 is 1 over 2³, which is 1/8.

  1. Sort out the question

    An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.

  2. Try one pair of numbers

    Try s = 2 and t = −3 in each expression:

    • I: 2−(−3)=2+3=52 - (-3) = 2 + 3 = 5
    • II: 2×(−3)2=2×9=182 \times (-3)^{2} = 2 \times 9 = 18
    • III: 2−3=123=182^{-3} = \frac{1}{2^{3}} = \frac{1}{8}

    All three are positive this time. Now see why that always happens.

  3. Explain expression I

    Use the rule for subtracting:

    s−t=s+(−t)s - t = s + (-t)

    t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.

    Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.

  4. Explain expression II

    t2t^{2} means t × t. A negative times a negative is positive, so t2t^{2} is always positive. Then s×t2s \times t^{2} is a positive times a positive. That is always positive too.

  5. Explain expression III

    Use the rule for a negative exponent:

    st=1s−ts^{t} = \frac{1}{s^{-t}}

    −t is a positive integer, so s−ts^{-t} is a positive integer. So sts^{t} is 1 over a positive number. That is a positive fraction.

    Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.

  6. Choose the answer

    Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.

    So the answer is D, I, II, and III.

Check your answer

Try the values closest to 0: s = 1 and t = −1. I gives 1−(−1)=21 - (-1) = 2. II gives 1×(−1)2=11 \times (-1)^{2} = 1. III gives 1−1=11=11^{-1} = \frac{1}{1} = 1. All three are still positive.

More questions like this

The original DOE question
Question 18 as shown on the DOE’s digital test, page 91 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 91, question 18. This is the DOE’s version, before the corrections listed in the analysis.

Sample #928575