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Reading a figure, table or graph

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Question 1 of 108

2026 Practice Test A Math, question 37

Question

When a student buys 4 notebooks, the total cost is $12. Each notebook costs the same amount. Select each point on the graph that best represents y, the total cost of x notebooks.

Select all the correct answers.

A graph titled BUYING NOTEBOOKS, with Number of Notebooks on the x-axis from 0 to 6 and Total Cost (dollars) on the y-axis from 0 to 15, with grid lines every 1 unit. Seven points are plotted: (1, 1), (1, 3), (2, 5), (3, 7), (3, 9), (4, 12) and (5, 14).

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

  • A(1, 1)
  • B(1, 3)
  • C(2, 5)
  • D(3, 7)
  • E(3, 9)
  • F(4, 12)
  • G(5, 14)
Show the answer

Correct answer: B, E, and F

Solution:

There are three correct points: (1, 3), (3, 9), and (4, 12).

Find the cost per notebook: 124=3\frac{12}{4} = 3.

The equation of the line is y=3xy = 3x, where y is the total cost of x notebooks.

Check each point by plugging the x-value into the equation y=3xy = 3x. If the y-value from the equation matches the y-value of the point, then the point is on the line y=3xy = 3x.

Check all seven points:

(1, 1) y=3(1)=3y = 3(1) = 3 y-values do not match; incorrect point

(1, 3) y=3(1)=3y = 3(1) = 3 y-values match; CORRECT point

(2, 5) y=3(2)=6y = 3(2) = 6 y-values do not match; incorrect point

(3, 7) y=3(3)=9y = 3(3) = 9 y-values do not match; incorrect point

(3, 9) y=3(3)=9y = 3(3) = 9 y-values match; CORRECT point

(4, 12) y=3(4)=12y = 3(4) = 12 y-values match; CORRECT point

(5, 14) y=3(5)=15y = 3(5) = 15 y-values do not match; incorrect point

Question analysis

A graph titled Buying Notebooks shows seven points, and students select every point that fits the total cost of buying notebooks, given the cost of a few notebooks and that each costs the same. It tests finding a unit rate and recognizing points of a proportional relationship. Some points line up in tempting patterns, and the item does not say how many to select.

NYS standards
  • 7.RP.2b constant of proportionality
  • 7.RP.2c proportional relationship equations
  • 6.RP.3a tables of equivalent ratios
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a unit rate and deciding which points fit a proportional relationship are grade 7 skills, and the arithmetic here is easy. What adds demand is testing all seven points with no count given, since one miss or extra pick loses the item.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 36 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

Finding the price is one division, 12 ÷ 4 = 3, so the question is really about checking: each chosen dot must fit y = 3x exactly, and the item does not say how many do. The graph punishes judging by eye, since a straight row of dots climbs from (1, 3) at $2 a notebook and two dots sit a single gridline off the true line. Multiplying each x by 3 and reading each y from the axis, not from the pattern, settles all seven.

  1. Every choice must be checked moderate Choosing the answer

    Seven dots must each be tested against $3 a notebook, and the item never says how many fit; one missed point or one extra loses the whole item, so no dot can be skipped. (Embretson & Wetzel, 1987)

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

    The graph stacks two dots above 1 notebook and two above 3, and each total cost has to be read off a grid ruled at every dollar up to 15. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. A very close wrong answer minor Choosing the answer

    (5, 14) sits one gridline below (5, 15), where 5 notebooks at $3 would be, and (2, 5) sits one below (2, 6), so judging by the line's direction cannot rule either out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the cost of one notebook from the total given in the question, then multiply each point's x-coordinate by that price and keep only the points whose y-coordinate matches exactly.

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.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit price
price of one=total cost÷number bought\text{price of one} = \text{total cost} \div \text{number bought}
The unit price is what one item costs. Divide the total cost by the number of items bought. This works because every notebook costs the same, so the total splits into equal parts.
A proportional relationship
y=kxy = kx
y is the total cost in dollars, and x is the number of notebooks. k is the price of one notebook, the number you always multiply by. A point fits only if its y is exactly k times its x. The graph is a straight line through (0, 0).

  1. Sort out the question

    Four notebooks cost $12, and each one costs the same. On the graph, x is the number of notebooks, and y is the total cost in dollars. The graph shows seven points. You must select every point that fits. The question does not say how many fit, so test all seven.

  2. Find the price of one notebook

    Split the $12 equally among the 4 notebooks:

    12÷4=312 \div 4 = 3

    Each notebook costs $3. So the total cost is 3 times the number of notebooks:

    y=3xy = 3x

  3. Test each point

    For each point, multiply its x by 3. Keep the point only if the result equals its y.

    ChoicePoint3 × xFits?
    A(1, 1)3No
    B(1, 3)3Yes
    C(2, 5)6No
    D(3, 7)9No
    E(3, 9)9Yes
    F(4, 12)12Yes
    G(5, 14)15No

  4. Watch for look-alike points

    Watch out: the dots (1, 3), (2, 5) and (3, 7) sit in a neat straight row. But that row climbs $2 per notebook, not $3. Lining up is not enough: (2, 5) and (3, 7), choices C and D, do not fit.

    Watch out: 5 notebooks cost 5×3=155 \times 3 = 15 dollars. The dot above 5 sits one gridline lower, at 14. Read the y-axis carefully, and you will skip choice G.

  5. Name the points

    Three points fit: (1, 3), (3, 9) and (4, 12). On the test, you click each of these dots on the graph. On this site, you check their boxes. So the answers are B, E and F.

Check your answer

All three points fit y=3xy = 3x: 3=3×13 = 3 \times 1, 9=3×39 = 3 \times 3 and 12=3×412 = 3 \times 4. They also line up with (0, 0), since 0 notebooks cost $0. And (4, 12) is the fact the question gave you.

More questions like this

The original DOE question
Question 37 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 A Math, page 108, question 37.

Question 2 of 108

2026 Practice Test A Math, question 43

Question

Graph the solution set for the inequality −3≤2x−5-3 \leq 2x - 5.

Select a ray. Then, move the endpoint of the ray to the correct position on the number line.

On the test you draw the ray on the number line; here, choose its endpoint, whether the endpoint is closed or open, and its direction. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

-5-4-3-2-1012345

Show the answer

Correct answer: a ray with a closed endpoint at 1, pointing right

Solution:

A number line from −5 to 5 with a tick at every integer, labeled at −5, 0 and 5. A solid dot at 1, with the line drawn thick from 1 to the right and ending in an arrow.

Solve the inequality for x, then graph the solution on the number line:

−3≤2x−5-3 \leq 2x - 5

Add 5 to both sides of the inequality: 2≤2x2 \leq 2x

Divide both sides by 2: 1≤x1 \leq x

Rewrite the inequality with x on the left side: x≥1x \geq 1

The graph of x≥1x \geq 1 is a closed ray (solid point) beginning at 1 and extending to the right on the number line.

Question analysis

Students graph the solution set of a two-step inequality on a number line by choosing a ray: where it starts, whether its endpoint is open or closed, and which way it points. It tests solving a linear inequality and reading its solution. The variable sits on the right of the inequality symbol, so reading the result the right way round is part of the challenge.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality and graphing its solution set is a named grade 7 skill. This one puts the variable on the right of the symbol and asks students to build the graph themselves rather than pick it from four.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 26 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

Solving takes two routine moves; the points are lost in turning 1 ≤ x into a picture. Because x sits on the right, the symbol seems to point left, and a student who shades the way it points draws the values below 1, which all fail. The dot must also be solid, since x = 1 makes both sides −3. Testing x = 2, where 2x − 5 = −1 and −3 ≤ −1 holds, confirms the direction.

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

    The solution comes out as 1 ≤ x, with x on the right, so the ≤ sign points away from the shaded side; it has to be read as x ≥ 1 before the ray is drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Two solving moves, adding 5 and then halving, come before three graph choices, the endpoint, a solid dot and a direction, and a mistake in any of the five loses the item. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The ray is built, by picking one of four ray shapes and dragging its endpoint, rather than chosen from four finished graphs, so nothing on screen flags a ray that starts at the right number but points the wrong way. (Enright, Morley & Sheehan, 2002)

Strategy: Undo the subtraction and then the multiplication on both sides, rewrite the result with x first, decide whether the endpoint is included, and test one value on your ray in the original inequality.

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.)
  • 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).
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a≤b, then a+c≤b+c\text{if } a \leq b \text{, then } a + c \leq b + c
a and b are the two sides of an inequality, and c is any number. Adding the same number to both sides keeps the inequality true. So does dividing both sides by the same positive number. Dividing by a negative number would flip the symbol, but that does not happen here.
Turning an inequality around
a≤b means the same as b≥aa \leq b \text{ means the same as } b \geq a
a and b are any two numbers or terms. You may swap the two sides if you also turn the symbol around. The pointed end of the symbol always faces the smaller side. So 1 ≤ x and x ≥ 1 say the same thing.

  1. Sort out the question

    The symbol ≤ means is less than or equal to. The solution set is every value of x that makes −3≤2x−5-3 \leq 2x - 5 true. You show it as a ray. A ray is a dot, called the endpoint, with an arrow going one way forever. You need three things: where the dot goes, whether it is closed or open, and which way the ray points.

  2. Add 5 to both sides

    The right side is 2x − 5. Adding 5 undoes subtracting 5, so add 5 to both sides:

    −3+5≤2x−5+5-3 + 5 \leq 2x - 5 + 5

    2≤2x2 \leq 2x

    Watch out: taking 5 away from −3 instead gives −8 ≤ 2x. That wrong move puts the dot at −4.

  3. Divide both sides by 2

    Now x is multiplied by 2, so divide both sides by 2. You divide by a positive number, so the symbol stays the same:

    22≤2x2\frac{2}{2} \leq \frac{2x}{2}

    1≤x1 \leq x

  4. Read it with x first

    1≤x1 \leq x says 1 is less than or equal to x. So x is 1 or any number bigger than 1. With x first, that is:

    x≥1x \geq 1

    Bigger numbers are to the right on a number line, so the ray points right.

    Watch out: in 1 ≤ x, the symbol seems to point left. Do not follow it. A ray pointing left from 1 shows numbers that do not work.

  5. Choose a closed dot

    The words or equal to mean 1 itself is a solution. Check it: 2×1−5=−32 \times 1 - 5 = -3, and −3 ≤ −3 is true. So the dot at 1 is closed, which means filled in. An open dot, an empty circle, leaves the endpoint out. Use it only for < or >.

    Watch out: an open dot at 1 would leave out a number that works.

  6. Draw the ray

    On the test, choose the ray with a filled dot and an arrow pointing right. Then drag its dot to 1. On this site, choose 1 for the endpoint, then closed (filled dot) and pointing right.

    So the answer is a closed dot at 1, with the ray pointing right.

Check your answer

Test a number on your ray, such as 3. 2×3−5=12 \times 3 - 5 = 1, and −3 ≤ 1 is true. Now test a number off the ray, such as 0. 2×0−5=−52 \times 0 - 5 = -5, and −3 ≤ −5 is false. So the ray points the right way.

More questions like this

The original DOE question
Question 43 as shown on the DOE’s digital test, page 114 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 114, question 43.

Question 3 of 108

2026 Practice Test B English, question 35
Passage Using Fire to Keep a Prairie Healthy

Using Fire to Keep a Prairie Healthy

Inside the 40,000-acre Joseph H. Williams Tallgrass Prairie Preserve in northern Oklahoma, herds of bison roam on the scenic land that is home to hundreds of different species of plants and animals. While one of the greatest threats to the prairie is wildfire, the use of controlled fires is actually an effective way to protect this idyllic landscape.

Fires that are started by lightning or other natural circumstances are inevitable, and the results can be disastrous. However, hundreds of years ago the American Indian people who inhabited the Great Plains area between Minnesota and Texas realized that these fires could also be helpful. The bison in the area seemed to prefer grazing on tender new grass on the recently burned land rather than on grass in the unburned areas. The American Indian people began to deliberately burn areas of land for bison to graze on, which enticed the herds away from the people’s crops.

Today conservationists at the Tallgrass Prairie Preserve and farmers in the Great Plains continue to use controlled burns for land management. Intentionally burning a portion of land can dramatically improve the quality of the vegetation that regrows there. The fires burn away weedy undergrowth and help limit the overcrowding of shrubs and trees in the burn area, creating less competition for water and nutrients. Additionally, the process of burning excess plant matter adds nitrogen, an essential element for plant growth, to the soil.

One common target of controlled burns at the preserve and the surrounding area is the invasive red cedar tree. A single red cedar tree can consume up to 40 gallons of water per day, taking this vital resource away from other plant life. These tall trees also cast shade that prevents sunlight from reaching the plants beneath them. The fast-growing red cedar trees tend to crowd out prairie grasses, the primary food source for wild and domestic animals that make the prairie their home. An imbalance in one component of the prairie’s ecosystem affects the entire web of life. Controlled burns help maintain this ecosystem’s delicate balance.

Of course, these controlled fires are intended to burn only a portion of an area. A total burn, which is a risk with an uncontrolled wildfire, would destroy all wildlife as well as the crops farmers plant for income. The key to using controlled fires is knowing which areas of land to burn and when. Conservation experts at the preserve employ the “patch-burn” approach, meaning they rotate which portion of land is burned each year. They study the land to find out which areas would most benefit from being burned, and then they arrange about a dozen burns over one-third of the land. This patch burning contains the fire within a specific area and allows animals in the burn area to safely relocate. The following year, conservationists will burn a different section of the preserve, while the land burned the previous year regrows healthier than before.

Researchers have tracked and studied the variety of plant species and animals that live on the preserve, and their studies show that the patch-burn approach has restored biodiversity to the area by promoting the growth of species that were at risk of being crowded out. The patch-burn system is so successful that the conservationists at the preserve provide training to prairie farmers about conducting controlled burns on their own land. Burning land to make it healthy may seem counterintuitive, but strategic controlled fires have helped the prairie sustain life for hundreds of years and, with careful management, will continue to do so.

Boxed diagram titled “OVERVIEW OF THE PATCH-BURN METHOD,” introduced by the text “By using a patch-burn method, conservationists and farmers can reduce the risk of unintentional fires and control where grazing animals, such as bison, roam.” Below it, a map of one stretch of land is divided by a branching stream, labeled Stream, into three shaded areas: a dark Unburned Area across the top, a medium gray Previously Burned Area on the left, and a light gray Recently Burned Area on the lower right, where eight bison stand. Three captions under the map read “The unburned area is overrun with weeds and has the greatest risk of catching fire,” “The previously burned area is cleared of weeds, and young trees and crops grow freely and thrive,” and “The recently burned area has been cleared of all plants. Bison prefer to eat the fresh grass that will begin to grow in the recently burned area, and they will generally remain in that area until newer, fresher grass is available elsewhere.”

Question

How do the diagram and its text provide additional support for the topic presented in the passage?

  • Aby depicting how the landscape in a given area changes as the patch-burn method is applied
  • Bby indicating that patch-burn fires are best suited for use in areas with certain features
  • Cby revealing that the patch-burn method is used primarily on uninhabited areas of land
  • Dby comparing the size of the area burned by the patch-burn method with that of unburned areas
Show the answer

Correct answer: A

The question asks how the diagram and its text provide additional support for the topic presented in the passage.

A. CORRECT. The text of the diagram states that “by using a patch-burn method, conservationists and farmers can reduce the risk of unintentional fires and control where grazing animals, such as bison, roam.” By showing the pattern of controlled burns and the resulting bison migration in a given area, the diagram depicts how the landscape changes as the patch-burn method is applied.

B. Incorrect. Although the text mentions areas “where grazing animals, such as bison, roam,” the diagram does not depict distinctive features of the land such as trees or brush and therefore does not indicate that patch-burn fires are best suited for use in areas with certain features.

C. Incorrect. While the text mentions “grazing animals, such as bison” and the diagram shows the bison grazing in only one area, the diagram and its text do not indicate that the patch-burn method is used on uninhabited land.

D. Incorrect. While the areas of land depicted in the diagram do indicate that the sizes of the areas differ, the text provided does not compare the sizes of the areas burned by the patch-burn method with those of unburned areas or indicate that this factor is important in reducing the risk of uncontrolled fires or controlling “where grazing animals, such as bison, roam.”

Question analysis

This question asks how a diagram titled “Overview of the Patch-Burn Method,” printed with the article on controlled burns at an Oklahoma prairie preserve, supports the passage's topic. The diagram shows a stretch of land divided into three areas, each with a caption, so students must take in what the picture and its text show together and tie that to the article.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R2 theme or central idea
Grade level
Grade 7 reading and task
Compared with grade 7
At grade 7 Explaining how a graphic adds to a text is a grade 7 skill. This diagram is simple and clearly labeled, but students must take in its three areas and captions together and match them to an abstract description, a typical demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Evidence
Answered from the whole text
Reading level
Passage: about grade 11, 598 words; question and choices: about grade 9, 79 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The diagram shows three areas side by side, and the key insight is that they are one piece of land at three moments in the patch-burn cycle. Students who read it only as a map look for spatial claims instead, about land features, people or the sizes of the areas. The caption calling the unburned area “overrun with weeds” is the likeliest snare, since it sounds like advice on which land to burn.

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

    The diagram’s meaning is split among a map of three shaded areas, an introduction and three captions, which must be read together to see what the picture adds to the passage. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Reading between the lines moderate Working it out

    Nothing in the diagram mentions the passing of time; students must infer that the unburned, recently burned and previously burned areas are stages the same land passes through as the burns rotate. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

  3. How, not what minor Working it out

    The stem asks how the diagram supports the passage, a question of function, so describing what is drawn is not enough; it must be tied to the yearly rotation of paragraph 5. (Anderson & Krathwohl, 2001)

Strategy: Read the diagram's title, introduction and three captions before the choices, sum up in one sentence what the diagram shows, and check each choice against something you can actually point to in it.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This question asks how the diagram helps the passage. A diagram is a drawing that explains how something works. You need to say what it adds, not just what it shows. So read its title, its short introduction and its three captions. Captions are the short notes under a picture.

  2. Read the diagram

    The title, printed in capital letters, is “Overview of the Patch-Burn Method.” The drawing shows one stretch of land. A stream splits it into three areas, labeled “Unburned Area,” “Recently Burned Area” and “Previously Burned Area.” A herd of bison stands in the recently burned area.

  3. Read the three captions

    • The unburned area “is overrun with weeds and has the greatest risk of catching fire.”
    • The recently burned area “has been cleared of all plants.” Bison come to eat the fresh grass that grows there.
    • The previously burned area “is cleared of weeds, and young trees and crops grow freely and thrive.”

    These are three stages the same land goes through: weedy, burned, then healthy. Paragraph 5 says the experts burn a different section each year, while “the land burned the previous year regrows healthier than before.”

  4. Check each choice

    • A is right. The diagram shows land changing from weedy, to burned, to healthy as the method is used.
    • B is wrong. The captions describe each stage. They don’t say some kinds of land suit burning best.
    • C is wrong. Bison live in the drawing, and a caption mentions crops. Uninhabited means with no one living there, and nothing says that.
    • D is wrong. The areas are different sizes in the drawing. But no caption compares sizes.

    Watch out: B is tempting because “overrun with weeds” sounds like a reason to burn. But that caption describes the first stage. It is not a rule for choosing land.

  5. Choose the answer

    Depicting means showing in a picture. So the answer is A: “by depicting how the landscape in a given area changes as the patch-burn method is applied.”

Why this answer and not the closest wrong one

B is the closest wrong answer, because the weedy unburned area does sound like land that needs a fire. But the diagram gives no rules for picking land. It shows one piece of land changing, stage by stage, as the burns rotate. That is what A says.

More questions like this

The original DOE question
Question 35 as shown on the DOE’s digital test, page 48 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B English, page 48, question 35.
The passage “Using Fire to Keep a Prairie Healthy” as printed (page 41)The passage “Using Fire to Keep a Prairie Healthy” as printed on page 41 of the DOE handbook.

Question 4 of 108

2026 Practice Test C Math, question 6

Question

A parallelogram that leans to the right. The angle at its upper left vertex is labeled 3x°, and the angle at its lower left vertex is labeled 2x°.

In the parallelogram above, what is the value of x?

  • A36
  • B18
  • C6
  • D5
Show the answer

Correct answer: A

In a parallelogram, any two adjacent angles are supplementary, so the sum of the two given angles is equal to 180°.

Write an equation showing that the sum of 3x∘3x^{\circ} and 2x∘2x^{\circ} is equal to 180°:

3x+2x=1803x + 2x = 180

Solve for x:

5x=1805x = 180

5x5=1805\frac{5x}{5} = \frac{180}{5}

x=36x = 36

Question analysis

A parallelogram is drawn with the two angles at the ends of its left side labeled 3x° and 2x°, and students find the value of x. It tests the angle facts of a parallelogram and solving a simple equation. Knowing what the two angles must add up to is the key fact, since the figure gives no other number.

NYS standards
  • 7.G.5 angle relationships and equations
  • 8.G.5 angle sums, transversals, angle-angle similarity
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 The angle fact needed, about consecutive angles of a parallelogram, rests on grade 8 parallel-line reasoning. Once it is known, the work is one short equation, like a typical grade 7 problem about angle sums.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 10 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

Once the angle fact is in place, 5x = 180 is a one-line equation, so the question tests recall more than computation. Students who use 90° for the pair land on half the right value, and the two smallest choices come only from combining the labels' 3 and 2. A sense check helps: x = 36 makes the angles 108° and 72°, one obtuse and one acute, just as the drawing shows.

  1. Something you must already know moderate Working it out

    The figure gives no number of degrees, only 3x° and 2x°, so everything rests on recalling that the two angles at the ends of one side of a parallelogram add to 180°. (Anderson & Krathwohl, 2001)

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

    The marked angles sit at the upper and lower ends of the left side, consecutive rather than opposite, which is what makes them supplementary instead of equal. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide what two angles that share a side of a parallelogram add up to, set 3x + 2x equal to that total, solve for x, and check that the two angles fit the figure.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Stops one step early: A choice is a number from partway through the solution.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Neighboring angles of a parallelogram
a+b=180∘a + b = 180^{\circ}
a and b are two angles of a parallelogram that share a side, like the top left and bottom left angles. They always add to 180°. Here is why. The four angles add to 360°, and opposite angles are equal. So each pair of neighbors is half of 360°.

  1. Sort out the question

    The figure is a parallelogram. That is a four-sided shape whose opposite sides are parallel, so they never meet. Two angles are marked: 3x° at the upper left and 2x° at the lower left. You need the value of x.

  2. Write an equation for the angles

    The two marked angles are at the two ends of the left side. So they are neighbors, not opposite angles. Use the rule that neighbors add to 180°:

    3x+2x=1803x + 2x = 180

    Watch out: the total is 180°, not 90°. Using 90 gives 5x = 90 and x = 18, choice B.

  3. Solve for x

    Three x’s plus two x’s make five x’s:

    5x=1805x = 180

    Divide both sides by 5 to get x alone:

    x=1805=36x = \frac{180}{5} = 36

    Watch out: stopping at 5, the number in front of x, gives choice D. Multiplying 3 × 2 = 6 gives choice C, but that uses no angle fact.

  4. Choose the answer

    The value of x is 36. It is a number, not the size of an angle. The angles themselves are 3 × 36 and 2 × 36 degrees.

    So the answer is A, 36.

Check your answer

Find the two angles. The top one is 3×36=1083 \times 36 = 108 degrees. The bottom one is 2×36=722 \times 36 = 72 degrees. They add to 180°. In the figure, the top angle is wide, more than 90°. The bottom one is narrow, less than 90°. That matches.

More questions like this

The original DOE question
Question 6 as shown on the DOE’s digital test, page 79 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 79, question 6.

Question 5 of 108

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 108

2026 Practice Test C Math, question 12

Question

A spinner has 5 congruent sections labeled as shown. One side of a fair coin is heads, and the other side is tails.

A spinner shaped like a regular pentagon, divided into 5 congruent triangular sections that meet at the center, where an arrow points straight up. Going clockwise from the upper left, the sections are labeled 1, 2, 1, 4 and 3. Beside the spinner is a coin (a dime) showing its heads side, captioned Heads.

The spinner is spun one time, and the coin is flipped. What is the probability that the spinner lands on 1 and the coin lands on tails?

Select the place on the number line that represents the probability.

On the test you click a point on the number line; here, choose it from the menu.

00.10.20.30.40.50.60.70.80.91

Show the answer

Correct answer: the point 0.2

The spinner has five equal-sized sections. Two of the sections are labeled with a 1. So, the probability of landing on a 1 is two out of five, or 2/5.

A coin has two sides, one heads and one tails. So, the probability of landing on tails is one out of two, or 1/2.

The compound probability of the spinner landing on a 1 and the coin landing on tails can be calculated by multiplying 2/5 and 1/2 as shown:

25⋅12=210=0.2\frac{2}{5} \cdot \frac{1}{2} = \frac{2}{10} = 0.2

Question analysis

A spinner with five equal sections is spun once and a fair coin is flipped. Students find the probability that the spinner lands on 1 and the coin shows tails, then select that point on a number line from 0 to 1. Counting the right sections and combining the two events correctly are the tricky parts.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 6.SP.6 probability from 0 to 1
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the chance of two independent events together is grade 7 compound probability, here with one spinner and one coin. Counting sections from the picture and placing the result on a number line add two small steps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 59 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The probability itself takes one multiplication of two simple fractions; the losses come before and after it. Before, the spinner has to be searched for every section labeled 1, and the coin's chance brought in as well. After, 1/5 has to be rewritten as 2/10 to find its point, since only 0, 1/2 and 1 are labeled. Checking that the result is smaller than both separate chances catches an addition.

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

    The two sections labeled 1, upper left and right, are not side by side, and the number line labels only 0, 1/2 and 1, so its nine points must be read as tenths. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Nothing says that the chance of two independent events both happening is the product of their chances, 2/5 × 1/2, rather than their sum. (Anderson & Krathwohl, 2001)

Strategy: Count the sections labeled 1 to get the spinner's chance, take the coin's chance of tails, multiply the two because both must happen, and write the result in tenths.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event
P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}
P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want. The bottom counts all possible results. It works when every result is equally likely, like equal sections of a spinner.
Probability that two things both happen
P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
A and B are two events that do not affect each other, like a spin and a coin flip. P(A) and P(B) are their chances. Multiply them to get the chance that both happen. Needing both makes the chance smaller, never bigger.

  1. Sort out the question

    The spinner has 5 congruent sections. Congruent means the same size and shape, so each is equally likely. The coin has a heads side and a tails side. You need the chance that the spinner lands on 1 and the coin lands on tails.

  2. Find the spinner’s chance

    Two sections show 1: the upper left one and the one on the right. So 2 of the 5 equal sections give a 1:

    P(1)=25P(1) = \frac{2}{5}

    Watch out: the two 1s are not next to each other. Counting only one gives 1/5 here, and 0.1 at the end.

  3. Find the coin’s chance

    A fair coin is equally likely to land on either side. Tails is 1 of its 2 sides:

    P(tails)=12P(\text{tails}) = \frac{1}{2}

  4. Multiply the two chances

    The spin and the flip do not affect each other, and both must happen. So multiply:

    25×12=210=0.2\frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = 0.2

    Watch out: stopping at 2/5 gives 0.4. Adding gives 25+12=910\frac{2}{5} + \frac{1}{2} = \frac{9}{10}, or 0.9. Both are wrong.

  5. Select the point

    On the test, the number line labels only 0, 1/2 and 1. Between 0 and 1 are 9 targets, one at each tenth. 0.2 is the second target to the right of 0. On this site, choose 0.2 from the Point menu.

    So the answer is the point at 0.2.

Check your answer

List every result. The spinner has 5 sections and the coin has 2 sides, so there are 5×2=105 \times 2 = 10 results, all equally likely. Just 2 of them are a 1 with tails, one for each section labeled 1. That is 2 out of 10, or 0.2.

More questions like this

The original DOE question
Question 12 as shown on the DOE’s digital test, page 85 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 85, question 12.

Question 7 of 108

2026 Practice Test C Math, question 17

Question

A number line from 0 to 8 with a tick and a label at every whole number. Four points are marked on it: M at 1, R at 3, S at 6 and N at 7.

What is the distance between the midpoints of MN\overline{MN} and RS\overline{RS}?

  • A12\frac{1}{2} unit
  • B1 unit
  • C1121\frac{1}{2} units
  • D2 units
Show the answer

Correct answer: A

MN\overline{MN} begins at 1 and ends at 7. Determine the length of MN\overline{MN}:

7−1=67 - 1 = 6

The midpoint of a line segment is located exactly halfway between the two endpoints. Since MN\overline{MN} is 6 units in length, the two endpoints are 6 units apart. Determine the location of the midpoint, in relation to each endpoint, by dividing 6 by 2:

6÷2=36 \div 2 = 3

The location of the midpoint of MN\overline{MN} is 3 units from either endpoint. Find the midpoint by either adding 3 to the location of M (1) or subtracting 3 from the location of N (7):

1+3=41 + 3 = 4

7−3=47 - 3 = 4

The midpoint of MN\overline{MN} is located at 4.

RS\overline{RS} begins at 3 and ends at 6. Determine the length of RS\overline{RS}:

6−3=36 - 3 = 3

Since RS\overline{RS} is 3 units in length, the two endpoints are 3 units apart. Determine the location of the midpoint, in relation to each endpoint, by dividing 3 by 2:

3÷2=1.53 \div 2 = 1.5

The location of the midpoint of RS\overline{RS} is 1.5 units from either endpoint. Find the midpoint by either adding 1.5 to the location of R (3) or subtracting 1.5 from the location of S (6):

3+1.5=4.53 + 1.5 = 4.5

6−1.5=4.56 - 1.5 = 4.5

The midpoint of RS\overline{RS} is located at 4.5.

Determine the distance between the two midpoints by subtracting:

4.5−4=0.54.5 - 4 = 0.5

0.5 is equivalent to 12\frac{1}{2}, making the distance between the two midpoints 12\frac{1}{2} unit.

Question analysis

A number line from 0 to 8 shows points M, R, S and N, and students find the distance between the midpoint of segment MN and the midpoint of segment RS. It tests locating points and finding distances on a number line. One segment sits inside the other, and several lengths in the figure can be mistaken for the distance wanted.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding distances between points on a number line is grade 7 work, here with whole-number positions. Two midpoints must be located before their distance can be found, so the task takes three short steps rather than one.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The positions are whole numbers, so the midpoints come from simple averages, (1 + 7) ÷ 2 and (3 + 6) ÷ 2. The difficulty is keeping several lengths apart: the half-lengths of the two segments, the gaps at their ends, and the distance actually asked for, half a unit. Marking both midpoints on the number line before subtracting makes the small gap visible.

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

    Segment RS lies inside MN, from 3 to 6, and its midpoint, 4 1/2, falls between two ticks of the number line rather than on a labeled point. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Two midpoints have to be found, at 4 and 4 1/2, before their distance is taken, and the half-length of RS met on the way, 1 1/2, is among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find each midpoint as the average of its segment's two endpoints, then subtract the smaller midpoint from the larger one.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line
midpoint=a+b2\text{midpoint} = \frac{a + b}{2}
a and b are the numbers at the two ends of a segment. The midpoint is the point exactly halfway between them. Adding the ends and dividing by 2 finds their average, which is the middle.
Distance on a number line
distance=larger−smaller\text{distance} = \text{larger} - \text{smaller}
The distance between two points on a number line is how many units apart they are. Subtract the smaller number from the larger one, so the distance is never negative.

  1. Sort out the question

    The number line shows M at 1, R at 3, S at 6 and N at 7. Segment MN goes from 1 to 7. Segment RS goes from 3 to 6. A midpoint is the point exactly halfway along a segment. You need the distance between the two midpoints.

  2. Find the midpoint of MN

    Use the midpoint formula with the ends 1 and 7:

    1+72=82=4\frac{1 + 7}{2} = \frac{8}{2} = 4

    The midpoint of MN is at 4, which is 3 units from each end.

  3. Find the midpoint of RS

    Use the same formula with the ends 3 and 6:

    3+62=92=4.5\frac{3 + 6}{2} = \frac{9}{2} = 4.5

    The midpoint of RS is at 4.5. That is halfway between the marks for 4 and 5, where no point is labeled.

  4. Subtract the two midpoints

    Use the distance formula:

    4.5−4=0.54.5 - 4 = 0.5

    0.5 is one half, so the midpoints are 1/2 unit apart.

    Watch out: half of MN is 3, and half of RS is 1 1/2. Subtracting those gives 1 1/2, choice C. But they are half-lengths, not positions. And 2, choice D, is only the distance from M to R.

  5. Choose the answer

    The midpoints, 4 and 4.5, are close together because RS sits near the middle of MN. So the answer is A, 1/2 unit.

Check your answer

Find each midpoint another way: walk half the length from the left end. MN is 7−1=67 - 1 = 6 units long, and 1+3=41 + 3 = 4. RS is 6−3=36 - 3 = 3 units long, and 3+1.5=4.53 + 1.5 = 4.5. Same midpoints, 1/2 unit apart.

More questions like this

The original DOE question
Question 17 as shown on the DOE’s digital test, page 90 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 90, question 17.

Question 8 of 108

2026 Practice Test C Math, question 20

Question

Two circles, with centers at A and C, are shown. Line segment AB is congruent to line segment DE.

Two circles. The larger circle has center A, and a radius is drawn from A to point B on the circle, labeled 5. The smaller circle has center C, and a segment is drawn from point D on the circle, through C, to point E on the circle.

Determine the area, A, and the circumference, C, of each circle to complete the table.

Select one answer in each row.

Choose one answer in each row, then check. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Circle ACircle CNeither
A=25πA = 25\pi
C=5πC = 5\pi
Show the answer

Correct answer: A = 25π: Circle A; C = 5π: Circle C

A=25πA = 25\pi: Circle A

C=5πC = 5\pi: Circle C

As shown in the illustration, AB (the radius of Circle A) is 5 units in length. Since DE (the diameter of Circle C) is congruent to AB, we know it is also 5 units in length.

Circle A:

A=πr2A = \pi r^{2}

A=25πA = 25\pi

r=5r = 5

We know the radius of Circle A is 5 so the answer to row 1 is A.

Circle C:

C=π⋅dC = \pi \cdot d

C=5πC = 5\pi

We have to look for the circle with a diameter of 5. That is circle C.

Question analysis

Two circles centered at A and C are shown: radius AB of Circle A is labeled 5, and segment DE, drawn through C, is congruent to AB. Students decide whether an area of 25π and a circumference of 5π each belong to Circle A, Circle C or neither. Keeping radius and diameter apart, and area and circumference, is the challenge.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying the area and circumference formulas is the grade 7 circle standard. This gives a radius for one circle and, for the other, a segment known only through congruence, and both rows of the table must be right.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 5, 40 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

Circle A is direct: its labeled radius of 5 gives an area of 25π. Circle C is where points are lost, since its size comes from a sentence rather than a label, and the segment through its center is a diameter, so its circumference is 5π while its area is only 6.25π. The row labels favor guessers, because the A and C in A = 25π and C = 5π match the circles' names, though they stand for area and circumference.

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

    Circle C's size is shown only by segment DE, drawn through center C with no label, and only the stem's sentence that DE is congruent to AB makes it a diameter of 5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know moderate Working it out

    Neither A = πr² nor C = πd is given, and one circle is measured by a radius and the other by a diameter, so each formula needs the right length. (Anderson & Krathwohl, 2001)

  3. Much to keep track of minor Working it out

    Four values are in play, areas of 25π and 6.25π and circumferences of 10π and 5π, and both rows must be matched correctly for the item to count. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write down the radius and the diameter of each circle, compute both circles' areas and circumferences in terms of π, then match each value in the table.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle
A=πr2A = \pi r^{2}
A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. Square the radius, then multiply by π.
Circumference of a circle
C=πdC = \pi d
C is the circumference, the distance around the circle. d is the diameter, the distance across the circle through its center. A diameter is twice as long as a radius. Multiply the diameter by π.

  1. Sort out the question

    Circle A has its center at A. Its radius, AB, is 5. Circle C has its center at C. Segment DE goes through C, so it is a diameter. Congruent means the same length, so DE is 5 too.

    Careful: in the table, A means area and C means circumference. Circle A and Circle C are just names.

  2. Find each radius and diameter

    Circle A: the radius is 5, so the diameter is 2×5=102 \times 5 = 10.

    Circle C: the diameter is 5, so the radius is 5÷2=2.55 \div 2 = 2.5.

    Watch out: without the sentence about DE, Circle C seems to have no size. Then you might pick Neither for the second row.

  3. Check the area row

    Use the area formula for each circle. For Circle A:

    A=π×52=25πA = \pi \times 5^{2} = 25\pi

    For Circle C:

    A=π×2.52=6.25πA = \pi \times 2.5^{2} = 6.25\pi

    Only Circle A has an area of 25π.

    Watch out: squaring Circle C’s diameter instead of its radius gives 25π by mistake.

  4. Check the circumference row

    Use the circumference formula for each circle. For Circle A:

    C=π×10=10πC = \pi \times 10 = 10\pi

    For Circle C:

    C=π×5=5πC = \pi \times 5 = 5\pi

    Only Circle C has a circumference of 5π.

    Watch out: π times the radius is not the circumference. It would give Circle A 5π.

  5. Fill in the table

    Choose one button in each row. In the row for A = 25π, choose the Circle A column. In the row for C = 5π, choose the Circle C column. Neither stays empty.

    So the answers are A = 25π: Circle A and C = 5π: Circle C.

Check your answer

Compare sizes. Circle A’s radius, 5, is twice Circle C’s radius, 2.5. So the distance around Circle C should be half of Circle A’s 10π. Half of 10π is 5π, which matches. And the bigger circle has the bigger area, 25π.

More questions like this

The original DOE question
Question 20 as shown on the DOE’s digital test, page 93 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 93, question 20.

Question 9 of 108

2026 Practice Test C Math, question 23

Question

Parallelogram PQRS, with Q and R at the top and P and S at the bottom; it leans to the right. A vertical segment QT is drawn from Q down to point T on side PS, with a right-angle mark at T, and labeled 18 cm. Side PS is labeled 34 cm, and side RS is labeled 22 cm.

On parallelogram PQRS above, the length of QT\overline{QT} is 18 centimeters. What is the area of the parallelogram?

  • A112 sq cm
  • B396 sq cm
  • C612 sq cm
  • D748 sq cm
Show the answer

Correct answer: C

PQRS is a parallelogram in which PS is a base and QT is an altitude. The base is 34 cm in length, and the height is the length of the altitude, so the height is 18 cm.

To determine the area of a parallelogram, multiply the base × the height:

34 cm×18 cm=612 cm234\text{ cm} \times 18\text{ cm} = 612\text{ cm}^{2}

Question analysis

Parallelogram PQRS is drawn with a 34 cm base PS, a 22 cm slanted side and an 18 cm segment QT at a right angle to the base, and students find its area. It tests the area formula for a parallelogram. One labeled length is not needed, so the base and height that go together have to be chosen.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 The area of a parallelogram is grade 6 content, used here once with the height already drawn and labeled. Grade 7 area tasks usually involve circles, composite shapes or scale drawings, so this is easier.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 5, 25 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 multiplication answers the question once the right pair of lengths is chosen. Each wrong choice pairs or combines the labeled lengths in another way: the two sides as for a rectangle, the height with the slanted side, or all four sides as a perimeter. Following the right-angle mark at T from QT down to PS identifies the base and height that belong together.

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

    Three lengths are labeled, 34 cm, 18 cm and 22 cm, but only the base PS and the height QT meet at the right-angle mark; the slanted side RS is not needed. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The formula, base times height, is not given, and the height must be the perpendicular distance between two sides, not the length of a slanted side. (Anderson & Krathwohl, 2001)

Strategy: Multiply the base by the height measured at a right angle to it, and check that the length you use as the height meets that base at a right-angle mark.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a parallelogram
A=b×hA = b \times h
A is the area. b is the base, the side the shape sits on. h is the height, the straight-up distance from the base to the opposite side, at a right angle to the base. It works because you can cut a triangle off one end, move it to the other end, and make a rectangle.

  1. Sort out the question

    PQRS is a parallelogram, a four-sided shape whose opposite sides are parallel. The figure shows three lengths. Side PS, the bottom, is 34 cm. Segment QT goes straight down from Q to PS and is 18 cm. Side RS is slanted and is 22 cm. You need the area, in square centimeters.

  2. Pick the base and the height

    The base is PS, 34 cm. The height must meet the base at a right angle. The small square at T shows that QT does. So the height is QT, 18 cm.

    Watch out: the 22 cm side is slanted, so it is not the height. Multiplying 22 × 18 gives 396, choice B.

  3. Multiply the base by the height

    Use the area formula:

    A=34×18A = 34 \times 18

    Break it up to multiply:

    34×18=34×20−34×2=680−68=61234 \times 18 = 34 \times 20 - 34 \times 2 = 680 - 68 = 612

    The area is 612 square centimeters.

  4. Choose the answer

    Watch out: 34 × 22 = 748, choice D, treats the shape like a rectangle. Adding all four sides gives 112, choice A. That is the perimeter, the distance around.

    So the answer is C, 612 sq cm.

Check your answer

Estimate: 34 is about 30 and 18 is about 20, and 30×20=60030 \times 20 = 600. 612 is close. Also, the area must be less than 34×22=74834 \times 22 = 748, because the height is shorter than the slanted side.

More questions like this

The original DOE question
Question 23 as shown on the DOE’s digital test, page 96 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 96, question 23.

Question 10 of 108

2026 Practice Test C Math, question 26

Question

A coordinate grid with the x-axis and y-axis each from 0 to 10, grid lines every 1 unit, labeled at 2, 4, 6, 8 and 10, and the origin labeled O. Four straight lines are drawn: one through M (0, 8) and S (8, 0); one through V (2, 3) and P (3, 6); a horizontal one through V (2, 3) and Q (6, 3); and one drawn from the origin O through T (4, 1) and R (8, 2).

Which straight line is the graph of a proportional relationship?

  • AMS↔\overleftrightarrow{\mathrm{MS}}
  • BVP↔\overleftrightarrow{\mathrm{VP}}
  • CVQ↔\overleftrightarrow{\mathrm{VQ}}
  • DTR↔\overleftrightarrow{\mathrm{TR}}
Show the answer

Correct answer: D

The graph of any proportional line must intersect the origin, (0,0)(0, 0).

The only line on the coordinate plane that contains (0,0)(0, 0) is TR↔\overleftrightarrow{\mathrm{TR}}.

Both point T(4,1)\mathrm{T}\,(4, 1) and point R(8,2)\mathrm{R}\,(8, 2) have a y-value that is 14\frac{1}{4} of the x-value, and every point on the line fits the proportional equation y=14xy = \frac{1}{4}x.

When x=0, y=14(0)x = 0,\ y = \frac{1}{4}(0). So the line contains the origin, (0,0)(0, 0).

Question analysis

A coordinate grid shows four straight lines, each named by two labeled points such as M and S or V and Q, and students choose the one that is the graph of a proportional relationship. It tests recognizing proportional relationships on a graph. Every line is straight, so being straight is not enough, and the lines must be told apart by what defines proportionality.

NYS standards
  • 7.RP.2a testing for proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Deciding from a graph whether a relationship is proportional is a named grade 7 skill, and this is a typical case: four lines on a clean grid, each checked against the same rule, with no calculation needed.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 9, 14 words.
Corrected from the DOE handbook
  • Coordinates and an equation in the explanation were corrected to match the graph.
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

Once the origin rule is recalled, the graph answers the question at a glance: only the line from O through T and R starts at (0, 0). Students who think of proportional as any steady pattern have three other candidates, from the falling line through M and S to the flat line at height 3. The steep line through V and P is the closest call, because its lower end disappears near the corner where the axes meet.

  1. Something you must already know minor Working it out

    The rule that a proportional graph is a straight line through the origin is not printed, and all four lines are straight, so straightness alone rules none of them out. (Anderson & Krathwohl, 2001)

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

    Line VP runs off the grid just right of O, crossing the x-axis at (1, 0), so whether it passes through the origin has to be read from the gridlines, not from its general direction. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Recall what every proportional graph must look like, then test each line: see where it would be when x is 0, and whether y divided by x stays the same at both labeled points.

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.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation says “Both point T (1, 4) and point R (2, 8) have a y-value that is 4 times the x-value, and every point on the line fits the proportional equation y = 4x”, and then “When x = 0, y = 4(0).” On the graph, T is at (4, 1) and R at (8, 2): each y-value is 1/4 of its x-value, and the line through the origin, T and R is y = (1/4)x. (The line y = 4x would pass through (1, 4) and (2, 8), which are not marked.) The 2022–23 guide, which first printed this question with the same graph, has the same slip. The site now gives T (4, 1) and R (8, 2), “a y-value that is 1/4 of the x-value”, y = (1/4)x, and y = (1/4)(0) at x = 0. The answer is unchanged: TR is still the only line through the origin.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship
y=kxy = kx
y and x are two amounts that change together. k is the constant of proportionality: the number you always multiply x by to get y. When x is 0, y is k times 0, which is 0. So the graph is a straight line through the origin, the point (0, 0) where the axes meet.

  1. Sort out the question

    The grid shows four straight lines. Each one is named by two labeled points on it: line MS, line VP, line VQ and line TR.

    You need the line that shows a proportional relationship. In that kind of relationship, y is always the same number times x. For example, y might always be 3 times x.

  2. Recall what a proportional graph looks like

    Use the rule y=kxy = kx. Put in 0 for x:

    y=k×0=0y = k \times 0 = 0

    So every proportional graph goes through the origin. That is the corner point (0, 0), labeled O, where the two axes meet. The graph must also be a straight line.

    All four lines here are straight. So being straight does not settle it. Look for the line through O.

  3. Check where each line starts

    • Line MS meets the y-axis at M (0, 8). When x is 0, y is 8, not 0.
    • Line VP is steep. Its lower end crosses the x-axis at (1, 0), one unit to the right of O.
    • Line VQ is flat at a height of 3. It never comes down to 0.
    • Line TR starts at O and runs up through T and R.

    Watch out: the lower end of line VP looks close to O. Follow it along the grid lines. It misses O by one unit, so choice B is wrong.

  4. Test the ratio on line TR

    In a proportional relationship, the ratio y ÷ x is the same at every point. T is at (4, 1) and R is at (8, 2):

    T: 14R: 28=14\text{T: } \frac{1}{4} \qquad \text{R: } \frac{2}{8} = \frac{1}{4}

    Both give 14\frac{1}{4}. So each y-value is 14\frac{1}{4} of its x-value, and the line is y=14xy = \frac{1}{4}x.

    Watch out: line VQ keeps the same y-value, 3, but not the same ratio. At V it is 32\frac{3}{2}, and at Q it is 36=12\frac{3}{6} = \frac{1}{2}. So choice C is wrong.

  5. Give the answer

    Line MS starts at 8 on the y-axis, so choice A is out too. Only line TR goes through the origin and keeps y ÷ x the same.

    So the answer is D, line TR.

Check your answer

Put the origin into the rule for line TR: y=14×0=0y = \frac{1}{4} \times 0 = 0. So (0, 0) is on the line. Then test R: 14×8=2\frac{1}{4} \times 8 = 2, which matches R (8, 2). The rule fits every labeled point on the line.

More questions like this

The original DOE question
Question 26 as shown on the DOE’s digital test, page 99 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 99, question 26. This is the DOE’s version, before the corrections listed in the analysis.

Question 11 of 108

2026 Practice Test C Math, question 27

Question

A graph titled ICE COSTS, with Weight of Ice (lb) on the x-axis from 0 to 20, labeled every 2, and Cost (dollars) on the y-axis from 0 to 3.0, labeled every 0.3, with a grid line at each label. A straight line starts at the origin, O, and rises 0.3 dollars for every 2 pounds, passing through (4, 0.6) and (20, 3.0).

Some friends are buying bags of ice for a picnic. The graph shows the cost of the ice, in dollars, in relation to the weight, in pounds. Which statement about a point on the graph is true?

  • AThe point (20, 3) means that 3 lbs of ice costs $20.00.
  • BThe point (1, 0.15) means that 1 bag of ice costs $0.15.
  • CThe point (10, 1.5) means that 10 lbs of ice costs $1.50.
  • DThe point (0, 0) means that 0 lbs of ice will be needed if 0 people go to the picnic.
Show the answer

Correct answer: C

The x-axis represents the weight, in pounds (lbs), of ice purchased, and the y-axis represents the cost of the ice, in dollars.

The x-value in the point (10, 1.5) represents 10 lbs of ice, and the y-value represents 1.5 dollars, or $1.50. Therefore, the cost of 10 lbs of ice is $1.50.

Question analysis

Some friends buying ice for a picnic use a graph of the ice's cost in dollars against its weight in pounds, a straight line from the origin. Students pick the statement that correctly explains what a point on the graph means. Each statement names a point that really is on the line, so the units and the order of the coordinates decide.

NYS standards
  • 7.RP.2d points on proportional graphs
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Explaining what a point on a proportional graph means in its situation is a named grade 7 skill. The graph is clear and starts at the origin, but each statement has to be read closely against the axis labels.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 2, 77 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Nothing needs computing, but each statement attaches words to two numbers, and three of them attach the wrong words. The point (1, 0.15) is the unit rate students are trained to look for, and the “bags” in the stem’s story make its statement sound right, though the graph measures pounds. Saying each point aloud as so many pounds for so many dollars exposes the swapped coordinates and the extra people as well.

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

    Each statement must be checked against the axis labels: x is the weight of ice in pounds and y the cost in dollars, on a grid ruled every 2 pounds and every $0.30. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Every choice must be checked minor Choosing the answer

    All four statements name a point that really lies on the line, (1, 0.15) included, so being on the graph rules nothing out; only the meaning given to each coordinate does. (Embretson & Wetzel, 1987)

Strategy: Read both axis labels and their units first, then for each choice say in words what its x-coordinate and its y-coordinate stand for, and compare that with the statement.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • 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

Unit rate
cost per pound=costweight\text{cost per pound} = \frac{\text{cost}}{\text{weight}}
The cost is in dollars, and the weight is in pounds. The unit rate is the cost of just 1 pound. On a straight line through the origin, it is the same at every point. So you can find it from any point you can read.

  1. Sort out the question

    The graph is called ICE COSTS. Read both axes first:

    • The x-axis, across the bottom, is the weight of the ice in pounds (lb).
    • The y-axis, up the side, is the cost in dollars.

    A point is written (x, y), weight first and cost second. So a point means: x pounds of ice cost y dollars. You need the statement that says this correctly.

  2. Find the cost of one pound

    The line reaches $3.00 at 20 pounds. Use the unit rate formula:

    cost per pound=320=0.15\text{cost per pound} = \frac{3}{20} = 0.15

    So 1 pound of ice costs $0.15. This helps you test each point.

  3. Test the point in choice C

    Find 10 on the weight axis and go straight up to the line. The line crosses there at 1.5 on the cost axis. The unit rate agrees:

    10×0.15=1.510 \times 0.15 = 1.5

    So the point (10, 1.5) means 10 pounds of ice cost $1.50. That is exactly what choice C says.

  4. Check the other statements

    Each one names a real point on the line. The problem is the meaning it gives the numbers.

    Watch out: choice A reads (20, 3) backward. It means 20 pounds cost $3, not 3 pounds cost $20.

    Watch out: choice B is tricky. The point (1, 0.15) is on the line, but the x-axis counts pounds, not bags. So it means 1 pound costs $0.15.

    Choice D talks about people, but the graph never counts people. The point (0, 0) means 0 pounds cost $0.

  5. Give the answer

    Only choice C gives the right meaning to both numbers.

    So the answer is C, 10 lbs of ice costs $1.50.

Check your answer

Use halves. The point (20, 3) says 20 pounds cost $3.00. Half as much ice, 10 pounds, should cost half as much. Half of $3.00 is $1.50. That matches choice C.

More questions like this

The original DOE question
Question 27 as shown on the DOE’s digital test, page 100 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 100, question 27.

Question 12 of 108

2026 Practice Test C Math, question 31

Question

NUMBER OF VISITORS
TO BROOKSIDE ZOO

Male Female
Children 360 340
Adults 210 290

The table above shows the number of people who visited Brookside Zoo in one day. The largest of the four groups made up what percentage of all visitors on that day?

  • A30%
  • B28%
  • C24%
  • D17%
Show the answer

Correct answer: A

First, determine the total number of visitors.

360+340+210+290=1200360 + 340 + 210 + 290 = 1200

The largest group is 360 (Male Children).

Find the percent that represents 3601200\frac{360}{1200}.

3601200×1001%=36012×11%=30%\frac{360}{1200} \times \frac{100}{1}\% = \frac{360}{12} \times \frac{1}{1}\% = 30\%

Question analysis

A two-way table counts one day's visitors to Brookside Zoo by age, children or adults, and by male or female. Students find what percent of all visitors the largest of the four groups made up. They must total every cell first, then make sure the count they divide is really the largest one.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding what percent one count is of a total is grade 6 work. A two-way table adds two quick steps, totaling the four groups and spotting the largest, but the numbers divide evenly, so it stays below grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is friendly, since 360 over 1,200 reduces to 3 over 10, so the work is in reading the table. Two of its cells are close, 340 just below 360, and every wrong choice is the share of one of the other groups. Adding all four counts first, and circling the largest before dividing, keeps the right cell in play.

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

    The two-way table gives four separate counts and no totals, so the 1,200 visitors must be added up from the cells and the largest group picked out of a grid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three short steps run in order: total the four cells, pick the largest, and divide; using the wrong cell in the last step still gives a believable percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add all four cells for the day's total, pick the biggest single cell, and divide it by the total; simplifying the fraction before dividing keeps the arithmetic short.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a whole
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the one group you care about. The whole is everyone. The fraction tells what share of everyone the group is. Multiplying by 100 turns that share into a percent, which means out of 100.

  1. Sort out the question

    The table splits one day’s visitors into four groups:

    MaleFemale
    Children360340
    Adults210290

    You need the largest group as a percent of all the visitors. The table gives no total, so you must find it.

  2. Add up all the visitors

    Add all four groups to get the whole:

    360+340+210+290=1,200360 + 340 + 210 + 290 = 1{,}200

    So 1,200 people visited the zoo that day.

  3. Find the largest group

    Compare the four numbers. The biggest is 360, the male children.

    Watch out: 340 is close to 360, but it is smaller. Using 340 gives about 28%, choice B. Using the 290 female adults gives about 24%, choice C. And 210 is the smallest group, which gives 17.5%, near choice D.

  4. Find the percent

    Put the part over the whole. Then simplify by dividing the top and bottom by 120:

    3601,200=310\frac{360}{1{,}200} = \frac{3}{10}

    Now multiply by 100 to get a percent:

    310×100=30\frac{3}{10} \times 100 = 30

    So the male children made up 30% of the visitors.

  5. Give the answer

    The largest group, 360 male children, is 30% of all 1,200 visitors.

    So the answer is A, 30%.

Check your answer

Work backward. 30% of 1,200 is 0.30×1,200=3600.30 \times 1{,}200 = 360, the male children. Also, with four groups an even split would be 25% each. The largest group should be a bit more than 25%, and 30% is.

More questions like this

The original DOE question
Question 31 as shown on the DOE’s digital test, page 104 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 104, question 31.

Question 13 of 108

2026 Practice Test C Math, question 32

Question

A box plot titled MONTHLY HIGH TEMPERATURES IN NEW YORK above a number line labeled Temperature (°F), from 36 to 88 with a labeled tick every 4. The left whisker starts at about 39, just left of the tick for 40. The box runs from 48 to about 78, between the ticks for 76 and 80, with the median line at 64. The right whisker ends at 84.

The box plot shows the average monthly high temperatures in New York City for 12 months. What is the difference between the range and the interquartile range of the temperature data?

Enter your answer in the space.

Fill-in question: work out the answer and type it as a number.

Show the answer

Correct answer: 15

Determine the range by subtracting the minimum value on the box plot from the maximum value on the box plot:

Maximum value =84∘ F= 84^{\circ}\text{ F}, Minimum value =39∘ F= 39^{\circ}\text{ F}

84∘ F−39∘ F=45∘ F84^{\circ}\text{ F} - 39^{\circ}\text{ F} = 45^{\circ}\text{ F}

Determine the interquartile range by subtracting the lower quartile value on the box plot from the upper quartile value on the box plot:

Upper Quartile Value =78∘ F= 78^{\circ}\text{ F}, Lower Quartile Value =48∘ F= 48^{\circ}\text{ F}

78∘ F−48∘ F=30∘ F78^{\circ}\text{ F} - 48^{\circ}\text{ F} = 30^{\circ}\text{ F}

Determine the difference between the range (45° F) and the interquartile range (30° F):

45∘ F−30∘ F=15∘ F45^{\circ}\text{ F} - 30^{\circ}\text{ F} = 15^{\circ}\text{ F}

Question analysis

A box plot of New York City's average monthly high temperatures for 12 months sits above a number line labeled every 4 degrees. Students type the difference between the range and the interquartile range of the data. Two of the plot's values fall between labeled ticks, so careful reading of the scale decides the answer.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Reading a box plot and finding its interquartile range are grade 7 skills. Here two of the five values fall between ticks spaced 4 apart, and the range, the IQR and their difference must all be found and typed with no choices.
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 7, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is standard, but the key rests on reading the scale between its labels. The axis is ticked only every 4 degrees, and the minimum and the upper quartile land off the ticks, so rounding either one to the nearest label moves the result by a degree or two with nothing to flag it. Students who find the range correctly can also stop there, entering 45, the larger of the two spreads, instead of the difference the question asks for.

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

    Two of the five values fall between ticks spaced 4 degrees apart: the left whisker ends just short of the 40 tick, at 39, and the box’s right edge sits midway between 76 and 80, at 78. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Four values are read and combined in three subtractions, the range, the interquartile range and then their difference, so one misread value changes the final entry. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    Neither measure is defined on the page: the range spans the whiskers, and the interquartile range is the width of the box, the upper quartile minus the lower. (Anderson & Krathwohl, 2001)

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

    The result is typed with no choices to compare, so an entry one or two degrees off, from rounding a value to the nearest tick, looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Read the minimum, both quartiles and the maximum, judging where each mark falls between the ticks, then subtract for the range and for the IQR, and subtract those two results.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range
range=maximum−minimum\text{range} = \text{maximum} - \text{minimum}
The maximum is the greatest value, at the right end of the right whisker. The minimum is the least value, at the left end of the left whisker. The range tells how far the whole set of data spreads.
Interquartile range (IQR)
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q3 is the upper quartile, the right edge of the box. Q1 is the lower quartile, the left edge of the box. The IQR is the width of the box. It tells how far the middle half of the data spreads.

  1. Sort out the question

    A box plot shows data with five numbers. The two whiskers are the lines on each side of the box. Their ends are the minimum, the least value, and the maximum, the greatest. The box runs from the lower quartile to the upper quartile. These are the middles of the lower half and the upper half of the data. The line inside the box is the median, the middle value.

    You need the range minus the interquartile range (IQR), in degrees.

  2. Read the values between the ticks

    The ticks on the number line are 4 degrees apart. So each degree is one quarter of the space between two ticks. Three values sit right on ticks: the box’s left edge at 48, the median at 64, and the right whisker’s end at 84.

    • The left whisker starts just left of the 40 tick. It sits about one quarter of the way back toward 36. So the minimum is 40−1=3940 - 1 = 39 degrees.
    • The box’s right edge is halfway between 76 and 80. Halfway is 78 degrees.

  3. Find the range

    Use the range formula:

    range=84−39=45\text{range} = 84 - 39 = 45

    The data spread over 45 degrees.

    Watch out: rounding the minimum to the 40 tick gives a range of 44. Then the final answer comes out 14, which is wrong.

  4. Find the IQR

    Use the IQR formula with the two edges of the box:

    IQR=78−48=30\text{IQR} = 78 - 48 = 30

    The middle half of the data spreads over 30 degrees.

    Watch out: reading the box’s right edge as 76 gives an IQR of 28. Then the final answer comes out 17.

  5. Subtract the IQR from the range

    The question asks for the difference between the two:

    45−30=1545 - 30 = 15

    Watch out: 45 is only the range. Do not stop there. Type only the number, with no degree sign.

    So you would enter 15.

Check your answer

The range covers the whole plot, and the IQR covers only the box. So their difference is the two whiskers added together. The left whisker is 48−39=948 - 39 = 9 degrees long. The right whisker is 84−78=684 - 78 = 6 degrees long. 9+6=159 + 6 = 15, which matches.

More questions like this

The original DOE question
Question 32 as shown on the DOE’s digital test, page 105 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 105, question 32.

Question 14 of 108

2026 Practice Test C Math, question 34

Question

A teacher will randomly select 3 students to be on a school committee. He will choose either a student from Classroom A or a student from Classroom B with each selection. This tree diagram shows the selection possibilities.

A tree diagram with three columns of boxes. From a starting point, two arrows lead to boxes A and B. From each of those, two arrows lead to boxes A and B, four boxes in all. From each of those four, two arrows lead to boxes A and B, so the last column has eight boxes: A, B, A, B, A, B, A, B.

What is the probability that the teacher will select 2 students from Classroom A and 1 student from Classroom B (in any order) for the committee?

  • A18\frac{1}{8}
  • B38\frac{3}{8}
  • C12\frac{1}{2}
  • D34\frac{3}{4}
Show the answer

Correct answer: B

The tree diagram shows that each of the 3 randomly selected students is from Classroom A or Classroom B. So the first chosen is either A or B, represented by the left column of the diagram.

The second student is either A or B, so if the first is A, the first two chosen will be either AA or AB. If the first is B, then the first two chosen will be either BA or BB, represented by the middle column.

The third student is either A or B, so if the first 2 were AA, the group of 3 would be either AAA or AAB. These outcomes are represented by the top two boxes of the right column. This pattern is followed so that 8 distinct outcomes can be visualized in order:

AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB

Since the order does not matter, there are 3 groups made up of 2 students from Classroom A and 1 student from Classroom B: AAB, ABA, and BAA.

Out of 8 possible outcomes, 3 outcomes have 2 students from Classroom A and 1 student from Classroom B. Therefore, the probability is 3 out of 8, or 38\frac{3}{8}.

Question analysis

A teacher picks 3 students for a committee, each from Classroom A or Classroom B, and a tree diagram shows every possible sequence of picks. Students find the probability of getting 2 students from A and 1 from B in any order. Reading the tree as a list of outcomes, and counting each order, is the key.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a compound probability from a tree diagram is a grade 7 standard, and here the tree is already drawn. Listing its paths and counting the ones that fit the event, in any order, is typical grade 7 classwork.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 7, 65 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tree does the hard part of the setup, but it shows each outcome only as a path through three columns, not as a list. The parenthesis “in any order” is the key phrase: missing it leaves a single path, while reading the event loosely, as at least two from Classroom A, adds A, A, A. Writing out all eight paths and marking those with exactly two A’s settles the count.

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

    The tree prints only A or B in each box, so the eight three-pick outcomes, such as A, B, A, have to be read by tracing each path across all three columns. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Much to keep track of minor Working it out

    All eight paths must be tracked while counting those with exactly two A’s and one B, and “in any order” means three of them count, not just A, A, B. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Follow each path through the tree to write out every three-letter outcome, mark the ones with exactly two A's and one B, and divide by the total number of outcomes.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=outcomes you wantall possible outcomesP = \frac{\text{outcomes you want}}{\text{all possible outcomes}}
P is the probability, the chance something happens. The top counts the outcomes that fit the event. The bottom counts every outcome that can happen. This works when every outcome is equally likely, like each path through this tree.

  1. Sort out the question

    The teacher picks 3 students, one at a time. Each pick is from Classroom A or Classroom B.

    You need the chance of getting 2 students from A and 1 from B. The words “in any order” matter. They mean the B can be the first, second or third pick.

  2. List every path in the tree

    Each path from left to right is one outcome, or possible result. The top path is A, then A, then A. Read all of them:

    AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB

    Each pick has 2 choices, so there are 2×2×2=82 \times 2 \times 2 = 8 outcomes in all.

  3. Mark the ones that fit

    Look for exactly two A’s and one B. Three outcomes fit: AAB, ABA and BAA.

    Watch out: counting only AAB misses “in any order” and gives 18\frac{1}{8}, choice A. AAA has three A’s, not two. Counting it too gives 48=12\frac{4}{8} = \frac{1}{2}, choice C. Counting every mixed outcome gives 68=34\frac{6}{8} = \frac{3}{4}, choice D, but three of those have two B’s.

  4. Find the probability

    Use the probability formula. 3 outcomes fit, out of 8:

    P=38P = \frac{3}{8}

    So the answer is B, 38\frac{3}{8}.

Check your answer

Count every outcome by type. Two A’s and one B: 3 outcomes. Two B’s and one A: 3 outcomes (ABB, BAB, BBA). All A’s: 1. All B’s: 1. 3+3+1+1=83 + 3 + 1 + 1 = 8, so each outcome was counted once.

More questions like this

The original DOE question
Question 34 as shown on the DOE’s digital test, page 107 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 107, question 34.

Question 15 of 108

2026 Practice Test C Math, question 42

Question

Tariq is making a clay block in the shape of a cube with a side length of 20 cm20\text{ cm}. He hollows out the center by removing clay to form a rectangular prism-shaped hole with a base of 5 cm⋅5 cm5\text{ cm} \cdot 5\text{ cm} and a height of 20 cm20\text{ cm}.

A cube labeled 20 cm along its front right edge. A tall rectangular prism-shaped hole runs from the top face of the cube down to the bottom face, the full height of the cube; its square opening on the top face is shaded gray and labeled 5 cm across. Hidden edges of the cube and the hole are dashed.

What is the volume, in cubic centimeters, of the block after the clay is removed from the center?

Enter your answer in the space.

Fill-in question: work out the answer and type it as a number.

Show the answer

Correct answer: 7,500

First, find the volume of the cube using the formula V=lwhV = lwh.

The length, width, and height of the cube are each equal to 20 cm.

Vc=20 cm×20 cm×20 cm=8,000 cm3V_c = 20\text{ cm} \times 20\text{ cm} \times 20\text{ cm} = 8{,}000\text{ cm}^3

Next, find the volume of the cut-out square prism, with equal length and width of 5 cm and height of 20 cm.

Vp=5 cm×5 cm×20 cm=500 cm3V_p = 5\text{ cm} \times 5\text{ cm} \times 20\text{ cm} = 500\text{ cm}^3

Finally, subtract the volume of the prism from the volume of the cube.

8,000 cm3−500 cm3=7,500 cm38{,}000\text{ cm}^3 - 500\text{ cm}^3 = 7{,}500\text{ cm}^3

Question analysis

Tariq makes a clay cube 20 centimeters on each side and removes a rectangular prism from its center, with a 5 cm by 5 cm base and a height of 20 cm. Students type the volume of the block that is left, in cubic centimeters. It tests the volume of rectangular prisms, and the hole's size must be read carefully from the words and figure.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.2 volume with fractional edge lengths
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the volume of a solid made from rectangular prisms, here by taking one away, is named in the grade 7 standards. This is a direct case with whole numbers, two volumes and one subtraction.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 8, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a direct composite-volume problem with whole numbers: one cube, one hole, one subtraction. What it checks is that both volumes are real volumes, the hole as tall as the cube, and that the subtraction is done at the end. Students who stop at the hole’s volume, or who subtract only the square face, give amounts that are easy to type and hard to doubt without choices.

  1. Something you must already know minor Working it out

    The volume of a rectangular prism, length times width times height, is not given, and it is needed twice, once for the 20 cm cube and once for the hole. (Anderson & Krathwohl, 2001)

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

    The figure labels only the 5 cm opening and the cube’s 20 cm edge; the hole’s 20 cm height comes from the words, while the drawing shows it reaching from top to bottom. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

    The volume is typed with no choices to compare, so the hole’s 500 cubic centimeters or the cube’s 8,000 can be entered without any warning. (Enright, Morley & Sheehan, 2002)

Strategy: Find the volume of the full cube, then the volume of the square hole that runs from the top face to the bottom, and subtract the hole from the cube.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a rectangular prism
V=l×w×hV = l \times w \times h
V is the volume, the space inside a solid. l is the length, w is the width and h is the height. A cube is a prism with all three the same. Volume is measured in cubic centimeters, tiny cubes 1 cm on each side.

  1. Sort out the question

    Tariq’s block starts as a cube 20 cm on each side. He removes a hole shaped like a rectangular prism, a box shape with flat rectangle sides. Its base is 5 cm by 5 cm, and its height is 20 cm. The figure shows it running from the top face to the bottom face.

    You need the volume of the clay that is left, in cubic centimeters.

  2. Find the volume of the cube

    Use the volume formula. The length, width and height are all 20 cm:

    V=20×20×20=8,000V = 20 \times 20 \times 20 = 8{,}000

    The full cube holds 8,000 cubic cm of clay.

  3. Find the volume of the hole

    The hole is a prism too. Its length and width are 5 cm, and its height is 20 cm:

    V=5×5×20=500V = 5 \times 5 \times 20 = 500

    So 500 cubic cm of clay are removed.

    Watch out: the hole is 20 cm tall, not 5 cm. Treating it as a small 5 cm cube gives 125 and the wrong answer 7,875. And 5 × 5 = 25 is only the area of the opening, not a volume. Subtracting it gives 7,975, also wrong.

  4. Subtract the hole from the cube

    The clay left is the cube minus the hole:

    8,000−500=7,5008{,}000 - 500 = 7{,}500

    The block has 7,500 cubic cm of clay left.

    Watch out: 500 is the clay taken out, not the clay that is left.

  5. Enter the number

    Type only the number, with no units. You can type 7500 or 7,500.

    So you would enter 7,500.

Check your answer

Think of the cube as 20 flat layers, each 1 cm thick. Each layer is a 20 by 20 square, 400 square cm, with a 5 by 5 hole, 25 square cm. So each layer holds 400−25=375400 - 25 = 375 cubic cm of clay. Then 20×375=7,50020 \times 375 = 7{,}500. It matches.

More questions like this

The original DOE question
Question 42 as shown on the DOE’s digital test, page 115 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 115, question 42.

Question 16 of 108

2026 Practice Test D English, question 5
Passage A Memory Revolution

A Memory Revolution

A high school senior logs on to a computer at the library to double-check the application due date listed on a college admissions website. Nearby, a librarian helps a group of biology students use a database to search for recent studies about mammals. In the past few decades, the Internet has become an integral component of daily life for many people. The seemingly limitless power of search engines made the Internet search extremely common, and today people increasingly rely on the Internet’s vast accumulation of sources to access all types of information. Scientists are beginning to examine how this reliance is modifying the strategies people use to store and prioritize information in their mind.

A Dependable and Valuable Asset

Psychologist Benjamin Storm from the University of California, Santa Cruz, and researchers Sean Stone and Aaron Benjamin devised an experiment to study students’ tendency to depend on the Internet for facts. To begin, the scientists divided sixty participants into multiple groups, including an “Internet” group and a “memory” group, and placed them in front of computers. The Internet group was required to use the search engine Google to answer eight challenging trivia questions. In contrast, participants in the memory group were permitted to use only their personal knowledge to answer the questions. In the second round, the researchers administered notably easier questions. This time, they allowed each group the option of using Google as they answered. Their results showed that 83 percent of the Internet group continued to consult Google in the second round, while only 63 percent of the memory group chose to do so.

People’s growing inclination to rely on the Internet in order to retrieve information, particularly facts and figures, is called cognitive offloading. Canadian researcher Evan F. Risko and British researcher Sam Gilbert, who have written extensively about the topic, say a similar process has been taking place for centuries. In the past, people used resources like encyclopedias to assist their memories; however, today the Internet, serving as a vast extended memory, allows people to digitally access and retrieve much larger volumes of information. Consequently, people’s minds are free for other cognitive feats, such as connecting data, learning new information, or solving problems.

Filing Information Away

The use of the Internet also appears to be modifying the strategies people use to store information in their mind. Researchers Betsy Sparrow from Columbia University, Jenny Liu from the University of Wisconsin-Madison, and Daniel Wegner from Harvard University conducted several studies to discover how people efficiently manage their information intake.

To begin, the researchers examined how people evaluate which information deserves their effort to remember. For this experiment, participants read forty trivia facts, such as “An ostrich’s eye is bigger than its brain,” and typed the statements into a computer file. Half the participants had been previously told the file would be saved, while half believed it would be erased. Next, the participants wrote down every fact they could recall. Those who believed the information would be erased and no longer available could recall 40 percent more facts than those who thought the information would be saved.

In another experiment, the same researchers tested the ability of study participants to remember where to access information. For this trial, participants read and typed trivia statements, which they saved in folders with generic names such as “Facts” and “Items.” After spending ten minutes writing down all the facts they could recall from memory, participants were asked which folder contained a particular fact based on a keyword. For example, “Which folder has the fact about ostriches?” Overall, participants recalled the information’s location more often than the content itself, correctly identifying 49 percent of the folders for specific facts while remembering only 23 percent of the actual trivia. The researchers concluded that our memory is adapting to the Internet age by prioritizing where to locate information even when the specific details are forgotten. According to Sparrow, the Internet has become an important form of transactive memory, an external source of the recollections and associative networks that constitute memory.

As the Internet’s resources continue to expand our “external” memory, some question whether the process may cause people to depend too heavily on technology. However, Steven Pinker, a professor of psychology at Harvard University, says, “Knowledge is increasing exponentially; human brainpower and waking hours are not. Fortunately, the Internet and information technologies are helping us manage, search and retrieve our collective intellectual output at different scales, from Twitter and previews to e-books and online encyclopedias. Far from making us stupid, these technologies are the only things that will keep us smart.”

Flow diagram titled “HOW MEMORY WORKS”: an arrow labeled “New Information” points down into a box labeled “Sensory Memory,” and an arrow labeled “Attention” leads down from that box to a box labeled “Short-Term Memory.” From Short-Term Memory, an arrow points right to the note “Unrehearsed information is forgotten,” and on its left a curved arrow labeled “Rehearsal” loops from the bottom of the box back into its top, beside the note “Information is maintained through repetition.” An arrow labeled “Storage” leads down from Short-Term Memory to a box labeled “Long-Term Memory,” a curved arrow labeled “Retrieval” leads from Long-Term Memory back up into Short-Term Memory, and an arrow points right from Long-Term Memory to the note “Some information may be forgotten over time.”

Question

How does the diagram provide additional support for the topic presented in the passage?

  • AIt reveals why human brains must adapt to obtaining information from the Internet as opposed to other sources.
  • BIt indicates how people can use the Internet to help improve their long-term recollection of information.
  • CIt shows how study participants’ brains distinguished between important and unimportant details.
  • DIt depicts the idea that repetition and rehearsal are necessary to recall information when tools such as search engines are unavailable.
Show the answer

Correct answer: D

The question asks how the diagram provides additional support for the topic presented in the passage.

A. Incorrect. While the diagram indicates that the human brain has the ability to maintain different types of memory, the diagram does not show the need for the brain to adapt to obtaining information from the Internet rather than from other sources.

B. Incorrect. The diagram enhances the reader’s understanding of how long-term memories are formed, but it does not provide details about how the Internet can be used to improve long-term recollection of information.

C. Incorrect. The diagram provides an understanding of how details that are important to people can become stored because of an effort made to recall them. However, the diagram does not show how people decide which information is important and which is not.

D. CORRECT. The diagram depicts the connections that the human brain maintains between the different forms of memory. It also illustrates the idea that repetition and rehearsal are important steps in creating memories, as unrehearsed information will be forgotten. This supports the passage by highlighting the idea that access to search engines and other tools leads people to skip rehearsal and repetition steps that create memories, because people know that the information is readily accessible.

Question analysis

This question pairs “A Memory Revolution” with its flow diagram, “How Memory Works,” which follows new information from sensory memory through short-term and long-term memory, and asks how the diagram supports the topic of the passage. Students must read the diagram’s boxes, arrows and notes and relate them to an article about the Internet that never refers to the diagram, so the connection has to be worked out.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R2 theme or central idea
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 students regularly connect a diagram to a text. This flow diagram of memory never mentions the Internet, so students must build the bridge to the article’s studies themselves, and every choice claims something the diagram alone cannot show.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from the whole text
Reading level
Passage: about grade 14, 763 words; question and choices: about grade 12, 81 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The diagram is a general model of memory, and every choice, even the right one, goes beyond it by bringing in the Internet, search engines or the study participants. The way through is to ask which choice rests on something the diagram really shows, and only one names its rehearsal loop and the forgetting of unrehearsed information. The strongest pull uses the diagram’s long-term memory and retrieval but claims the Internet improves recollection, the opposite of what paragraph 5’s experiment found.

  1. Combines ideas moderate Working it out

    The diagram never mentions the Internet and the article never mentions the diagram, so students must join its rehearsal loop to paragraph 5’s finding that people who expected the facts to be saved remembered fewer. (Embretson, 1998; Daniel & Embretson, 2010)

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

    Students must follow the arrows from “Attention” into short-term memory and find the “Rehearsal” loop beside “Information is maintained through repetition,” with “Unrehearsed information is forgotten” on the other side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Every choice must be checked minor Choosing the answer

    Three choices tie the diagram to the Internet, adaptation or the study participants, and even the key mentions search engines, so each claim must be held against the boxes and arrows actually drawn. (Embretson & Wetzel, 1987)

Strategy: Trace the diagram’s arrows from new information to long-term memory and say what each label shows, then ask which choice links that process to the studies without crediting the diagram with ideas it never shows.

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 how the diagram supports the article. The diagram is the flow chart called “HOW MEMORY WORKS” at the end of the passage. It never mentions the Internet. So you must find what it shows, then link that to the article.

  2. Follow the arrows in the diagram

    Start at the top. An arrow labeled “New Information” goes into the box “Sensory Memory.” The “Attention” arrow leads to “Short-Term Memory.” From there, two things can happen. On the left, a loop labeled “Rehearsal” sits beside the note “Information is maintained through repetition.” On the right, a note says “Unrehearsed information is forgotten.” A “Storage” arrow leads down to “Long-Term Memory,” and a “Retrieval” arrow leads back up.

  3. Explain what the diagram means

    To rehearse information is to repeat it, the way you practice lines for a play. To maintain something is to keep it. So the diagram says that information you repeat stays with you. Information you don’t repeat is lost. Repeating takes effort.

  4. Link the diagram to the passage

    Paragraph 5 describes people who typed forty facts. Some thought the file would be saved. Others thought it would be erased. Those who thought it would be erased “could recall 40 percent more facts.” They knew they could not look the facts up later. So they made the effort to remember. The diagram shows that effort: rehearsal. When search engines are at hand, people may skip it.

  5. Check each choice

    • A does not fit. The diagram never mentions the Internet. It does not say brains must adapt.
    • B does not fit. The diagram shows nothing about using the Internet. Paragraph 5 even found that people remembered less when they expected the facts to be saved.
    • C does not fit. The diagram shows memory in general. It shows no study participants and no sorting of important facts.
    • D fits. The “Rehearsal” loop shows that repeating information keeps it. Without a tool to look things up, that repeating is what lets you recall.

    Watch out: B is tempting because the diagram has “Long-Term Memory” and “Retrieval,” and the article calls the Internet an outside memory. But the diagram is about the brain, not the Internet.

  6. Choose the answer

    The diagram shows rehearsal keeping information and unrehearsed information being forgotten. That explains why people who can’t look facts up must work to remember them. So the answer is D, “It depicts the idea that repetition and rehearsal are necessary to recall information when tools such as search engines are unavailable.”

Why this answer and not the closest wrong one

B is the closest wrong choice. It uses the diagram’s “Long-Term Memory” and “Retrieval” labels. But the diagram never shows the Internet improving recall. The passage says the opposite: expecting the facts to be saved made people remember fewer. D rests on what the diagram really shows, the “Rehearsal” loop and the note “Unrehearsed information is forgotten.”

More questions like this

The original DOE question
Question 5 as shown on the DOE’s digital test, page 9 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D English, page 9, question 5.
The passage “A Memory Revolution” as printed (pages 3–4)The passage “A Memory Revolution” as printed on page 3 of the DOE handbook.The passage “A Memory Revolution” as printed on page 4 of the DOE handbook.

Question 17 of 108

2026 Practice Test D Math, question 12

Question

A circle with center P. Points R and T are on the circle at the ends of a horizontal diameter through P, and point S is on the circle above it. Segment SP is labeled 12. A vertical segment from S meets RT at a right angle, dividing RT into a part labeled x, from R to that point, and a part labeled 5, from that point to T.

P is the center of the circle above, and R, S, and T are points on the circle. What is the value of x?

  • A12
  • B15
  • C17
  • D19
Show the answer

Correct answer: D

SP\overline{\mathrm{SP}} represents the radius of the circle and has length 12.

Both RP\overline{\mathrm{RP}} and PT\overline{\mathrm{PT}} are radii and also have length 12.

Therefore, the length of RT\overline{\mathrm{RT}}, which is a diameter, is 24 and is also represented by x+5x + 5.

x+5=24x + 5 = 24

x=19x = 19

Question analysis

A circle with center P has points R, S and T on it, a segment SP labeled 12, and a vertical segment from S that splits RT into a piece labeled x and a piece labeled 5. Students find the value of x. It tests basic circle facts, and the challenge is seeing which lengths in the figure must be equal.

NYS standards
  • 7.G.4 circle area and circumference
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the fact that every radius has the same length, but spotting which segments are radii in a figure with extra markings is less routine than using a formula.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 24 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure gives one measurement on a slanted radius and two labels on the diameter, and the right angle at the foot of the vertical segment is a decoy. Once RP and PT are seen as radii, RT is 24, and x is that diameter less the 5. Students who never connect the slanted 12 to the horizontal line fall back on the labeled numbers, adding 12 and 5, or take x to end at the center.

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

    The 12 labels the slanted segment SP, not the horizontal one, and the arrow for x runs from R past the center P to the foot of the vertical segment, so x is neither a radius nor a whole diameter. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. No obvious method minor Working it out

    The right-angle mark and the triangle under S suggest the Pythagorean theorem, which is no help; the way in is seeing that RP and PT are radii as well. (Schoenfeld, 1985)

Strategy: Mark every segment from P to the circle as 12, notice that R and T are the ends of a diameter, then write an equation for that diameter using x and the 5.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius and diameter
d=2rd = 2r
r is the radius, the distance from the center to any point on the circle. d is the diameter, the distance straight across the circle through the center. A diameter is two radii laid end to end, one on each side of the center, so it is twice as long.

  1. Sort out the question

    P is the center of the circle, and R, S and T are on the circle. Segment SP is 12. Segment RT goes straight across through P. A short line from S meets RT at a right angle. It splits RT into x, from R to that point, and 5, from that point to T. You need x.

  2. See that SP is a radius

    SP runs from the center P to the point S on the circle. So SP is a radius, and the radius is 12. Every radius of a circle has the same length. RP and PT also run from the center to the circle, so RP = 12 and PT = 12.

  3. Find the diameter RT

    RT passes through the center, so it is a diameter. Use the formula with r = 12:

    RT=2r=2×12=24\mathrm{RT} = 2r = 2 \times 12 = 24

    RT is 24 long. It is the radius RP plus the radius PT.

  4. Solve for x

    RT is also split into x and 5, so:

    x+5=24x + 5 = 24

    Subtract 5 from both sides:

    x=24−5=19x = 24 - 5 = 19

    Watch out: x runs from R past the center P, so it is longer than a radius. Reading it as 12, choice A, stops at P. Adding 12 + 5 = 17, choice C, puts the 12 in the wrong place.

  5. Work from the numbers, not the picture

    The figure is not drawn to scale. In it, x looks about three times as long as 5, which suggests 15, choice B. Trust the labels instead. The right triangle under S is not needed at all.

    So the answer is D, 19.

Check your answer

The point where the short line meets RT is 19 from R. P is 12 from R, so that point is 19 − 12 = 7 past the center. Then 7 + 5 = 12, which is PT, a radius. Everything fits.

More questions like this

The original DOE question
Question 12 as shown on the DOE’s digital test, page 85 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 85, question 12.

Question 18 of 108

2026 Practice Test D Math, question 28

Question

A graph shows the relationship between the number of gallons of water, y, that has been added to a tank and the number of hours, x, that water has been added to the tank at a constant rate. What does the ordered pair (3, 24) represent?

  • AThree gallons of water was added in 24 hours.
  • BThree gallons of water was added per hour.
  • CTwenty-four gallons of water was added in 3 hours.
  • DTwenty-four gallons of water was added per hour.
Show the answer

Correct answer: C

It is given that x-values on the graph represent the numbers of hours for which water was added to the tank, and the y-values represent the numbers of gallons of water added in x hours. In the ordered pair (3, 24), 3 represents an x-value and 24 represents a y-value.

Therefore, (3, 24) represents a data point for 24 gallons of water being added to the tank in a 3-hour period.

Question analysis

A graph relates the gallons of water added to a tank, y, to the hours it has been filling at a constant rate, x. Students choose what the ordered pair (3, 24) means in the situation. It tests interpreting coordinates in context, and the four statements mix totals with rates and switch the two numbers around.

NYS standards
  • 7.RP.2d points on proportional graphs
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Explaining what a point on the graph of a proportional relationship means is named in the grade 7 standards. This is a direct case with no graph to read, though the stem introduces the two quantities in an order that needs care.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
1 step
Reading level
Question and choices: about grade 5, 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

One fact decides the question: an ordered pair lists x first. The choices then test two confusions, which number is the gallons and whether a point is a rate. Reading the pair aloud as “after 3 hours, 24 gallons”, and noting that 24 ÷ 3 gives a rate of 8 gallons an hour that no choice states, rules out all three wrong statements.

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

    No graph is drawn, and the stem names y, the gallons, before x, the hours, so the pair (3, 24) has to be matched to its letters, not to the order the quantities were mentioned. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer minor Choosing the answer

    The four statements reuse 3 and 24 in two frames, a total added “in” some hours and an amount “per hour”, so the right statement and its swapped twin differ only in which number counts gallons. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Recall that an ordered pair lists x first and y second, write down what x and y measure here, and read the pair as one sentence about hours and gallons.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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

An ordered pair
(x,y)(x, y)
An ordered pair names one point on a graph. The first number is always the x-value. The second number is always the y-value. Here x counts hours and y counts gallons. So a pair tells how many gallons were added after that many hours.
Rate from a point
rate=yx\text{rate} = \frac{y}{x}
When water is added at a constant rate, divide the y-value by the x-value. The result is the rate: gallons added in 1 hour. A point itself shows a total, not a rate.

  1. Sort out the question

    The graph links two amounts:

    • x is the number of hours water has been added.
    • y is the number of gallons added.

    You need to say what the point (3, 24) means. No graph is drawn, so you work from the words.

  2. Match each number to its letter

    In an ordered pair, x always comes first and y second. So in (3, 24):

    • x = 3, which means 3 hours.
    • y = 24, which means 24 gallons.

    Watch out: the question names gallons first and hours second. Do not match the numbers to the words in that order. That mistake reads 3 as gallons and 24 as hours, choice A.

  3. Read the pair as a sentence

    Put the two meanings together: after 3 hours, 24 gallons have been added in all. The 24 is a total for the whole 3 hours.

    Watch out: 24 is not the amount for each hour. The rate per hour is 24÷3=824 \div 3 = 8 gallons. So choice D, 24 gallons per hour, is wrong. Choice B, 3 gallons per hour, is also not the rate.

    So the answer is C, twenty-four gallons of water was added in 3 hours.

Check your answer

Use the rate of 8 gallons each hour. In 3 hours, 3×8=243 \times 8 = 24 gallons go in. That matches choice C. No choice says 8 gallons per hour, so both “per hour” choices must be wrong.

More questions like this

The original DOE question
Question 28 as shown on the DOE’s digital test, page 101 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 101, question 28.

Question 19 of 108

2026 Practice Test D Math, question 39

Question

The table shows the probabilities of a hockey team scoring different numbers of goals in a game.

HOCKEY GOAL PROBABILITIES

Number of Goals Probability
0 0.16
1 0.21
2 0.29
3 0.21
4 0.09
5 0.04

What is the probability that the team will score three or more goals in the next game?

  • A0.13
  • B0.21
  • C0.34
  • D0.50
Show the answer

Correct answer: C

To find the probability that any one of several mutually exclusive events occurs, use the addition rule, and add the probabilities of each event:

P(A or B)=P(A)+P(B)\mathrm{P}(\mathrm{A} \text{ or } \mathrm{B}) = \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B})

Using the data in the table, the probability that the team will score 3 or more goals in the next game is represented by

P(3)+P(4)+P(5)=\mathrm{P}(3) + \mathrm{P}(4) + \mathrm{P}(5) =

0.21+0.09+0.04=0.34.0.21 + 0.09 + 0.04 = 0.34.

Question analysis

A table gives the probabilities of a hockey team scoring 0 to 5 goals in a game, and students find the chance that it scores three or more goals in the next one. This probability question tests adding the probabilities of separate outcomes. Reading exactly which rows the phrase covers is the main point of care.

NYS standards
  • 6.SP.8 probability models for simple events
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 8, 40 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

Adding three decimals is easy, so the question turns on which rows the words cover. Including the 3-goal row, and not stopping at it, are the two decisions, and each misreading gives a listed choice. Checking the complement, 1 − (0.16 + 0.21 + 0.29), confirms both the sum and the choice of rows.

  1. Turning words into math minor Understanding the question

    The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.

Traps

  • 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).
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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

Adding probabilities
P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)
A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
The chance it does not happen
P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.

  1. Sort out the question

    The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.

  2. Pick the right rows

    Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:

    • 3 goals: 0.21
    • 4 goals: 0.09
    • 5 goals: 0.04

    Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.

  3. Add the chances

    A game can end with only one number of goals. So add the three chances:

    0.21+0.09+0.04=0.340.21 + 0.09 + 0.04 = 0.34

    Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.

    So the answer is C, 0.34.

Check your answer

Find the chance of fewer than 3 goals, the other rows: 0.16+0.21+0.29=0.660.16 + 0.21 + 0.29 = 0.66. Then the chance of 3 or more is 1−0.66=0.341 - 0.66 = 0.34. Both ways give 0.34.

More questions like this

The original DOE question
Question 39 as shown on the DOE’s digital test, page 112 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 112, question 39.

Question 20 of 108

2026 Practice Test D Math, question 47

Question

A tall right rectangular prism standing on end. The bottom front edge is labeled 2 in., the bottom edge running back on the right is labeled 2 in., and the height is labeled 6 in. Right-angle marks are drawn at the lower left corner of the front face and at the front left corner of the top face.

What shape will result from a horizontal slice of the figure above?

  • Aa square that has 2-inch sides
  • Ba triangle that has 2-inch sides
  • Ca rectangle that has a 2-inch side and a 6-inch side
  • Da triangle that has one 2-inch side and two 6-inch sides
Show the answer

Correct answer: A

Based on the diagram, the figure shown is a right rectangular prism with a square base. A horizontal slice of the figure will be a plane parallel to the square base, as in this example.

The same prism, 2 in. by 2 in. by 6 in., drawn with its hidden edges dashed and a horizontal slice partway up: a square parallel to the base and the same size as the base.

Since the horizontal slice is parallel to the base, it will be congruent to the base, which is a square with 2-inch sides.

Question analysis

A figure shows a right rectangular prism measuring 2 inches by 2 inches by 6 inches, standing on end. This geometry question asks what two-dimensional shape a horizontal slice of the prism makes. It tests visualizing cross-sections of solids, using the labeled lengths to name both the shape and its size.

NYS standards
  • 7.G.3 slicing three-dimensional solids
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Describing the shape made by slicing a prism parallel to its base is exactly the grade 7 standard. This is its simplest case, a single slice of a labeled right prism, typical of a classroom example.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 2, 46 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A horizontal cut runs parallel to the ground, and this prism stands on a square base, so the slice copies the base. The main danger is picturing the cut along the tall front face, which gives a 2-by-6 rectangle, the shape a vertical slice makes. The two triangles need a slanted cut through a corner and catch only students guessing.

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

    The prism is drawn standing on end, 6 inches tall on a 2-by-2 base, so the cut has to be pictured in a flat drawing where the tall front face is the most visible shape. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Students must know that a slice parallel to the base of a prism has the same shape and size as the base, which here is the square at the bottom. (Anderson & Krathwohl, 2001)

Strategy: Picture a flat cut across the prism parallel to the ground, ask which face of the prism that cut runs parallel to, and read that face's edge lengths from the labels.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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

A slice parallel to the base
slice parallel to the base≅base\text{slice parallel to the base} \cong \text{base}
The sign ≅ means congruent: the same shape and the same size. A prism has the same shape all the way up, like a stack of matching cards. So a cut parallel to its base, at any height, is a copy of the base.

  1. Sort out the question

    The figure is a right rectangular prism, a box shape whose faces are all rectangles. It stands on end. The two edges along the bottom are labeled 2 in. The tall edge is labeled 6 in.

    You need the shape you get from a horizontal slice. Horizontal means flat, like the floor or a tabletop.

  2. Find the base

    The base is the face the prism stands on. The two bottom edges you can see are both labeled 2 in. The right-angle mark on the top face shows its corners are square. The top and the bottom match. So the base is a square with 2-inch sides.

  3. Picture the slice

    A horizontal slice cuts straight across the prism, parallel to the base. The walls of the prism go straight up, so the slice is a copy of the base. It is a square with 2-inch sides, at whatever height you cut.

    The 6-inch height does not matter. It only tells how tall the stack of squares is.

  4. Rule out the other shapes

    Watch out: a 2-inch by 6-inch rectangle, choice C, comes from a vertical slice. That cut goes straight down, through the 6-inch height. A triangle, choice B or D, needs a slanted cut through a corner.

    So the answer is A, a square that has 2-inch sides.

Check your answer

Picture the slice sliding up and down. At the bottom, it matches the base, 2 in. by 2 in. At the top, it matches the top face, the same square. In between, the walls are straight, so its size never changes.

More questions like this

The original DOE question
Question 47 as shown on the DOE’s digital test, page 120 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 120, question 47.

Question 21 of 108

2025 Practice Test A English, question 14
Passage Excerpt from “Scribe Like an Egyptian”

Excerpt from “Scribe Like an Egyptian”

by Hilary Wilson

In ancient Egypt, literacy was the key to success. However, contrary to popular belief, not all Egyptian scribes understood hieroglyphs. Many relied instead on the simpler hieratic script for the multitude of everyday documents generated by the Egyptian bureaucracy.

Hieroglyphs—“the Words of God”—compose a writing system with more than 1,000 distinct characters, the meanings of which were lost for 1,500 years before they were deciphered by Jean-François Champollion in 1822. Including both ideograms (which convey a whole word or idea, either concrete or abstract, in a single sign) and phonograms (representing either an alphabetic sound or a group of consonants), [the writing system] was used in formal inscriptions on tomb and temple walls as well as on elaborate funerary papyri. For everyday purposes, however, scribes used a shorthand version of the hieroglyphic script known as hieratic, which was quicker to write and more economical of space. The two writings existed side by side for at least 2,500 years.

Scraps of ancient hieratic writing, mostly penned by student scribes on limestone flakes called ostraca, suggest that no matter how humble his origins, an educated Egyptian could achieve almost anything. Horemheb (d.1292 B.C.) is a good example. Born of middle-ranking parents, his scribal training led to an army career. From Scribe of Recruits, during the reign of Akhenaten (1353–1336 B.C.), Horemheb rose through the military ranks and, by the rule of Tutankhamun (1332–1323 B.C.), he was commander in chief of the Egyptian forces. As a close adviser of the young Pharaoh, Horemheb was appointed “Deputy of the King throughout the Two Lands,” and might have expected to succeed to the throne should the king die childless. He had to wait a few years, but eventually Horemheb achieved the pinnacle of his career by becoming the last king of the 18th Dynasty, making his mark by instituting dramatic reforms to the organisation of the army, the judiciary and administration in general. The lasting success of these changes owed much to his scribal background. . . .

But education was not available to all. Government departments and major temples supported schools, where boys commenced their training at six or seven, sometimes earlier. To these boarding establishments . . . family or household servants delivered the students’ food and drink rations daily for several years, during which time the student was not contributing to the family’s income. Boys from poorer families could only hope to be educated with support from a wealthier relative or patron, or through apprenticeship to an older scribe, perhaps the local clerk or land agent, who would teach them the basics of the scribe’s craft. This limited the scope for employment but such “on the job” training allowed apprentices to help out at home while learning. . . .

Scribal education began with the elementary principles of the hieratic script. The lowliest scribes, who trained for just five or six years, probably learned only the rudiments of the hieroglyphic script. Students were set exemplar documents and extracts from popular texts to copy, to practice their hieratic handwriting on basic-format letters, reports and contracts, while absorbing the good advice contained in the texts. Surviving examples of copy-work sometimes include tutors’ corrections added in red. Some significant Egyptian literary works survive almost exclusively from student copies.

A schoolboy “dictionary” of hieroglyphs with their hieratic equivalents shows that a knowledge of more than 450 signs was required for everyday writing purposes. Lessons in record-keeping and filing and labelling enabled any half-competent scribe to perform that most essential of all scribal functions: the making and updating of lists. For professions such as those of government official, priest or lawyer, a scribe would train for several more years, increasing his vocabulary to perhaps a thousand or more signs. Those with the best handwriting or drawing skills might follow the craft of creating beautifully illustrated copies of funerary texts, commonly called Books of the Dead. Others could become draughtsmen, artists or architects. Doctors compiled their own collections of medication recipes, treatments and associated incantations, many copied from texts found in the House of Life, the temple library. Lawyers had to be familiar with the corpus of civil and religious laws and precedents found in the official records, which were administered by archivists. Egypt’s bureaucratic society depended on the skills of an army of scribes of all ranks from filing clerk to tax assessor. For young Egyptians, “be a scribe” was the best of career advice.

From “Scribe Like an Egyptian” by Hilary Wilson from HISTORY TODAY, August 8, 2019. Copyright © 2019 by History Today Ltd. Company.

A pyramid diagram titled “ANCIENT EGYPTIAN SOCIAL CLASS STRUCTURE,” divided into five horizontal levels, each with a small drawing of figures in ancient Egyptian style and a labeled box. From top to bottom the levels are labeled: Pharaoh (a single standing figure); Nobility and priests (a kneeling figure and a standing figure); Traders, shopkeepers, scribes, and artisans (two seated figures working); Animal herders and farmers (a man standing beside a group of cattle); and Unskilled workers (three figures bending forward).

bureaucracy: an administrative staff of government officials
funerary papyri: a sheet or scroll of papyrus containing religious images and hieroglyphs meant to help the deceased be reborn in the afterlife
rudiments: basics
schoolboy: slang that refers to materials used during the course of receiving an education
draughtsmen: people who draw plans of machinery or structures
corpus: a collection of writings

Question

The diagram after paragraph 6 best provides additional support for the topic of the excerpt by

  • Ademonstrating the place of scribes in ancient Egyptian social structure.
  • Bdepicting the types of tasks performed by each social class.
  • Cindicating that scribes were able to easily improve their social ranking.
  • Dcomparing the social positions of each social class in ancient Egypt.
Show the answer

Correct answer: A

The question asks how the diagram of ancient Egyptian class structure supports the topic of the excerpt.

A. CORRECT. The focus of the excerpt is the opportunities available to scribes in different professions and how people became scribes and therefore improved their position. By illustrating the position of the scribes in the class structure of ancient Egypt, the diagram supports the idea that scribes were valuable members of society.

B. Incorrect. The professions of the different social classes are shown in the diagram, but it does not address specific tasks. Furthermore, the main topic of the excerpt is the role of the scribes in society rather than the tasks performed by the different social classes.

C. Incorrect. While the diagram does show the position of scribes in the social hierarchy, it does not provide an illustration of movement between the different classes or rankings, so the diagram does not support the idea that scribes were able to improve their social standing.

D. Incorrect. The diagram shows the class structure as hierarchical and can support readers’ efforts to compare the different social positions. However, the role of scribes in society, rather than a general comparison of different classes, is the main focus of the excerpt.

Question analysis

This question on “Scribe Like an Egyptian” asks how the diagram after paragraph 6, a pyramid of ancient Egyptian social classes from the pharaoh at the top to unskilled workers at the base, gives additional support to the excerpt’s topic. Students must decide what the excerpt is mainly about and then judge what the graphic adds, and several choices describe something the pyramid really shows.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R2 theme or central idea
Grade level
Grade 7 reading and task
Compared with grade 7
At grade 7 Linking a graphic to a text is regular grade 7 work, and this five-level pyramid is simple and clearly labeled. The demand lies in naming the excerpt’s topic first and judging how the diagram bears on it.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Evidence
Answered from the whole text
Reading level
Passage: about grade 12, 810 words; question and choices: about grade 9, 57 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

More than one choice is true of the pyramid, so the question is decided by the excerpt’s topic, not by the picture alone. Students who look only at the diagram see a ranking of all the classes; students who remember that the article is about scribes look for where scribes stand. The choice about rising easily borrows Horemheb’s story from the text and credits it to a drawing that shows no movement.

  1. A very close wrong answer moderate Choosing the answer

    The choice about comparing every class’s position differs from the right one in a single detail, all classes rather than scribes, and both are true of the pyramid. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Scribes are not a tier of their own: they share the third of five tiers with merchants and craftsmen, below the pharaoh, nobles and priests and above farmers, herders and laborers. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. How, not what minor Working it out

    The stem asks how the diagram supports the excerpt’s topic, so students must first decide the topic is scribes, then judge what the picture adds to it. (Anderson & Krathwohl, 2001)

Strategy: State the excerpt’s topic in a few words, then test each choice twice: is it true of the diagram, and does it connect the diagram to that topic?

Traps

  • True but not the answer: A choice says something accurate about the text that does not answer this question.
  • Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
  • Wrong purpose or structure: Names a purpose, structure or effect that the part does not serve, or describes what a part says when the question asks what it does.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This asks how the diagram after paragraph 6 adds to the topic of the excerpt. First name the topic. Then ask two things about each choice. Is it true of the diagram? And does it connect the diagram to that topic?

  2. Name the excerpt’s topic

    The excerpt is about scribes, people trained to read and write for others. It tells how they learned, who could study, and what jobs they could get. Every paragraph comes back to scribes. So the diagram should add something about scribes.

  3. Read the diagram

    The diagram is a pyramid with five levels. From the top down, they are the Pharaoh, nobles and priests, then “Traders, shopkeepers, scribes, and artisans,” then herders and farmers, and last unskilled workers. Artisans are skilled craftspeople. Scribes sit in the third level, the middle one. They are below the ruler and the nobles, and above farmers and workers.

  4. Check each choice

    • A fits. The diagram shows where scribes stood in Egyptian society, which ties it to the topic.
    • B does not fit. The labels name groups of people, not the tasks each group did.
    • C does not fit. The levels are fixed, and nothing in the diagram shows anyone moving up. Horemheb’s rise comes from paragraph 3, not the diagram.
    • D does not fit. The pyramid does rank all the classes. But that alone does not connect it to an excerpt about scribes.

    Watch out: More than one choice is true of the picture. The right one must also fit the topic.

  5. Choose the answer

    The diagram supports the topic by showing where scribes fit among the classes of ancient Egypt. So the answer is A.

Why this answer and not the closest wrong one

D is the closest wrong choice, because the pyramid really does compare every class. But the excerpt is about scribes, not about every class. A is true of the diagram too, and it also links the diagram to the topic. It points to the place of scribes in the middle level.

More questions like this

The original DOE question
Question 14 as shown on the screen of the DOE’s digital test.
As shown on the screen of the DOE’s digital test: 2025 Practice Test A English, question 14.
The passage “Excerpt from “Scribe Like an Egyptian”” as shown on the testThe passage “Excerpt from “Scribe Like an Egyptian”” as shown on the DOE’s digital test, screen 1 of 6.The passage “Excerpt from “Scribe Like an Egyptian”” as shown on the DOE’s digital test, screen 2 of 6.The passage “Excerpt from “Scribe Like an Egyptian”” as shown on the DOE’s digital test, screen 3 of 6.The passage “Excerpt from “Scribe Like an Egyptian”” as shown on the DOE’s digital test, screen 4 of 6.The passage “Excerpt from “Scribe Like an Egyptian”” as shown on the DOE’s digital test, screen 5 of 6.The passage “Excerpt from “Scribe Like an Egyptian”” as shown on the DOE’s digital test, screen 6 of 6.

Question 22 of 108

2024 Test A English, question 18
Passage Excerpt from “Impressions of an Indian Childhood”

Excerpt from “Impressions of an Indian Childhood”

by Zitkala-Sa

Zitkala-Sa (Gertrude Simmons Bonnin) was a Native American writer, musician, teacher, and political activist who was raised on the Yankton Sioux Reservation in South Dakota. In 1900 she published “Impressions of an Indian Childhood” (the term Indian was commonly used at the time to refer to Native American people) to expose readers to what life is like on a reservation.

Soon after breakfast Mother sometimes began her beadwork. On a bright, clear day, she pulled out the wooden pegs that pinned the skirt of our wigwam1 to the ground, and rolled the canvas part way up on its frame of slender poles. Then the cool morning breezes swept freely through our dwelling, now and then wafting the perfume of sweet grasses from newly burnt prairie.

Untying the long tasseled strings that bound a small brown buckskin2 bag, my mother spread upon a mat beside her bunches of colored beads, just as an artist arranges the paints upon his palette. On a lapboard she smoothed out a double sheet of soft white buckskin; and drawing from a beaded case that hung on the left of her wide belt a long, narrow blade, she trimmed the buckskin into shape. Often she worked upon small moccasins for her small daughter. Then I became intensely interested in her designing. With a proud, beaming face, I watched her work. In [my] imagination, I saw myself walking in a new pair of snugly fitting moccasins. I felt the envious eyes of my playmates upon the pretty red beads decorating my feet.

Close beside my mother I sat on a rug, with a scrap of buckskin in one hand and an awl in the other. This was the beginning of my practical observation lessons in the art of beadwork. From a skein3 of finely twisted threads of silvery sinews my mother pulled out a single one. With an awl she pierced the buckskin, and skillfully threaded it with the white sinew. Picking up the tiny beads one by one, she strung them with the point of her thread, always twisting it carefully after every stitch.

It took many trials before I learned how to knot my sinew thread on the point of my finger, as I saw her do. Then the next difficulty was in keeping my thread stiffly twisted, so that I could easily string my beads upon it. My mother required of me original designs for my lessons in beading. At first I frequently ensnared many a sunny hour into working a long design. Soon I learned from self-inflicted punishment to refrain from drawing complex patterns, for I had to finish whatever I began.

After some experience I usually drew easy and simple crosses and squares. These were some of the set forms. My original designs were not always symmetrical nor sufficiently characteristic, two faults with which my mother had little patience. The quietness of her oversight made me feel strongly responsible and dependent upon my own judgment. She treated me as a dignified little individual as long as I was on my good behavior; and how humiliated I was when some boldness of mine drew forth a rebuke from her!

In the choice of colors she left me to my own taste. I was pleased with an outline of yellow upon a background of dark blue, or a combination of red and myrtle-green. There was another of red with a bluish-gray that was more conventionally used. When I became a little familiar with designing and the various pleasing combinations of color, a harder lesson was given me. It was the sewing on, instead of beads, some tinted porcupine quills, moistened and flattened between the nails of the thumb and forefinger. My mother cut off the prickly ends and burned them at once in the centre fire. These sharp points were poisonous, and worked into the flesh wherever they lodged. For this reason, my mother said, I should not do much alone in quills until I was as tall as my cousin Warca-Ziwin.

Always after these confining lessons I was wild with surplus spirits, and found joyous relief in running loose in the open again. Many a summer afternoon a party of four or five of my playmates roamed over the hills with me. We each carried a light sharpened rod about four feet long, with which we pried up certain sweet roots. When we had eaten all the choice roots we chanced upon, we shouldered our rods and strayed off into patches of a stalky plant under whose yellow blossoms we found little crystal drops of gum. Drop by drop we gathered this nature’s rock-candy, until each of us could boast of a lump the size of a small bird’s egg. Soon satiated with its woody flavor, we tossed away our gum, to return again to the sweet roots.

From “Impressions of an Indian Childhood” by Zitkala-Sa—Public Domain

A BRIEF HISTORY OF BEADWORK IN SOUTH DAKOTA

DateEvent
1500sThe Dakota use beads made from bones, shells, stones, and animal teeth. European traders bring glass beads to North America.
Late 1600sThe Dakota begin trading with the French in Minnesota.
1830sThe Dakota begin using glass beads in clothing, artwork, and decorations in place of Native-made beads.
1862–1865The Dakota are expelled from their homelands in Minnesota as a result of the U.S.-Dakota War.
1900s–1920sThe Dakota create items to sell outside their community.

Source: MNopedia

1wigwam: hut with an arched framework of poles covered with bark, mats, or animal hides

2buckskin: leather made from the skin of a male deer

3skein: coiled length of yarn or other thread loosely wound on a reel

Question

The table after paragraph 7 expands on a central idea in the excerpt because it shows that

  • Athe craft the author was learning was a tradition that endured through many generations and changes.
  • Bthe author’s family incorporated traditional materials into their craft as a way of resisting the influence from European traders.
  • Cthe uniqueness of the cultural tradition that the author learned as a child was eventually recognized in Europe.
  • Dthe author was able to incorporate color into her craftwork as a result of trade with other peoples.
Show the answer

Correct answer: A

The question asks how the table after paragraph 7 expands on a central idea in the excerpt.

A. CORRECT. The table provides the chronology from the 1500s, when the Dakota created beadwork using “beads made from bones, shells, stones, and animal teeth,” to the 1830s, when the Dakota began “using glass beads,” and the 1900s, when they created items to “sell outside their community.” Thus, the table helps expand on a central idea by showing that the craft the author was learning is a tradition that has endured through many generations.

B. Incorrect. Although the table shows that the Dakota used “beads made from bones, shells, stones, and animal teeth” in the 1500s and that by the 1830s began “using glass beads in clothing, artwork, and decorations in place of Native-made beads,” it does not provide any information about the type of materials used by the author’s family.

C. Incorrect. While the information included in the table states that “European traders [brought] glass beads to North America” in the 1500s and that by the 1920s the Dakota created “items to sell outside their community,” there is no mention of whether the uniqueness of Dakota beadwork was recognized in Europe.

D. Incorrect. The table indicates that by the late 1600s the Dakota began trading with French traders from Europe who brought glass beads to the Americas and that by the 1830s the Dakota were “using glass beads in clothing, artwork, and decorations in place of Native-made beads.” However, the table provides no information about whether this work incorporated color. Therefore, the table does not show that the author was able to incorporate color into her craftwork as a result of trade with other peoples.

Question analysis

This reading question asks how a table titled A Brief History of Beadwork in South Dakota, printed after Zitkala-Sa's excerpt, expands on one of its central ideas. Students must read a short timeline of beads, trade and war and connect it to a girl learning beadwork, while avoiding choices that sound historically likely but go beyond the table.

NYS standards
  • 7R5 structure and genre features
  • 7R2 theme or central idea
Grade level
Grade 7 reading and task
Compared with grade 7
At grade 7 Connecting a table to a text's central idea is grade 7 work. The table is short and plainly worded, and linking a people's history to one girl's lessons is a typical grade 7 step.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Evidence
Answered from several paragraphs or stanzas
Reading level
Passage: about grade 8, 946 words; question and choices: about grade 11, 87 words.
Corrected from the DOE handbook
  • A date in the explanation was corrected to match the table.
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 table holds plenty of tempting raw material: European traders, glass beads, sales outside the community. Wrong answers build unsupported stories out of it, about resistance, European recognition or the arrival of color, and the last is the hardest to resist because the excerpt is full of colored beads. Students who sum up the table in one phrase, a craft carried on through centuries of change, can check each choice against what the rows actually say.

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

    The table runs from the 1500s to the 1920s, and its point appears only when the five rows are read as a timeline: new beads, trade, war and expulsion, yet the beadwork goes on. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Students must join the table's history of the Dakota to one girl's lessons in paragraphs 3–6 and see that the craft she learns belongs to that long line. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Reading between the lines minor Working it out

    Endurance is never stated; it has to be drawn from a beadwork history that spans four centuries, from bone and shell beads to items made for sale, through every change between. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

Strategy: Read the table from top to bottom and say in a few words what it shows over time, then pick the choice that fits that pattern and also connects to the excerpt.

Traps

  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
  • Outside knowledge: A choice is plausible in the real world but not supported by this text.
  • Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: In rejecting choice D, the handbook says that “by 1830” the Dakota were using glass beads in clothing, artwork and decorations. The table places that change in the 1830s (“The Dakota begin using glass beads …”), and the handbook’s own reason for rejecting choice B says “by the 1830s.” The explanation now reads “by the 1830s.”

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    The excerpt ends with a table, a chart of facts in rows. You need to see how the table adds to a central idea of the excerpt. A central idea is an important message.

  2. Read the table as a timeline

    • 1500s: the Dakota make beads from bones, shells, stones and teeth.
    • Late 1600s: they begin trading with the French.
    • 1830s: they use glass beads “in place of Native-made beads.”
    • 1862–1865: a war forces them out of their homelands.
    • 1900s–1920s: they make items to sell outside their community.

  3. Find the pattern

    The beads change. Trade begins. A war drives the people from their land. But through all of it, the beadwork goes on for hundreds of years. The girl in the excerpt is learning this same craft from her mother. So the table shows her craft is a tradition, a custom passed down through many generations.

  4. Check each choice

    • A is right. The craft lasted through many changes over hundreds of years.
    • B is wrong. The table shows the Dakota took up glass beads, not that they resisted traders.
    • C is wrong. No row says Europe ever praised the craft.
    • D is wrong. The table never mentions color.

    Watch out: D is tempting because the excerpt is full of colored beads, and traders brought glass beads. But the link to color is a guess. The table doesn’t say it.

  5. Choose the answer

    The table shows beadwork lasting through centuries of change, just like the craft the girl learns. So the answer is A.

Why this answer and not the closest wrong one

D is the closest wrong answer. Traders did bring glass beads, and the girl’s beads are colorful. But no row of the table says trade gave the Dakota color. Nothing ties it to the author’s own work either. A sums up what the rows actually show: a craft that went on through change.

More questions like this

The original DOE question
Question 18 as printed on page 46 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 46, question 18. Open page 46 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 “Excerpt from “Impressions of an Indian Childhood”” as printed (pages 42–43)The passage “Excerpt from “Impressions of an Indian Childhood”” as printed on page 42 of the DOE handbook.The passage “Excerpt from “Impressions of an Indian Childhood”” as printed on page 43 of the DOE handbook.

Question 23 of 108

2024 Test A English, question 34
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

The table builds on the information in the passage mainly by

  • Aemphasizing that people have continued to swim across the channel and have significantly improved the speed record.
  • Bsuggesting that Ederle inspired many women to swim across the channel to attempt to break the current speed record.
  • Crevealing that other women competed with Ederle to set a record time for swimming across the English Channel.
  • Dproviding a comparison between the earliest channel-swimming records and the current record times.
Show the answer

Correct answer: A

The question asks how the table mainly builds on the information in the passage.

A. CORRECT. The table supports the information in the passage by showing that people have continued to swim the channel and have significantly improved on past records. The information in the table about records set by those who came after Ederle builds on the idea that “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” (paragraph 9).

B. Incorrect. Even though paragraph 9 states that Ederle “inspired [everyday American citizens] to be more active” and the table presents the time for the current female record holder, there is no indication in the passage or in the table that the subsequent female record holders were inspired to swim across the channel by Ederle.

C. Incorrect. Although paragraph 3 mentions that Ederle had a female rival and the table includes the speed record of another female swimmer, the passage does not provide information about other female swimmers competing with Ederle to set the channel-swim record.

D. Incorrect. Although paragraph 1 mentions that Ederle “broke the men’s record by two hours” and the table allows for comparisons between the earliest speed records and those of the present day, the passage mentions no channel-swim records set after Ederle’s swim, so it says nothing about current record times. The table allows for comparison, but this aspect of the table does not build on the information presented in the passage.

Question analysis

This question asks how a table of English Channel speed records, from Matthew Webb's 1875 crossing to a 2012 record, builds on the passage about Gertrude Ederle. Students must read times from two eras and decide what the table adds to the text, while the choices mix accurate observations about the table with claims neither source supports.

NYS standards
  • 7R5 structure and genre features
  • 7R1 textual evidence and inferences
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Relating a table to a passage is within grade 7 standards, but here the student must compare records from two eras, connect them to ideas in the text and choose among closely worded statements, well above typical grade 7 demand.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Evidence
Answered from the whole text
Reading level
Passage: about grade 9, 790 words; question and choices: about grade 11, 78 words.
Corrected from the DOE handbook
  • A self-contradictory statement 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

This is one of the hardest questions on the test because a wrong choice describes the table exactly, in its own words. The passage stops with Ederle's 1926 record, and what the table adds is the rest of the story: swimmers kept crossing, and the record fell to under seven hours. Students who describe what the table contains, rather than what it tells a reader who has finished the passage, lose the point.

  1. A very close wrong answer major Choosing the answer

    The comparison choice repeats the table's own headings, “Earliest Speed Records” and “Current Speed Records”, and is literally true; only the stem's “builds on the information in the passage” and “mainly” separate it from the choice about the falling record. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The student must compare times across the table's two sections, from Webb's 21 hours 45 minutes in 1875 and Ederle's 14 hours 31 minutes to Grimsey's 6 hours 55 minutes in 2012. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. How, not what moderate Working it out

    The question asks what the table adds to a passage that ends in 1926, so the student must say what its later rows extend, not what its layout holds. (Anderson & Krathwohl, 2001)

Strategy: Read both sections of the table and say what it adds to Ederle's story, then keep only a choice that the table shows and that connects to an idea the passage itself develops.

Traps

  • Wrong purpose or structure: Names a purpose, structure or effect that the part does not serve, or describes what a part says when the question asks what it does.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
  • Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: In rejecting choice D, the handbook notes that paragraph 1 says Ederle “broke the men’s record by two hours,” and then says “there is no mention in the passage of past, present, or future channel-swim records.” Paragraph 1 does mention a past record: the one Ederle broke. What the passage never mentions is a record set after hers, which is what the table’s “Current Speed Records” add. The explanation now says the passage mentions no channel-swim records set after Ederle’s swim, so it says nothing about current record times. The reason for rejecting D is otherwise unchanged.

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Understand the question

    To build on something means to add to it. So this asks what the table adds to what the passage already tells you. Don’t just describe what is in the table. Ask what a reader learns from it after finishing the passage.

  2. Read the table’s two parts

    The table has two parts, “Earliest Speed Records” and “Current Speed Records.” These rows matter most:

    YearSwimmerTime
    1875Matthew Webb21 hours, 45 minutes
    1926Gertrude Ederle14 hours, 31 minutes
    2012Trent Grimsey6 hours, 55 minutes

  3. Compare the times

    Ederle swam for 14 hours and 31 minutes. Half of her time would be about 7 hours and 15 minutes. Grimsey’s 2012 time, 6 hours and 55 minutes, is even faster than that. So people kept swimming the channel after 1926, and the record got much faster.

  4. See what the passage leaves out

    The passage does mention one record. In paragraph 1, Ederle “broke the men’s record by two hours.” But the passage stops in 1926. It never tells of any record set after her swim. The table adds that later part of the story: swimmers kept crossing, and the record kept falling.

  5. Check each choice

    • A fits: people kept crossing and “significantly improved the speed record,” and significantly means by a lot.
    • B is wrong because nothing links any later woman to Ederle, and the table lists only one later woman.
    • C is wrong because Ederle is the only woman of her time in the table, and Hlavacova swam 80 years later, in 2006.
    • D is wrong because it only names what the table holds, not what the comparison adds to Ederle’s story.

    Watch out: D uses the table’s own headings, so it sounds right. But the question asks what the table adds, not how it is laid out.

  6. Choose the answer

    The table carries the story past 1926 and shows the record falling far below Ederle’s time. So the answer is A.

Why this answer and not the closest wrong one

D comes closest, because the table really does set the earliest and current records side by side. But D stops at describing the table. A says what that comparison shows a reader who has finished the passage: the record kept falling, from Ederle’s 14 hours and 31 minutes to under 7 hours.

More questions like this

The original DOE question
Question 34 as printed on page 56 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 56, question 34. Open page 56 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 “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 24 of 108

2024 Test A English, question 57
Passage Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”

Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”

by Emma Marris

The author of this opinion article attended a symposium, or conference, in Missoula, Montana, at which biologists and other scientists discussed species of plants and animals that are not native to the areas in which they live.

Invasive species are scary. It was ecologist Charles Elton, back in the 1950s, who introduced the militaristic “invasion” metaphor to describe exotic plants and animals—but there’s no question some can be extremely destructive.

The brown tree snake has eaten a dozen kinds of forest birds in Guam to extinction; zebra mussels clog pipes around the Great Lakes; the common house cat turns out to be, in Australia, a mercilessly effective killer of cute, fluffy marsupials like the bilby and the numbat.

As scientists have sounded the alarm about these pests, the public has gotten the message. Citizen groups rip out non-native plants. Native gardens have become increasingly popular, both as ways to celebrate the unique flora of each region and as tiny hot spots of diversity. Native trees provide food for native bugs, which feed native birds. Food chains developed over thousands of years of co-evolution unfold in our backyards. . . .

So we’ve learned, scientists and laypeople alike, that native species are good and non-natives are bad.

Julian Olden, a biologist at the University of Washington, Seattle, who co-organized the symposium, recently polled nearly 2,000 ecologists. Among his findings: a substantial number of them said they would immediately eradicate a hypothetical non-native forest plant, even if it were shown to have no effect on the forest. Olden calls this the “guilty even when proven innocent” approach.

That kind of approach is not very useful on a rapidly changing planet.

Exotics Are Everywhere

Climate change is making it harder even to decide who the invaders are.

How, scientists at the symposium wondered, do you define “native” on a warming planet, when plants and animals are already moving toward the poles or up mountainsides in search of climate conditions they can tolerate? Should we consider them “invasive” in their new homes? Regardless of what we label them, conservationists will be reluctant to remove them from their new environs—to do so would stymie1 their chances of adapting to the warmer future we’re creating.

And then there are the non-natives that we actually like. Most domestic crops are exotic in most of the places they’re grown, but there are even wild exotics that “do good,” forming useful relationships with native species.

Edwin Grosholz of the University of California, Davis, told the recent symposium about one such relationship. On beaches in his state, non-native spartina grass has become important habitat for the endangered California clapper rail, a plump shorebird with a downward curving bill more at home on land than in the air. A project to rip out and poison the spartina—which grows in dense swaths that exclude many other shorebirds—saw clapper rail numbers go tumbling downward.

There are other examples like that. The endangered southwestern willow flycatcher nests in “invasive” tamarisk shrubs. Many native (and beautiful) Hawaiian flowers are now pollinated by the Japanese white-eye bird—because the native pollinators have been driven extinct by other non-native species.

Should we impose further risk on already endangered natives by severing these relationships? Or should we admire the resilience of nature and let such “well-behaved” exotics stay? . . .

Leave them alone, more and more conservationists are arguing, and stop focusing obsessively on categorizing species as native or non-native. Mark Davis, an ecologist at Macalester College in St. Paul, Minnesota, once considered himself an “invasion biologist”—but not anymore. “I am actively trying to get the field to retire the invader narrative,” he said in Missoula.

A Good Thing, Not the Only Thing

After all, nativeness is just one environmental value, and arguably not as important as preventing extinctions and preserving biodiversity. In some cases we can best serve biodiversity by leaving the non-natives alone or even—brace yourself, now—introducing them on purpose.

This is the thinking behind, for example, installing the Aldabra tortoise on the islands of Mauritius. The islands lost their own large tortoises, and the fruiting plants that formerly had their seeds moved around by these fruit-loving reptiles have been on the decline. A tortoise that’s related to the island’s large tortoises—a non-native from the Seychelles in the Indian Ocean that was intentionally introduced in 2004—is now handling some of that work.

Most of the time, for the time being, conserving species still means focusing on supporting them in their historical habitats, planting natives and removing non-natives. We can and should do that in places where it is feasible and important to us.

From “It’s Time to Stop Thinking That All Non-Native Species Are Evil” by Emma Marris from NATIONAL GEOGRAPHIC MAGAZINE, July 24, 2014. Copyright © 2014 by National Geographic Society.

Chart titled “Possible Management of Non-Native Species, Based on Extent of Invasion,” plotting Area Affected by Non-Native Species (vertical axis) against Time (horizontal axis) as an S-shaped curve that stays low, then rises steeply and levels off high, where it is labeled “Widespread and abundant throughout range.” The time line is divided into four shaded stages, left to right: Prevention, “Species absent”; Eradication, “Few, localized populations”; Containment, “Rapid increase in populations”; and Local Control and Management Only, “Eradication no longer feasible; can only protect small areas based on economic and ecological need.” Markers on the curve show “Species introduced” at the end of Prevention, “Detection” within Eradication and “Public awareness begins” near the end of Containment, and two large arrows read “Management = simple; eradication likely” across Prevention and Eradication and “Management = difficult and costly; eradication unlikely or impossible” across Containment and Local Control and Management Only.

1stymie: hinder or interfere with

Question

How does the chart provide additional support for a central idea of the excerpt?

  • Aby implying that the public becomes aware of non-native species only after removal is impossible
  • Bby showing that the complete removal of non-native species is usually not achievable after a certain period of time
  • Cby demonstrating why non-native species can be destructive if they are not removed from an area immediately
  • Dby explaining how some conservation groups have approached the removal of non-native species
Show the answer

Correct answer: B

The question asks how the chart provides additional support for a central idea of the excerpt.

A. Incorrect. The chart indicates that the public tends to become aware of non-native species once eradication is “unlikely or impossible,” not strictly impossible. Furthermore, this idea is not a central idea of the excerpt.

B. CORRECT. The chart strongly suggests that once “rapid increase in populations” of non-native species has occurred, eradication will be “unlikely or impossible.”

C. Incorrect. While the chart does show that non-native species can eventually be impossible to eradicate, it never addresses the issue of how the non-native species might be destructive to their new environments.

D. Incorrect. The chart shows how non-native species become harder and harder to remove as their population grows but does not explain how any conservation groups have approached their removal.

Question analysis

This question asks how a chart titled “Possible Management of Non-Native Species, Based on Extent of Invasion” supports a central idea of an article arguing that not all non-native species are harmful. Students must read the chart's stages, labels and curve, then decide which statement the chart actually shows and which also fits the article's argument.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R2 theme or central idea
Grade level
Grade 8 reading and task
Compared with grade 7
At grade 7 Grade 7 students connect graphics to a text, but this chart packs four stages, two arrows, a curve and several labels into one figure, and its pattern must be linked to an argument made in words, demanding but within upper grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Evidence
Answered from the whole text
Reading level
Passage: about grade 11, 815 words; question and choices: about grade 12, 78 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The chart is easy to read in outline, simple management on the left and costly, likely failed eradication on the right, and the right choice restates that contrast. The pull comes from the “Public awareness begins” marker, placed at the border of the stage where eradication is no longer feasible, which makes a statement about public awareness look like a faithful reading. It is nearly faithful, but it describes one marker, not a central idea the article develops.

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

    The chart packs four shaded stages, two arrows, an S-shaped curve and three markers into one figure, and its message lies in the arrows: eradication “likely” early, “unlikely or impossible” later. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer moderate Choosing the answer

    The right choice and the public-awareness choice both read the late stages where removal fails; the awareness marker even sits on the border of “Eradication no longer feasible”, so only centrality separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Combines ideas minor Working it out

    The chart's pattern has to be matched to the article's argument that nativeness is one value among several and that removal makes sense only where “feasible and important to us”. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Read the chart's title, axes and stage labels, say in one sentence what it shows happening over time, then ask which part of the article's argument that pattern supports.

Traps

  • Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
  • Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Know what the question asks

    This asks how the chart supports a central idea, a main point of the article. First read the chart and say in your own words what it shows. Then link that to the article’s argument.

  2. Read the chart from left to right

    Time moves from left to right. The curve shows how much area a non-native species covers. Eradication means wiping a species out completely. Early on, the arrow says management is “simple” and eradication is “likely.” Then the species spreads fast. From there on, management is “difficult and costly,” and eradication is “unlikely or impossible.” In the last stage it is “no longer feasible,” or no longer possible.

  3. Link the chart to the article

    The article argues that we shouldn’t treat every non-native species the same way. Paragraph 16 says we should remove non-natives “in places where it is feasible and important to us.” The chart shows when removal is possible: mostly early, before the species spreads. After that, wiping it out usually can’t be done.

  4. Check each choice

    • B fits: complete removal is usually not possible after a certain time, which is the chart’s main message.
    • A is wrong because the “Public awareness begins” marker is one small label, and the article never discusses when the public notices.
    • C is wrong because the chart shows how far a species spreads, not how much damage it does.
    • D is wrong because the chart shows general stages, not what any real group has done.

    Watch out: A reads the marker correctly, since it does sit late, where removal is already hard. But it picks a detail, not the chart’s main point.

  5. Choose the answer

    The chart shows that removal works only early, which backs up the article’s argument. So the answer is B.

Why this answer and not the closest wrong one

A comes closest, since the chart does place “Public awareness begins” late, when removal is already hard. But that is one small marker, and the article never says when the public first notices a species. B states the chart’s main message, that removal works only early. That backs up the article’s point that removing non-natives isn’t always possible.

More questions like this

The original DOE question
Question 57 as printed on page 72 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 72, question 57. Open page 72 of the handbook (PDF)
The passage “Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”” as printed (pages 68–70)The passage “Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”” as printed on page 68 of the DOE handbook.The passage “Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”” as printed on page 69 of the DOE handbook.The passage “Excerpt from “It’s Time to Stop Thinking That All Non-Native Species Are Evil”” as printed on page 70 of the DOE handbook.

Question 25 of 108

2024 Test A Math, question 5

Question

RESULTS FROM SURVEY OF 110 FAMILIES

Number of Children in the Family Number of Families
0 45
1 32
2 19
3 8
4 6

The table above shows the number of children in each of 110 families. What is the median number of children in these families?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 1

Since 110 families were surveyed, the median will have 55 values less than or equal to it, and 55 values greater than or equal to it.

If the values in the table were listed in order, they would look like this:

0,…,0⏞45,1,…,1⏞32,2,…,2⏞19,3,…,3⏞8,4,…,4⏞6\overbrace{0,\ldots,0}^{45},\overbrace{1,\ldots,1}^{32},\overbrace{2,\ldots,2}^{19},\overbrace{3,\ldots,3}^{8},\overbrace{4,\ldots,4}^{6}

The first row of the table indicates that 45 families had 0 children.

The second row of the table indicates that 32 families had 1 child.

So 77 families had either 0 children or 1 child, and 45 families had 0 children.

Since 55 is between 45 and 77, the median must be a value in the second row of the table.

So the median is 1.

Question analysis

This statistics grid-in gives a frequency table from a survey of 110 families, showing how many families have each number of children from none to four. Students find the median number of children per family. The table lists counts rather than the 110 data values themselves, so locating the middle value takes a running total of the families.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 median problems usually list a dozen or so values. Here 110 values are hidden in a frequency table, so students must reason about positions with a running count, a less familiar and more error-prone task.
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 7, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Students know a median as the middle of a list, but here the list is never written out. Treating the table's columns as the data gives quick, confident answers, 2 from the middle of the children column or 19 from the middle of the family counts, and neither can be checked against choices. The question rewards counting positions: 45 families have no children, so the 55th and 56th families both fall among the 32 with one child.

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

    The table hides 110 data values in five rows: the 0 to 4 column lists the possible values, and 45, 32, 19, 8 and 6 say how many families have each. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. No obvious method moderate Working it out

    The median of 110 values lies between the 55th and 56th; finding them takes a running count of families, 45 and then 77, a method seldom practiced with tables. (Schoenfeld, 1985)

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

    As a grid-in there are no choices to compare, so a student who takes the middle row's 2 has nothing on the page to suggest it is wrong. (Enright, Morley & Sheehan, 2002)

Strategy: With 110 values the median sits between the 55th and 56th; add the family counts row by row until the running total passes 55, and read that row's number of children.

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.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Spot of the median
middle spot=n+12\text{middle spot} = \frac{n + 1}{2}
The median is the middle value when all the values are put in order. n is how many values there are. This gives the spot of the middle value, with the same number of values before and after it. If it ends in a half, the median is halfway between the two values on either side.

  1. Sort out the table

    The table covers 110 families. The left column is a number of children. The right column says how many families have that many children. For example, 45 families have 0 children. You need the median number of children. The median is the middle value when all the values are in order.

  2. Picture the values in order

    Think of a list with one number for each family: how many children it has. The list has 45 zeros, then 32 ones, then 19 twos, then 8 threes, then 6 fours. It is already in order, from least to most.

    45+32+19+8+6=11045 + 32 + 19 + 8 + 6 = 110

    There are 110 values. Watch out: the median is not the middle row, 2. Each row stands for a different number of families.

  3. Find the middle spot

    Use the median rule with n = 110.

    middle spot=n+12\text{middle spot} = \frac{n + 1}{2}

    110+12=55.5\frac{110 + 1}{2} = 55.5

    The middle is halfway between the 55th and 56th values. So you need those two values in the list.

  4. Count down the list

    Keep a running total of families, row by row. The zeros fill spots 1 to 45. Now add the 32 families with 1 child.

    45+32=7745 + 32 = 77

    So the ones fill spots 46 to 77. Spot 55 and spot 56 are both in that range. So both values are 1.

  5. Give the median

    The 55th and 56th values are both 1. Halfway between 1 and 1 is 1. So the median is 1 child. Watch out: the median is a number of children, not of families. Gridding 19, from the families column, is wrong. So is the mean, or average, which is about 1.07. So you would grid 1.

Check your answer

Half of 110 is 55. Families with 0 children: 45, which is less than half. So the median is not 0. Families with 0 or 1 child: 77, which is more than half. So the median is not more than 1. It must be 1.

More questions like this

The original DOE question
Question 62 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 26 of 108

2024 Test A Math, question 8

Question

Square PQRS, with Q at the top left, R at the top right, P at the bottom left and S at the bottom right, and a right-angle mark at S. Point S is marked as the center of a circle that passes through P and R. The quarter of the circle inside the square, bounded by sides SP and SR and the arc from P to R, is shaded.

In the figure above, PQRS is a square. Point S is the center of the circle, and points P and R are on the circle. If the area of the square is 4 square centimeters, what is the area, in square centimeters, of the shaded quarter of the circle?

  • Aπ4\frac{\pi}{4}
  • Bπ\pi
  • C2π2\pi
  • D4π4\pi
Show the answer

Correct answer: B

The area of square PQRS is 4 square centimeters. So

s2=4s^{2} = 4

s2=4\sqrt{s^{2}} = \sqrt{4}

s=2 cms = 2\text{ cm}

Sides PS and RS are each a radius of the circle. So the radius of the circle is 2 cm.

The shaded region is 14\frac{1}{4} of the circle. The area of a circle is A=πr2A = \pi r^{2}. So the area of the shaded region, in square centimeters, is

14A=14πr2=\frac{1}{4}A = \frac{1}{4}\pi r^{2} =

14π(22)=\frac{1}{4}\pi(2^{2}) =

14π(4)=π\frac{1}{4}\pi(4) = \pi

Question analysis

A geometry question with a figure: square PQRS has one corner at the center of a circle, and the quarter of the circle inside the square is shaded. Given the area of the square, students find the area of the shaded quarter circle. The key is connecting the square's side to the circle's radius.

NYS standards
  • 7.G.4 circle area and circumference
  • 6.G.5 perfect squares and perfect cubes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Circle area is grade 7 content, and this question adds only one linking step from the square's area to the radius. Typical grade 7 circle work gives the radius directly; here it takes one short extra inference.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 4, 52 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains three short facts: from the square's area to its side, from the side to the radius, and from the whole circle to a quarter of it. None is hard, but each wrong choice marks a broken link, most often the whole circle's area, which looks like a finished answer. Pointing to S as the center before computing anything settles the radius.

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

    The radius is never labeled: students must see that S is both a corner of the square and the circle's center, so sides PS and RS are radii. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Neither the circle area formula, π times the radius squared, nor the step from a square's area to its side length is given. (Anderson & Krathwohl, 2001)

  3. Combines ideas minor Working it out

    Square and circle are linked in one chain: the area 4 gives a side of 2, the side is the radius, and the shaded part is a quarter of π × 2². (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Take the square root of the square's area to get its side, which is also the radius, then take one quarter of pi times the radius squared.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Stops one step early: A choice is a number from partway through the solution.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square
A=s2A = s^2
A is the area, the space inside. s is the length of one side, and s² means s × s. All four sides of a square are equal, so its area is side times side.
Area of a circle
A=πr2A = \pi r^2
A is the area of the whole circle. r is the radius: the distance from the center to the edge. π, called pi, is a number a little more than 3, about 3.14. Multiply π by r × r. This works for every circle.

  1. Sort out the figure

    PQRS is a square with an area of 4 sq cm. Point S is the center of the circle. Points P and R are on the circle. The shaded part is a quarter of the circle, inside the square. You need the shaded area, in square centimeters. The answer will have π in it.

  2. Find the side of the square

    Use the square formula. The area is 4, so find the number that times itself gives 4.

    A=s2A = s^2

    4=s24 = s^2

    s=2s = 2

    Each side is 2 cm long, because 2×2=42 \times 2 = 4.

  3. See that the side is the radius

    Side SP goes from the center, S, to point P on the circle. So SP is a radius. So is side SR. The radius is 2 cm. Watch out: the side is a radius, not a diameter, the full width of the circle. Using a radius of 1 leads to π4\frac{\pi}{4}, choice A.

  4. Find the area of the whole circle

    A=πr2A = \pi r^2

    A=π×22=π×4=4πA = \pi \times 2^2 = \pi \times 4 = 4\pi

    The whole circle has an area of 4π sq cm. Watch out: this is the whole circle. Stopping here gives choice D.

  5. Take one quarter

    The shaded part is one quarter of the circle, so take one quarter of 4π.

    14×4π=π\frac{1}{4} \times 4\pi = \pi

    The shaded area is π sq cm. So the answer is B, π\pi.

Check your answer

π is about 3.14, so the shaded part is about 3.14 sq cm. The whole square is 4 sq cm. In the figure, the shaded part fills most of the square, but not all. So a bit less than 4 makes sense. The choice 4π is about 12.6, far too big.

More questions like this

The original DOE question
Question 65 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 27 of 108

2024 Test A Math, question 21

Question

A coordinate grid with origin O, the x-axis and y-axis each marked 5, 10, 15 and 20, and grid lines every 5 units. One line segment joins the points (0, 20) and (20, 0), and another joins (0, 15) and (10, 0); all four points are marked with dots. The region between the two segments, with corners at (0, 15), (0, 20), (20, 0) and (10, 0), is shaded.

What is the area, in square units, of the shaded region shown in the figure above?

  • A75
  • B125
  • C150
  • D200
Show the answer

Correct answer: B

The area of the shaded region, in square units, is the area of the triangle created by the shaded and unshaded regions minus the area of the triangle created by the unshaded region.

The area of the triangle created by the shaded and unshaded regions is 200 square units.

12bh=12(20)(20)=12(400)=200\frac{1}{2}bh = \frac{1}{2}(20)(20) = \frac{1}{2}(400) = 200

The area of the triangle created by the unshaded region is 75 square units.

12bh=12(10)(15)=12(150)=75\frac{1}{2}bh = \frac{1}{2}(10)(15) = \frac{1}{2}(150) = 75

Subtract to find the area of the shaded region.

200−75=125200 - 75 = 125

Question analysis

A coordinate geometry question: two line segments cross a grid, one from (0, 20) to (20, 0) and one from (0, 15) to (10, 0), and the region between them is shaded. Students find the shaded area. The shape is not a standard figure, so they need to see it as one triangle with another removed.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Area of triangles is grade 6 content; finding a composite area by subtraction is the kind of task grade 7 students meet regularly. Points on the axes make every length easy to read.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The shaded shape has no name, so the question is about seeing it as a difference of two right triangles whose legs lie on the axes. Once that is seen, both areas are one-line calculations. The wrong choices come from stopping after either triangle or from pairing a base from one segment with a height from the other.

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

    All four lengths come from where the two segments meet the axes, at 20 and 15 on the y-axis and at 10 and 20 on the x-axis, read against grid lines 5 units apart. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. No obvious method minor Working it out

    The shaded region is a four-sided shape with no formula of its own; it has to be seen as the large triangle with the small one removed. (Schoenfeld, 1985)

  3. Something you must already know minor Working it out

    The area of a triangle, one half of base times height, is assumed and not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of the large right triangle under the outer segment and subtract the area of the smaller triangle under the inner segment, reading the legs from the axes.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

  1. Sort out the figure

    The grid lines are 5 units apart. Two slanted segments run from the y-axis down to the x-axis. The outer one goes from 20 on the y-axis to 20 on the x-axis. The inner one goes from 15 on the y-axis to 10 on the x-axis. The shaded region is the strip between them. You need its area, in square units.

  2. See the shape as two triangles

    The shaded region has four sides, and it has no simple formula. But look at the big triangle under the outer segment, with its corner at O. It is the shaded strip plus the small white triangle. So the shaded area is the big triangle minus the small one.

  3. Find the area of the big triangle

    The big triangle has a right angle at O. Its base runs along the x-axis from 0 to 20, so it is 20 units. Its height runs up the y-axis from 0 to 20, so it is 20 units.

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×20×20=200A = \frac{1}{2} \times 20 \times 20 = 200

    Watch out: this is the whole big triangle. Stopping here gives 200, choice D.

  4. Find the area of the small triangle

    The small triangle also has its right angle at O. Its base is 10 units, from 0 to 10 on the x-axis. Its height is 15 units, from 0 to 15 on the y-axis.

    A=12×10×15=75A = \frac{1}{2} \times 10 \times 15 = 75

    Watch out: 75 is the white triangle, choice A, not the shaded part. Also, pair the base and height of the same triangle. Mixing 20 and 15 gives 150, choice C.

  5. Subtract to find the shaded area

    Take the small white triangle away from the big triangle.

    200−75=125200 - 75 = 125

    The shaded area is 125 square units. So the answer is B, 125.

Check your answer

In the figure, the shaded strip looks bigger than the white triangle. So the shaded area should be more than half of the big triangle, which is 200÷2=100200 \div 2 = 100. 125 is more than 100. And 125+75=200125 + 75 = 200, the whole big triangle.

More questions like this

The original DOE question
Question 78 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 81, question 78. Open page 81 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 28 of 108

2024 Test A Math, question 25

Question

Points M, N, P and Q, each marked with a dot, lie in that order from left to right on a line. A dimension arrow from M to N is labeled x cm, and a longer dimension arrow from M to P is labeled y cm. No lengths are marked for NP, PQ or MQ.

In the figure above, N is the midpoint of MQ\overline{MQ}. Which segment has length (2x−y)(2x - y) centimeters?

  • APQ\overline{PQ}
  • BNP\overline{NP}
  • CMQ\overline{MQ}
  • DMP\overline{MP}
Show the answer

Correct answer: A

These facts are given:

  • N is the midpoint of MQ\overline{MQ}.
  • MN\overline{MN} has a length of x centimeters.
  • MP\overline{MP} has a length of y centimeters.

Use the first two facts to determine that MQ\overline{MQ} has a length of 2x2x centimeters. So (2x−y)(2x - y) can be represented by MQ−MPMQ - MP, which is PQ\overline{PQ}.

Question analysis

A geometry and algebra question with a labeled figure: points M, N, P and Q lie on a line, MN is x cm, MP is y cm, and N is the midpoint of MQ. Students find which segment has length 2x − y. They must turn the midpoint fact into an expression and read a difference of lengths on the figure.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students evaluate and rewrite expressions but rarely read one as a length on a figure. Combining the midpoint idea from geometry with a difference of expressions, and no numbers, makes this more abstract than typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 9, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question reverses the usual task: instead of finding a length, students match an expression to a segment. The midpoint fact turns into MQ = 2x, and taking away the y of MP leaves the piece from P to Q. The wrong choices reward stopping at 2x, reading only the labeled arrows, or matching the letter y; picking x = 5 and y = 7 lets students check with numbers.

  1. Variables instead of numbers moderate Understanding the question

    Every length is a letter: MN is x, MP is y, and the target is the expression 2x − y, so the matching segment cannot be found by measuring. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The figure marks only MN and MP with arrows; the length of MQ comes from the stem's midpoint fact, not from the drawing, and has to be written as 2x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the midpoint to write the length of the whole segment in terms of x, or pick sample values for x and y, then see which segment's length matches 2x − y.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint
MQ=2×MNMQ = 2 \times MN
N is the midpoint of segment MQ, the point exactly in the middle. MN is the half from M to N, and MQ is the whole segment. The whole is two equal halves, so it is 2 times MN.
Adding segments
MP+PQ=MQMP + PQ = MQ
When points lie in order on a line, the short segments add up to the long one. MP runs from M to P, and PQ runs from P to Q. Put end to end, they make MQ.

  1. Sort out the figure

    Points M, N, P and Q lie in that order on a line. Segment MN is x cm long, and segment MP is y cm long. N is the midpoint of MQ\overline{MQ}, the point exactly halfway from M to Q. You need the segment that is 2x−y2x - y cm long.

  2. Find the length of MQ

    MN is half of MQ, so MQ is twice as long.

    MQ=2×MN=2xMQ = 2 \times MN = 2x

    MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.

  3. Take MP away from MQ

    Subtracting y from 2x means taking MP away from MQ. MP covers the part from M to P. What is left of MQ is the piece from P to Q.

    MP+PQ=MQMP + PQ = MQ

    PQ=MQ−MP=2x−yPQ = MQ - MP = 2x - y

    So PQ is 2x−y2x - y cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is y−xy - x, choice B. So the answer is A, PQ\overline{PQ}.

Check your answer

Try numbers. Say x = 5, so MN = 5 and MQ = 10. Say y = 7, so MP = 7. Then PQ is 10−7=310 - 7 = 3. And 2x−y=10−7=32x - y = 10 - 7 = 3. They match. NP would be 7−5=27 - 5 = 2, which does not.

More questions like this

The original DOE question
Question 82 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 82. Open page 82 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 29 of 108

2024 Test A Math, question 26

Question

A parallelogram with horizontal top and bottom sides; its left and right sides both slant down to the right. The top side is labeled 30 ft and the slanted left side is labeled 25 ft. A vertical double-headed arrow from the top side down to the bottom side, with a right-angle mark where it meets the bottom side, is labeled 20 ft.

What is the area of the parallelogram shown above?

  • A750 sq ft
  • B600 sq ft
  • C500 sq ft
  • D300 sq ft
Show the answer

Correct answer: B

To find the area of a parallelogram, multiply the length of a side by the altitude for that side.

One property of parallelograms is that opposite sides are the same length.

So the bottom side of the parallelogram is 30 feet long and has an altitude of 20 feet.

Therefore, the area of the parallelogram, in square feet, is A=bh=30×20=600A = bh = 30 \times 20 = 600.

Question analysis

A geometry question with a figure: a parallelogram is labeled with a 30-foot top side, a 25-foot slanted side and a 20-foot height drawn at a right angle. Students find its area. The figure gives one length that is not needed, so students must pick the base and height that belong together.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Area of a parallelogram is grade 6 content, and this is a one-step use of the formula. It is easier than typical grade 7 area work, which involves composite figures, circles or scale drawings.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 2, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-step formula question with whole numbers, made slightly harder by one length too many. A student who multiplies the two side lengths, as for a rectangle, or brings in a triangle's one half, finds a listed choice. Matching the base to the height marked with a right angle is all it takes.

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

    The figure labels three lengths, but only the 30-foot side and the 20-foot height drawn at a right angle to it belong together; the slanted 25-foot side is extra. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The parallelogram area formula, base times height, is not given, and its height must be the perpendicular distance, not the slanted side. (Anderson & Krathwohl, 2001)

Strategy: Use base times height, where the height is the perpendicular distance between the two parallel sides chosen as bases.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a parallelogram
A=b×hA = b \times h
A is the area. b is the base, one of the two parallel sides. h is the height: the straight distance between the two parallel sides, at a right angle to the base. It works because you can cut a triangle off one end, move it to the other end and make a rectangle.

  1. Sort out the figure

    A parallelogram is a four-sided shape whose opposite sides are parallel. Parallel means they never meet. The figure shows three lengths. The top side is 30 ft. The slanted left side is 25 ft. The arrow from top to bottom is 20 ft, and the small square shows a right angle. You need the area, in square feet.

  2. Choose the base and height

    The height must make a right angle with the base. The 20 ft arrow meets the bottom side at a right angle. Opposite sides of a parallelogram are equal, so the bottom side is 30 ft too. So the base is 30 ft and the height is 20 ft. Watch out: the slanted 25 ft side is not needed. Using 25×20=50025 \times 20 = 500 gives choice C.

  3. Find the area

    A=b×hA = b \times h

    A=30×20=600A = 30 \times 20 = 600

    The area is 600 sq ft. Watch out: do not multiply the two sides, 30×25=75030 \times 25 = 750. That is choice A. Do not take half either, which is for triangles and gives 300, choice D. So the answer is B, 600 sq ft.

Check your answer

Picture cutting the triangle off the left end and sliding it to the right end. The shape becomes a rectangle 30 ft long and 20 ft tall, with an area of 600 sq ft. 750 is too big, because the shape is only 20 ft tall, not 25 ft.

More questions like this

The original DOE question
Question 83 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 30 of 108

2024 Test A Math, question 28

Question

QUIZ SCORES IN
MRS. ARCH’S CLASS

Quiz Score Number of Students
60 9
70 7
80 4
90 5
100 3

In the table above, what is the mean quiz score?

  • A60
  • B70
  • C75
  • D80
Show the answer

Correct answer: C

First, find the total number of students who took the quiz.

9+7+4+5+3=289 + 7 + 4 + 5 + 3 = 28

Multiply each quiz score by the number of students who received that score. Add those products together to find the total number of points scored.

(60×9)+(70×7)+(80×4)+(90×5)+(100×3)=2,100(60 \times 9) + (70 \times 7) + (80 \times 4) + (90 \times 5) + (100 \times 3) = 2{,}100

To find the mean quiz score, divide the total number of points by the number of students who took the quiz.

2,100÷28=752{,}100 \div 28 = 75

Question analysis

A frequency table question: the table lists five quiz scores from 60 to 100 and how many students in Mrs. Arch's class earned each. Students find the mean quiz score for the whole class. Each score has to count once for every student who earned it, and the class size has to be added up from the table.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 A mean from a frequency table is grade 6 content in a three-step form that grade 7 students meet regularly. Five rows of products make it slightly longer than a typical exercise, but the arithmetic is friendly.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 2, 19 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The table invites a shortcut: average the five scores in the first column and get a round number that is a choice. That ignores the fact that more students earned the low scores than the high ones, which pulls the true mean down. Weighting each score by its count is a five-row calculation with friendly numbers, and the other choices catch students who confuse the mean with the most common or the middle score.

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

    Each row of the table stands for several students, 9 with 60, 7 with 70 and so on, so a score counts once per student and the class size, 28, must be added up. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Five products, 540, 490, 320, 450 and 300, must be found and added to 2,100 before dividing by the 28 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    The last step, 2,100 ÷ 28, is not a familiar fact; noticing that 28 × 75 = 2,100, or testing the choices, keeps it quick. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply each score by its number of students, add the products for the total points, add the student counts, and divide the points by the students.

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.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean from a table
mean=total pointsnumber of students\text{mean} = \frac{\text{total points}}{\text{number of students}}
The mean is the average score. Total points is every student’s score added up. Each score in the table counts once for every student who got it. Number of students is how many took the quiz.

  1. Sort out the table

    The left column lists quiz scores. The right column says how many students got each score. For example, 9 students scored 60. You need the mean quiz score, which is the average.

  2. Count the students

    9+7+4+5+3=289 + 7 + 4 + 5 + 3 = 28

    28 students took the quiz. So there are 28 scores in all, not just 5.

  3. Find the total points

    Each score counts once for each student who got it. So multiply each score by its number of students.

    ScoreStudentsPoints
    60960×9=54060 \times 9 = 540
    70770×7=49070 \times 7 = 490
    80480×4=32080 \times 4 = 320
    90590×5=45090 \times 5 = 450
    1003100×3=300100 \times 3 = 300

    540+490+320+450+300=2,100540 + 490 + 320 + 450 + 300 = 2{,}100

    The class scored 2,100 points in all.

  4. Divide to find the mean

    mean=total pointsnumber of students\text{mean} = \frac{\text{total points}}{\text{number of students}}

    mean=2,10028=75\text{mean} = \frac{2{,}100}{28} = 75

    To divide, try 28×70=1,96028 \times 70 = 1{,}960. That leaves 2,100−1,960=1402{,}100 - 1{,}960 = 140, and 140=28×5140 = 28 \times 5. So the mean is 70+5=7570 + 5 = 75. Watch out: do not just average the five scores in the table. That gives 80, choice D. Also, 70 is the median and 60 is the most common score, choices B and A. So the answer is C, 75.

Check your answer

More students scored low, 60 or 70, than high. So the mean should be below 80, the middle of the listed scores. 75 is below 80. Also, 28×75=2,10028 \times 75 = 2{,}100, the total points.

More questions like this

The original DOE question
Question 85 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 85. Open page 83 of the handbook (PDF)

Question 31 of 108

2024 Test A Math, question 29

Question

Which graph represents the solution to the inequality x+4≥3x + 4 \geq 3?

  • ANumber line from −4 to 7 with a tick at every integer. A closed dot at −1, with the line shaded from −1 to the right, ending in an arrow.
  • BNumber line from −4 to 7 with a tick at every integer. A closed dot at 1, with the line shaded from 1 to the right, ending in an arrow.
  • CNumber line from −4 to 7 with a tick at every integer. A closed dot at 3, with the line shaded from 3 to the right, ending in an arrow.
  • DNumber line from −4 to 7 with a tick at every integer. A closed dot at 4, with the line shaded from 4 to the right, ending in an arrow.
Show the answer

Correct answer: A

Use the properties of equations to solve the inequality for x.

x+4≥3x + 4 \geq 3

Subtract 4 from both sides of the inequality.

x+4−4≥3−4x + 4 - 4 \geq 3 - 4

x≥−1x \geq -1

The graph of x≥−1x \geq -1 will have a point at −1 and will include all values that are greater than, or to the right of, −1. The greater than or equal sign is used, which indicates that the solution includes −1. So the point at −1 will be solid. Option A is the correct answer.

Question analysis

This inequality question asks which number line graph shows the solution set of x + 4 ≥ 3. Students solve a one-step inequality by undoing the addition, then match the result to a graph. All four graphs use a solid dot and an arrow pointing right, so the whole question rests on where the shaded ray begins.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 inequality work is usually two-step, such as 3x − 5 ≥ 10, and asks students to choose an open or closed dot and a direction. This one needs a single subtraction, and all four graphs share a closed dot and a rightward arrow.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 13, 9 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

Solving takes a single subtraction, and the graphs differ only in where the solid dot sits, so the question comes down to 3 − 4. Students who lose the negative sign land on the graph that starts at 1, while those who never isolate x pick a graph that starts at a number printed in the inequality. Testing 0, which satisfies 0 + 4 ≥ 3, shows that the shading must include 0.

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

    All four graphs share a closed dot and a rightward arrow, so the only thing to read is where the dot sits, against tick labels printed with raised minus signs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic minor Working it out

    Undoing the + 4 means computing 3 − 4, the one step that crosses zero; a dropped minus sign moves the dot from −1 to 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 4 from both sides to get x alone, then find the graph whose solid dot sits at that number; test one shaded value in the original inequality to confirm.

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.)
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an inequality balanced
if a≥b, then a−c≥b−c\text{if } a \geq b \text{, then } a - c \geq b - c
a and b are the two sides of an inequality, and c is any number. The sign ≥ stands for greater than or equal to. Taking the same number away from both sides keeps the comparison true, like taking the same weight off both pans of a scale. Use it to get x by itself.

  1. Sort out the question

    An inequality is a math sentence that compares two amounts. The sign ≥\geq means is greater than or equal to. So x+4≥3x + 4 \geq 3 says that x plus 4 is 3 or more.

    You need the graph that shows every number x that makes this true. All four graphs have a solid dot and an arrow to the right. Only the place where the dot sits is different.

  2. Get x by itself

    Right now 4 is added to x. To undo adding 4, subtract 4 from both sides.

    if a≥b, then a−c≥b−c\text{if } a \geq b \text{, then } a - c \geq b - c

    x+4−4≥3−4x + 4 - 4 \geq 3 - 4

    x≥−1x \geq -1

    So x is −1 or any number greater than −1. Watch out: 3 − 4 is −1, not 1. Dropping the minus sign puts the dot at 1, choice B. Putting the dot at 3 or 4 means you read a number off the inequality without solving. That leads to C or D.

  3. Match the solution to a graph

    On a number line, greater numbers are to the right. So the shading starts at −1 and goes right. The arrow shows that it keeps going.

    The dot is solid, or filled in, because of the or equal to part. A solid dot means −1 itself is a solution, a number that makes the inequality true.

    Graph A has a solid dot at −1 and is shaded to the right. So the answer is A, the graph with a solid dot at −1.

Check your answer

Test a number from the shaded part, such as 0. 0+4=40 + 4 = 4, and 4 is more than 3. Now test the dot: −1+4=3-1 + 4 = 3, and 3 is equal to 3, so −1 works. A number left of the dot, −2, gives −2+4=2-2 + 4 = 2. That is less than 3, so −2 should not be shaded, and it isn’t.

More questions like this

The original DOE question
Question 86 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 86. Open page 83 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 32 of 108

2024 Test A Math, question 32

Question

INGREDIENTS FOR 4 SERVINGS OF OATMEAL

Ingredient Cups
Oats 23\frac{2}{3}
Water 3143\frac{1}{4}

The table shows the cups of ingredients used to make 4 servings of oatmeal. What is the unit rate for cups of oats per cup of water?

  • A16\frac{1}{6}
  • B839\frac{8}{39}
  • C1316\frac{13}{16}
  • D138\frac{13}{8}
Show the answer

Correct answer: B

The unit rate for cups of oats per cup of water can be found by dividing the number of cups of oats by the number of cups of water:

23314\frac{\frac{2}{3}}{3\frac{1}{4}}

The denominator can be rewritten as 134\frac{13}{4}.

The quotient 23134\frac{\frac{2}{3}}{\frac{13}{4}} can be rewritten as:

23×413=839.\frac{2}{3} \times \frac{4}{13} = \frac{8}{39}.

Therefore, the unit rate for cups of oats per cup of water is 839\frac{8}{39}.

Question analysis

This unit rate question uses a recipe table for 4 servings of oatmeal, with 2/3 cup of oats and 3 1/4 cups of water, and asks for cups of oats per cup of water. Students divide one fraction by another, including a mixed number. Deciding which quantity goes on top, and whether the number of servings matters, is the tricky part.

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Unit rates from fractions, such as 1/2 mile in 1/4 hour, are exactly grade 7 work. This one adds a mixed number, a fraction divided by a fraction, and a table titled for 4 servings, so students must decide which numbers the rate uses.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards reading the rate's wording exactly and then carrying out a fraction division without slips. The 4 servings in the title is the hook: dividing by 4 gives a per-serving amount that sounds like a unit rate and is listed. Students who write oats over water as one fraction, convert the mixed number and multiply by the reciprocal finish in a few lines.

  1. Error-prone arithmetic moderate Working it out

    The rate is a fraction over a mixed number, 2/3 ÷ 3 1/4, so 3 1/4 must become 13/4 and the division a multiplication by 4/13. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    In "cups of oats per cup of water" the quantity before "per" goes on top; reading the phrase loosely puts water over oats or brings in servings. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table's title, "for 4 servings", supplies the only other number on the page, and it plays no part in a rate of oats to water. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Write the rate as cups of oats over cups of water, change the mixed number to an improper fraction, and divide by multiplying by its reciprocal.

Traps

  • Information you don't need: The question gives numbers or facts that play no part in the answer.
  • 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.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=first amountsecond amount\text{unit rate} = \frac{\text{first amount}}{\text{second amount}}
A unit rate tells how much of the first amount goes with 1 of the second amount. The word per means for each. The amount named before per goes on top, and the amount after it goes on the bottom.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
Dividing by a fraction
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. This works because dividing by a number undoes multiplying by it.

  1. Sort out the question

    The table shows 23\frac{2}{3} cup of oats and 3143\frac{1}{4} cups of water. You need the unit rate for “cups of oats per cup of water.” That is how many cups of oats go with 1 cup of water.

    Watch out: the 4 servings play no part here. Dividing the oats by 4 servings gives 16\frac{1}{6} cup for each serving, choice A.

  2. Set up the rate

    Oats comes before per, so oats goes on top. Water goes on the bottom.

    unit rate=first amountsecond amount\text{unit rate} = \frac{\text{first amount}}{\text{second amount}}

    unit rate=23÷314\text{unit rate} = \frac{2}{3} \div 3\frac{1}{4}

    Watch out: water over oats flips the rate. That gives a number bigger than 1, like choice D.

  3. Change the mixed number

    A mixed number is a whole number plus a fraction. Change 3143\frac{1}{4} to a fraction so you can divide.

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    314=3×4+14=1343\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{13}{4}

    So 3 and one quarter cups is 13 quarters of a cup.

  4. Divide by multiplying

    To divide by 134\frac{13}{4}, multiply by its reciprocal, 413\frac{4}{13}.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    23÷134=23×413=2×43×13=839\frac{2}{3} \div \frac{13}{4} = \frac{2}{3} \times \frac{4}{13} = \frac{2 \times 4}{3 \times 13} = \frac{8}{39}

    No whole number bigger than 1 divides both 8 and 39, so the fraction can’t be made simpler. The rate is 839\frac{8}{39} cup of oats per cup of water.

    So the answer is B, 839\frac{8}{39}.

Check your answer

There is much more water than oats. So the oats for 1 cup of water should be small, and 839\frac{8}{39} is about 0.2. Also multiply back: 839×134=104156=23\frac{8}{39} \times \frac{13}{4} = \frac{104}{156} = \frac{2}{3}. That is the cups of oats in the table.

More questions like this

The original DOE question
Question 89 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 84, question 89. Open page 84 of the handbook (PDF)

Question 33 of 108

2024 Test A Math, question 36

Question

CELL PHONE SALES BY COLOR

Color Percentage of Cell Phones Sold
White 8k8k
Black 30
Blue 30−2k30 - 2k
Red k+5k + 5
Total 100

The table above shows cell phone sales by color. What percentage of the cell phones sold were blue?

  • A18%
  • B20%
  • C22%
  • D28%
Show the answer

Correct answer: B

First, use the information given in the table and properties of equations to determine the value of k:

8k+30+30−2k+k+5=1008k + 30 + 30 - 2k + k + 5 = 100

8k−2k+k+30+30+5=1008k - 2k + k + 30 + 30 + 5 = 100

7k+65=1007k + 65 = 100

7k+(65−65)=100−657k + (65 - 65) = 100 - 65

7k=357k = 35

k=5k = 5

Then, determine the percentage of cell phones sold that were blue by finding the value of the expression 30−2k30 - 2k (given in the table), when k=5k = 5:

30−2k=30−2(5)=30 - 2k = 30 - 2(5) =

30−10=2030 - 10 = 20

The percentage of cell phones sold that were blue is 20%.

Question analysis

A table of cell phone sales by color gives each color's percentage as an expression in k, such as 8k for white and 30 − 2k for blue, with a total of 100. Students write and solve an equation for k, then find the percentage for blue. Combining terms with a negative coefficient, and remembering that k itself is not the answer, both need care.

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve two-step equations that are written out for them. Here the equation must be built from a table whose percentages total 100, with like terms to combine, and the value found must then be substituted back.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 4, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hides an equation in a table, and the key insight is that the four percentages must add to 100. After that the algebra is routine, and no listed choice rewards stopping early, since neither k nor any other color's percentage appears. The wrong choices sit near the answer or use the blue expression with k ignored, so students who cannot set up the equation are left guessing.

  1. Many steps moderate Working it out

    The answer needs a chain: add the four expressions, set the sum to 100, collect 8k − 2k + k into 7k, solve for k, then substitute into 30 − 2k. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    No equation is printed; the table's Total row of 100 is what turns the four percentages written in k into one equation to solve. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Error-prone arithmetic minor Working it out

    Collecting like terms includes the negative coefficient in the blue row, −2k, and the constants 30, 30 and 5 have to be gathered too. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four color expressions and set the sum equal to 100, combine like terms to solve for k, then substitute k into the blue row's expression.

Traps

  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms
ak+bk=(a+b)kak + bk = (a + b)k
k is the unknown number. a and b are the numbers in front of k, called coefficients. Terms with the same letter are called like terms. You can add their coefficients, because a groups of k plus b groups of k make a + b groups of k.
Keeping an equation balanced
if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c
a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it balanced. Use it to undo each step and get k alone.

  1. Sort out the table

    The table gives the percent of phones sold in each color. Some entries use k, a letter that stands for an unknown number. 8k means 8 times k. You need the percent that were blue, 30−2k30 - 2k. To find it, you first need k.

  2. Write an equation for the total

    Every phone sold is one of the four colors. So the four percents add up to the total, 100, as the last row of the table shows.

    8k+30+(30−2k)+(k+5)=1008k + 30 + (30 - 2k) + (k + 5) = 100

  3. Combine like terms

    Terms with k in them are called like terms, so they can be combined. Group the k terms together and the plain numbers together. A k on its own means 1k.

    ak+bk=(a+b)kak + bk = (a + b)k

    8k−2k+k=(8−2+1)k=7k8k - 2k + k = (8 - 2 + 1)k = 7k

    30+30+5=6530 + 30 + 5 = 65

    So the equation becomes 7k+65=1007k + 65 = 100. Watch out: the blue row takes away 2k. Keep its minus sign, or the k terms will be wrong.

  4. Solve for k

    Undo the adding first, then the multiplying.

    if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c

    7k+65−65=100−657k + 65 - 65 = 100 - 65

    7k=357k = 35

    k=35÷7=5k = 35 \div 7 = 5

    So k is 5. But k is not the answer. You still need the blue row.

  5. Find the blue percent

    Put 5 in for k in the blue row.

    30−2k=30−2×5=30−10=2030 - 2k = 30 - 2 \times 5 = 30 - 10 = 20

    So 20% of the phones sold were blue. Watch out: 2k means 2 times k, not 2. Reading 30−2k30 - 2k as 30−230 - 2 gives 28%, choice D.

    So the answer is B, 20%.

Check your answer

Put k = 5 into every row. White: 8×5=408 \times 5 = 40. Black: 30. Blue: 20. Red: 5+5=105 + 5 = 10. Now add them: 40+30+20+10=10040 + 30 + 20 + 10 = 100. The total matches the table, so k = 5 is right.

More questions like this

The original DOE question
Question 93 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 93. Open page 85 of the handbook (PDF)

Question 34 of 108

2024 Test A Math, question 37

Question

Number line with points X, Y and Z marked from left to right; Y is closer to X than to Z. Above the line, a dimension arrow from X to Y is labeled 1 and one from Y to Z is labeled 2.

On the number line above, the distance between X and Y is 1 unit, and the distance between Y and Z is 2 units. What is the distance, in units, between Y and the midpoint of X and Z?

  • A12\frac{1}{2}
  • B1
  • C1121\frac{1}{2}
  • D3
Show the answer

Correct answer: A

First, find the distance between X and Z by adding the distance between X and Y to the distance between Y and Z.

1+2=31 + 2 = 3

Since the distance between X and Z is 3, the midpoint between X and Z is 1121\frac{1}{2} units from X.

The midpoint between X and Z is 1121\frac{1}{2} units from X, and the distance between X and Y is 1. The distance between Y and the midpoint of X and Z is 112−1=121\frac{1}{2} - 1 = \frac{1}{2}.

Question analysis

This number line question shows points X, Y and Z, with 1 unit from X to Y and 2 units from Y to Z, and asks how far Y is from the midpoint of X and Z. Students find the length of XZ, locate its midpoint, and compare it with Y. Each step produces a tempting number, so stopping early is the usual mistake.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 number line problems usually involve negative numbers and fractions as coordinates. This one has no coordinates, only two whole-number distances, and needs three short steps: a total length, a midpoint and a difference.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 39 words.
Corrected from the DOE handbook
  • An imprecise statement about the midpoint 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 arithmetic is light, but every intermediate result is waiting among the choices, so the risk is stopping before the last subtraction. Students also have to keep two midpoints apart: the one asked about lies between X and Z, not halfway along the 2-unit stretch from Y. Marking X at 0 and the midpoint at 1.5 on the drawing makes the half-unit gap visible.

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

    The number line has no coordinates, only dimension arrows labeled 1 and 2, so students must assign positions themselves, such as X at 0, Y at 1 and Z at 3. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three short results come in sequence, the length XZ, the midpoint's distance from X, then its distance from Y, and each is also a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Place X at 0 so that Y and Z get positions from the given distances, find the midpoint of X and Z as half their distance, then compare it with Y's position.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “the midpoint between X and Z is 1½” (twice). A midpoint is a point, and this number line has no numbers on it, so the 1½ is the midpoint’s distance from X (half of XZ = 3). Both sentences now say the midpoint is “1½ units from X.” The rest is unchanged: Y is 1 unit from X, so the distance between Y and the midpoint is 1½ − 1 = ½.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding lengths on a line
XZ=XY+YZXZ = XY + YZ
X, Y and Z are points on one line, with Y between X and Z. XY and YZ are the lengths of the two short pieces. XZ is the length of the whole. The two pieces join end to end to make the whole.
Midpoint of a segment
distance from an end to the midpoint=length of the segment2\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}
A segment is the straight piece of a line between two points, its ends. The midpoint is the point exactly halfway between the ends. It splits the segment into two equal halves. So it is half the length away from each end.

  1. Sort out the picture

    The points go X, then Y, then Z, from left to right. X to Y is 1 unit, and Y to Z is 2 units. You need the midpoint of X and Z, the point halfway between them. Then you need how far Y is from that midpoint. The line has no numbers, so measure everything from X.

  2. Find the length from X to Z

    XZ=XY+YZXZ = XY + YZ

    XZ=1+2=3XZ = 1 + 2 = 3

    X to Z is 3 units. Watch out: 3 is the whole length, choice D. It is not a distance to the midpoint.

  3. Place the midpoint

    The midpoint of X and Z is half of 3 units from X.

    distance from an end to the midpoint=length of the segment2\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}

    32=112\frac{3}{2} = 1\frac{1}{2}

    So the midpoint is 1121\frac{1}{2} units to the right of X. Watch out: this is its distance from X, not from Y. Stopping here gives choice C.

  4. Measure from Y to the midpoint

    Y is 1 unit to the right of X. The midpoint is 1121\frac{1}{2} units to the right of X. Subtract to find the gap between them.

    112−1=121\frac{1}{2} - 1 = \frac{1}{2}

    Watch out: use the midpoint of X and Z, not of Y and Z. Halving the 2 units from Y to Z gives 1, choice B.

    So the answer is A, 12\frac{1}{2} unit.

Check your answer

Give each point a number, with X at 0. Then Y is at 1 and Z is at 3. Halfway between 0 and 3 is 1121\frac{1}{2}. From Y at 1 to 1121\frac{1}{2} is 12\frac{1}{2}. This fits the picture too: Y sits a little left of the middle of XZ.

More questions like this

The original DOE question
Question 94 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 94. Open page 86 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.

Question 35 of 108

2024 Test A Math, question 39

Question

Triangle with its two slanted sides each labeled 13 in. and its base labeled 10 in. Beside it, a vertical arrow as tall as the triangle, from its base to its top vertex, is labeled 12 in.

Raquel is cutting out pieces of cardboard to make a pyramid. She will use a square piece for the base and identical triangular pieces for the sides. The figure above shows the dimensions of the triangle for each side. What will be the total surface area, in square inches, of the pyramid, including the square base?

  • A280
  • B295
  • C340
  • D360
Show the answer

Correct answer: C

Surface area is found by calculating the area of each side, including the base, and adding them all together.

The base of the pyramid measures 10 inches on each side, so the area of the base is

10×10=10010 \times 10 = 100

Each triangular face has a base of 10 inches and a height of 12 inches, so the area of each face is

12×10×12=60\frac{1}{2} \times 10 \times 12 = 60

There are 4 triangular faces (because the square base has 4 edges, each of which will have a triangular face). So the total surface area of the pyramid is the sum of the areas of 4 triangles and 1 square.

(4×60)+100=240+100(4 \times 60) + 100 = 240 + 100

So the total surface area is 340 square inches.

Question analysis

This surface area word problem shows one triangular face, with sides of 13 inches, a base of 10 inches and a height of 12 inches, that Raquel will use to build a cardboard pyramid with a square base. Students find the total area of the base and all the triangular sides. Picturing the solid and choosing the right measurements from the figure are the tricky parts.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.4 nets and surface area
  • 6.G.1 area of polygons
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students find the surface area of a pyramid from a drawn net or solid. Here only one triangular face is shown, so students must picture the whole pyramid, count its faces, and choose which labeled length belongs in the area formula.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 6, 56 words.
Corrected from the DOE handbook
  • A wrong unit 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

Students have to build the pyramid in their heads from one flat triangle, then choose the right measurement from three labels. The 13-inch sides look like the natural height, and using them gives a listed total; picturing a pyramid with three sloping faces gives another. Sketching the square base with a triangle on each edge, and marking 12 as the height, settles both the face count and the formula.

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

    The figure gives one triangle with three length labels, 13, 13 and 10, and puts its height, 12, on a separate arrow beside it, where it is easy to overlook. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    The solid is never drawn: students must see that the square's side equals the triangle's 10-inch base, count four triangles, and add triangle and square areas. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Something you must already know minor Working it out

    A triangle's area uses its perpendicular height, not a slanted side, and a pyramid on a square base has four triangular faces; neither fact is given. (Anderson & Krathwohl, 2001)

Strategy: Sketch the pyramid: a square base whose side is the triangle's base, and one triangular face on each side of the square. Find each area with the matching formula, then add them all.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Information you don't need: The question gives numbers or facts that play no part in the answer.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends “So the total surface area is 340 square feet.” The triangle’s dimensions are in inches and the question asks for square inches, so the site now reads “340 square inches.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.
Area of a square
A=s×sA = s \times s
A is the area and s is the length of one side. All four sides of a square are equal, so the area is side times side.
Surface area
surface area=sum of the areas of all the faces\text{surface area} = \text{sum of the areas of all the faces}
A face is one flat side of a solid shape. The surface area is the total area of the outside. Picture unfolding the solid into flat pieces. Then add up the area of every piece.

  1. Sort out the question

    Raquel’s pyramid has a square base. Each side is a triangle like the one in the figure. The triangle’s base is 10 in. Its two slanted sides are 13 in. Its height, the arrow beside it, is 12 in. You need the total surface area, including the square base, in square inches.

  2. Picture the pyramid

    Surface area is the total area of all the flat sides, called faces. Each triangle stands on one edge of the square. So each edge of the square is 10 inches, the same as the triangle’s base. A square has 4 edges, so the pyramid has 4 triangles.

    surface area=sum of the areas of all the faces\text{surface area} = \text{sum of the areas of all the faces}

    Here that is 4 triangles plus 1 square. Watch out: counting only 3 triangles gives 280, choice A.

  3. Find the area of one triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×10×12=60A = \frac{1}{2} \times 10 \times 12 = 60

    Each triangle is 60 square inches. Watch out: the height is 12 in., the straight-up arrow. It is not 13 in., the slanted side. Using 13 gives 65 for each triangle and 360 in all, choice D.

  4. Find the area of the base

    The base is a square with sides of 10 inches, the same as each triangle’s base.

    A=s×sA = s \times s

    A=10×10=100A = 10 \times 10 = 100

    The square base is 100 square inches.

  5. Add all five faces

    surface area=4×60+100=240+100=340\text{surface area} = 4 \times 60 + 100 = 240 + 100 = 340

    The four triangles and the square together cover 340 square inches.

    So the answer is C, 340 square inches.

Check your answer

Each triangle is half of a 10 by 12 rectangle, which has an area of 120. So two triangles make 120, and four make 240. Add the base: 240+100=340240 + 100 = 340. The units fit too: inches times inches gives square inches.

More questions like this

The original DOE question
Question 96 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 96. Open page 86 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.

Question 36 of 108

2024 Test A Math, question 55

Question

The table below shows the number of cups of red paint and blue paint used to make a purple paint mixture.

PURPLE PAINT

Cups of Red Paint Cups of Blue Paint
1 1.5
4 6
11 y

Based on the relationship between the number of cups of red paint and the number of cups of blue paint, what is the value of y?

  • A10.5
  • B13
  • C16.5
  • D24
Show the answer

Correct answer: C

Divide the number of cups of blue paint by the number of cups of red paint to find the constant of proportionality.

1.51=64=1.5\frac{1.5}{1} = \frac{6}{4} = 1.5

Multiply to find the value of y.

11×1.5=16.511 \times 1.5 = 16.5

Question analysis

A table gives cups of red and blue paint for a purple mixture: 1 and 1.5, 4 and 6, and 11 and an unknown y. This ratio table question asks for y based on the relationship between the two colors. The tricky part is noticing whether the relationship works by multiplying or by adding.

NYS standards
  • 7.RP.2b constant of proportionality
  • 6.RP.3a tables of equivalent ratios
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a constant of proportionality from a table is grade 7 work, but here one row already pairs 1 cup of red with its cups of blue, so a single multiplication finishes it, lighter than typical grade 7 tasks.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 57 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The relationship is multiplicative, 1.5 cups of blue for every cup of red, and one multiplication finishes the question. The trap is additive thinking: red grows by 7 cups from the second row to the third, and adding the same 7 to the blue column gives a listed answer. Testing a rule on every row exposes the additive one, since blue is 0.5 more than red in the first row but 2 more in the second.

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

    The table's first row, 1 cup of red to 1.5 cups of blue, hands over the constant of proportionality; the other two rows only confirm it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the row with 1 cup of red paint to read the cups of blue per cup of red, multiply that rate by 11, and check the rate against the row with 4 cups.

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.
  • 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

Constant of proportionality
k=yxk = \frac{y}{x}
x and y are two amounts that change together. k is the constant of proportionality: how much y goes with 1 of x. In a proportional relationship, y divided by x gives the same k every time.
Proportional relationship
y=k×xy = k \times x
x is the cups of red paint, and y is the cups of blue paint. k is the cups of blue for each cup of red. In a proportional relationship, y is always k times x. Use it to find y for any amount of red.

  1. Sort out the table

    The table shows cups of red and blue paint for the same purple. 1 cup of red goes with 1.5 cups of blue. 4 cups of red go with 6 cups of blue. You need y, the cups of blue for 11 cups of red. The same color means the paints mix at the same rate in every row.

  2. Find the blue for each cup of red

    k=yxk = \frac{y}{x}

    k=1.51=1.5andk=64=1.5k = \frac{1.5}{1} = 1.5 \quad \text{and} \quad k = \frac{6}{4} = 1.5

    Both rows give 1.5. So there are always 1.5 cups of blue for each cup of red. This number is the constant of proportionality.

  3. Use the rate for 11 cups

    y=k×xy = k \times x

    y=1.5×11=16.5y = 1.5 \times 11 = 16.5

    Think of it as 11 cups plus half of 11 cups: 11+5.5=16.511 + 5.5 = 16.5. So y is 16.5 cups of blue.

  4. Avoid the traps and answer

    Watch out: red goes up 7 cups from 4 to 11. Adding the same 7 cups to the 6 cups of blue gives 13, choice B. Paint mixes by multiplying, not by adding. Watch out: 7×1.5=10.57 \times 1.5 = 10.5, choice A, is only the extra blue for the 7 extra cups of red.

    So the answer is C, 16.5.

Check your answer

Divide to check: 16.5÷11=1.516.5 \div 11 = 1.5. That matches the other rows. Also, every row has more blue than red, so y must be more than 11. That rules out 10.5.

More questions like this

The original DOE question
Question 112 as printed on page 90 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 90, question 112. Open page 90 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 37 of 108

2024 Test B English, question 40
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

How does the map provide additional support for a central idea of the passage?

  • Aby demonstrating that Morse’s telegraph system greatly exceeded the limitations of previous long-distance communication systems
  • Bby indicating that Morse was lacking foresight by seeking a patent only in North America
  • Cby proving that Gale’s advice helped Morse extend the telegraph’s range much farther than first thought possible
  • Dby showing that the Morse telegraph had a surprisingly immense influence across the world
Show the answer

Correct answer: A

The question asks how the map provides additional support for a central idea of the passage.

A. CORRECT. The map shows the telegraph line crossing continents in 1902, just forty-four years after the first transatlantic message was sent in 1858. This is a much greater distance than the first line Morse built between “the Supreme Court chamber of the Capitol building and the railroad station in Baltimore” (paragraph 7).

B. Incorrect. The map shows the distance the line covers; it does not indicate Morse’s lack of foresight in seeking patents only in North America.

C. Incorrect. While Gale’s advice was crucial to Morse’s success, the map shows the physical line stretching across continents many years later, after their initial work.

D. Incorrect. While it can be inferred that the telegraph had influence across the world, the amount of influence cannot be determined.

Question analysis

This question pairs “Invention of the Telegraph” with a map of the British All-Red Telegraph Line in 1902 and asks how the map adds support for a central idea of the passage. Students must read a world map with a key and a caption, connect what it shows to the story of Samuel Morse’s invention, and sort through four claims about the map that vary in how far they go.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R2 theme or central idea
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Linking a graphic to a text is familiar grade 7 work. Here students must pin down a central idea of the passage and work out how a 1902 world map, made decades after Morse’s first line, bears on it, a step beyond typical demand.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from the whole text
Reading level
Passage: about grade 11, 910 words; question and choices: about grade 11, 75 words.
Corrected from the DOE handbook
  • An inaccurate detail 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

Lines on a map can prove how far the telegraph reached, and the question turns on how much more than that each choice claims. The key depends on a link back to paragraph 1, where a message went only as far as the next hilltop tower, read many minutes before the map. The influence choice skips that link and states the map’s message in bigger terms, so students who do not ask what a map of routes can show are drawn to it.

  1. A very close wrong answer major Choosing the answer

    The influence choice, a telegraph with “a surprisingly immense influence across the world,” sounds like a fair reading of lines circling the globe; it differs from the key in claiming impact and surprise rather than reach beyond earlier systems. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Evidence in several places moderate Working it out

    The earlier systems the key measures against are in paragraph 1, towers on distant hills passing flag and light signals from station to station, far from the map at the end of the passage. (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)

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

    The map shows a 1902 British line from London to Canada and Vancouver, south through Bermuda and Barbados to Cape Town, and on to India, Australia and New Zealand, with Washington left unconnected. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Look at what the map and its caption actually show, sum up a central idea of the passage in your own words, and keep only a choice that both of them back up.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says the 1902 line came “forty-four years after the first message was sent in 1858.” The first telegraph message was Morse’s in 1844 (paragraph 1); 1858 is the year of the first transatlantic message, as the map’s caption says. The site now says “the first transatlantic message.”

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This question asks how a graphic adds to the passage. You need two things. First, what does the map really show? Second, which central idea, a main point the whole passage builds, does it back up?

  2. Read the map and its caption

    This map is a picture of one British network of telegraph wires, as it stood in 1902. In its key, a thick black line stands for an electric telegraph line. From London, a line crosses the Atlantic Ocean and Canada to Vancouver. Other lines reach Bermuda, Barbados, Cape Town, India, Australia and New Zealand. According to the caption, by 1902 this network “connected most of the world.”

  3. Recall the older ways of signaling

    Next, compare that with paragraph 1. Before electricity, people signaled with flags or lights. An observer would “decipher a signal from a high tower on a distant hill and then send it on to the next station.” So a message could only jump as far as a person could see. Even Morse’s first line, in paragraph 6, reached Baltimore, just “forty miles away.”

  4. Connect the map to that idea

    The caption says the first message across the Atlantic was sent in 1858. Only 44 years later, in 1902, lines crossed several oceans and linked much of the globe. Paragraph 8 adds that telegraph lines “connected the continents of Europe and America” before Morse died. So the map supports a central idea: electric telegraph lines went far beyond the reach of older signal systems.

  5. Check each choice

    • A fits. Lines that span oceans go far past the limits of hilltop towers.
    • B does not fit. A map of lines shows nothing about patents. Also, paragraph 5 says Morse sought patent protection in Europe, too.
    • C does not fit. The map names no inventors, so it cannot prove what Gale’s advice did decades before.
    • D does not fit. The map shows where lines went, not how much they changed the world, and nothing on it shows surprise.

  6. Choose the answer

    The map shows telegraph lines reaching around the globe, far past what flags and towers could do. So the answer is A.

Why this answer and not the closest wrong one

D is the closest trap, because lines around the world seem to prove a huge effect. But a map of routes shows only how far messages could travel. It cannot measure influence, and it cannot show that anyone was surprised. A sticks to what a map of lines can prove: the telegraph reached much farther than earlier systems.

More questions like this

The original DOE question
Question 40 as printed on page 188 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 188, question 40. Open page 188 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 38 of 108

2024 Test B Math, question 3

Question

Scale drawing of a garden shaped like a rectangle with its lower right corner cut out; all six sides are horizontal or vertical. The top side is labeled 13 cm and the left side 6 cm. From the top right corner, the right side, labeled 3 cm, goes down to an unlabeled horizontal side running left, then an unlabeled vertical side runs down to the bottom side, which is labeled 8 cm.

The figure above shows a scale drawing of a garden, where 1 centimeter represents 2.5 meters. What is the perimeter of the actual garden in meters?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 95

First, find the perimeter of the figure.

Two sides do not have labels. The top of the figure measures 13 cm and the bottom shows a length measuring 8 cm, so the horizontal side without a label measures 5 cm (13−8)(13-8).

The left side of the figure measures 6 cm and the right side shows a length measuring 3 cm, so the vertical side without a label measures 3 cm (6−3)(6-3).

So the total perimeter of the scale drawing is

13+3+5+3+8+6=38 cm13+3+5+3+8+6=38\text{ cm}

Use the scale to find the perimeter of the actual garden:

1 cm2.5 m=38 cmp m\frac{1\text{ cm}}{2.5\text{ m}}=\frac{38\text{ cm}}{p\text{ m}}

p=2.5×38=95p=2.5\times 38=95

Question analysis

A grid-in on a scale drawing of an L-shaped garden, where each centimeter stands for 2.5 meters and two sides carry no labels. Students find the perimeter of the drawing, then convert it to the real garden's perimeter in meters. The unlabeled sides are easy to leave out.

NYS standards
  • 7.G.1 scale drawings
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings are grade 7 content, and a typical task converts one or two given lengths. Here two missing sides must be worked out and the perimeter assembled before the scale applies, more steps but nothing beyond grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 9, 24 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure hides the work: it labels four of the six sides, and the missing two are short pieces of the notch that are easy to overlook. Once all six are found, the scale step is a single multiplication, but it must go the right way, since the real garden is larger than the drawing. Tracing the outline and counting six sides guards against the most common loss.

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

    Two sides of the notch carry no labels, so students must find them from the opposite sides: 13 − 8 = 5 cm across and 6 − 3 = 3 cm down. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas minor Working it out

    The perimeter of an L-shaped drawing has to be joined to a scale of 2.5 meters per centimeter, two separate skills in one answer. (Embretson, 1998; Daniel & Embretson, 2010)

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

    With no choices, nothing flags a perimeter that is missing two sides or still in centimeters; the student must check the six-sided total alone. (Enright, Morley & Sheehan, 2002)

Strategy: Use the opposite sides to work out the two unlabeled lengths, add all six sides of the drawing, then multiply the total in centimeters by 2.5 meters per centimeter.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter
P=sum of all the side lengthsP = \text{sum of all the side lengths}
P is the perimeter, the distance all the way around a shape. Add the length of every side, once each. It works for any shape with straight sides, even one with a corner cut out.
Using a scale
actual length=drawing length×s\text{actual length} = \text{drawing length} \times s
s is the scale, the number of meters that 1 centimeter on the drawing stands for. Here s is 2.5. Each centimeter on the drawing is 2.5 meters in real life, so you multiply.

  1. Sort out what you know

    The figure is a scale drawing, a small copy of the real garden. Each 1 cm on the drawing stands for 2.5 m in the real garden. You need the perimeter of the real garden in meters. The perimeter is the distance all the way around.

    The shape has 6 sides. Four are labeled: 13 cm, 3 cm, 8 cm and 6 cm. The two sides around the cut-out corner have no labels.

  2. Find the missing side across

    Look at the top side, 13 cm. Below it, the bottom side (8 cm) and the short unlabeled side across the notch together reach just as far. So the missing side is the difference.

    13−8=513 - 8 = 5

    The missing side across is 5 cm.

  3. Find the missing side going down

    Now look at the left side, 6 cm. On the right, the 3 cm side and the unlabeled side going down together make the same height.

    6−3=36 - 3 = 3

    The missing side going down is 3 cm.

  4. Add all six sides

    P=sum of all the side lengthsP = \text{sum of all the side lengths}

    P=13+3+5+3+8+6=38P = 13 + 3 + 5 + 3 + 8 + 6 = 38

    The perimeter of the drawing is 38 cm. Watch out: adding only the four labeled sides gives 30 cm. That leads to the wrong answer 75.

  5. Change to real meters

    Each centimeter stands for 2.5 meters, so multiply by 2.5.

    actual length=drawing length×s\text{actual length} = \text{drawing length} \times s

    actual length=38×2.5=76+19=95\text{actual length} = 38 \times 2.5 = 76 + 19 = 95

    Here 38×2=7638 \times 2 = 76 and 38×0.5=1938 \times 0.5 = 19. The real perimeter is 95 meters. Watch out: 38 is still in centimeters, so do not grid it. Dividing by 2.5 gives 15.2, which makes the garden smaller than its drawing.

    So you would grid 95.

Check your answer

Slide the two notch sides out to the corner. The distance around stays the same, so it equals a 13 cm by 6 cm rectangle: 2×(13+6)=382 \times (13 + 6) = 38. Also, 95 meters is more than 38, which fits, because each centimeter stands for more than 1 meter.

More questions like this

The original DOE question
Question 60 as printed on page 198 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 198, question 60. Open page 198 of the handbook (PDF)

Question 39 of 108

2024 Test B Math, question 5

Question

The figure below is a triangular prism. The lateral sides are rectangles.

A triangular prism with a triangle on the top and bottom and three rectangular sides. On the top triangle, a dashed segment labeled 8 in. runs from the far vertex to the opposite side and meets it at a right angle. The bottom edge below that opposite side is labeled 4 in., and the vertical edge at its left end is labeled 2 in.

What is the volume of the figure, in cubic inches?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 32

The volume of a triangular prism is found by multiplying the area of the triangular base by the height of the prism.

The area of the triangular base of the prism is 12bh=12(4)(8)=16\frac{1}{2}bh=\frac{1}{2}(4)(8)=16 square inches.

The height of the triangular prism is 2 inches.

So the volume of the triangular prism, in cubic inches, is 16∙2=3216\bullet 2=32.

Question analysis

A geometry grid-in: find the volume, in cubic inches, of a triangular prism drawn with its triangular faces on top and bottom. The drawing gives a 4-inch side of the triangle, a dashed 8-inch height drawn to that side, and a 2-inch edge. Students must recall the volume formula and match each length to its role.

NYS standards
  • 7.G.6 area, surface area, and volume
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the volume of a right triangular prism is named in the grade 7 standards, and this is a direct two-step case. The only extra demand is reading a slanted drawing to see which length plays which role.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards knowing that a prism's volume is its base area times its height, then matching each length to its job. The drawing lies on its side, with a short 2-inch height and the triangle flat on top, which makes the base harder to spot. Most lost points come from the one half: leaving it out doubles the volume, and stopping at the triangle's area gives half of it.

  1. Something you must already know moderate Working it out

    Neither formula is given: the volume is the area of the triangular base times the prism's height, and that area needs the one half in 1/2 × base × height. (Anderson & Krathwohl, 2001)

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

    The prism lies with its triangles on top and bottom, so the 2-inch vertical edge is its height, while the dashed 8-inch line is the triangle's height to the 4-inch side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

    There are no choices, so a product that forgets the one half, or one that stops at the area of the triangle, gives no warning. (Enright, Morley & Sheehan, 2002)

Strategy: Find the area of the triangular face as one half times base times height, using the 4-inch side and the dashed 8-inch height, then multiply by the 2-inch edge of the prism.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Stops one step early: A choice is a number from partway through the solution.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, one side of the triangle. h is the height, the straight distance from that side to the opposite corner, at a right angle to the side. You take half because a triangle is half of a rectangle with the same base and height.
Volume of a prism
V=B×HV = B \times H
V is the volume, the space inside. B is the area of the base, the shape at each end. H is the height of the prism, the distance between the two ends. It works because the prism is like a stack of layers, each the shape of the base.

  1. Sort out the figure

    A prism is a solid with two matching ends joined by flat sides. Here the ends are triangles, on the top and bottom. The other sides are rectangles. You need the volume, the space inside, in cubic inches.

    The dashed line on the top triangle is 8 in. It meets one side of the triangle at a right angle (the small square). The bottom edge under that side is 4 in. The edge going straight up is 2 in.

  2. Find the triangle’s base and height

    The side that the dashed line meets is the top edge of a rectangle face. Opposite sides of a rectangle are equal. So that side is 4 in., the same as the bottom edge under it.

    So the triangle has a base of 4 in. Its height is the dashed line, 8 in., because it meets the base at a right angle.

  3. Find the area of the triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×4×8=16A = \frac{1}{2} \times 4 \times 8 = 16

    The triangle’s area is 16 square inches. Watch out: 16 is only the area of one end. Stopping here gives the wrong answer 16.

  4. Multiply by the prism’s height

    The height of the prism is the distance between the two triangles. That is the 2 in. edge.

    V=B×HV = B \times H

    V=16×2=32V = 16 \times 2 = 32

    The volume is 32 cubic inches. Watch out: do not forget the one half. 4×8×2=644 \times 8 \times 2 = 64 is the volume of a box, not of this prism.

    So you would grid 32.

Check your answer

Picture a box that is 4 in. by 8 in. by 2 in. Its volume is 4×8×2=644 \times 8 \times 2 = 64 cubic inches. The triangle has half the area of a 4 by 8 rectangle, so the prism holds half as much as the box. Half of 64 is 32.

More questions like this

The original DOE question
Question 62 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 199, question 62. Open page 199 of the handbook (PDF)

Question 40 of 108

2024 Test B Math, question 6

Question

This graph shows y, the weight of a pole of length x.

Graph titled WEIGHT OF POLE. The x-axis, Length (feet), is marked 1, 2, 3 and 4, and the y-axis, Weight (ounces), is marked 4, 8, 12 and 16, with grid lines at each mark; the origin is labeled O. A straight line starts at O and rises through the grid corners (1, 4), (2, 8), (3, 12) and (4, 16), ending in an arrow, and the point (1, 4) is marked with a dot and labeled (1, 4).

What is represented by the point with coordinates (1, 4)?

  • AThe unit rate is 4 ounces per foot.
  • BThe y-intercept is 4.
  • CA pole 4 feet long weighs 1 ounce.
  • DThe length increases 4 feet for each 1 ounce of weight.
Show the answer

Correct answer: A

The x-axis represents the length of the pole, in feet, and the y-axis represents the weight of the pole, in ounces.

The point (1, 4) means that a 1-foot pole would weigh 4 ounces.

Since the graph is linear, the pole weighs 4 ounces for every additional 1 foot of length.

So the unit rate is 4 ounces per foot.

Question analysis

A graph of a pole's weight in ounces against its length in feet is a straight line through the origin, with the point (1, 4) marked. Students decide what that point says about poles, choosing among statements about rates, intercepts and particular poles, so the meaning of each coordinate has to be pinned down.

NYS standards
  • 7.RP.2d points on proportional graphs
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Explaining what a point on a proportional graph means in the situation is a named grade 7 skill. This is a direct, one-step example with a clear graph, typical of grade 7 classwork.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
1 step
Reading level
Question and choices: about grade 2, 45 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 needs computing; the question tests whether a point on a graph can be put into words. Each wrong statement misuses the same two numbers, as an intercept, with the coordinates swapped, or as a rate turned upside down. Saying the point aloud with its units, a 1-foot pole weighs 4 ounces, and comparing that sentence with each statement settles it quickly.

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

    The axes carry the meaning: length in feet runs along x and weight in ounces up y, so (1, 4) describes a 1-foot pole, and the line through O makes that a rate. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer minor Choosing the answer

    Both rate statements pair 4 with 1, one as ounces per foot and the other as feet per ounce; only the axis labels say which way the rate runs. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Something you must already know minor Working it out

    The choices use the terms unit rate and y-intercept, which the question does not define, and two statements hang on knowing what each one means. (Anderson & Krathwohl, 2001)

Strategy: Note which quantity each axis measures, say in words what the x-coordinate and the y-coordinate of the marked point stand for, then compare that sentence with each choice.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • 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

Unit rate
unit rate=yx\text{unit rate} = \frac{y}{x}
For a point on the line, y is the weight in ounces and x is the length in feet. The unit rate is the weight of 1 foot of pole. This works for a straight line that starts at the origin, the point (0, 0).

  1. Read the axes

    The graph has two axes (number lines). The x-axis, across the bottom, shows length in feet. The y-axis, going up, shows weight in ounces.

    A point is written (x, y). The across number comes first, then the up number. So in (1, 4), x is 1 and y is 4.

  2. Say what the point means

    x = 1 means a length of 1 foot. y = 4 means a weight of 4 ounces. So a pole 1 foot long weighs 4 ounces.

    Watch out: do not swap the numbers. “A pole 4 feet long weighs 1 ounce” reads them the wrong way round. That is choice C.

  3. See that this is a unit rate

    A unit rate is an amount for 1 unit, here the ounces in 1 foot. The line starts at O, the origin (0, 0), and it is straight. So each extra foot adds the same weight.

    unit rate=yx=41=4\text{unit rate} = \frac{y}{x} = \frac{4}{1} = 4

    The unit rate is 4 ounces per foot. Watch out: choice D turns this upside down. It gives feet for each ounce, not ounces for each foot.

  4. Rule out the y-intercept

    The y-intercept is where the line crosses the y-axis. This line crosses it at O, where y is 0. So the y-intercept is 0, not 4, and choice B is wrong.

    So the answer is A, the unit rate is 4 ounces per foot.

Check your answer

Try another point on the line, such as (3, 12). A pole 3 feet long weighs 12 ounces. 12÷3=412 \div 3 = 4, so it is still 4 ounces per foot. The rate stays the same all along the line.

More questions like this

The original DOE question
Question 63 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 63. Open page 200 of the handbook (PDF)

Question 41 of 108

2024 Test B Math, question 7

Question

The number line shows points E, F, G, and H.

Number line from −7 to 7 with a tick at every integer and arrows at both ends; the ticks are labeled −7, −5, −3, −1, 0, 1, 3, 5 and 7. Point E is on −7, point F is on the unlabeled tick between −3 and −1, point G is on the unlabeled tick between 1 and 3, and point H is on 7.

Which point represents the sum of 2.5+(−4.5)2.5+(-4.5)?

  • Apoint E
  • Bpoint F
  • Cpoint G
  • Dpoint H
Show the answer

Correct answer: B

Find the sum of 2.5+(−4.5)2.5+(-4.5):

2.5+(−4.5)=2.5−4.5=−22.5+(-4.5)=2.5-4.5=-2

Point F is located at −2.

Question analysis

A number line question with points E, F, G and H marked on a line labeled from −7 to 7. Students add 2.5 and negative 4.5 and find which point shows the sum. It tests adding a positive and a negative decimal, then reading positions on a line whose labels skip some integers.

NYS standards
  • 7.NS.1b rational number addition
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding rational numbers with different signs and showing the sum on a number line is exactly the grade 7 standard. This is one routine sum with simple decimals, a typical classroom exercise.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 1, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-line sum followed by a lookup, and most students finish quickly. The four points are placed so that each common sign mistake lands on one of them: −7, 2 and 7 all come from mishandling the negative. Picturing the sum as a move of 4.5 units to the left from 2.5 shows both the size and the direction at once.

  1. Error-prone arithmetic minor Working it out

    Adding −4.5 to 2.5 means taking 4.5 away from 2.5, and the larger size of −4.5 makes the sum negative, a sign rule that is easy to apply backwards. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    Only 0 and the odd numbers are labeled, so point F sits at an unlabeled tick between −3 and −1 and must be found by counting. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Rewrite adding −4.5 as subtracting 4.5, work out the sum, then find it on the line by counting tick marks from a labeled number, one tick per integer.

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.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding a negative number
a+(−b)=a−ba + (-b) = a - b
a is any number, and −b is a negative number. Adding a negative number is the same as subtracting its size. On a number line, it means moving b steps to the left.

  1. Sort out the question

    You need to find 2.5+(−4.5)2.5 + (-4.5). Then find that sum (the answer to the addition) on the number line.

    The number line has a tick at every whole number from −7 to 7. Only 0 and the odd numbers have labels.

  2. Turn the addition into subtraction

    a+(−b)=a−ba + (-b) = a - b

    2.5+(−4.5)=2.5−4.52.5 + (-4.5) = 2.5 - 4.5

    You are taking 4.5 away from 2.5. You take away more than you have, so the answer will be negative.

  3. Do the subtraction

    The two sizes differ by 2, because 4.5−2.5=24.5 - 2.5 = 2. The larger size, 4.5, belongs to the negative number. So the sum is negative.

    2.5−4.5=−22.5 - 4.5 = -2

    Watch out: the answer is −2, not 2. Point G is at 2.

  4. Find −2 on the line

    −2 has no label. It is the tick between −3 and −1. Point F sits on that tick.

    Watch out: adding 2.5 + 4.5 = 7 lands on point H. Making both numbers negative gives −7, point E.

    So the answer is B, point F.

Check your answer

Start at 2.5 on the number line and move 4.5 steps to the left. The first 2.5 steps take you to 0. You still have 2 steps to go. Two more steps left of 0 lands on −2, which is point F.

More questions like this

The original DOE question
Question 64 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 64. Open page 200 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 42 of 108

2024 Test B Math, question 18

Question

A taxi driver recorded the number of fares each day for one month. The box plot displays the data.

Box plot titled DAILY FARES above a number line labeled Number of Fares, with arrows at both ends and a tick labeled at every even number from 0 to 32. The left whisker starts at 2, the box runs from 12 to 24 with the median line at 16, and the right whisker ends at 32.

Based on the box plot, which statement is a correct interpretation of the data?

  • AThe data point 2 is an outlier.
  • BThe mean of the data is 16.
  • CThe interquartile range (IQR) of the data is 30.
  • DOn about 25% of the days, there were between 12 and 16 fares.
Show the answer

Correct answer: D

The amount of data in a box plot is broken into 4 parts, or quartiles, which each represent 25% of the data. So the amount of data represented between Quartile 1 (12) and the median (16) is about 25% of the data.

Question analysis

A statistics question on a box plot of the number of fares a taxi driver recorded each day for one month. Students decide which of four statements is a correct interpretation of the data, using what a box plot does and does not show. The statements draw on several different box plot ideas, so each one must be checked.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Reading box plots, finding the interquartile range and testing for outliers are all grade 7 standards in New York. This item gathers several of them in one question, but each check is short, typical of grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 63 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

No calculation here is long; the challenge is knowing what a box plot does and does not show. Two of the wrong statements sound reasonable because they borrow real features of the plot, its long left whisker and its middle line, and give them the wrong meaning. Remembering that the plot splits the days into four groups of about equal size makes the true statement easy to confirm.

  1. Something you must already know moderate Working it out

    Each statement draws on a box plot fact the test does not supply: what the middle line marks, how the IQR is found, and when a value counts as an outlier. (Anderson & Krathwohl, 2001)

  2. Every choice must be checked moderate Choosing the answer

    The four statements test four separate ideas, an outlier, the mean, the IQR and a quartile's share, so ruling out one says nothing about the next. (Embretson & Wetzel, 1987)

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

    The scale is labeled every 2 fares with the numbers crowded together, so the five values 2, 12, 16, 24 and 32 must be read off with care. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the minimum, quartiles, median and maximum off the plot, then test each statement in turn, computing the IQR and the outlier cutoffs rather than judging by eye.

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.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range (IQR)
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q1 is the first quartile, the left edge of the box. Q3 is the third quartile, the right edge. The IQR is the length of the box. It tells how spread out the middle half of the data is.
Low cutoff for an outlier
low cutoff=Q1−1.5×IQR\text{low cutoff} = Q_1 - 1.5 \times \text{IQR}
Q1 and the IQR are found from the box. An outlier is a value far away from the rest. A value below this cutoff counts as an outlier on the low side. A value at or above it does not.

  1. Read the five numbers

    A box plot sums up a set of data with five numbers. Here each piece of data is the number of fares on one day.

    • Minimum, the smallest value: 2
    • First quartile, Q1, the left edge of the box: 12
    • Median, the middle value, the line inside the box: 16
    • Third quartile, Q3, the right edge of the box: 24
    • Maximum, the largest value: 32

  2. Learn what each part holds

    These five numbers split the days into four parts. Each part holds about one fourth of the days, or about 25%. For example, about 25% of the days fall between the minimum and Q1. Another 25% fall between Q1 and the median.

  3. Check choice D

    Choice D talks about days with between 12 and 16 fares. 12 is Q1, and 16 is the median. That stretch is one of the four parts. So on about 25% of the days, there were between 12 and 16 fares. Choice D is correct.

  4. Rule out choices B and C

    Choice B: a box plot shows the median, 16, not the mean (the average). You cannot find the mean from it. Watch out: do not mix up the median and the mean.

    IQR=Q3−Q1=24−12=12\text{IQR} = Q_3 - Q_1 = 24 - 12 = 12

    Choice C says the IQR is 30. But 30 is the range, 32−232 - 2. Watch out: the IQR uses the edges of the box, not the ends of the whiskers.

  5. Rule out choice A and answer

    A long whisker does not make 2 an outlier. Use the cutoff rule.

    low cutoff=Q1−1.5×IQR\text{low cutoff} = Q_1 - 1.5 \times \text{IQR}

    low cutoff=12−1.5×12=12−18=−6\text{low cutoff} = 12 - 1.5 \times 12 = 12 - 18 = -6

    2 is above −6, so 2 is not an outlier. The long whisker just shows that the lowest fourth of the days is spread out.

    So the answer is D.

Check your answer

Go back over the four choices. Only D uses what a box plot really shows: four parts of about 25% each. Choice B needs the mean, which the plot does not show. Choice C gives the range, not the IQR. And 2 is not far enough out to be an outlier, so A is wrong.

More questions like this

The original DOE question
Question 75 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 75. Open page 203 of the handbook (PDF)

Question 43 of 108

2024 Test B Math, question 19

Question

Serena was studying the number of water stations for different lengths of races. The table below shows the data she collected.

WATER STATIONS IN RACES

Length of Race (mi) Number of Water Stations
3 2
12 8
18 12

Which equation represents the relationship between the length of the race, x miles, and the number of water stations, y?

  • Ay=23xy=\frac{2}{3}x
  • By=32xy=\frac{3}{2}x
  • Cy=x+6y=x+6
  • Dy=x−1y=x-1
Show the answer

Correct answer: A

There is a proportional relationship between the length of the race, x miles, and the number of water stations, y. The constant of proportionality is

yx=23\frac{y}{x}=\frac{2}{3}

y=23xy=\frac{2}{3}x

Therefore, the equation y=23xy=\frac{2}{3}x represents this relationship.

Question analysis

A table from a study of races shows the number of water stations for races of 3, 12 and 18 miles. Students choose the equation that links the race length x to the number of stations y. It tests reading a relationship from a table and expressing it as an equation in x and y.

NYS standards
  • 7.RP.2c proportional relationship equations
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an equation for a relationship given in a table is a central grade 7 skill. With three clean rows and whole numbers, this is a typical grade 7 exercise.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 57 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The table is small and its numbers are friendly, so the question comes down to which way the ratio runs and whether every row is checked. One wrong equation fits only the first row and another fits none, so testing a single row is not enough. Putting the 12-mile race into each equation, which should give 8 stations, leaves only one rule standing.

  1. A very close wrong answer moderate Choosing the answer

    The coefficients 2/3 and 3/2 use the same two digits, and only dividing stations by miles, 2 ÷ 3, rather than miles by stations, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The table lists miles first and stations second, while each equation puts stations, y, alone on the left, so every row must be read as an (x, y) pair. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Test each equation on all three rows of the table, not just the first; only a rule that fits every row represents the relationship.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Checks only one number: Tries one value in each choice and takes the first that works, when that value happens to work for a wrong choice too; a second value would rule it out.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship
y=kxy = kx
x is the race length in miles, and y is the number of water stations. k is the constant of proportionality, the number of stations for each mile. Find it with y ÷ x. If every row gives the same k, the table is proportional.

  1. Sort out the table

    Each row of the table is a pair of numbers. The first is the race length, x, in miles. The second is the number of water stations, y. You need the equation that works for every row.

  2. Divide y by x in each row

    See if every row has the same ratio of stations to miles. Divide y by x.

    Miles (x)Stations (y)y ÷ x
    3223\frac{2}{3}
    128812=23\frac{8}{12} = \frac{2}{3}
    18121218=23\frac{12}{18} = \frac{2}{3}

    Every row gives 2/3. So the relationship is proportional: y is always the same number times x.

  3. Write the equation

    y=kxy = kx

    Here k is the ratio you found, 23\frac{2}{3}.

    y=23xy = \frac{2}{3}x

    Watch out: divide stations by miles, not miles by stations. 3 ÷ 2 gives 3/2, choice B. That rule would give more stations than miles.

  4. Test the other choices and answer

    Watch out: test every row, not just the first one. Choice D, y = x − 1, works for 3 miles. But it gives 11 stations for 12 miles, not 8. Choice C, y = x + 6, gives 9 stations for 3 miles, not 2.

    So the answer is A, y=23xy = \frac{2}{3}x.

Check your answer

Put each race length into y=23xy = \frac{2}{3}x. 23×3=2\frac{2}{3} \times 3 = 2, 23×12=8\frac{2}{3} \times 12 = 8 and 23×18=12\frac{2}{3} \times 18 = 12. All three match the table.

More questions like this

The original DOE question
Question 76 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 204, question 76. Open page 204 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 44 of 108

2024 Test B Math, question 20

Question

Centerville will hold a vote on whether to build a new park. The town council surveyed a random sample of 200 residents to determine whether they would be in favor of building this park. The table shows the results of the survey.

SURVEY RESULTS

Opinion Number of Residents
Yes 142
No 38
Undecided 20

The mayor estimates that half the undecided voters will vote in favor of building the park. Based on the survey results, if the mayor is correct, how many of the 24,100 voters will be in favor of building the park?

  • A17,111
  • B18,316
  • C19,401
  • D19,521
Show the answer

Correct answer: B

If half the undecided voters (half of 20) vote in favor of building the park, the total number of “yes” votes from the survey will be 152.

142+10=152142+10=152

There are 24,100 voters, and 200 residents were surveyed. Set up a proportion to solve:

152200=x24,100\frac{152}{200}=\frac{x}{24{,}100}

1925=x24,100\frac{19}{25}=\frac{x}{24{,}100}

(24,100)(1925)=x(24{,}100)\left(\frac{19}{25}\right)=x

18,316=x18{,}316=x

Question analysis

A survey question: a random sample of 200 Centerville residents answered yes, no or undecided about building a park, and the mayor expects half the undecided to vote yes. Students use the sample to predict how many of 24,100 voters will favor the park. It joins a proportion from a sample with large-number arithmetic.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.SP.1b samples and populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using a random sample to estimate a count in a whole population is a typical grade 7 proportion task. The extra step for the undecided group and the large numbers add care without going beyond grade 7.
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, 88 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup is a sample-to-population proportion, and the trap is the mayor's assumption in the middle of the stem. Students who scale the 142 yes answers directly, or who count every undecided resident, reach believable totals that sit right among the choices. Once the count of 152 is settled, a rough estimate, about three quarters of 24,000, is enough to confirm the exact product.

  1. Error-prone arithmetic moderate Working it out

    Scaling 152 out of 200 up to 24,100 voters means computing 24,100 × 0.76, or 120.5 × 152, by hand with five-digit numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    The mayor's sentence adds a step: “half the undecided voters” turns the 20 undecided residents into 10 more yes votes, giving 152 before any scaling. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table gives Yes, No and Undecided counts, and the half must be taken of the Undecided row, not of the No row just above it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Count the sample's expected yes votes, including half of the undecided, write them as a fraction of 200, and multiply that fraction by 24,100; estimate first to check the choice.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling up a sample
yes in samplesample size=yes in townvoters in town\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}
A random sample is a smaller group picked by chance. Its share of yes votes should be about the same as the whole town’s share. So the two fractions are set equal. Use it to scale a survey up to a whole group.

  1. Sort out what you know

    The town surveyed 200 people. 142 said Yes, 38 said No, and 20 were Undecided (not sure yet). The mayor thinks half of the undecided people will vote yes. There are 24,100 voters in all. You need how many of them will vote yes.

  2. Count the yes votes in the survey

    Half of the 20 undecided people is 20÷2=1020 \div 2 = 10. Add them to the yes votes.

    142+10=152142 + 10 = 152

    So 152 of the 200 people surveyed would vote yes. Watch out: take half of the Undecided row, not the No row. Half of 38 leads to 19,401, choice C.

  3. Find the share who vote yes

    152200=0.76\frac{152}{200} = 0.76

    That is 76% of the people surveyed. The sample was random, so about 76% of all the voters should vote yes too. Watch out: using only the 142 yes votes gives 17,111, choice A. Counting all 20 undecided people gives 19,521, choice D.

  4. Scale up to all the voters

    yes in samplesample size=yes in townvoters in town\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}

    152200=x24,100\frac{152}{200} = \frac{x}{24{,}100}

    x=0.76×24,100x = 0.76 \times 24{,}100

    Break it into easy parts. 75% is 34×24,100=18,075\frac{3}{4} \times 24{,}100 = 18{,}075, and 1% is 24,100÷100=24124{,}100 \div 100 = 241.

    x=18,075+241=18,316x = 18{,}075 + 241 = 18{,}316

    So the answer is B, 18,316.

Check your answer

Estimate. 76% is a little more than three fourths. Three fourths of 24,000 is 18,000. So the answer should be a little more than 18,000, and 18,316 fits. Also, 18,316÷24,100=0.7618{,}316 \div 24{,}100 = 0.76, the same share as the survey.

More questions like this

The original DOE question
Question 77 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 204, question 77. Open page 204 of the handbook (PDF)

Question 45 of 108

2024 Test B Math, question 25

Question

Which number line represents the values of x that satisfy the inequality −5≤1−3x≤4-5\leq 1-3x\leq 4?

  • ANumber line from −3 to 3 with a tick at every integer. Open circles at −1 and 2, with the segment between them shaded.
  • BNumber line from −3 to 3 with a tick at every integer. Closed circles at −1 and 2, with the segment between them shaded.
  • CNumber line from −3 to 3 with a tick at every integer. Open circles at −1 and 2, with the line shaded from −1 to the left and from 2 to the right.
  • DNumber line from −3 to 3 with a tick at every integer. A closed circle at −1 and an open circle at 2, with the line shaded from −1 to the left and from 2 to the right.
Show the answer

Correct answer: B

Split the compound inequality into two inequalities.

−5≤1−3x-5\leq 1-3x and 1−3x≤41-3x\leq 4

For each inequality, subtract 1 from each side.

−6≤−3x-6\leq -3x and −3x≤3-3x\leq 3

For each inequality, divide both sides by −3. Note that dividing each side of an inequality by a negative number changes the direction of the inequality.

2≥x2\geq x and x≥−1x\geq -1

Combining the pieces gives the compound inequality −1≤x≤2-1\leq x\leq 2.

The solution is that x is greater than or equal to −1 and less than or equal to 2, so choose the number line with filled circles at −1 and 2, with the filled segment between the circles.

Question analysis

An inequality question: choose the number line that shows the solutions of the compound inequality −5 ≤ 1 − 3x ≤ 4. Students solve an inequality whose x term has a negative coefficient, then match the endpoints, open or closed circles, and shading to one of four graphs.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph two-step inequalities one at a time. This compound inequality needs two solved together and a division by a negative number, which reverses the signs, a step many grade 7 classes only touch on.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 12, 13 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question stacks two classic traps: reversing the inequality signs when dividing by −3, and matching filled or open circles to ≤. All four graphs use the endpoints −1 and 2, so finding those numbers proves little; the choice turns on the circles and the direction of the shading. Testing x = 0, x = 3 and the endpoint x = 2 in the original inequality checks both decisions in under a minute.

  1. Error-prone arithmetic moderate Working it out

    Dividing all three parts of −6 ≤ −3x ≤ 3 by −3 reverses both signs and the order, giving 2 ≥ x ≥ −1, which must be rewritten as −1 ≤ x ≤ 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    Two graphs share the endpoints −1 and 2 and the shading between them; only the circles, filled for ≤ and open for <, separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Each graph must be read for three things at once: the endpoints, the fill of each circle, and whether the shading runs between them or outward. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Subtract 1 from all three parts, divide all three by −3 while reversing both inequality signs, then rewrite the result from smallest to largest before matching circles and shading.

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.)
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting from every part
if a≤b, then a−c≤b−c\text{if } a \leq b \text{, then } a - c \leq b - c
a and b are the two sides of an inequality, and c is any number. Subtracting the same number from every part keeps the inequality true. The sign does not change.
Dividing by a negative number
if a≤b, then a÷(−c)≥b÷(−c)\text{if } a \leq b \text{, then } a \div (-c) \geq b \div (-c)
a and b are numbers, and c is a positive number, so −c is negative. When you divide both sides by a negative number, the inequality sign flips. For example, 2 is less than 3, but −2 is greater than −3.

  1. Sort out the question

    The inequality −5≤1−3x≤4-5 \leq 1 - 3x \leq 4 says that 1−3x1 - 3x is between −5 and 4. The sign ≤ means less than or equal to. So −5 and 4 themselves are allowed. You need the number line that shows every x that works. To find them, get x alone in the middle.

  2. Subtract 1 from every part

    First take away the 1 in the middle. Subtract 1 from all three parts.

    if a≤b, then a−c≤b−c\text{if } a \leq b \text{, then } a - c \leq b - c

    −5−1≤1−3x−1≤4−1-5 - 1 \leq 1 - 3x - 1 \leq 4 - 1

    −6≤−3x≤3-6 \leq -3x \leq 3

  3. Divide by −3 and flip

    Now x is multiplied by −3. Divide all three parts by −3. Dividing by a negative number flips each sign.

    if a≤b, then a÷(−c)≥b÷(−c)\text{if } a \leq b \text{, then } a \div (-c) \geq b \div (-c)

    −6−3≥−3x−3≥3−3\frac{-6}{-3} \geq \frac{-3x}{-3} \geq \frac{3}{-3}

    2≥x≥−12 \geq x \geq -1

    Watch out: if you forget to flip, x seems to lie outside −1 and 2. That leads to choices C and D, which are shaded outward.

  4. Read the result

    Turn it around so the smaller number is on the left. It means the same thing.

    −1≤x≤2-1 \leq x \leq 2

    So x can be any number from −1 to 2. The ends, −1 and 2, are included, because the signs are ≤.

  5. Pick the number line

    On a number line, a closed (filled-in) circle means the end number is included. An open circle means it is not. So you need closed circles at −1 and 2, with the part between them shaded. Watch out: choice A has the right shading but open circles, which leave out −1 and 2.

    So the answer is B.

Check your answer

Test a few numbers in 1−3x1 - 3x. For x = 0 you get 1, which is between −5 and 4, and 0 is shaded. For x = 2 you get −5, which is allowed, so the circle at 2 is filled in. For x = 3 you get −8, which is too small, so 3 is not shaded.

More questions like this

The original DOE question
Question 82 as printed on page 206 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 82. Open page 206 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 46 of 108

2024 Test B Math, question 28

Question

A community center has two 30-member exercise groups. The director recorded the ages of each group’s members in the box plots.

Two box plots titled AGES OF GROUPS over one number line from 15 to 33, with a tick at every whole number and labels at 15, 18, 21, 24, 27, 30 and 33. Group P: whiskers from 17 to 33, box from 19 to 27, median line at 24. Group Q: whiskers from 15 to 29, box from 17 to 26, median line at 21.

Which statement about the two groups can be verified based on the box plots?

  • ABoth groups have at least one member who is exactly 29 years old.
  • BGroup P has more members who are at least 17 years old than Group Q has.
  • CBoth groups have an equal number of members who are between 21 and 24 years old.
  • DGroup Q has fewer members who are between 19 and 27 years old than Group P has.
Show the answer

Correct answer: B

Groups P and Q have an equal number of members. All the members of Group P are 17 or older, but only about 34\frac{3}{4} of the members of Group Q are 17 or older.

Question analysis

Box plots show the ages of the members of two 30-member exercise groups at a community center, Group P and Group Q. Students decide which statement comparing the groups can be verified from the plots alone. The challenge is knowing what a box plot guarantees about individual ages and counts, and what it leaves unknown.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 7.SP.4 comparative inferences about populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students read box plots and compare medians and spreads. Deciding which claims two box plots can prove, from what the five key values guarantee about counts, is a harder reasoning task 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 5, 91 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a question about evidence: the right statement is the only one the box plots guarantee, while the others may or may not be true. Students who treat the plots as pictures of every member, assuming ages fill each whisker or spread evenly through each box, find support for claims that cannot be checked. Asking what must hold for any 30 ages with these summaries separates the verifiable claim from the merely plausible ones.

  1. Every choice must be checked major Choosing the answer

    Three of the statements are not false but unverifiable, so each must be tested against what the five-number summaries guarantee, and none can be dismissed at a glance. (Embretson & Wetzel, 1987)

  2. Something you must already know moderate Working it out

    Students must know that the minimum and maximum are actual ages, that each quarter of a plot holds about a quarter of the 30 members, and that a whisker shows nothing in between. (Anderson & Krathwohl, 2001)

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

    The scale is labeled only every 3 years, so values such as 17, 19, 26 and 29 have to be read from unlabeled ticks on two separate plots. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List each group's minimum, first quartile, median, third quartile and maximum, then test every statement against only what those five values guarantee about the 30 ages.

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.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Members in each part of a box plot
members in each part≈14×n\text{members in each part} \approx \frac{1}{4} \times n
n is the number of members in a group, here 30. A box plot splits the data into four parts at the minimum, Q1, the median, Q3 and the maximum. Each part holds about one fourth of the data. It does not show how the ages spread out inside a part.

  1. Read the two box plots

    A box plot shows five numbers. The minimum is the youngest age, and the maximum is the oldest. Q1 is the left edge of the box, and Q3 is the right edge. The median is the line inside the box.

    GroupMinQ1MedianQ3Max
    P1719242733
    Q1517212629

    Each group has 30 members.

  2. Know what can be verified

    To verify a statement means to know for sure that it is true. A box plot shows only five numbers, not every age. The minimum and maximum are real ages of real members. The other ages are hidden inside the box and whiskers.

  3. Check choice B

    In Group P, the youngest member is 17. So all 30 members are at least 17. In Group Q, the youngest member is 15, and 15 is under 17. So at most 29 members of Group Q are at least 17.

    Group P has 30 such members, and Group Q has 29 or fewer. So choice B is true for sure.

  4. Rule out choices A and C

    A: Group Q’s oldest member is 29, so Group Q has a 29-year-old. But in Group P, 29 is inside the right whisker. A whisker does not show single ages. Watch out: 29 is in Group P’s range, but maybe no one in Group P is exactly 29.

    C: The plots do not show how many members are between 21 and 24 in either group.

  5. Rule out choice D and answer

    Group P’s box, from 19 to 27, is two of its four parts.

    members in each part≈14×30=7.5\text{members in each part} \approx \frac{1}{4} \times 30 = 7.5

    So about 15 members of Group P are between 19 and 27. But 19 and 27 fall in the middle of Group Q’s parts, and ages inside a part can bunch up in any way. Group Q could have more or fewer. Watch out: D assumes the ages are spread out evenly, and a box plot never tells you that.

    So the answer is B.

Check your answer

Choice B uses only the two minimums, which are real ages. Group P’s youngest member is 17, and Group Q’s is 15. That alone proves it. Each other choice needs ages inside a box or whisker, which a box plot does not show.

More questions like this

The original DOE question
Question 85 as printed on page 207 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 85. Open page 207 of the handbook (PDF)

Question 47 of 108

2024 Test B Math, question 30

Question

GIA’S MIX

Snack Number in Bag
Pretzels 12
Raisins 9

Gia and her friends counted the numbers of pretzels and raisins in their snack mix. The table above shows the numbers in Gia’s mix. Which other mix has a ratio of pretzels to raisins in the same proportional relationship as Gia’s mix?

  • A

    ADELE’S MIX

    SnackNumber in Bag
    Pretzels6
    Raisins18
  • B

    CARL’S MIX

    SnackNumber in Bag
    Pretzels15
    Raisins12
  • C

    ISABEL’S MIX

    SnackNumber in Bag
    Pretzels18
    Raisins15
  • D

    TREVON’S MIX

    SnackNumber in Bag
    Pretzels16
    Raisins12
Show the answer

Correct answer: D

The ratio of pretzels to raisins in Gia’s bag of snack mix, 12:912:9, is equivalent to the ratio of pretzels to raisins in Trevon’s bag, 16:1216:12.

129=1612=43\frac{12}{9}=\frac{16}{12}=\frac{4}{3}

Question analysis

Gia's snack mix table lists how many pretzels and raisins are in her bag, and four friends' tables list theirs. Students pick the mix whose ratio of pretzels to raisins matches Gia's. It tests equivalent ratios and proportional relationships, and some mixes look close at a glance, so each ratio has to be tested exactly.

NYS standards
  • 7.RP.2a testing for proportionality
  • 6.RP.3a tables of equivalent ratios
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Testing whether two ratios are equivalent is core grade 7 proportional reasoning. The numbers are small whole numbers in four short tables, so the work is routine, though every table has to be checked against Gia's.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 77 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A short ratio check made harder only by an additive pull: Carl's and Isabel's mixes keep the gap of 3 between pretzels and raisins that Gia's mix has, so students who compare differences instead of ratios see two matches. Reducing every table to lowest terms, Gia's 12:9 to 4:3, shows that exactly one mix keeps the ratio.

  1. Every choice must be checked minor Choosing the answer

    Each of the four tables has to be reduced in turn, to 1:3, 5:4, 6:5 and 4:3, because none of them can be ruled out at a glance. (Embretson & Wetzel, 1987)

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

    The counts sit in five separate tables, and pretzels must stay first in every ratio, in the same order as Gia's 12 pretzels to 9 raisins. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Simplify the pretzels-to-raisins ratio in Gia's table to lowest terms, then simplify each other mix's ratio in the same order and compare.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent ratios
a:b=(a÷k):(b÷k)a : b = (a \div k) : (b \div k)
a and b are the two amounts being compared, in order. k is any number that divides both evenly. Dividing both amounts by the same number keeps the ratio the same. When no number but 1 divides both, the ratio is in simplest form.

  1. Sort out the question

    A ratio compares two amounts. Gia’s mix has 12 pretzels and 9 raisins. So her ratio of pretzels to raisins is 12 : 9.

    The “same proportional relationship” means the ratios are equal once you simplify them. For example, 2 : 1 and 6 : 3 are equal. Always keep pretzels first and raisins second.

  2. Simplify Gia’s ratio

    Both 12 and 9 can be divided by 3. So divide both by 3.

    12:9=(12÷3):(9÷3)=4:312 : 9 = (12 \div 3) : (9 \div 3) = 4 : 3

    Gia’s mix has 4 pretzels for every 3 raisins.

  3. Simplify each other mix

    Do the same for each choice. Divide both numbers by the biggest number that goes into both.

    ChoicePretzels : raisinsDivide bySimplest form
    A, Adele6 : 1861 : 3
    B, Carl15 : 1235 : 4
    C, Isabel18 : 1536 : 5
    D, Trevon16 : 1244 : 3

  4. Pick the matching mix

    Only Trevon’s mix simplifies to 4 : 3, the same as Gia’s.

    Watch out: Carl’s and Isabel’s mixes each have 3 more pretzels than raisins, just like Gia’s. But adding the same number to both amounts changes the ratio. Equal ratios come from multiplying or dividing. That trap leads to choice B or C.

    So the answer is D, Trevon’s mix.

Check your answer

Go the other way. Multiply both parts of 4 : 3 by 4. You get 16 : 12, which are Trevon’s exact numbers. You can also divide pretzels by raisins. Both 12÷912 \div 9 and 16÷1216 \div 12 equal 43\frac{4}{3}.

More questions like this

The original DOE question
Question 87 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 87. Open page 208 of the handbook (PDF)

Question 48 of 108

2024 Test B Math, question 32

Question

Two circular plates have the dimensions shown.

Two circles side by side. The larger circle has a dashed diameter labeled 18 in., and the smaller circle has a dashed diameter labeled 12 in.

What is the difference between the areas of the two plates, in square inches?

  • A6π6\pi
  • B9π9\pi
  • C45π45\pi
  • D180π180\pi
Show the answer

Correct answer: C

First, determine the area of each circle using the formula A=πr2A=\pi r^{2}:

A circle with a diameter of 18 inches has a radius of 9 inches, so the area is π(9)2=81π\pi(9)^{2}=81\pi.

A circle with a diameter of 12 inches has a radius of 6 inches, so the area is π(6)2=36π\pi(6)^{2}=36\pi.

The difference between the two areas is 81π−36π=45π81\pi-36\pi=45\pi.

Question analysis

Two circular plates are drawn with their diameters marked in inches, and students find the difference between their areas in square inches, in terms of pi. It tests the area of a circle. The diameters, not the radii, are marked, so a step comes before the area formula, which the test does not provide.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a circle's area is a grade 7 standard, and this asks for two areas and their difference. It matches typical classwork: the formula must be recalled and each diameter halved first, with no further twist.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two area calculations and a subtraction, kept in terms of π, make this routine for students who know the area formula. Points are lost by using the labeled diameters as radii, by subtracting circumferences, or by subtracting the radii before squaring, and each of these has a listed answer. Writing both areas, 81π and 36π, before subtracting avoids all three.

  1. Something you must already know moderate Working it out

    The area formula A = πr² is not given, and the figure labels each plate by its diameter, so 18 and 12 must be halved to radii of 9 and 6 first. (Anderson & Krathwohl, 2001)

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

    The dashed lines run all the way across each circle, which marks them as diameters; reading the 18 in. and 12 in. as radii leads to a listed choice. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Halve each diameter to get the radius, find each plate's area as pi times the radius squared, and subtract the smaller area from the larger.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle
A=πr2A = \pi r^2
A is the area, the flat space inside the circle. r is the radius, the distance from the center to the edge. It is half the diameter. π (pi) is a fixed number, about 3.14. r² means r × r. Use it for the space inside any circle.

  1. Sort out the question

    Each dashed line goes all the way across its circle, through the center. A line like that is a diameter. The big plate has a diameter of 18 inches. The small plate has a diameter of 12 inches.

    You need the difference between the two areas. That means the big area minus the small area, in square inches.

  2. Find each radius

    The area formula uses the radius, not the diameter. The radius is half the diameter.

    18÷2=918 \div 2 = 9

    12÷2=612 \div 2 = 6

    The radii are 9 inches and 6 inches. Watch out: If you use 18 and 12 as the radii, you get 180π180\pi, choice D.

  3. Find the big plate’s area

    A=πr2=π×92=π×81=81πA = \pi r^2 = \pi \times 9^2 = \pi \times 81 = 81\pi

    The big plate’s area is 81π81\pi square inches. Keep π as a symbol, since all the choices use π.

  4. Find the small plate’s area

    A=π×62=π×36=36πA = \pi \times 6^2 = \pi \times 36 = 36\pi

    The small plate’s area is 36π36\pi square inches. Squaring means multiplying a number by itself, so 62=6×6=366^2 = 6 \times 6 = 36.

  5. Subtract the two areas

    81π−36π=45π81\pi - 36\pi = 45\pi

    Subtract the numbers in front of π, just as 81 apples minus 36 apples is 45 apples. The difference is 45π45\pi square inches.

    Watch out: Don’t subtract the radii first. Squaring 9 − 6 = 3 gives 9π9\pi, choice B. Subtracting the distances around the plates, 18π−12π18\pi - 12\pi, gives 6π6\pi, choice A. That uses the wrong formula.

    So the answer is C, 45π45\pi.

Check your answer

Estimate with π about 3. The big plate is about 3×81=2433 \times 81 = 243 square inches. The small one is about 3×36=1083 \times 36 = 108. The difference is about 135, and 45×3=13545 \times 3 = 135. It matches.

More questions like this

The original DOE question
Question 89 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 89. Open page 208 of the handbook (PDF)

Question 49 of 108

2024 Test B Math, question 39

Question

A closed rectangular box drawn in three dimensions. Its height is labeled 20 cm, the bottom edge running back on the left is labeled 40 cm, and the bottom front edge is labeled 50 cm.

What is the total surface area, in square centimeters, of the closed box shown above?

  • A3,800
  • B5,600
  • C7,600
  • D40,000
Show the answer

Correct answer: C

To find surface area, use the formula A=2(hw+hl+lw)A=2(hw+hl+lw):

A=2[(20)(40)+(20)(50)+(50)(40)]A=2[(20)(40)+(20)(50)+(50)(40)]

A=2(800+1,000+2,000)A=2(800+1{,}000+2{,}000)

A=2(3,800)=7,600A=2(3{,}800)=7{,}600

Question analysis

A drawing of a closed box gives its length, width and height in centimeters, and students find its total surface area in square centimeters. It tests surface area of a rectangular prism. The work is short, but every face of the box has to be accounted for, and surface area must be kept apart from volume.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.4 nets and surface area
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Surface area of a rectangular prism is a grade 7 standard, and this box has whole-number dimensions shown in a drawing. Three face areas, a sum and the opposite faces make it typical practice rather than a stretch.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 8, 15 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

Finding the three face areas, 800, 1,000 and 2,000, is quick; the question tests whether students account for all six faces. The wrong choices are what the box would give with only three faces, with the top left off, or with the dimensions multiplied into a volume. Listing the faces in pairs before adding is the safest route.

  1. Something you must already know minor Working it out

    No surface-area formula is given; students must know that a box's six faces come in three matching pairs, 20 × 40, 20 × 50 and 40 × 50. (Anderson & Krathwohl, 2001)

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

    The drawing shows only three faces and labels 20, 40 and 50 cm on separate edges, so the hidden faces and their dimensions have to be pictured. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the areas of the three different faces from the labeled length, width and height, add them, and double the sum to cover the opposite faces.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Surface area of a box
SA=2(lw+lh+wh)SA = 2(lw + lh + wh)
SA is the surface area, the total area of all six faces. l is the length, w is the width and h is the height. lw, lh and wh are the areas of the three different faces. You double because each face has a matching face on the opposite side.

  1. Sort out the question

    The box is a rectangular prism, a box shape with six rectangle faces. Its length is 50 cm, its width is 40 cm and its height is 20 cm.

    The box is closed, so it has a top. Surface area is the total area of all the outside faces, in square centimeters.

  2. Find the three different faces

    The faces come in matching pairs. Find the area of one face from each pair. Multiply its two edges.

    FacesEdgesArea of one (sq cm)
    Front and back50 × 201,000
    Left and right40 × 20800
    Top and bottom50 × 402,000

  3. Add the faces and double

    Use the formula. Add the three areas. Then double the sum to cover the matching faces you can’t see.

    SA=2(50×40+50×20+40×20)SA = 2(50 \times 40 + 50 \times 20 + 40 \times 20)

    SA=2(2,000+1,000+800)=2×3,800=7,600SA = 2(2{,}000 + 1{,}000 + 800) = 2 \times 3{,}800 = 7{,}600

    The surface area is 7,600 square centimeters.

  4. Avoid the traps

    Watch out: Stopping at 3,800 counts only three faces, choice A. Leaving off the top, as for an open box, gives 5,600, choice B. Multiplying all three edges gives 40,000, choice D. That is the volume, in cubic centimeters.

    So the answer is C, 7,600.

Check your answer

Add all six faces one by one: 1,000+1,000+800+800+2,000+2,000=7,6001{,}000 + 1{,}000 + 800 + 800 + 2{,}000 + 2{,}000 = 7{,}600. Also check the units. Area multiplies two lengths, so it is in square centimeters. Multiplying three lengths gives cubic units, which is volume.

More questions like this

The original DOE question
Question 96 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 96. Open page 210 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 50 of 108

2024 Test B Math, question 48

Question

Two triangles. Triangle MNP has N at the top and base MP labeled 10 cm; sides MN and NP each have a tick mark, showing they are equal. The smaller triangle XYZ has Y at the top, base XZ labeled 6 cm, and sides XY and YZ each labeled 9 cm.

Triangle MNP is similar to triangle XYZ, and MN=NP\mathrm{MN}=\mathrm{NP}. What is the perimeter of triangle MNP?

  • A25 cm
  • B30 cm
  • C36 cm
  • D40 cm
Show the answer

Correct answer: D

Similar geometric figures have sides that are proportional.

106=x9\frac{10}{6}=\frac{x}{9}

53=x9\frac{5}{3}=\frac{x}{9}

x=(5)(9)3=15x=\frac{(5)(9)}{3}=15

Because MN=NP\mathrm{MN}=\mathrm{NP}, both unknown sides are equal to 15. Therefore, the three sides of MNP are 15, 15, and 10. The perimeter of a triangle is the sum of its three side lengths.

15+15+10=4015+15+10=40

Question analysis

Two isosceles triangles are drawn, one with all its side lengths labeled and the other with only its base labeled, and the triangles are similar. Students find the perimeter of triangle MNP in centimeters. It tests similar figures and proportional sides, and the catch is that two sides must be found from the scale and the equal-sides marks.

NYS standards
  • 8.G.4 similarity and dilations
  • 7.G.1 scale drawings
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Similar triangles are grade 8 geometry, though grade 7 scale drawings use the same proportional reasoning. Students also need the equal-sides marks to find a missing side, a combined step beyond routine grade 7 scale work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains a scale factor to a perimeter, and both links have a trap. Some students scale by adding, since the base grew by 4, and others find the 15 cm leg correctly but then add only some of the sides. Writing all three sides of the large triangle, 15, 15 and 10, before adding is what the question rewards.

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

    The figure carries every number: tick marks show MN = NP, the base MP is 10 cm, and triangle XYZ has 9 cm legs on a 6 cm base. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Students must know that corresponding sides of similar triangles share one scale factor, here 10/6 = 5/3, though the stem never says so. (Anderson & Krathwohl, 2001)

  3. Error-prone arithmetic minor Working it out

    The scale factor 5/3 is a fraction, and 9 × 5/3 = 15 is easier to get right by dividing 9 by 3 first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Compare the two bases to get the scale factor, multiply a leg of the small triangle by it, then use the equal sides to add up the perimeter.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Similar triangles
MNXY=MPXZ\frac{\mathrm{MN}}{\mathrm{XY}} = \frac{\mathrm{MP}}{\mathrm{XZ}}
Similar triangles have the same shape but can differ in size. MN matches XY, and MP matches XZ. Each side of the big triangle is the same number of times as long as its matching side. So these two fractions are equal.
Perimeter of a triangle
P=a+b+cP = a + b + c
P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. Add them to go once around the shape.

  1. Sort out the question

    In triangle MNP, the base MP is 10 cm. The tick marks on MN and NP show those two sides are equal. In triangle XYZ, the base XZ is 6 cm, and XY and YZ are each 9 cm.

    Similar means the same shape, with every side scaled by the same amount. You need the perimeter of MNP, the sum of its three sides.

  2. Match the sides

    The letters line up in order: M with X, N with Y, and P with Z. So the base MP matches the base XZ. The side MN matches XY, and NP matches YZ.

  3. Find the scale factor

    The scale factor is how many times bigger MNP is. Compare the two bases.

    MPXZ=106=53\frac{\mathrm{MP}}{\mathrm{XZ}} = \frac{10}{6} = \frac{5}{3}

    Every side of MNP is 53\frac{5}{3} times as long as its matching side in XYZ.

  4. Find the equal sides

    Use the similar triangles formula with XY = 9.

    MN9=53\frac{\mathrm{MN}}{9} = \frac{5}{3}

    MN=9×53=3×5=15\mathrm{MN} = 9 \times \frac{5}{3} = 3 \times 5 = 15

    Divide 9 by 3 first to keep the numbers small. MN is 15 cm. Since MN = NP, NP is 15 cm too.

    Watch out: The base grew by 4 cm, but adding 4 cm to each side is wrong. That gives 13 cm sides and 36, choice C. Similar shapes grow by multiplying.

  5. Add the three sides

    P=15+15+10=40P = 15 + 15 + 10 = 40

    The perimeter is 40 cm. Watch out: Leaving out one 15 cm side gives 25, choice A. Leaving out the base gives 30, choice B.

    So the answer is D, 40 cm.

Check your answer

The small triangle’s perimeter is 9 + 9 + 6 = 24 cm. Perimeters grow by the same scale factor as the sides. So 24×53=4024 \times \frac{5}{3} = 40. It matches.

More questions like this

The original DOE question
Question 105 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 105. Open page 212 of the handbook (PDF)

Question 51 of 108

2024 Test B Math, question 52

Question

Two open boxes, labeled A and B. Box A holds two balls, numbered 2 and 6. Box B holds three balls, numbered 3, 4 and 5.

Five numbered balls are placed in two boxes as shown in the figure above. Davina picks one ball at random from Box B and places it in Box A. Then Yusuf picks one ball from Box A without looking. What is the probability that Yusuf will pick a ball with an odd number?

  • A29\frac{2}{9}
  • B13\frac{1}{3}
  • C25\frac{2}{5}
  • D23\frac{2}{3}
Show the answer

Correct answer: A

The probability of Davina picking an odd-numbered ball from Box B is 23\frac{2}{3}. If Davina places an odd-numbered ball in Box A, then the probability of Yusuf picking an odd-numbered ball from Box A is 13\frac{1}{3}. The probability of both of these events happening is the product of the two probabilities.

23×13=29\frac{2}{3}\times\frac{1}{3}=\frac{2}{9}

Question analysis

Two boxes hold five numbered balls, and Davina moves one ball at random from Box B into Box A before Yusuf draws one from Box A. Students find the probability that Yusuf's ball shows an odd number. It tests two-stage probability, and the tricky part is that the second draw depends on which ball was moved.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 compound probability lists outcomes for independent stages such as coins or spinners. Here the first random move changes what is in the second box, so students must follow two linked stages, a dependence typical grade 7 work does not include.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
3 steps
Reading level
Question and choices: about grade 4, 57 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two chance events are hidden in what reads like a question about a single draw. Every listed wrong answer is a reasonable single-stage answer, either one of the two factors on its own or a count of odd balls across both boxes. Listing the nine equally likely pairs, Davina's ball and then Yusuf's, shows that only two of them end with an odd ball and makes the structure visible without any formula.

  1. No obvious method major Working it out

    The stem asks one question about Yusuf's draw, but the answer depends on Davina's earlier random pick; the key is that Box A starts with only even balls, so Yusuf can draw odd only by drawing the ball she moved. (Schoenfeld, 1985)

  2. Something you must already know minor Working it out

    Multiplying the chances of the two stages, 2/3 for an odd ball moving and 1/3 for Yusuf then drawing it, is a rule students must bring themselves. (Anderson & Krathwohl, 2001)

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

    The numbers appear only in the figure, Box A holding 2 and 6 and Box B holding 3, 4 and 5, so the odd balls have to be spotted there. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List every equally likely pair of a ball moved by Davina and a ball picked by Yusuf, and count the pairs that end with an odd ball.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Read the figure

    Box A holds balls numbered 2 and 6. Box B holds balls numbered 3, 4 and 5.

    An odd number does not divide evenly by 2. So the only odd balls are 3 and 5, and both start in Box B. Box A starts with no odd balls at all.

  2. Spot the two chance events

    Two things happen by chance, one after the other:

    1. Davina picks one of the 3 balls in Box B and moves it to Box A.
    2. Then Yusuf picks one of the 3 balls now in Box A.

    Yusuf gets an odd ball only if both go right. Davina must move an odd ball. Then Yusuf must pick that very ball.

    The question asks only about Yusuf. But his chance depends on Davina’s pick, which comes first.

  3. Find the chance for Davina’s pick

    Box B has 3 balls, and 2 of them, the 3 and the 5, are odd.

    P(odd ball moved)=23P(\text{odd ball moved}) = \frac{2}{3}

    If she moves the 4 instead, Box A holds only even balls. Then Yusuf can’t get an odd ball at all.

  4. Find the chance for Yusuf’s pick

    Say Davina moved an odd ball. Box A now holds 3 balls: the 2, the 6 and that one odd ball.

    P(Yusuf picks it)=13P(\text{Yusuf picks it}) = \frac{1}{3}

    Watch out: 13\frac{1}{3} alone is choice B. It forgets that Davina might move the 4.

  5. Multiply the two chances

    Both events must happen, so use the rule for events in a row.

    P(odd)=23×13=29P(\text{odd}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}

    Watch out: 23\frac{2}{3} alone, choice D, stops after Davina’s pick. 25\frac{2}{5}, choice C, counts 2 odd balls out of all 5. But Yusuf picks only from Box A.

    So the answer is A, 29\frac{2}{9}.

Check your answer

List all 9 equally likely outcomes. Davina can move 3, 4 or 5. Then Yusuf picks one of 3 balls.

Ball movedBox A thenOdd picks
32, 6, 31
42, 6, 40
52, 6, 51

That is 2 odd picks out of 9, or 29\frac{2}{9}.

More questions like this

The original DOE question
Question 109 as printed on page 214 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 109. Open page 214 of the handbook (PDF)

Question 52 of 108

2024 Test B Math, question 53

Question

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

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

Correct answer: A

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

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

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

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

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 9, 7 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know moderate Working it out

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

  2. A very close wrong answer moderate Choosing the answer

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Read it as a distance

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

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

  3. Find the two boundary points

    Start at −1 and go 2 units each way.

    −1−2=−3-1 - 2 = -3

    −1+2=1-1 + 2 = 1

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

  4. Decide which side to shade

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

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

  5. Match the graph

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

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

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

Check your answer

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

More questions like this

The original DOE question
Question 110 as printed on page 214 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 110. Open page 214 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 53 of 108

2023 Test A Math, question 4

Question

Line l is vertical and line n is horizontal, and a small square marks the right angle where they meet. Line m rises from the lower left to the upper right; it crosses line l below line n and crosses line n to the right of line l. Where m crosses l, the angle between the lower part of l and the lower-left part of m is marked 35°. Where m crosses n, the angle between the right part of n and the upper-right part of m is marked x°.

In the figure above, line l is perpendicular to line n. What is the value of x?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 55

Since vertical angles are congruent, the right triangle in the figure has acute angles with measures of 35° and x°. The interior angles of a triangle sum to 180°, so set up an equation to solve for x:

90+35+x=18090 + 35 + x = 180

125+x=180125 + x = 180

x=55x = 55

Question analysis

Three lines cross in a figure: line l is perpendicular to line n, and a slanted line m meets them at a marked 35° angle and an unknown angle x. This geometry grid-in asks for the value of x. Students must see the right triangle the lines form and carry the marked angles into it using vertical angles.

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 vertical, supplementary and complementary angles. This one also needs the angle sum of a triangle, which New York places in grade 8, and the triangle is not drawn as a separate shape, so students must spot it.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure hides a small right triangle between the three lines, and the question is easy once students see it. Its angles are the right angle, the vertical partner of the 35° angle and the vertical partner of x, so x is what the right angle and 35° leave of 180°. Students who never spot the triangle tend to guess a direct relationship between the two marked angles, which no angle fact supports.

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

    Neither marked angle is inside the right triangle that lines l, m and n form: the 35° opens below the point where m crosses l, and x opens above line n, so each must be carried in as a vertical angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know moderate Working it out

    The stem states only that l is perpendicular to n; students must supply that vertical angles are equal and that the three angles of a triangle add to 180°. (Anderson & Krathwohl, 2001)

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

    As a grid-in it offers no choices, so 125, the right angle plus the 35° angle, can be gridded with nothing to signal the missing subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Find the triangle formed by the three lines, use vertical angles to move the 35° and x° angles inside it, and apply the triangle's angle sum with the right angle.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Vertical angles
angle a=angle b\text{angle } a = \text{angle } b
When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle
A+B+C=180A + B + C = 180
A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

  1. Sort out the figure

    Line l is perpendicular to line n. Perpendicular means they meet at a right angle, 90°. The small square in the figure marks it. Line m crosses both lines. The three lines make a triangle with a right angle. You need x, the angle marked x° where m crosses n.

  2. Move the 35° angle into the triangle

    The 35° angle is below the point where m crosses l. So it is outside the triangle. Lines l and m make an X there. The triangle’s angle is straight across from the 35° angle. They are vertical angles, so they are equal. The triangle’s angle at that corner is 35°.

  3. Move the x° angle into the triangle

    The x° angle is above line n, so it is outside the triangle too. Lines m and n make another X. The triangle’s angle at that crossing is straight across from x°. So the third angle of the triangle is also x°. Watch out: line m does not cross l and n at the same angle. Lines l and n are not parallel, so x is not 35.

  4. Use the triangle’s angle sum

    The triangle’s three angles are 90°, 35° and x°.

    A+B+C=180A + B + C = 180

    90+35+x=18090 + 35 + x = 180

    125+x=180125 + x = 180

    x=180−125=55x = 180 - 125 = 55

    Watch out: 125 is only the two known angles added together. You still subtract it from 180. Gridding 125 is wrong. So you would grid 55.

Check your answer

In a right triangle, the two smaller angles add up to 90°. That is because the right angle uses 90° of the 180°. Check: 35+55=9035 + 55 = 90. Also, in the figure the x° angle looks a bit bigger than the 35° angle, and 55 is.

More questions like this

The original DOE question
Question 61 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 76, question 61. Open page 76 of the handbook (PDF)

Question 54 of 108

2023 Test A Math, question 10

Question

A number line from 0 to 10 with a tick and a label at every whole number and arrows at both ends. Point W is marked with a dot at 0, and point X is marked with a dot at 2.

Points Y and Z are not shown on the number line above. If X is the midpoint of WY\overline{WY}, and Y is the midpoint of WZ\overline{WZ}, where on the number line would point Z be located?

  • A2
  • B4
  • C6
  • D8
Show the answer

Correct answer: D

First, find where Y is located on the number line. Point X is the midpoint of WY\overline{WY}. Since there are 2 units between W and X, there must also be 2 units between X and Y. Thus, Y is located at 4 on the number line.

Point Y is the midpoint of WZ\overline{WZ}. Use the same reasoning as above. Since there are 4 units between W and Y, there must be 4 units between Y and Z. So, Z is located at 8 on the number line.

Question analysis

A number line shows point W at 0 and point X at 2, and two more points, Y and Z, are described but not drawn. Students use two midpoint facts, one after the other, to find where Z lies. Each midpoint fact is about a different segment, so keeping track of which point is the middle one matters.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Midpoints on a number line rest on distance, which grade 7 students find as a difference. Here every point lies at a whole number of zero or more, so the work is two short steps with no negative numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 36 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 midpoint statement doubles a distance from W, so the problem is two short doublings, 2 to 4 and then 4 to 8. The risk is stopping halfway, since Y's position is found first and appears among the choices, or reusing X as an endpoint in the second statement. Rereading which point the question asks about, and checking that Y is halfway between W and Z, catches both.

  1. Many steps minor Working it out

    The two midpoint statements must be used in order: the first places Y at 4, and only then can the second, about segment WZ, place Z. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    Only W at 0 and X at 2 are marked on the 0-to-10 line, and the overbars name WY and WZ as segments, so Y and Z have to be placed from the statements alone. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the first midpoint to place Y as far beyond X as X is from W, then use the second midpoint the same way to place Z beyond Y.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint of a segment
AM=MBAM = MB
A and B are the ends of a segment, a straight piece of a line. M is its midpoint, the point exactly halfway. AM is the distance from A to M, and MB is the distance from M to B. A midpoint splits a segment into two equal parts, so these are equal.

  1. Sort out the question

    The number line shows W at 0 and X at 2. Points Y and Z are not shown. X is the midpoint of segment WY. Y is the midpoint of segment WZ. A midpoint is the point exactly halfway along a segment. You need the number where Z sits.

  2. Place point Y

    X is halfway from W to Y. So X to Y is as long as W to X.

    WX=2−0=2WX = 2 - 0 = 2

    XY=WX=2XY = WX = 2

    So Y is 2 units past X.

    2+2=42 + 2 = 4

    Point Y is at 4. Watch out: this is Y, not Z. Stopping here gives 4, choice B.

  3. Place point Z

    Now Y is halfway from W to Z. W to Y is 4 units, so Y to Z is also 4 units.

    YZ=WY=4−0=4YZ = WY = 4 - 0 = 4

    4+4=84 + 4 = 8

    Watch out: the second midpoint is for segment WZ, not XZ. Taking just 2 more units from Y puts Z at 6, choice C. So the answer is D, 8.

Check your answer

Check both midpoints. Halfway from 0 to 4 is (0+4)÷2=2(0 + 4) \div 2 = 2, which is X. Halfway from 0 to 8 is (0+8)÷2=4(0 + 8) \div 2 = 4, which is Y. Both facts in the question are true.

More questions like this

The original DOE question
Question 67 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 67. Open page 78 of the handbook (PDF)

Question 55 of 108

2023 Test A Math, question 15

Question

A swim instructor used the box plot below to display the distribution of the ages of students who signed up for swim lessons.

Box plot titled STUDENTS’ AGES above a number line from 5 to 13 with a tick and a label at every whole number and arrows at both ends. The left whisker starts at 7, the box runs from 8 to 10 with the median line at 9, and the right whisker ends at 12.

Which statement about the distribution of ages is true?

  • AThe data contain an outlier.
  • BApproximately half the students are exactly 9 years old.
  • CApproximately one-fourth of the students are at least 10 years old.
  • DThe interquartile range is 3 greater than the range of the data.
Show the answer

Correct answer: C

The data points in the box plot are as follows:

  • minimum: 7
  • first quartile: 8
  • median: 9
  • third quartile: 10
  • maximum: 12

Since the value of the third quartile is 10, approximately one-fourth of the students are 10–12 years old.

Question analysis

A swim instructor's box plot shows the ages of students who signed up for lessons. Students decide which of four statements about the distribution is true, using the minimum, quartiles, median and maximum. The statements test different box plot ideas, including outliers, the median, the quartiles and the range, so each needs its own check.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 New York places box plots, the interquartile range and outliers in grade 7. Here four statements test those ideas one at a time; each check is short, so the item matches typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Reading the five values off the plot is quick; the work is knowing what a box plot can and cannot say. It shows where quarters of the data begin and end, not how many students share any one age, and a long whisker is not an outlier unless it passes the 1.5 × IQR cutoff, here 13. The true statement uses the third quartile: about a quarter of the students are 10 or older.

  1. Something you must already know moderate Working it out

    The statements rely on box plot facts the question does not give: each section holds about a quarter of the data, the middle line is the median, and outliers lie beyond 1.5 IQRs. (Anderson & Krathwohl, 2001)

  2. Every choice must be checked moderate Choosing the answer

    Four statements test four separate ideas, an outlier, the share at one age, a quarter of the data and two spreads, so ruling one out says nothing about the rest. (Embretson & Wetzel, 1987)

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

    The five values, 7, 8, 9, 10 and 12, are read off a 5-to-13 scale, where the longer right whisker draws the eye more than the box does. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the five key values off the plot, then test each statement: compute the range, the interquartile range and the outlier cutoffs, and recall what share of the data each part holds.

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.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range
range=maximum−minimum\text{range} = \text{maximum} - \text{minimum}
The maximum is the largest value, the right end of a box plot. The minimum is the smallest, the left end. The range is how far apart they are. It shows how spread out all the data are.
Interquartile range
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
IQR is short for interquartile range. Q1, the first quartile, is the left edge of the box. Q3, the third quartile, is the right edge. The IQR is the width of the box. It is the spread of the middle half of the data. The four parts of a box plot each hold about one-fourth of the data.
Outlier test
below Q1−1.5×IQR or above Q3+1.5×IQR\text{below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}
An outlier is a value far from the rest of the data. A common rule says a value is an outlier if it lies more than 1.5 IQRs below Q1 or above Q3.

  1. Read the five values off the plot

    A box plot shows five numbers. The median is the middle value when the ages are in order.

    PartAge
    Minimum, the left whisker’s end7
    First quartile, Q1, the box’s left edge8
    Median, the line in the box9
    Third quartile, Q3, the box’s right edge10
    Maximum, the right whisker’s end12

  2. Check choice A, an outlier

    IQR=10−8=2\text{IQR} = 10 - 8 = 2

    Q3+1.5×2=13 and Q1−1.5×2=5Q_3 + 1.5 \times 2 = 13 \text{ and } Q_1 - 1.5 \times 2 = 5

    An outlier would be above 13 or below 5. The ages only run from 7 to 12. So there is no outlier. Watch out: a long whisker is not an outlier. It just means the top ages are spread out. Choice A is false.

  3. Check choice B, the median

    The median, 9, is the middle age. About half the students are 9 or younger. About half are 9 or older. It does not tell how many are exactly 9. Watch out: the median line does not mean half the students are 9. Choice B is false.

  4. Check choice C, the top fourth

    The box plot splits the students into four groups of about the same size. Each group holds about one-fourth of them. The last group runs from Q3, 10, to the maximum, 12. At least 10 means 10 or more. So about one-fourth of the students are at least 10 years old. Choice C is true.

  5. Check choice D, the two spreads

    range=12−7=5\text{range} = 12 - 7 = 5

    IQR=2\text{IQR} = 2

    The IQR is 3 less than the range, not 3 greater. The box always fits inside the whiskers, so the IQR can never be bigger. Watch out: swapping the two measures gives choice D. So the answer is C: about one-fourth of the students are at least 10 years old.

Check your answer

Picture 20 students. About 5 would be in each fourth of the plot. The top 5 would be ages 10 to 12. That is about one-fourth, just as choice C says. Each other choice fails a fact you can read straight off the plot.

More questions like this

The original DOE question
Question 72 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 72. Open page 79 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 56 of 108

2023 Test A Math, question 24

Question

A student uses Spinner R and Spinner T to generate a list of two-digit numbers.

Two spinners, each with an arrow pointing out from its center. Spinner R is a circle divided into 4 equal sections, labeled 1 and 2 across the top and 3 and 4 across the bottom. Spinner T is a circle divided into 6 equal sections, labeled, clockwise from the top left, 1, 1, 2, 3, 4 and 4.

Spinner R determines the digit in the tens place, and Spinner T determines the digit in the ones place. What is the probability that the two-digit number determined by spinning each spinner one time is a prime number?

  • A38\frac{3}{8}
  • B512\frac{5}{12}
  • C35\frac{3}{5}
  • D58\frac{5}{8}
Show the answer

Correct answer: A

Spinner R has 4 equal sections and Spinner T has 6 equal sections, so the number of possible two-digit numbers that can be created is found by multiplying 4 and 6, which is 24. List the possible outcomes:

11 11 12 13 14 14
21 21 22 23 24 24
31 31 32 33 34 34
41 41 42 43 44 44

There are 9 prime numbers out of 24 possible outcomes: 11, 11, 13, 23, 31, 31, 41, 41, and 43.

So the probability that the two-digit number created is a prime number is 924\frac{9}{24}, which is equivalent to 38\frac{3}{8}.

Question analysis

Spinner R, with four equal sections, gives the tens digit and Spinner T, with six equal sections and some labels repeated, gives the ones digit of a two-digit number. Students find the probability that the number is prime. Building the full table of outcomes, counting repeated sections separately, and checking whether each number is prime all take care.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Listing a sample space for two spinners is grade 7 work, usually with distinct labels and a simple event. Here one spinner repeats two numbers, the space has 24 outcomes, and each two-digit result must be tested for being prime.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 7, 57 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question combines a compound sample space with a test for primes, and its figure adds a twist: two labels repeat on Spinner T. Counting the 24 section pairs rather than the 16 different numbers is the careful route, though here both happen to give the same fraction. The points are lost in the prime test, where 33 is easy to count as prime, and in writing the count as a share of all 24 outcomes.

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

    Spinner T has six equal sections but only four labels, with 1 and 4 each appearing twice, so the 24 equally likely outcomes include repeats such as 11 and 11. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Much to keep track of moderate Working it out

    Twenty-four two-digit numbers have to be built and each tested for primality while the repeated sections are kept track of; a 4-by-6 table is the reliable way to hold them. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Something you must already know minor Working it out

    Which two-digit numbers are prime is not given; students must test 21, 33, 43 and the rest themselves, and 21 = 3 × 7 and 33 = 3 × 11 look prime at a glance. (Anderson & Krathwohl, 2001)

Strategy: Make a 4-by-6 table with Spinner R's digits as tens and Spinner T's six sections as ones, mark each prime, and count every section, including repeated labels.

Traps

  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
  • 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.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting outcomes
outcomes=a×b\text{outcomes} = a \times b
a is the number of equal sections on the first spinner, and b is the number on the second. Each section of the first can go with each section of the second. So the number of pairs is a times b.
Probability
P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}
P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.

  1. Sort out the question

    Spinner R gives the tens digit. Spinner T gives the ones digit. Together they make a two-digit number, like 23. You need the chance that the number is prime. A prime number has exactly two factors, 1 and itself. For example, 13 is prime, but 21 = 3 × 7 is not.

  2. Count all the outcomes

    Spinner R has 4 equal sections. Spinner T has 6 equal sections. Look closely: T has two 1s and two 4s. Each section counts as its own outcome.

    outcomes=4×6=24\text{outcomes} = 4 \times 6 = 24

    There are 24 outcomes, all equally likely.

  3. Make a table of the numbers

    Each row is a tens digit from R. Each column is a section of T. Prime numbers are in bold.

    R112344
    1111112131414
    2212122232424
    3313132333434
    4414142434444

  4. Count the prime numbers

    Numbers ending in 2 or 4 are even, so 2 divides them. They are not prime. Check the rest. 11, 13, 23, 31, 41 and 43 are prime. But 21 = 3 × 7 and 33 = 3 × 11 are not. Count every section, so 11, 31 and 41 each count twice. That makes 3 + 1 + 2 + 3 = 9 primes. Watch out: counting 33 as prime gives 1024=512\frac{10}{24} = \frac{5}{12}, choice B.

  5. Find the probability

    P=924=9÷324÷3=38P = \frac{9}{24} = \frac{9 \div 3}{24 \div 3} = \frac{3}{8}

    Watch out: divide by all 24 outcomes. Comparing the 9 primes with the 15 others gives 915=35\frac{9}{15} = \frac{3}{5}, choice C. So the answer is A, 38\frac{3}{8}.

Check your answer

There are 24−9=1524 - 9 = 15 numbers that are not prime. Their chance is 1524=58\frac{15}{24} = \frac{5}{8}. And 38+58=1\frac{3}{8} + \frac{5}{8} = 1, as it should be. Notice that 58\frac{5}{8}, choice D, is the chance of not prime.

More questions like this

The original DOE question
Question 81 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 81. Open page 81 of the handbook (PDF)

Question 57 of 108

2023 Test A Math, question 25

Question

Line h is horizontal, with arrows at both ends. Line j rises from the lower left to the upper right, and line k rises steeply from the bottom, crossing line h to the right of line j; lines j and k meet at point Q above line h. Where j crosses h, the angle above h and to the left of j is marked 135°. Where k crosses h, the angle below h and to the right of k is marked 100°. At Q, the angle between the parts of lines k and j above Q is marked x°.

Lines j and k intersect at point Q, and h is a straight line. What is the value of x?

  • A55
  • B45
  • C35
  • D30
Show the answer

Correct answer: C

First, find the measures of the three angles inside the triangle. The lower left angle is supplementary to the one marked 135°, so the measure of that angle is 180−135=45180 - 135 = 45. The angle on top is a vertical angle with the one marked x°, so the measure of that angle is also x°. The angle on the lower right is a vertical angle with the one marked 100°, so the measure of that angle is also 100°. Since the measures of the three interior angles of a triangle sum to 180°, set up an equation and solve for x:

45+x+100=18045 + x + 100 = 180

145+x=180145 + x = 180

x=35x = 35

Question analysis

Lines j and k cross at point Q above a straight line h, forming a triangle, with a 135° angle marked where j meets h and a 100° angle where k meets h. Students find the angle x at Q. Each marked angle lies outside the triangle, so it has to be turned into an inside angle before the triangle's angle sum can be used.

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 students use supplementary and vertical angles to find unknown angles. This figure needs both, applied at three different vertices, plus the angle sum of a triangle, which New York teaches in grade 8.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 2, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every angle the solution needs has to be carried into the triangle from outside, and the two marked angles need different treatment. The 135° is supplementary to the triangle's angle, but the 100° is vertical to it, so subtracting both from 180° out of habit builds the wrong triangle. Checking which side of each line a marked angle sits on, not just how near the triangle it is, separates the two cases.

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

    None of the three marked angles is inside the triangle formed by h, j and k: the 135° is outside on the left, the 100° is below line h, and x is above Q. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Two different angle facts are needed: the 135° gives a supplementary interior angle of 45°, while the 100° and x carry into the triangle unchanged as vertical angles. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Something you must already know minor Working it out

    Supplementary angles, vertical angles and the triangle's 180° sum are all assumed, and the question names none of them. (Anderson & Krathwohl, 2001)

Strategy: Mark the three angles of the triangle formed by lines h, j and k: one is supplementary to 135°, one is vertical to 100°, and the one at Q is vertical to x°.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b=180a + b = 180
a and b are two angles side by side on a straight line. Together they make a half turn, a straight angle. So they add up to 180°. Angles like this are called supplementary.
Vertical angles
angle a=angle b\text{angle } a = \text{angle } b
When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle
A+B+C=180A + B + C = 180
A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

  1. Sort out the figure

    Lines h, j and k make a triangle. Its corners are where j crosses h, where k crosses h, and point Q. The figure marks two angles outside the triangle, 135° and 100°. You need x, the angle marked above Q.

  2. Find the angle where j crosses h

    The 135° angle is above h and to the left of j. The triangle’s angle is right next to it, on the other side of j. Together they make the straight line h. Call the triangle’s angle a.

    a+b=180a + b = 180

    135+a=180135 + a = 180

    a=180−135=45a = 180 - 135 = 45

    So that corner of the triangle is 45°. Watch out: this is only one corner, not x. Stopping here gives 45, choice B.

  3. Find the angle where k crosses h

    The 100° angle is below h and to the right of k. The triangle’s angle is above h and to the left of k. Lines h and k make an X there, and these two angles are straight across from each other. They are vertical angles, so the triangle’s angle is 100°. Watch out: it is not 180 − 100 = 80. That angle is beside the 100°, not across from it. Using 80 gives 55, choice A.

  4. Find the angle at Q

    The x° angle is above Q, between the top parts of lines j and k. The triangle’s angle at Q is between the parts of j and k below Q. These two angles are straight across from each other. So they are vertical angles, and the triangle’s angle at Q is also x°.

  5. Use the triangle’s angle sum

    A+B+C=180A + B + C = 180

    45+100+x=18045 + 100 + x = 180

    145+x=180145 + x = 180

    x=180−145=35x = 180 - 145 = 35

    The three angles of the triangle must add up to 180°. Only x is unknown. So the answer is C, 35.

Check your answer

Add the triangle’s three angles: 45+100+35=18045 + 100 + 35 = 180. Also, the angle at Q looks narrow in the figure, much less than a right angle of 90°. 35° fits that.

More questions like this

The original DOE question
Question 82 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 82. Open page 82 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 58 of 108

2023 Test A Math, question 31

Question

A circle divided into 6 equal spaces, labeled clockwise from the top right P, Q, R, S, T and U. An arrow from the center of the circle points into Space P.

The arrow starts on Space P and moves clockwise around the circle. It moves through one space each minute. What space will the arrow point to in 100 minutes?

  • AR
  • BS
  • CT
  • DU
Show the answer

Correct answer: C

There are 6 spaces, so first divide 100 by 6: 100÷6=16100 \div 6 = 16, remainder 4. So, the arrow goes around 16 full times and then 4 more spaces. Four spaces from Space P is Space T.

Question analysis

This patterns question shows a circle split into six lettered spaces, with an arrow that starts on P and moves one space clockwise each minute, and asks where it points after 100 minutes. Students use division with a remainder to skip the full trips around. Counting the leftover moves without including the starting space, and in the right direction, is the tricky part.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Division with a remainder is grade 6 arithmetic, but using that remainder to step around a six-part circle is a pattern task that grade 7 classes seldom practice. It needs careful counting rather than new content.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 1, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The division is short; the points are lost in turning the remainder into a space. The arrow starts on P, so the 4 leftover moves land on Q, R, S and T, and counting P itself as a move stops one space short. Going around the wrong way, or one move too far, gives the other two choices. Checking a small case first, such as 7 minutes landing on Q, fixes the rule before it is used.

  1. No obvious method minor Working it out

    The stem never mentions division; the step that unlocks it is seeing that every 6 minutes brings the arrow back to P, so only the remainder of 100 ÷ 6 matters. (Schoenfeld, 1985)

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

    The six spaces and their clockwise order, P, Q, R, S, T, U, appear only in the circle diagram, so the cycle length and the direction of counting both come from the figure. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Divide 100 by 6: every full trip around the circle brings the arrow back to P, so only the remainder matters; count that many moves clockwise, starting with the space after P.

Traps

  • 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).
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder
a=b×q+ra = b \times q + r
a is the number you divide, here the minutes. b is the number you divide by, here the spaces in one trip around. q is how many whole groups of b fit into a. r is the remainder, the amount left over. The remainder is always less than b.

  1. Sort out the question

    The circle has 6 spaces. Going clockwise, the way a clock’s hands turn, they are P, Q, R, S, T and U. The arrow starts on P and moves one space each minute. You need the space it points to after 100 minutes.

  2. Notice that the pattern repeats

    After 6 minutes, the arrow has gone all the way around and points to P again. After 12 minutes it is back on P again, and so on. So full trips of 6 minutes don’t change where it points. Only the leftover minutes matter.

  3. Divide 100 by 6

    a=b×q+ra = b \times q + r

    100=6×16+4100 = 6 \times 16 + 4

    The arrow makes 16 full trips. That takes 6×16=966 \times 16 = 96 minutes, and it is back on P. There are 4 minutes left over.

  4. Count the last 4 moves

    From P, move one space each minute: Q (minute 97), R (minute 98), S (minute 99), T (minute 100).

    Watch out: don’t count P itself as the first move. That stops on S, choice B. Also count clockwise. Counting the other way lands on R, choice A.

    So the answer is C, T.

Check your answer

Use a smaller number with the same remainder. 100 − 96 = 4, and 96 minutes is a whole number of trips. So after 100 minutes the arrow points where it does after 4 minutes. Four moves from P are Q, R, S, T. That is T again.

More questions like this

The original DOE question
Question 88 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 88. Open page 83 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 59 of 108

2023 Test A Math, question 34

Question

Parallelogram PQRS, with P at the bottom left, Q at the top left, R at the top right and S at the bottom right. The angle at P is marked 56°, the angle at Q is marked x°, and the angle at S is marked y°. The angle at R is not marked.

In the parallelogram above, what is the value of x+yx + y?

  • A112
  • B124
  • C148
  • D248
Show the answer

Correct answer: D

In a parallelogram, opposite angles are congruent, so the measure of the unmarked angle is 56°. The sum of the interior angles of a parallelogram is equal to 360°. Use that information to solve the problem:

x+y+56+56=360x + y + 56 + 56 = 360

x+y+112=360x + y + 112 = 360

x+y=248x + y = 248

Question analysis

This geometry question shows parallelogram PQRS with a 56° angle at P and unknown angles x and y at Q and S, and asks for the sum x + y. Students need angle facts about parallelograms: opposite angles match, and the four angles total 360°. Stopping after finding only one angle, or adding the wrong pair, are the common mistakes.

NYS standards
  • 7.G.5 angle relationships and equations
  • 8.G.5 angle sums, transversals, angle-angle similarity
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 The parallelogram facts needed, equal opposite angles and a 360° angle sum, rest on grade 8 reasoning about parallel lines and triangles. Once they are known, the arithmetic is short, about the size of a typical grade 7 angle problem.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 10, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two short angle facts finish the question, but the figure marks three of the four corners, and the unmarked one at R is easy to forget. Students who find one obtuse angle stop with half the total, and those who add the two 56° angles give the part already known. Checking that the four angles, 56, 124, 56 and 124, add to 360 confirms the sum.

  1. Something you must already know moderate Working it out

    The figure gives one angle and no rules: students must bring both that opposite angles of a parallelogram are equal and that a quadrilateral's angles total 360°, or that neighboring angles add to 180°. (Anderson & Krathwohl, 2001)

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

    The labels x and y sit at Q and S, opposite corners, while the corner R across from the 56° angle at P carries no label at all. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Mark the angle at R equal to the angle at P, since opposite angles of a parallelogram are equal, then take both away from the total of a quadrilateral's four angles.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Opposite angles of a parallelogram
∠P=∠R\angle P = \angle R
A parallelogram is a four-sided shape whose opposite sides are parallel. P and R are opposite corners: they don’t share a side. Opposite angles of a parallelogram are always equal.
Angles of a four-sided shape
∠P+∠Q+∠R+∠S=360\angle P + \angle Q + \angle R + \angle S = 360
P, Q, R and S are the four corners. Their angles always add up to 360 degrees. A line from one corner to the opposite corner cuts the shape into two triangles, and each triangle’s angles add up to 180 degrees.
Neighboring angles of a parallelogram
∠P+∠Q=180\angle P + \angle Q = 180
P and Q are corners next to each other, joined by one side. In a parallelogram, their angles add up to 180 degrees. The same is true of any two neighboring corners.

  1. Sort out the figure

    PQRS is a parallelogram. The angle at corner P is 56°. The angle at Q is x°, and the angle at S is y°. The angle at R has no label. You need x + y, the angles at Q and S added together.

  2. Find the angle at R

    R is across from P, so they are opposite corners.

    ∠P=∠R\angle P = \angle R

    ∠R=56\angle R = 56

    So the angle at R is 56°, the same as at P.

  3. Use the 360° total

    ∠P+∠Q+∠R+∠S=360\angle P + \angle Q + \angle R + \angle S = 360

    56+x+56+y=36056 + x + 56 + y = 360

    x+y+112=360x + y + 112 = 360

    x+y=360−112=248x + y = 360 - 112 = 248

    Watch out: 112 is the two angles you already know, not x + y. Answering 112 is choice A.

  4. Find x and y one at a time

    Q is next to P, so use the neighbor rule.

    ∠P+∠Q=180\angle P + \angle Q = 180

    56+x=180,x=12456 + x = 180, \quad x = 124

    S is also next to P, so y = 124 too. Then 124+124=248124 + 124 = 248, the same total. Watch out: stopping at one angle, 124, gives choice B. The question asks for both added together.

    So the answer is D, 248.

Check your answer

Add all four angles: 56+124+56+124=36056 + 124 + 56 + 124 = 360. That is the right total for a four-sided shape. Also, the angles at Q and S look wider than a right angle in the figure, and 124° is wider than 90°.

More questions like this

The original DOE question
Question 91 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 91. Open page 84 of the handbook (PDF)

Question 60 of 108

2023 Test A Math, question 35

Question

A number line with tick marks every 2 units, labeled −6, −4, −2, 0, 2, 4 and 6. Point J is the dot at −4 and point K is the dot at 6.

On the number line above, point L (not shown) is located on line segment JK so that JL=23LK\mathrm{JL} = \frac{2}{3}\mathrm{LK}. What is the position of point L?

  • A−2
  • B0
  • C2
  • D4
Show the answer

Correct answer: B

JK=6−(−4)=10\mathrm{JK} = 6 - (-4) = 10 units. Let x=JLx = \mathrm{JL}. Then LK=10−x\mathrm{LK} = 10 - x. Substitute these expressions in the given equation:

x=23(10−x)x = \frac{2}{3}(10 - x)

Solve for x:

x=23(10−x)x = \frac{2}{3}(10 - x) Apply the multiplicative inverse property; multiply both sides of the equation by 3.
3x=2(10−x)3x = 2(10 - x) Apply the distributive property; multiply the 2 by 10 and x.
3x=20−2x3x = 20 - 2x Apply the additive inverse property; add 2x2x to both sides of the equation.
5x=205x = 20 Apply the multiplicative inverse property; divide both sides of the equation by 5.
x=4x = 4

So, point L is 4 units to the right of point J: −4+4=0-4 + 4 = 0. Point L is located at 0 on the number line.

Question analysis

This number line question puts J at −4 and K at 6 and asks for the position of a point L on segment JK where JL is two thirds of LK. Students find the segment's length, divide it in a 2-to-3 ratio, and turn a distance into a coordinate. Reading the ratio backwards, or giving a distance instead of a position, leads to wrong answers.

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
At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The ratio is stated between two unknown pieces, which is less familiar than a fraction of a whole, and the final step turns a length into a position. Students who stop at JL's length, read 2/3 the wrong way round, or use its 2 as a distance all reach listed points. Sketching five equal shares of 2 units from −4 to 6, two for JL and three for LK, makes the position plain.

  1. Turning words into math moderate Understanding the question

    JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length of JK, split it into five equal shares with two for JL and three for LK, then count JL's length to the right from J.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
Subtract the position on the left from the position on the right. The answer is how many units apart the two points are. A distance is never negative.
Splitting a length in a ratio
one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}
A ratio a : b compares two pieces. If a length is split in the ratio a : b, think of it as a + b equal parts. The first piece gets a of them and the second gets b. Divide the whole length by a + b to find one part.

  1. Sort out the question

    Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is 23\frac{2}{3} of LK. You need the position of L.

  2. Find the length of JK

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    K is on the right, so subtract J’s position from K’s.

    JK=6−(−4)=6+4=10\mathrm{JK} = 6 - (-4) = 6 + 4 = 10

    Taking away −4 is the same as adding 4. J and K are 10 units apart.

  3. Turn the rule into equal parts

    JL is 23\frac{2}{3} of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.

    one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}

    one part=102+3=2\text{one part} = \frac{10}{2 + 3} = 2

    Each part is 2 units long.

  4. Find the two lengths

    JL=2×2=4\mathrm{JL} = 2 \times 2 = 4

    LK=3×2=6\mathrm{LK} = 3 \times 2 = 6

    Watch out: JL is 23\frac{2}{3} of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.

  5. Find the position of L

    L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.

    −4+4=0-4 + 4 = 0

    Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

Put L at 0 and test the rule. JL = 0 − (−4) = 4, and LK = 6 − 0 = 6. Then 23×6=4\frac{2}{3} \times 6 = 4, which is JL. Also 4 + 6 = 10, the whole length JK.

More questions like this

The original DOE question
Question 92 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 92. Open page 84 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 61 of 108

2023 Test A Math, question 37

Question

A circle with a shaded sector between two radii, one pointing straight up and one pointing up and to the right. The central angle of the shaded sector is marked 45°.

The shaded sector of the circle shown above has an area of 18π18\pi square feet. What is the circumference of the circle?

  • A144π144\pi ft
  • B24π24\pi ft
  • C18π18\pi ft
  • D9π9\pi ft
Show the answer

Correct answer: B

The area of the entire circle is πr2\pi r^{2}.

The fraction of the circle that is shaded is 45360=18\frac{45}{360} = \frac{1}{8}. So, the area of the sector is 18πr2\frac{1}{8}\pi r^{2}. Use that information to find r:

18πr2=18π\frac{1}{8}\pi r^{2} = 18\pi

r2=144r^{2} = 144

r=12r = 12

Use r to calculate the circumference of the circle:

c=2πr=2π(12)=24πc = 2\pi r = 2\pi(12) = 24\pi

The circumference is 24π24\pi feet.

Question analysis

This circle question shows a shaded 45° sector with an area of 18π square feet and asks for the circumference of the whole circle. Students work backward: from the sector to the full area, from the area to the radius, and from the radius to the circumference. Mixing up area and circumference, or stopping partway, gives wrong choices.

NYS standards
  • 7.G.4 circle area and circumference
  • 6.G.5 perfect squares and perfect cubes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 circle problems usually give the radius and ask for area or circumference. This one runs backward from the area of a 45° sector, needing the sector's share of the circle, a square root and a switch to circumference: four linked steps.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 5, 30 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question moves from a part of the circle's area to the whole circle's length, so area and circumference both appear and are easy to swap. Many students scale the sector up to 144π and stop, giving an area when a length was asked. Solving r² = 144 also needs a square root, after which 2πr is routine. Asking at the end whether the result is a length, measured in feet rather than square feet, catches the main trap.

  1. Combines ideas moderate Working it out

    Three ideas chain together: the 45° sector is 1/8 of the circle, so the whole area is 8 × 18π = 144π, then the radius comes from the area, and the circumference from the radius. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know moderate Working it out

    Neither πr² nor 2πr is given, nor that a full turn is 360°, so every link of the chain rests on formulas students must already know. (Anderson & Krathwohl, 2001)

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

    The stem says only “shaded sector”; its 45° angle is marked at the center in the figure, the one place the fraction of the circle comes from. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Work out what fraction of the circle a 45° sector is, scale its area up to the whole circle, solve for the radius, and only then use the circumference formula.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a circle in a sector
fraction of the circle=angle360\text{fraction of the circle} = \frac{\text{angle}}{360}
A sector is a slice of a circle, like a slice of pizza. Its angle is the angle at the center, in degrees. A full turn around the center is 360 degrees. So the slice is that fraction of the whole circle.
Area of a circle
A=πr2A = \pi r^{2}
A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π, called pi, is a number a little more than 3. r² means r times r.
Circumference of a circle
C=2πrC = 2\pi r
C is the circumference, the distance around the circle. r is the radius and π is pi. Two radii make the diameter, the distance across, so this is the same as π times the diameter.

  1. Sort out the question

    The shaded slice has a 45° angle at the center. Its area is 18π18\pi square feet. You need the circumference, the distance around the whole circle. For that you need the radius, so work back to it.

  2. Find what fraction is shaded

    fraction of the circle=angle360\text{fraction of the circle} = \frac{\text{angle}}{360}

    45360=18\frac{45}{360} = \frac{1}{8}

    The shaded slice is one eighth of the circle. Eight slices like it would fill the circle.

  3. Find the area of the whole circle

    8×18π=144π8 \times 18\pi = 144\pi

    The whole circle has an area of 144π144\pi square feet. Watch out: this is an area, not the circumference. Stopping here gives choice A.

  4. Find the radius

    A=πr2A = \pi r^{2}

    πr2=144π\pi r^{2} = 144\pi

    r2=144r^{2} = 144

    r=12r = 12

    Divide both sides by π. Then ask what number times itself makes 144. It is 12, so the radius is 12 feet.

  5. Find the circumference

    C=2πrC = 2\pi r

    C=2×π×12=24πC = 2 \times \pi \times 12 = 24\pi

    Leave π as it is, because the choices are written with π. The circumference is 24π24\pi feet. So the answer is B, 24π24\pi ft.

Check your answer

Work forward. A radius of 12 feet gives an area of π×12×12=144π\pi \times 12 \times 12 = 144\pi square feet. One eighth of that is 144π÷8=18π144\pi \div 8 = 18\pi square feet, the shaded area in the question.

More questions like this

The original DOE question
Question 94 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 94. Open page 85 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 62 of 108

2023 Test A Math, question 38

Question

Which graph represents the solution to −3x−7>−4-3x - 7 > -4?

  • ANumber line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the left, ending in an arrow.
  • BNumber line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the left, ending in an arrow.
  • CNumber line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the right, ending in an arrow.
  • DNumber line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the right, ending in an arrow.
Show the answer

Correct answer: A

−3x−7>−4-3x - 7 > -4 Apply the additive inverse property; add 7 to both sides of the inequality.
−3x>3-3x > 3 Apply the multiplicative inverse property; divide both sides of the inequality by −3. Note: when dividing an inequality by a negative number, the inequality sign is reversed.
x<−1x < -1

The solution is that x is less than −1, so choose the graph that starts at −1 and continues to the left.

Question analysis

This inequality question asks which number line graph shows the solution of −3x − 7 > −4. Students solve a two-step inequality whose x term has a negative coefficient, then match the boundary point and the direction of the shading. Handling the inequality symbol at the division step is where most mistakes happen, and each slip has a matching graph.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving and graphing a two-step inequality is a core grade 7 standard. This one has a negative coefficient on x, the classic grade 7 inequality trap, and four similar graphs test whether the sign and the direction were both handled.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 7 words.
Corrected from the DOE handbook
  • A wrong term 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 solving takes two moves, but the second carries the rule the question tests: dividing an inequality by a negative number flips its direction. The four graphs cross two features, boundary at −1 or 1 and shading left or right, so forgetting the flip, dropping a sign, or both each land on a graph. Testing a shaded value such as x = −2 in the original inequality, where 6 − 7 = −1 is greater than −4, confirms the choice.

  1. Something you must already know moderate Working it out

    The step that decides the question, reversing the symbol when dividing both sides by −3, is a rule students must bring themselves; −3x > 3 becomes x < −1, not x > −1. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The graph shaded to the right from an open circle at −1 matches the right graph's boundary exactly and differs only in direction, the result of skipping the reversal. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Each graph must be read for two things, where its open circle sits, at −1 or 1, and which way the arrow runs, so a boundary alone does not identify the answer. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add 7 to both sides, divide by −3 and reverse the inequality symbol, then match both the boundary point and the shading direction; test one shaded value in the original.

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.)

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s steps say “add 7 to both sides of the equation” and “divide both sides of the equation by −3” (p. 142, question 95). But −3x − 7 > −4 is an inequality, not an equation, and the step’s own note is the point of the question: dividing an inequality by a negative number reverses the sign. The site now says “both sides of the inequality” in both steps, as the DOE’s later guides word this step (“Subtract 4 from both sides of the inequality” in the 2024–2025 guide). The working and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding to both sides of an inequality
if a>b, then a+c>b+c\text{if } a > b\text{, then } a + c > b + c
a and b are the two sides, and c is any number. Adding the same amount to both sides keeps the comparison true, like adding the same weight to both pans of a scale.
Dividing an inequality by a negative number
if a>b and c<0, then ac<bc\text{if } a > b \text{ and } c < 0\text{, then } \frac{a}{c} < \frac{b}{c}
a and b are the two sides, and c is a negative number. When you divide both sides by a negative number, the greater-than sign flips to less-than. That is because negatives reverse order: 2 is more than 1, but −2 is less than −1.

  1. Sort out the question

    An inequality compares two amounts. The sign > means is greater than. You need the graph of every x that makes −3x−7>−4-3x - 7 > -4 true. The graphs differ in two ways: where the open circle sits, at −1 or 1, and which way the shading goes.

  2. Add 7 to both sides

    if a>b, then a+c>b+c\text{if } a > b\text{, then } a + c > b + c

    −3x−7+7>−4+7-3x - 7 + 7 > -4 + 7

    −3x>3-3x > 3

    Adding 7 undoes the −7, so the x term is alone on the left.

  3. Divide by −3 and flip the sign

    if a>b and c<0, then ac<bc\text{if } a > b \text{ and } c < 0\text{, then } \frac{a}{c} < \frac{b}{c}

    −3x−3<3−3\frac{-3x}{-3} < \frac{3}{-3}

    x<−1x < -1

    So x is any number less than −1. Watch out: if you forget to flip the sign, you get x > −1, and that graph is choice C. Also, 3 ÷ (−3) is −1, not 1. Dropping that minus sign leads to B or D.

  4. Match the solution to a graph

    Less than −1 means to the left of −1 on the number line. The circle at −1 is open, or empty, because −1 itself does not work. The sign means greater than, not greater than or equal to.

    Graph A has an open circle at −1 and is shaded to the left. So the answer is A, the graph shaded left from an open circle at −1.

Check your answer

Test a number on each side of −1. For x = −2: −3(−2)−7=6−7=−1-3(-2) - 7 = 6 - 7 = -1, and −1 is greater than −4, so −2 works. For x = 0: −3(0)−7=−7-3(0) - 7 = -7, and −7 is less than −4, so 0 does not work. The shading must go left.

More questions like this

The original DOE question
Question 95 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 95. Open page 85 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 63 of 108

2023 Test A Math, question 45

Question

A rectangle. Point R is on the top side, near its right end. Line segment RS runs down and to the right from R, crosses the right side of the rectangle, and ends at point S outside it. At R, the angle between the top side, to the left of R, and segment RS is marked 130°. Where RS crosses the right side, the angle outside the rectangle between the right side, above the crossing, and the part of RS going on to S is marked y.

The diagram above shows line segment RS intersecting a rectangle. What is the measure of angle y?

  • A140°
  • B130°
  • C50°
  • D40°
Show the answer

Correct answer: A

First, find the measures of the interior angles of the triangle formed by the line segment and the rectangle. The top angle on the right is a vertex of the rectangle, so the measure of that angle is 90°. The top angle on the left is supplementary to the given angle of 130°. So, the measure of that angle is 180−130=50180 - 130 = 50. The measure of the lower angle is 40°, since 180−90−50=40180 - 90 - 50 = 40. Angle y is supplementary to the lower angle of 40°, so the measure of angle y is 180−40=140180 - 40 = 140.

Question analysis

This geometry question shows line segment RS cutting across a corner of a rectangle, with a 130° angle marked where RS meets the top side, and asks for angle y where it crosses the right side. Students combine supplementary angles, the right angle of a rectangle and the angle sum of a triangle. Stopping at an angle inside the triangle is the main trap.

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 adds the angle sum of a triangle, which New York places in grade 8, and chains three facts around a small triangle that students must first spot in the figure.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The angle facts are basic, but the route runs through a triangle the figure does not point out, and the answer lies outside it. Students who stop inside the triangle give its 40° angle or its 50° angle, and those who assume y matches the marked 130° skip the work entirely. Checking that y and the triangle's angle beside it form a straight line, 140 + 40 = 180, confirms the result.

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

    The small triangle cut off at the rectangle's top right corner is never outlined or labeled, its 90° corner is unmarked, and y sits outside it, between RS and the rectangle's right side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three steps follow in order, a straight angle at R, the triangle's angle sum, and a straight angle at the right side, and the first two results are both listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    The rectangle's corner must be taken as 90° although no right-angle mark is drawn, and the 180° totals for a straight line and a triangle are assumed. (Anderson & Krathwohl, 2001)

Strategy: Look at the small triangle cut off at the rectangle's corner: get its angle at R from the straight top side, use the 90° corner and the triangle's angle sum, then use the straight right side.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b=180a + b = 180
a and b are two angles side by side that together make a straight line. A straight line makes an angle of 180 degrees. So if you know one of them, subtract it from 180 to get the other.
Angles in a triangle
a+b+c=180a + b + c = 180
a, b and c are the three angles inside any triangle. They always add up to 180 degrees. If you know two of them, subtract both from 180 to find the third.
Outside angle of a triangle
outside angle=a+b\text{outside angle} = a + b
An outside angle sits next to a triangle’s angle, along a straight line. a and b are the two inside angles at the other corners. The outside angle equals their sum.

  1. Sort out the figure

    Segment RS starts at R on the top side of the rectangle. It goes down to the right, crosses the right side, and ends at S. Call the rectangle’s top right corner P. Call the point where RS crosses the right side Q. Then R, P and Q make a small triangle. Angle y is at Q, outside the rectangle.

  2. Find the triangle’s angle at R

    The 130° angle and the triangle’s angle at R sit side by side along the top side, a straight line.

    a+b=180a + b = 180

    130+∠R=180130 + \angle \mathrm{R} = 180

    ∠R=180−130=50\angle \mathrm{R} = 180 - 130 = 50

    Watch out: 50° is only the first step. Stopping here gives choice C.

  3. Find the triangle’s angle at Q

    P is a corner of the rectangle, so its angle is 90°.

    a+b+c=180a + b + c = 180

    50+90+∠Q=18050 + 90 + \angle \mathrm{Q} = 180

    ∠Q=180−140=40\angle \mathrm{Q} = 180 - 140 = 40

    Watch out: y is outside the triangle, so this 40° angle, choice D, is not y.

  4. Find angle y

    Angle y and the triangle’s 40° angle at Q sit side by side along RS, a straight line.

    a+b=180a + b = 180

    y+40=180y + 40 = 180

    y=140y = 140

    So the answer is A, 140°.

Check your answer

Use the outside angle rule. Angle y is outside the triangle at Q. The two inside angles at the other corners are 50° and 90°, so y=50+90=140y = 50 + 90 = 140. Also, y looks wider than a right angle in the figure, so it must be more than 90°.

More questions like this

The original DOE question
Question 102 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 102. Open page 87 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 64 of 108

2023 Test A Math, question 46

Question

ESSAY LENGTH

Number of Words Number of Essays
<100 6
100–250 4
251–500 11
>500 9

All 150 students in Grade 8 at a school are assigned to write an essay on the same topic. A teacher records the number of words in a random sample of the essays, as shown in the table above. Based on this sample, how many students in the entire grade would be expected to write essays with at least 100 words?

  • A20
  • B30
  • C100
  • D120
Show the answer

Correct answer: D

The number of students in the sample who wrote an essay with at least 100 words is 24 out of the 30 students. Write it as a fraction: 2430=45\frac{24}{30} = \frac{4}{5}. To predict the number of students in the entire grade who would write an essay with at least 100 words, find 45\frac{4}{5} of 150, which is 45(150)=120\frac{4}{5}(150) = 120.

Question analysis

This statistics question gives a table of word counts for a random sample of Grade 8 essays and asks how many of all 150 students would be expected to write at least 100 words. Students find the sample size, the share meeting the condition, and scale it to the whole grade. Matching the words at least to the table's ranges is the tricky part.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a random sample to predict a count for a whole population is typical grade 7 statistics and ratio work. Students total the sample from a table, read at least 100 against the table's ranges, and scale by a whole-number factor.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 65 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The proportion at the end is routine; the question is decided at the table. Students must turn “at least 100 words” into the right rows and also find the sample size, which is printed nowhere. Each listed wrong answer scales up a different group of rows, so a careless reading still produces a believable number. Noting that only 6 of 30 essays fall short, one fifth, gives a quick check.

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

    The table never prints the sample size; it has to be built by adding all four rows, 6 + 4 + 11 + 9 = 30, and the qualifying essays sit in three rows, not one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Turning words into math minor Understanding the question

    The bold “at least 100 words” must be matched to the rows 100–250, 251–500 and >500, keeping the row that begins at 100 and leaving out only the <100 row. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps minor Working it out

    Four steps follow in order: the sample total, the qualifying count of 24, the fraction 24/30 = 4/5, and that fraction of all 150 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the table's counts to find the sample size, add the rows with 100 or more words, and scale that fraction of the sample up to all 150 students.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from a sample
expected number=count in samplesample size×whole group\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}
A sample is a smaller group picked at random from the whole group. The fraction of the sample with some feature is your best guess for the fraction of the whole group. So multiply that fraction by the size of the whole group.

  1. Sort out the question

    The table shows a random sample of the essays from a grade of 150 students. You need to predict how many of the 150 students wrote at least 100 words. At least 100 means 100 or more.

  2. Find the sample size

    The table doesn’t give the total, so add all four rows.

    6+4+11+9=306 + 4 + 11 + 9 = 30

    The sample has 30 essays.

  3. Count the essays with enough words

    Three rows count: 100–250, 251–500 and >500. Only the <100 row is left out.

    4+11+9=244 + 11 + 9 = 24

    Watch out: the 100–250 row counts, because 100 is at least 100. Leaving it out gives 100, choice C. Using the 6 short essays instead gives 30, choice B.

  4. Scale up to the whole grade

    expected number=count in samplesample size×whole group\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}

    2430×150=45×150=120\frac{24}{30} \times 150 = \frac{4}{5} \times 150 = 120

    About 4 out of every 5 students write at least 100 words. So the answer is D, 120 students.

Check your answer

The whole grade, 150, is 5 times the sample of 30. So multiply the sample count by 5: 24×5=12024 \times 5 = 120. The short essays scale to 6×5=306 \times 5 = 30, and 120 + 30 = 150, the whole grade.

More questions like this

The original DOE question
Question 103 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 103. Open page 87 of the handbook (PDF)

Question 65 of 108

2023 Test A Math, question 49

Question

A shaded rectangle inside a larger, unshaded outer rectangle. The bottom side of the outer rectangle is labeled 4 ft. The left side of the shaded rectangle is labeled 2 ft. The shaded rectangle is wider than it is tall.

In the figure above, the shaded rectangle is similar to the outer rectangle. The length of the outer rectangle is 4 feet, and the perimeter of the outer rectangle is 14 feet. If the width of the shaded rectangle is 2 feet, what is the area of the shaded rectangle?

  • A5135\frac{1}{3} sq ft
  • B6 sq ft
  • C9139\frac{1}{3} sq ft
  • D12 sq ft
Show the answer

Correct answer: A

For the outer rectangle, the length is 4 feet and the perimeter is 14 feet. Write and solve an equation to find the width:

2(4)+2w=142(4) + 2w = 14 Apply the additive inverse property; subtract 8 from both sides of the equation.
2w=62w = 6 Apply the multiplicative inverse property; divide both sides of the equation by 2.
w=3w = 3

Use that information in a proportion to find the length of the shaded rectangle:

x4=23\frac{x}{4} = \frac{2}{3} Cross multiply.
3x=4(2)3x = 4(2) Apply the multiplicative inverse property; divide both sides of the equation by 3.
x=83x = \frac{8}{3}

Now find the area of the shaded rectangle:

2×83=163=5132 \times \frac{8}{3} = \frac{16}{3} = 5\frac{1}{3}

Question analysis

This geometry question shows a shaded rectangle similar to an outer rectangle that is 4 feet long with a perimeter of 14 feet; the shaded rectangle is 2 feet wide. Students find the shaded area. They must find a missing side, match corresponding sides, and scale by a factor rather than a fixed amount, which makes it a multistep similar-figures problem.

NYS standards
  • 8.G.4 similarity and dilations
  • 7.G.1 scale drawings
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Similar figures are grade 8 geometry, though grade 7 scale drawings use the same proportions. Students must first find a missing side from the perimeter, match corresponding sides, solve a proportion with a fractional result, and then find an area.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 5, 57 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is familiar, but the three together, perimeter to width, width to scale factor, and sides to area, demand a plan. The strongest pull is scaling by subtraction, shrinking each side by 1 foot because the width drops from 3 to 2, which gives a neat whole-number area. Others stop at the outer rectangle or add the sides instead of multiplying. Checking that 8/3 to 2 matches 4 to 3 confirms the shape is similar.

  1. Combines ideas moderate Working it out

    Perimeter, similarity and area are chained: 14 feet gives the outer width of 3 feet, the 2-foot side sets a scale factor of 2/3, and only then can the area be found. (Embretson, 1998; Daniel & Embretson, 2010)

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

    The figure marks the shaded rectangle's 2-foot side running up and down and the outer 4-foot length along the bottom, so the 2 feet must be paired with the outer 3-foot side, which is not labeled. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Error-prone arithmetic minor Working it out

    The scale factor 2/3 makes the shaded length 8/3 feet, and the area 2 × 8/3 = 16/3 has to be written as the mixed number 5 1/3. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Use the perimeter to find the outer rectangle's width, match the shaded width of 2 feet to it, scale the 4-foot length by the same factor, then multiply length by width.

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.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a rectangle
P=2l+2wP = 2l + 2w
P is the perimeter, the distance all the way around. l is the length and w is the width. A rectangle has two lengths and two widths, so you add each one twice.
Scale factor of similar shapes
scale factor=small sidematching large side\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}
Similar shapes have the same shape but different sizes. Each side of the small shape equals its matching side on the large shape times one number, the scale factor. So one pair of matching sides tells you the scale factor.
Area of a rectangle
A=l×wA = l \times w
A is the area, the space inside. l is the length and w is the width. You multiply because the rectangle is made of rows of unit squares: w rows with l squares in each.

  1. Sort out the question

    The outer rectangle is 4 feet long, and its perimeter is 14 feet. The shaded rectangle is similar to it, so it has the same shape. The shaded rectangle’s width, its left side, is 2 feet. You need the shaded area, so you also need the shaded length.

  2. Find the outer rectangle’s width

    P=2l+2wP = 2l + 2w

    14=2×4+2w14 = 2 \times 4 + 2w

    14=8+2w14 = 8 + 2w

    2w=6,w=32w = 6, \quad w = 3

    The outer rectangle is 3 feet wide. That is its up-and-down side.

  3. Find the scale factor

    The shaded width, 2 feet, matches the outer width, 3 feet.

    scale factor=small sidematching large side\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}

    scale factor=23\text{scale factor} = \frac{2}{3}

    Every shaded side is 23\frac{2}{3} of the matching outer side. Watch out: similar shapes shrink by the same factor, not by the same number of feet. Taking 1 foot off each side gives 2 by 3, an area of 6, choice B.

  4. Find the shaded length

    Multiply the outer length, 4 feet, by the scale factor.

    l=23×4=83=223l = \frac{2}{3} \times 4 = \frac{8}{3} = 2\frac{2}{3}

    The shaded rectangle is 2232\frac{2}{3} feet long.

  5. Find the shaded area

    A=l×wA = l \times w

    A=83×2=163=513A = \frac{8}{3} \times 2 = \frac{16}{3} = 5\frac{1}{3}

    Watch out: 4 × 3 = 12 is the outer area, choice D. So the answer is A, 5135\frac{1}{3} sq ft.

Check your answer

Check the shape. For the outer rectangle, length ÷ width is 4÷3=434 \div 3 = \frac{4}{3}. For the shaded one, it is 83÷2=43\frac{8}{3} \div 2 = \frac{4}{3}. They match, so the shapes are similar. Also, 5135\frac{1}{3} square feet is less than the outer 12, as it should be.

More questions like this

The original DOE question
Question 106 as printed on page 88 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 88, question 106. Open page 88 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 66 of 108

2023 Test A Math, question 51

Question

At a carnival, visitors can win a prize if an employee cannot correctly guess their age. The employee records the first 15 ages that he guesses one Saturday. The box plot shown below displays data about the ages he recorded.

A box plot above a number line from 7 to 18 with a tick at every integer. The left whisker starts at 7, the box runs from 8 to 17 with the median line at 10, and the right whisker ends at 18.

What is the interquartile range of the box plot?

  • A7
  • B9
  • C10
  • D11
Show the answer

Correct answer: B

Quartile 1 is the median of the lower half of the data, 8.

Quartile 3 is the median of the upper half of the data, 17.

IQR=Q3−Q1=17−8=9\text{IQR} = \text{Q3} - \text{Q1} = 17 - 8 = 9

The interquartile range is 9.

Question analysis

This statistics question shows a box plot of 15 ages that a carnival employee guessed and asks for its interquartile range. Students read the quartiles from the edges of the box and subtract. Confusing the interquartile range with the full range, or using the median line in place of a quartile, gives the wrong choices.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Reading a box plot and finding its interquartile range is a grade 7 standard in New York. The values sit on labeled whole-number ticks and one subtraction is needed, a routine grade 7 task once the box's edges are known as the quartiles.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is one subtraction, so the question tests whether students know which parts of a box plot to use. The whiskers' ends give the range, the line inside the box gives the median, and both are easy to grab in place of the quartiles. The story about the carnival and the 15 ages plays no part in the answer.

  1. Something you must already know minor Working it out

    The plot shows five values with no labels, so students must know that the box's ends mark the quartiles and that the interquartile range is the box's width, not the whiskers'. (Anderson & Krathwohl, 2001)

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

    The values 7, 8, 10, 17 and 18 are read off a scale ticked at every whole number, with the box's ends at 8 and 17 and the median line at 10. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the first and third quartiles on the plot by recalling which parts of a box plot mark them, then subtract the smaller quartile from the larger.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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.
  • Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q1, the first quartile, is the middle of the lower half of the data. Q3, the third quartile, is the middle of the upper half. In a box plot, Q1 is the left end of the box and Q3 is the right end. So the IQR is the width of the box.

  1. Sort out the question

    A box plot shows five numbers. The whiskers, the lines sticking out, end at the smallest and largest values. The box covers the middle half of the data. The line inside the box is the median, the middle value. You need the interquartile range of the ages.

  2. Read the five numbers

    Read each one off the number line below the plot.

    • Smallest age, the left whisker: 7
    • Q1, the left end of the box: 8
    • Median, the line inside the box: 10
    • Q3, the right end of the box: 17
    • Largest age, the right whisker: 18

  3. Find the interquartile range

    IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

    IQR=17−8=9\text{IQR} = 17 - 8 = 9

    The interquartile range is 9 years. Watch out: 18 − 7 = 11 is the range, the whole spread from whisker to whisker. That is choice D. The median, 10, is choice C.

    So the answer is B, 9.

Check your answer

Count the spaces across the box. From 8 to 17 there are 9 spaces of 1 year each, so the box is 9 years wide. The answer is also less than the whole range, 11, as it must be.

More questions like this

The original DOE question
Question 108 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 108. Open page 89 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 67 of 108

2023 Test A Math, question 56

Question

Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. The line is shaded from the unlabeled tick halfway between 0 and 1 to the unlabeled tick halfway between 1 and 2, with closed dots at both ends.

If all possible values of x are indicated by the shaded part of the number line above, which number line best shows all possible values of 1x\frac{1}{x}?

  • ANumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 1 and 2 to the tick halfway between 2 and 3, with closed dots at both ends.
  • BNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from 1 to 2, with closed dots at both ends.
  • CNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 0 and 1 to 2, with closed dots at both ends.
  • DNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from a closed dot between the tick halfway between 0 and 1 and the tick at 1, about two-thirds of the way from 0 to 1, to a closed dot at 2.
Show the answer

Correct answer: D

The shaded region goes from 12\frac{1}{2} to 32\frac{3}{2}. So, the possible values of 1x\frac{1}{x} would be between the reciprocal of 12\frac{1}{2} and the reciprocal of 32\frac{3}{2}, which are 2 and 23\frac{2}{3}, respectively. Find the number line in the answer options that has a shaded region that ranges from 23\frac{2}{3} to 2, which is option D.

Question analysis

This number line question shades all possible values of x from one half to three halves and asks which number line shows all possible values of 1/x. Students find reciprocals of fractions and work out how a whole range of values changes. Keeping track of what happens to each end of the range, and placing fractions accurately on the line, makes it tricky.

NYS standards
  • 6.NS.1 dividing fractions by fractions
  • 6.NS.7a inequality as relative position
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Reciprocals of single fractions are grade 6 work. Applying them to every value in a shaded range at once, and working out where the new range begins and ends, is reasoning about a whole set of numbers that grade 7 classes rarely practice.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 12, 27 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

Few students have flipped an interval before, so the question rewards testing its two ends and noticing that the order reverses. The harder part comes at the end: 2/3 falls between the tick marks, and a graph ending at the nearby 1/2 tick looks right at a glance. Students who believe a reciprocal is always at least 1, or who shift the interval instead of flipping it, reach the other graphs.

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

    The shaded segment's ends sit on unlabeled half-unit ticks, 1/2 and 3/2, and each choice must be read the same way, with 2/3 falling between ticks rather than on one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer moderate Choosing the answer

    The graph from 1/2 to 2 matches the right graph's right end and differs at the left end by only a sixth of a unit, 1/2 against 2/3. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The question asks about 1/x for every x in a range, so a whole interval must be transformed, and its ends swap: the smallest x gives the largest 1/x. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Take the reciprocal of each end of the shaded part, ask whether a bigger x gives a bigger or a smaller 1/x, and look for the graph whose endpoints are exactly those values.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • 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.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal
reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}
a and b are the top and bottom of a fraction. The reciprocal flips the fraction upside down. The number 1 over x is the reciprocal of x. A number times its reciprocal always makes 1.

  1. Sort out the question

    x can be any number in the shaded part of the first number line. You need the number line that shows every possible value of 1x\frac{1}{x}, the reciprocal of x. The reciprocal is the number flipped upside down, like 12\frac{1}{2} for 2.

  2. Read the shaded part

    The ticks come every half unit. The shading starts at the tick halfway between 0 and 1, which is 12\frac{1}{2}. It ends halfway between 1 and 2, which is 32\frac{3}{2}, or 1121\frac{1}{2}. Both dots are filled in, so both ends count.

  3. Find the reciprocals of the ends

    Flip each end of the shaded part.

    reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}

    reciprocal of 12=21=2\text{reciprocal of } \frac{1}{2} = \frac{2}{1} = 2

    reciprocal of 32=23\text{reciprocal of } \frac{3}{2} = \frac{2}{3}

    So the ends of the new shading are 2 and 23\frac{2}{3}.

  4. Notice that the ends swap

    When x gets bigger, 1x\frac{1}{x} gets smaller. So the smallest x, 12\frac{1}{2}, gives the biggest value, 2. The biggest x, 32\frac{3}{2}, gives the smallest, 23\frac{2}{3}. So 1x\frac{1}{x} runs from 23\frac{2}{3} to 2, ends included.

  5. Find the matching number line

    23\frac{2}{3} lies between the half tick and 1, a third of the way from 12\frac{1}{2} to 1. Line D starts there and ends at 2. Watch out: line C starts at the half tick, 12\frac{1}{2}, not at 23\frac{2}{3}. Line B starts at 1.

    So the answer is D, the line shaded from 23\frac{2}{3} to 2.

Check your answer

Test a value in the middle. x = 1 is shaded, and 11=1\frac{1}{1} = 1. On line D, 1 is shaded too. Test an end: 32×23=1\frac{3}{2} \times \frac{2}{3} = 1, so 23\frac{2}{3} really is the reciprocal of 32\frac{3}{2}.

More questions like this

The original DOE question
Question 113 as printed on page 91 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 113. Open page 91 of the handbook (PDF)

Question 68 of 108

2023 Test B English, question 17
Passage Excerpt from “The Food Business Incubator”

Excerpt from “The Food Business Incubator”

by Mandalit Del Barco

“La Cocina” means “the kitchen” in Spanish. It’s also the name of a business incubator1 based in San Francisco’s Mission District. Since it began in 2005, it’s been helping local food entrepreneurs, many of whom are low-income immigrant women, develop their small businesses.

Over the years, many of its alumni have found success: more than 50 chefs in its program have become self-sufficient business owners, and many of them have opened their own brick-and-mortar restaurants. Two alumnae of its culinary program . . . were even recognized as semi-finalists for prestigious James Beard awards.2

A new cookbook, We Are La Cocina: Recipes in Pursuit of the American Dream, tells some of their stories.

Executive director Caleb Zigas says the nonprofit La Cocina grew out of two grassroots economic development organizations3 who found many people cooking at their homes and selling food on the streets. The vendors needed an affordable commercial kitchen space and technical assistance in order for their businesses to be legally viable. La Cocina provided just such a space, in addition to helping them develop business plans, pull city permits and more.

Zigas says as many as eight businesses can work in the kitchen space at La Cocina at any one time. Some can prep for a farmers’ market sale, corporate catering gigs or weddings, while others might be making and packaging their food products.

“It’s just an incredible and exciting range of techniques, flavors, perspectives, age, language. And that’s a really beautiful thing,” he says. . . .

Twice a year, La Cocina hosts “F&B: Voices from the Kitchen,” a storytelling project where their chefs can tell their own stories, as they do in the new cookbook. Some of the chefs will be on tour to promote the book, whose proceeds will go to support La Cocina entrepreneurs.

NPR4 caught up with . . . La Cocina chefs, who shared their stories:

Mariko Grady, Aedan Fermented Foods

At La Cocina, you can often hear Mariko Grady singing or humming as she prepares miso, koji, and amasake. Her fermented products come in four different flavors, including mushroom and chicken, to be used in soups and sauces. She originally brought the fermenting rice and barley koji seeds from Japan, where she had a 30-year career as a singer and dancer with the prestigious modern theatrical dance group she founded, Pappa Tarahumara. They performed around the world, and 16 years ago, had a one-night show in San Francisco. The man who would become her husband was in the audience. She soon joined him in San Francisco, often returning to Tokyo to rehearse. But after the earthquake and tsunami hit Japan in 2011, her company disbanded. . . . Grady focused on nourishing her family and creating a line of fermented products that she sells online, at local Bay Area stores and at the Ferry Plaza Farmers Market. She began at La Cocina in 2012 and named her business after her son, Aedan. (Written in Kanji characters, the name means “wisdom” and “handed down from generation to generation.”) . . . Her fermented products are “also full of wisdom about how to relate to nature and how to create a healthy life,” she says. . . .

Shani Jones, Peaches Patties

Shani Jones is a native San Franciscan. Her father was born in New Orleans, her mother, in Jamaica. She says their home was always filled with a variety of spices and dishes like jerk chicken and Jamaican patties—savory pastries filled with beef or chicken. Jones says she learned to cook from her mother, whose nickname was Peaches. After returning from college in Atlanta, Jones worked on her doctorate in organizational leadership and management while driving a Lyft5 car. She often told passengers about her idea of opening a catering company with her mom’s recipes. They steered her to La Cocina, where she developed her own business, named after her mother. Five years later, she caters and runs a kiosk6 at a small food cooperative in Bernal Heights, where some of her handmade patties have an Ethiopian twist, “because my husband is Ethiopian.” Jones has big aspirations for Peaches Patties: “The ultimate goal,” she says, “is to be the patty kingpin7 of the West Coast.”

From “The Food Business Incubator That Helps Immigrant Women Pursue The American Dream” by Mandalit Del Barco from WEEKEND EDITION SUNDAY, June 19, 2019. Copyright © 2019 NPR.

QUALIFICATIONS FOR LA COCINA APPLICANTS

Level of IncomeLa Cocina considers assets and access to opportunity when screening applicants. La Cocina’s mission is to support entrepreneurs who face barriers to becoming successful in the food industry. Applicants must be classified as low- or very-low-income earners according to the Housing and Urban Development (HUD) Guidelines for San Francisco.
Business PlanApplicants must have a business plan that is ready to be put into action. La Cocina will help applicants refine an existing business plan. Applicants without a business plan should contact one of La Cocina’s partner organizations for guidance in developing one.
Viable ProductThe applicant’s food must not only be delicious but also stand out. For a business to make a profit, there also has to be a market for the food.
Entrepreneurial Work EthicLa Cocina is excited to work with entrepreneurs who are motivated and passionate about growing their business. Succeeding in the food industry requires a lot of heart and hard work.
Community SpiritThe 30+ entrepreneurs who share La Cocina’s kitchen are from many different backgrounds. La Cocina values the community it has created and expects each entrepreneur to be active in bringing the spirit of La Cocina to the wider world.
In order to succeed in the La Cocina kitchen, accepted applicants must be flexible and willing to work closely with others.

Source: La Cocina

1business incubator: company that helps new and startup companies develop by providing services such as management training and office space

2James Beard awards: an award named after American cook and author James Beard, considered one of the highest honors in the culinary industry

3economic development organizations: organizations whose mission is to promote economic improvement within a specific geographical area

4NPR: National Public Radio, a nonprofit media organization that produces and distributes news and cultural programming

5Lyft: ride-sharing company where people use their own cars to provide rides to customers

6kiosk: cart or small structure with one or more open sides that is used to vend merchandise or services

7kingpin: the leader in a group or undertaking

Question

The table provides additional support for a central idea of the excerpt by

  • Ashowing how La Cocina works with partner organizations to further educate and advance successful applicants within the food industry.
  • Bexplaining how applicants are prepared to develop their own businesses once they leave La Cocina.
  • Cestablishing that La Cocina is dedicated to its mission to provide delicious food to different communities.
  • Demphasizing that applicants must be willing to dedicate time to promoting and developing their business with La Cocina.
Show the answer

Correct answer: D

The question asks how the table provides additional support for a central idea of the excerpt.

A. Incorrect. The table describes the requirements that allow food entrepreneurs to qualify to participate in La Cocina. While the table does mention that applicants without a business plan should contact one of La Cocina’s partner organizations, it does not specifically explain how La Cocina works with these partner organizations to help educate successful applicants.

B. Incorrect. The table describes the ways entrepreneur chefs must qualify in order to be accepted into La Cocina, not how their businesses can be successful after they leave.

C. Incorrect. While the table mentions community spirit, this is in reference to the type of energy La Cocina wishes to establish within its group kitchen space. The excerpt and table do not discuss providing food to different communities within or outside San Francisco as specific goals of La Cocina.

D. CORRECT. A central idea of the excerpt is that La Cocina helps local food entrepreneurs develop their businesses. The table lists requirements for people who want to participate in La Cocina. They must be willing to work hard to promote their own businesses as they use the resources that La Cocina provides.

Question analysis

This question asks how the table “Qualifications for La Cocina Applicants,” printed after the NPR article, provides additional support for a central idea of the excerpt. The table’s five rows, from level of income to community spirit, are written in full sentences, so students must read it as closely as the article and decide how the two connect.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R2 theme or central idea
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 readers often connect a table to a text. This one holds five rows of requirements written as sentences, not numbers, and students must tie it to a central idea through long choices, more than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from the whole text
Reading level
Passage: about grade 10, 1057 words; question and choices: about grade 13, 81 words.
Corrected from the DOE handbook
  • A detail of the table was restated accurately in the explanation.
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 table is easy to read row by row but hard to connect, because it lists entry requirements while the article describes help and success. Each wrong choice grabs a real phrase from the table or an idea from the article, partner organizations, delicious food, alumni starting their own businesses, and attaches it to a claim the table does not make. The right choice is the only one that sums up several rows at once, the work ethic and community spirit expected of applicants.

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

    The table holds five rows of full sentences, and the telling details sit in single phrases: “should contact one of La Cocina’s partner organizations,” “a lot of heart and hard work,” “active in bringing the spirit of La Cocina to the wider world.” (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. The right answer is reworded moderate Choosing the answer

    The right choice speaks of dedicating time to “promoting and developing” a business, words the table never uses, while wrong choices lift its own “partner organizations,” “mission,” “delicious” and “community.” (Freedle & Kostin, 1993; Kirsch & Mosenthal, 1990)

  3. Combines ideas minor Working it out

    Students must join two sources, a central idea of the article, that La Cocina helps cooks develop businesses, and a table about who may apply, then say what the second adds to the first. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Sum up each row of the table in a few words, state a central idea of the article in a sentence, and keep only a choice that the rows actually show and that fits that idea.

Traps

  • Word match: Repeats striking words or phrases from the text, so it looks supported, but its meaning does not fit the question.
  • Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
  • Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s explanation of Option A says the table states that applicants without a business plan “must contact” one of La Cocina’s partner organizations. The table’s Business Plan row says they “should contact” one: it is advice, while the row’s requirement is that applicants “must have a business plan that is ready to be put into action.” The site now says “should contact.” The answer is unchanged.

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. See how the table connects

    This question links the table to the article. The table, “Qualifications for La Cocina Applicants,” lists what people need to join. Qualifications are the skills or traits you must have. You must decide how the table supports a central idea of the article.

  2. Sum up each row

    RowWhat it asks of applicants
    Level of Incomeearn little money
    Business Planhave a plan ready to use
    Viable Productmake food that tastes great, stands out and can sell
    Entrepreneurial Work Ethicbe eager and work hard
    Community Spiritwork well with others and spread La Cocina’s spirit

    Most rows describe people ready to put in real effort.

  3. Connect the table to the article

    A central idea of the article is that La Cocina helps cooks develop their businesses. The table shows what La Cocina expects in return. It wants people “motivated and passionate about growing their business.” It says success “requires a lot of heart and hard work.” It expects each cook to be “active in bringing the spirit of La Cocina to the wider world.” So the help goes to people who will give time to their business.

  4. Check each choice

    • A is wrong. The table says applicants without a plan “should contact” a partner group for advice. That happens before applying. It doesn’t show La Cocina working with partners to train accepted cooks.
    • B is wrong. The table is about getting in, not about leaving La Cocina later.
    • C is wrong. The table uses the words “delicious” and “community.” But its community is the cooks who share the kitchen, not neighborhoods La Cocina feeds.
    • D is right. The work-ethic row calls for hard work at growing a business. The community row asks cooks to spread La Cocina’s spirit. Both take time and effort.

  5. Choose the answer

    The table shows that La Cocina wants cooks ready to work hard at building their businesses. So the answer is D.

Why this answer and not the closest wrong one

A is the closest wrong answer, because “partner organizations” really is in the table. But the table only suggests that people without a business plan get advice from one. It never shows La Cocina teaming up with partners to train its members. D sums up several rows at once.

More questions like this

The original DOE question
Question 17 as printed on page 162 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 162, question 17. Open page 162 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.
The passage “Excerpt from “The Food Business Incubator”” as printed (pages 158–160)The passage “Excerpt from “The Food Business Incubator”” as printed on page 158 of the DOE handbook.The passage “Excerpt from “The Food Business Incubator”” as printed on page 159 of the DOE handbook.The passage “Excerpt from “The Food Business Incubator”” as printed on page 160 of the DOE handbook.

Question 69 of 108

2023 Test B Math, question 1

Question

LAST YEAR’S TOTAL SALES

Vehicles Number
Four-door cars 5,850
Two-door cars 2,250
Trucks 900
Total 9,000

The table above shows the number of each type of vehicle sold at a dealership last year. If 1,000 vehicles will be sold next month, what is the best estimate (based on last year’s sales) of the number of two-door cars that will be sold?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 250

Let x represent the number of two-door cars that will be sold next month. Use the information in the table to set up a proportion:

x1,000=2,2509,000\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000}

x=(2,250)(1,000)9,000=2,2509=250x = \frac{(2,250)(1,000)}{9{,}000} = \frac{2{,}250}{9} = 250

Question analysis

A table shows last year's sales at a car dealership: four-door cars, two-door cars and trucks, 9,000 vehicles in all. This grid-in asks for the best estimate of how many two-door cars will be sold next month if 1,000 vehicles are sold. Students set up a proportion from the table; most of its rows are not needed.

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.SP.7 long-run relative frequency
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 proportion problems often take several steps or awkward numbers. This one scales a single share from a table to a new total, with numbers that divide cleanly, which is closer to grade 6 ratio work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Once the share is seen as 2,250 out of 9,000, a quarter, the arithmetic is light, which makes this one of the easier grid-ins. Points are lost by entering the share itself as a percent, by reading the wrong row of the table, or by building a monthly average when the question supplies next month’s total. With no choices to compare, checking that the entry is a quarter of 1,000 is the only safeguard.

  1. Turning words into math minor Understanding the question

    The phrase “best estimate (based on last year’s sales)” has to become a proportion: two-door cars keep the same share of next month’s 1,000 vehicles as of last year’s 9,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table lists three vehicle types and a total; only the two-door row, 2,250, and the 9,000 total are needed, so the four-door and truck rows are distractions. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find what fraction of last year's 9,000 vehicles were two-door cars, simplify it, then take that fraction of the 1,000 vehicles expected next month.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportion
partwhole=partwhole\frac{\text{part}}{\text{whole}} = \frac{\text{part}}{\text{whole}}
A proportion says two fractions are equal. Each fraction here is a part over its whole: two-door cars over all vehicles sold. Use it when a share stays the same while the total changes. Then you can find the missing part.

  1. Sort out the question

    The table shows last year’s sales. In all, 9,000 vehicles were sold, and 2,250 of them were two-door cars. Next month, 1,000 vehicles will be sold. You need about how many of them will be two-door cars.

    “Based on last year’s sales” means two-door cars keep the same share of all sales. A share is a part out of a whole. The four-door and truck rows don’t matter here.

  2. Find last year’s share of two-door cars

    Last year, 2,250 out of 9,000 vehicles were two-door cars. Write that as a fraction and simplify it.

    2,2509,000=14\frac{2{,}250}{9{,}000} = \frac{1}{4}

    This works because 2,250×4=9,0002{,}250 \times 4 = 9{,}000. So 1 out of every 4 vehicles sold was a two-door car.

  3. Set up the proportion and solve

    Let x be the number of two-door cars next month. Next month’s share must match last year’s share.

    x1,000=2,2509,000=14\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000} = \frac{1}{4}

    To get x alone, multiply both sides by 1,000.

    x=14×1,000=250x = \frac{1}{4} \times 1{,}000 = 250

    So about 250 two-door cars will be sold.

  4. Give the answer

    Watch out: one fourth is the same as 25%. Gridding 25 gives the percent, not the number of cars. Also be sure to use the two-door row. The four-door row would give 650. So you would grid 250.

Check your answer

Try it another way. Next month’s 1,000 vehicles are 19\frac{1}{9} of last year’s 9,000. So take 19\frac{1}{9} of the 2,250 two-door cars: 2,250÷9=2502{,}250 \div 9 = 250. Both ways give 250 cars.

More questions like this

The original DOE question
Question 58 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 58. Open page 191 of the handbook (PDF)

Question 70 of 108

2023 Test B Math, question 10

Question

A total of 24 students in a math class took a 100-point test:

  • None of the students scored 100 points.
  • 75% of the students scored less than 82 points.
  • The lowest score in the class was 54 points.

Which box plot best represents the test scores from this math class?

  • ABox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from 60 to about 72 with the median line at about 63, and the right whisker ends at about 82.5, halfway between 80 and 85.
  • BBox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from about 62.5 to about 82.5 with the median line at about 71, and the right whisker ends at about 98.
  • CBox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 51, the box runs from 60 to about 83 with the median line at about 74, and the right whisker ends at about 99.
  • DBox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from about 63 to about 83.5 with the median line at 70, and the right whisker ends at 100.
Show the answer

Correct answer: B

Use the information in the bullet points to reason which box plot could represent the data set of test scores.

Since none of the students scored 100, the maximum value on the box plot must be less than 100.

If 75% of the students scored less than 82, then 25% scored 82 points or more. Since the third quartile (Q3) indicates 34\frac{3}{4} (75%) of the data is to the left and 14\frac{1}{4} (25%) of the data is to the right, it is reasonable that Q3 would be at or about 82.

Given that the lowest score in the class was 54, the minimum point on the box plot must be 54.

The box plot graph in option B is the only one that meets all three criteria.

Question analysis

Twenty-four students took a 100-point test: nobody scored 100, 75% scored below 82, and the lowest score was 54. Students choose the box plot that fits all three facts. The skill is linking each statement to a part of the plot, the whiskers, the box and the quartiles, and checking every plot closely.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students usually read values off a box plot. This item runs the other way, turning a percent statement into a quartile and checking the whiskers against two more facts, but the content stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 2, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns three sentences into three features of a box plot, and a plot that matches two of them is easy to accept. The upper-quartile fact is the conceptual one: students who do not know that the box’s right edge marks 75 percent put 82 at the end of the whisker. The other two facts are quick visual checks, but only if every plot is actually checked.

  1. Every choice must be checked moderate Choosing the answer

    Each wrong plot fails a different fact, so all four must be tested against all three bullets: the left whisker against 54, the right whisker against 100, and the box’s right edge against 82. (Embretson & Wetzel, 1987)

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

    The axis is labeled every 5 points, and the deciding features fall between labels: one left whisker starts near 51 instead of 54, and box edges sit near 82.5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Something you must already know minor Working it out

    Nothing explains that “75% of the students scored less than 82” locates the upper quartile, the right edge of the box, rather than the highest score. (Anderson & Krathwohl, 2001)

Strategy: Turn each fact into a feature to check: where the left whisker starts, where the right whisker ends, and where the box ends. Rule out any plot that breaks one.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a box plot
minimum, Q1, median, Q3, maximum\text{minimum},\ Q_1,\ \text{median},\ Q_3,\ \text{maximum}
A box plot shows five numbers. The left whisker starts at the minimum, the lowest value. The box starts at Q1, the first quartile, and ends at Q3, the third quartile. The line inside is the median, the middle value. The right whisker ends at the maximum. About a quarter of the data falls in each of the four sections.

  1. Sort out the question

    24 students took a test. You know three facts. Nobody scored 100. 75% scored less than 82. The lowest score was 54. You need the box plot that fits all three facts.

  2. Turn each fact into a feature

    Match each fact to a part of the box plot.

    • The lowest score is 54, so the left whisker starts at 54.
    • Nobody scored 100, so the right whisker ends before 100.
    • 75% is three quarters. Three quarters of the scores are below Q3. So the box should end at about 82.

  3. Read the number line

    The ticks are 5 points apart: 50, 55, 60 and so on. A score of 54 sits just left of the 55 tick. A score of 82 sits a little less than halfway from 80 to 85. Look closely, because these points fall between labels.

  4. Check plot A

    In plot A, the right whisker ends near 82, and the box ends near 72. That would make 82 the highest score. Watch out: 75% below 82 does not make 82 the top score. It tells you where the box ends. So A is out.

  5. Check plot C

    In plot C, the left whisker starts near 51, just right of the 50 tick. But the lowest score was 54. So C is out.

  6. Check plot D

    Plot D starts at 54, and its box ends near 82. But its right whisker reaches 100. That means someone scored 100, and nobody did. Watch out: D fits two facts, so it is easy to stop early. Check all three. So D is out.

  7. Check plot B and answer

    Plot B starts at 54. Its box ends near 82. Its right whisker stops near 98, below 100. It fits all three facts. So the answer is B.

Check your answer

Go over B once more with all three facts. Its lowest point is 54. Its box ends near 82. Its highest point is under 100. Each other plot breaks one fact: A’s box, C’s left whisker and D’s right whisker.

More questions like this

The original DOE question
Question 67 as printed on page 194 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 67. Open page 194 of the handbook (PDF)

Question 71 of 108

2023 Test B Math, question 18

Question

A rectangle whose bottom side is labeled 16 ft and whose left side is labeled 8 ft. It is shaded gray except for a white triangle inside it: the triangle’s vertex is at the middle of the left side, and its base lies along the right side and is labeled 4 ft.

The diagram above shows a rectangle with an inscribed triangle. The triangle’s vertex and base touch the sides of the rectangle as shown. What is the total area of the shaded portion?

  • A32 sq ft
  • B64 sq ft
  • C96 sq ft
  • D128 sq ft
Show the answer

Correct answer: C

First, find the area of the rectangle, in square feet:

16×8=12816 \times 8 = 128

Then, find the area of the triangle:

12×4×16=32\frac{1}{2} \times 4 \times 16 = 32

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle:

128−32=96 sq ft128 - 32 = 96\text{ sq ft}

Question analysis

A 16-foot by 8-foot rectangle has a white triangle inside it, with one vertex at the middle of the left side and a 4-foot base along the right side. Students find the total area of the shaded part. They subtract the triangle's area, whose height is not labeled and must be read from the figure.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Composite areas built from rectangles and triangles appear in grades 6 and 7. This one is typical, but the triangle lies on its side, so students must see which length is its height rather than read it from a label.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The shaded region is a rectangle with a triangle cut out, so the plan is standard, but the triangle is turned so that its height runs across the figure. Students who see that the height is the 16-foot side have two short areas to find and one subtraction. The listed wrong answers stop at either area or leave out the one half.

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

    The triangle lies on its side: its 4-foot base runs along the right edge, so its height is the rectangle’s full 16-foot length, which the figure labels only as the bottom side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    The rectangle’s area, the triangle’s area and their difference are three separate results, and the first two, 128 and 32, are themselves listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the rectangle's area, find the triangle's area from its base on the right side and its height across the rectangle, then subtract.

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.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, the space inside a flat shape, in square units. l is the length and w is the width. It works because the rectangle is filled by w rows of unit squares, with l squares in each row.
Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side you measure from. h is the height, the distance from the base to the opposite corner, at a right angle to the base. You take half, because a triangle is half of a rectangle with the same base and height.

  1. Sort out the figure

    The figure is a rectangle 16 feet long and 8 feet tall. Inside it is a white triangle. Its tip touches the middle of the left side. Its base lies along the right side and is 4 feet long. The shaded part is the gray area around the triangle. You need its area.

    Plan: find the rectangle’s area, then take away the triangle’s area.

  2. Find the area of the rectangle

    A=l×wA = l \times w

    A=16×8=128A = 16 \times 8 = 128

    The rectangle’s area is 128 square feet. Watch out: this includes the white triangle. Stopping here gives choice D.

  3. Find the triangle’s height

    The triangle lies on its side. Its base is the 4-foot piece on the right side, which runs up and down. The height must meet the base at a right angle, a square corner. So the height runs straight across, from the base to the tip on the left side. That is the full length of the rectangle, 16 feet.

  4. Find the area of the triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×4×16=32A = \frac{1}{2} \times 4 \times 16 = 32

    The triangle’s area is 32 square feet. Watch out: don’t forget the 12\frac{1}{2}. Using 4×16=644 \times 16 = 64 leaves 64 for the shaded part, choice B.

  5. Subtract to find the shaded area

    128−32=96128 - 32 = 96

    The shaded part is 96 square feet. Watch out: 32 is the white triangle, not the shaded part, so choice A is wrong. So the answer is C, 96 sq ft.

Check your answer

The triangle’s base, 4 feet, is half the rectangle’s height of 8 feet. Its height is the full 16 feet. So the triangle is half of half the rectangle, or one fourth of it. One fourth of 128 is 32, and 128−32=96128 - 32 = 96.

More questions like this

The original DOE question
Question 75 as printed on page 196 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 75. Open page 196 of the handbook (PDF)

Question 72 of 108

2023 Test B Math, question 22

Question

A vendor is selling 10 packages of homemade fudge at a bake sale. The vendor weighed and labeled each package. The package weights are given in ounces:

4.6 6.0 3.8 3.5 4.5
5.4 3.9 6.5 4.8 6.2

The vendor created a box plot of the package weights. Which box plot best represents the data?

  • ABox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at about 4.1, the box runs from 5 to 6 with the median line at 5.5, and the right whisker ends at about 6.3.
  • BBox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 4.5, and the right whisker ends at 6.5.
  • CBox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from about 3.9, just left of 4, to 6, with the median line halfway between 4.5 and 5, and the right whisker ends at 6.5.
  • DBox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 5, and the right whisker ends at 6.5.
Show the answer

Correct answer: C

A set of 10 data points is given. The first step for determining the quartiles and minimum and maximum values of the box plot is to order the data from smallest to greatest value. For this data set the order is:

3.5 3.8 3.9 4.5 4.6
4.8 5.4 6.0 6.2 6.5

The minimum value is the smallest decimal, 3.5, and the maximum value is the greatest decimal, 6.5.

To determine the median in a set with an even number of values, locate the two middle values (5th and 6th in this case): 4.6 and 4.8. Determine the mean, or average, of the two middle numbers by adding them together and dividing the sum by 2:

4.6+4.82=9.42=4.7\frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7

To determine the value of the first quartile (Q1), locate the middle value of the 5 terms to the left of the median:

3.5 3.8 3.9 4.5 4.6

To determine the value of the third quartile (Q3), locate the middle value of the 5 terms to the right of the median:

4.8 5.4 6.0 6.2 6.5

Box plot C identifies the correct minimum, maximum, median, and quartiles of the data set.

Question analysis

A vendor weighs 10 packages of homemade fudge, from 3.5 to 6.5 ounces, and makes a box plot of the weights. Students choose the box plot that matches the data. They must sort the decimals, find the median of an even number of values and both quartiles, then compare plots that differ only slightly.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Building a box plot from data is grade 7 content. This version needs all five summary values from ten unsorted decimals, including a median between two values, then a close comparison of plots, longer and finer than typical.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 4, 43 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

Most of the work happens before the plots are even read: the weights must be sorted, and the median of ten values is the average of two of them. After that, three plots look almost identical, so a careless median, taken from the unsorted list or placed halfway between the extremes, settles on a wrong one. The careful student checks the median line and the box’s left edge against values that fall between the tick marks.

  1. Many steps moderate Working it out

    Ten unsorted decimals must be ordered, then five values found: the median as the mean of the fifth and sixth, 4.6 and 4.8, and each quartile as the middle of its half. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    Three plots share whiskers at 3.5 and 6.5 and a box ending at 6; they differ only in the median line, at 4.5, near 4.75 or at 5, and in where the box starts. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The axis is ticked every half ounce, and the correct plot’s median, 4.7, and lower quartile, 3.9, both fall between ticks, so each line’s position must be judged by eye. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. Slow to do carefully minor Choosing the answer

    Sorting ten decimals, computing a five-number summary and comparing four plots takes well over a minute and a half when done carefully. (Lu & Sireci, 2007)

Strategy: Sort the ten weights, then find the least and greatest, the median as the mean of the fifth and sixth values, and each quartile as the middle of its half.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Median of an even number of values
median=first middle value+second middle value2\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}
First put the values in order. With an even number of values, two sit in the middle. The median is their mean, or average: add them and divide by 2. The median splits the data in half.
Quartiles
Q1=middle of the lower halfQ3=middle of the upper halfQ_1 = \text{middle of the lower half} \qquad Q_3 = \text{middle of the upper half}
Q1 is the first quartile and Q3 is the third quartile. Split the ordered data in half at the median. Q1 is the middle of the lower half, and Q3 is the middle of the upper half. A box plot’s box runs from Q1 to Q3.

  1. Sort out the question

    The vendor weighed 10 packages of fudge, in ounces. You need the box plot that matches the weights. A box plot is built from five numbers. They are the minimum (least), the first quartile Q1, the median, the third quartile Q3 and the maximum (greatest). So find those five numbers first.

  2. Put the weights in order

    List the 10 weights from least to greatest.

    3.53.83.94.54.6
    4.85.46.06.26.5

    Watch out: you must sort first. The fifth weight as printed is 4.5. Using it as the median points to choice B.

  3. Find the least and greatest

    The minimum is 3.5 ounces and the maximum is 6.5 ounces. So the whiskers must run from 3.5 to 6.5. Plot A’s whiskers run from about 4.1 to about 6.3, so A is out.

  4. Find the median

    With 10 weights, the two middle ones are the 5th and 6th: 4.6 and 4.8.

    median=first middle value+second middle value2\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}

    median=4.6+4.82=9.42=4.7\text{median} = \frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7

    The median is 4.7 ounces. Watch out: the median is not halfway between the least and greatest. That gives (3.5+6.5)÷2=5(3.5 + 6.5) \div 2 = 5 and points to choice D.

  5. Find the two quartiles

    The lower half is the 5 weights below the median: 3.5, 3.8, 3.9, 4.5 and 4.6. Its middle value is Q1=3.9Q_1 = 3.9.

    The upper half is the 5 weights above the median: 4.8, 5.4, 6.0, 6.2 and 6.5. Its middle value is Q3=6.0Q_3 = 6.0.

  6. Match the summary to a plot

    The five numbers are 3.5, 3.9, 4.7, 6.0 and 6.5. Plots B, C and D all run from 3.5 to 6.5, and their boxes end at 6. They differ in the median line and where the box starts.

    • B: the box starts at 4, and the median line is at 4.5.
    • D: the box starts at 4, and the median line is at 5.
    • C: the box starts just left of 4, and the median line is between 4.5 and 5.

    C is the closest match to a Q1 of 3.9 and a median of 4.7. So the answer is C.

Check your answer

A median splits the data in half. Five weights, 3.5 to 4.6, are below 4.7. Five, 4.8 to 6.5, are above it. A median line at 5, as in D, would have six weights below it. A line at 4.5, as in B, would have only three below it.

More questions like this

The original DOE question
Question 79 as printed on page 198 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 198, question 79. Open page 198 of the handbook (PDF)

Question 73 of 108

2023 Test B Math, question 23

Question

FRUIT SOLD
AT STORE XYZ

Fruit Number Sold
Apples 8
Bananas 6
Cherries 12
Oranges 6
Peaches ?
Total 50

Peaches made up what percentage of the total number of pieces of fruit sold at Store XYZ, according to the table above?

  • A18%
  • B28%
  • C36%
  • D64%
Show the answer

Correct answer: C

First, find the number of peaches sold:

50−(8+6+12+6)=50 - (8 + 6 + 12 + 6) =

50−32=1850 - 32 = 18

Then, find the percentage:

1850=0.36=36%\frac{18}{50} = 0.36 = 36\%

Question analysis

A table lists the fruit sold at Store XYZ, apples, bananas, cherries and oranges, with the number of peaches shown as a question mark and a total of 50. Students find what percentage of the fruit sold were peaches. They first work out the missing count, then turn it into a percent of the total.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a missing value and writing a part as a percent of a whole are grade 6 skills. With a total of 50, the percent step is quick, so the task is lighter than typical grade 7 percent problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 34 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The total of 50 makes the percent step easy, since each piece of fruit is 2 percent, but the question first hides the count it needs. Students who find the 18 peaches and stop, or who give the share of the four listed fruits, reach listed choices. Checking that the peaches’ percent and the other fruits’ percent add to 100 confirms the answer.

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

    The table gives four counts, a question mark for peaches and a total of 50, so the missing count must be found by subtraction before any percent. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Adding 8, 6, 12 and 6, subtracting from 50, and turning 18 out of 50 into a percent are three steps, and the count of 18 is itself a listed percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the four known counts, subtract from 50 to get the peaches, then double that count, since each piece out of 50 is 2 percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a total
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the amount you are asking about, here the peaches. The whole is the total, here all 50 pieces of fruit. Dividing gives the fraction of the whole. Multiplying by 100 turns it into a percent, which means out of 100.

  1. Sort out the question

    The table shows how many of each fruit Store XYZ sold. The number of peaches is missing, shown by a question mark. The total is 50 pieces of fruit. You need the percent of the total that were peaches.

  2. Add the fruits you know

    8+6+12+6=328 + 6 + 12 + 6 = 32

    Apples, bananas, cherries and oranges add up to 32 pieces of fruit. That is every fruit except the peaches.

  3. Find the number of peaches

    The peaches are whatever is left of the 50.

    50−32=1850 - 32 = 18

    So 18 peaches were sold. Watch out: 18 is a count, not a percent. Choosing 18% is choice A.

  4. Turn the count into a percent

    percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100

    1850×100=0.36×100=36\frac{18}{50} \times 100 = 0.36 \times 100 = 36

    Peaches were 36% of the fruit sold. Watch out: 32 out of 50 is 64%, choice D. That is the other four fruits, not the peaches. So the answer is C, 36%.

Check your answer

Out of 50 pieces, each piece is 2% of the total, because 100÷50=2100 \div 50 = 2. So 18 peaches are 18×2=3618 \times 2 = 36 percent. Also, 36% for peaches plus 64% for the rest makes 100%.

More questions like this

The original DOE question
Question 80 as printed on page 198 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 198, question 80. Open page 198 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 74 of 108

2023 Test B Math, question 26

Question

The box plot below represents a set of data.

Box plot above a number line with arrows at both ends, a tick every 3 units from 0 to 30, and labels at 0, 6, 12, 18, 24 and 30. The left whisker starts at 3, the box runs from 6 to the unlabeled tick for 27 with the median line at 15, and the right whisker ends at 30.

What is the interquartile range of the set of data?

  • A9
  • B12
  • C21
  • D27
Show the answer

Correct answer: C

First, determine the values of the upper and lower quartiles of the set of data:

The same box plot, with an arrow labeled lower quartile pointing down to the left edge of the box, at 6, and an arrow labeled upper quartile pointing down to the right edge of the box, at the unlabeled tick for 27.

  • The upper quartile value is indicated by the vertical line segment at the far right edge of the box plot above the hash mark representing 27.
  • The lower quartile value is indicated by the vertical line segment at the far left edge of the box plot above the hash mark representing 6.

To determine the interquartile range, subtract the lower quartile value (6) from the upper quartile value (27):

27−6=2127 - 6 = 21

Question analysis

A box plot on a number line marked every 3 units, with labels every 6, represents a set of data. Students find the interquartile range. They have to read the two ends of the box, one of which falls on an unlabeled tick, and remember exactly which part of the plot the interquartile range measures.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the interquartile range from a box plot is a grade 7 skill, and this is a typical example. Read two values and subtract; the only extra care is a box edge that falls on a tick with no label.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 19 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

Once the two edges of the box are read, one subtraction finishes the question, so the work is reading the scale and knowing which part of the plot is meant. The upper quartile falls on an unlabeled tick, which costs care. The wrong answers measure the whole spread or only half of the box, all common mix-ups of what the interquartile range covers.

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

    The axis is labeled only every 6 units but ticked every 3, and the box’s right edge sits on the unlabeled tick for 27, so it has to be counted from 24. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Nothing defines the interquartile range; it is the width of the box, the upper quartile minus the lower, not the span of the whiskers. (Anderson & Krathwohl, 2001)

Strategy: Find the values at the left and right edges of the box, counting tick marks in steps of 3 where there is no label, and subtract the smaller from the larger.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
IQR stands for interquartile range. Q3, the third quartile, is the right edge of a box plot’s box. Q1, the first quartile, is the left edge. Subtracting gives the width of the box, where the middle half of the data lies.

  1. Sort out the question

    You see a box plot and need the interquartile range, or IQR. That is the width of the box: the upper quartile minus the lower quartile. The whiskers don’t matter here.

  2. Read the number line

    The labels are 0, 6, 12, 18, 24 and 30. But there is a tick mark halfway between each pair of labels. So the ticks go up by 3: 0, 3, 6, 9 and so on. The tick between 24 and 30 stands for 27.

  3. Find the edges of the box

    The left edge of the box is at 6. That is the lower quartile, Q1. The right edge is at the unlabeled tick after 24, which is 27. That is the upper quartile, Q3.

  4. Subtract to find the IQR

    IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

    IQR=27−6=21\text{IQR} = 27 - 6 = 21

    Watch out: 30−3=2730 - 3 = 27 uses the ends of the whiskers. That is the range, choice D. Also, the median line at 15 splits the box. 15−6=915 - 6 = 9 and 27−15=1227 - 15 = 12 are each only part of the box. So the answer is C, 21.

Check your answer

Count the tick spaces across the box, from 6 to 27. There are 7 spaces, and each is worth 3. So the box is 7×3=217 \times 3 = 21 units wide. That matches.

More questions like this

The original DOE question
Question 83 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 83. Open page 199 of the handbook (PDF)

Question 75 of 108

2023 Test B Math, question 27

Question

N={6,7,…,21,22}\mathrm{N} = \{6, 7, \ldots, 21, 22\}

Set L (not shown) is a list of 25 consecutive integers. The median of the integers in set N above is equal to the least integer in set L. What is the median of the integers in set L?

  • A25
  • B26
  • C27
  • D28
Show the answer

Correct answer: B

First, find the median of set N. There are 17 integers in set N, so the 9th integer is the median. The 9th integer is 14.

The least integer in set L is equal to the median of set N, so the least integer in set L is 14.

There are 25 integers in set L, so the median is the 13th integer. That integer is 26.

Question analysis

Set N lists the consecutive integers from 6 to 22, and set L is a list of 25 consecutive integers whose least member equals the median of N. Students find the median of set L. It chains two median questions together, and counting positions in a run of consecutive integers invites off-by-one slips.

NYS standards
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Medians of short lists are grade 6 content. This item hides both lists, one behind set notation with an ellipsis and one not shown at all, and links them, so it needs more planning and careful counting 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 5, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains two medians, and both depend on counting members correctly in runs of consecutive integers. The easy slip is between position and distance: the 13th member of L is 12 steps above its first, not 13. With four consecutive choices, every counting error has a home, so writing out the first and last members of L, 14 and 38, and averaging them is the safest check.

  1. A very close wrong answer moderate Choosing the answer

    The choices 25, 26, 27 and 28 are consecutive, so a count that is off by one position lands on a neighboring choice with nothing to flag it. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. No obvious method moderate Working it out

    Set L is never listed; its median has to be found from its least member, 14, and its size, 25, by working out that the middle is the 13th member. (Schoenfeld, 1985)

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

    Set N is written with an ellipsis, {6, 7, …, 21, 22}, so its 17 members must be counted, and 22 − 6 = 16 undercounts by one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find N's median by averaging its least and greatest members. L's median is its middle member, so work out that member's position and count up from L's least member.

Traps

  • 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).
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting the integers in a list
count=last−first+1\text{count} = \text{last} - \text{first} + 1
This counts the whole numbers in a list that goes up by 1, like 6, 7, 8. Subtracting gives the number of steps from the first number to the last. You add 1 because the first number counts too.
Position of the median
position=count+12\text{position} = \frac{\text{count} + 1}{2}
The count is how many numbers are in a list that is in order. When the count is odd, the median is at this position. It has the same number of values before it and after it.

  1. Sort out the question

    Set N is 6, 7, and so on up to 22. The dots stand for every integer in between. Integers are whole numbers. Set L has 25 consecutive integers, one right after another. Its least number equals the median of N. You need the median of L.

  2. Count the numbers in set N

    count=last−first+1\text{count} = \text{last} - \text{first} + 1

    22−6+1=1722 - 6 + 1 = 17

    Set N has 17 integers. Watch out: 22−6=1622 - 6 = 16 misses one, because the first number counts too.

  3. Find the median of set N

    position=count+12\text{position} = \frac{\text{count} + 1}{2}

    position=17+12=9\text{position} = \frac{17 + 1}{2} = 9

    The median is the 9th number. 6 is the 1st, so the 9th is 8 steps up: 6 + 8 = 14. The median of N is 14.

  4. Find the middle place in set L

    So set L starts at 14 and has 25 integers. Find the position of its median.

    position=25+12=13\text{position} = \frac{25 + 1}{2} = 13

    The median of L is its 13th number.

  5. Count up to the 13th number

    14 is the 1st number. The 13th is 12 steps after it.

    14+12=2614 + 12 = 26

    Watch out: the 13th number is 12 steps up, not 13. Adding 13 gives 27, choice C. And 25 is how many numbers L has, not its median. So the answer is B, 26.

Check your answer

Set L runs from 14 to 14+24=3814 + 24 = 38. For consecutive integers, the median is halfway between the ends: (14+38)÷2=26(14 + 38) \div 2 = 26. That matches. The same trick gives N’s median: (6+22)÷2=14(6 + 22) \div 2 = 14.

More questions like this

The original DOE question
Question 84 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 84. Open page 199 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 76 of 108

2023 Test B Math, question 30

Question

PRICE OF A USED VIDEO GAME AT VARIOUS STORES

Price of Used Video Game Number of Stores
$10.99 5
$11.99 2
$12.99 1
$13.99 4

The table above shows the price of a certain used video game at 12 different stores. What is the median price of this video game?

  • A$10.99
  • B$11.49
  • C$11.99
  • D$12.32
Show the answer

Correct answer: C

There are 5 stores where the video game is priced at $10.99 and a total of 5 stores where it is priced at either $12.99 or $13.99, so the median price must be $11.99.

Question analysis

A statistics question with a table showing the price of a used video game at 12 stores, grouped by price, and asking for the median price. Students put the prices in order using the number of stores at each price and find the middle. The table gives counts rather than a list, so the position of the middle value has to be worked out from the running totals.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a median is grade 6 work that grade 7 students meet often. The twist is the frequency table: the 12 prices are not listed one by one, so students count positions instead of reading down a list.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is trivial once the positions are found: counting down 5 stores at $10.99 and then 2 at $11.99 puts both the 6th and 7th prices at $11.99. The wrong choices reward three different misreadings of a frequency table: the most common price, a middle pair one place too early, and the mean, which takes a long computation to reach. A student who writes the running totals 5, 7, 8 and 12 beside the table avoids all three.

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

    The 12 prices are never listed; the table gives four prices with store counts of 5, 2, 1 and 4, so the 6th and 7th prices must be found by a running count. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    With 12 values the median is the mean of the 6th and 7th in order, a rule the question assumes; here both fall in the $11.99 row, so no averaging is needed. (Anderson & Krathwohl, 2001)

Strategy: Keep a running total of the store counts down the table; with 12 prices in order, the median is the average of the 6th and 7th, so find the rows where they fall.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Median of an even number of values
median=first middle value+second middle value2\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}
The median is the middle value when all the values are listed from least to greatest. With an even number of values, no single value sits in the middle. So take the two middle values and find their average: add them and divide by 2.

  1. Sort out the table

    Each row gives a price and how many stores charge it. For example, 5 stores charge $10.99. So $10.99 shows up 5 times in the list of prices. Add the stores to check the count.

    5+2+1+4=125 + 2 + 1 + 4 = 12

    There are 12 prices. You need the median. That is the middle price when the prices are in order from least to greatest.

  2. Find the middle places

    12 is an even number, so no single price sits in the middle. The two middle prices are the 6th and the 7th. Five prices come before the 6th one. Five prices come after the 7th one.

  3. Count down the table

    The table already lists the prices from least to greatest. Keep a running total of the stores. It shows which places in the list each price fills.

    PriceStoresPlaces in the list
    $10.9951st to 5th
    $11.9926th and 7th
    $12.9918th
    $13.9949th to 12th

    The 6th and 7th prices are both $11.99.

  4. Find the median

    Use the formula with the 6th and 7th prices.

    median=11.99+11.992=23.982=11.99\text{median} = \frac{11.99 + 11.99}{2} = \frac{23.98}{2} = 11.99

    The median price is $11.99.

    Watch out: $10.99 is the mode, the price that shows up most often. That is choice A, not the median. Averaging the 5th and 6th prices gives $11.49, choice B. Adding all 12 prices and dividing by 12 gives the mean, about $12.32. That is choice D.

    So the answer is C, $11.99.

Check your answer

Count the prices on each side of $11.99. Five stores charge less: the five at $10.99. Five stores charge more: one at $12.99 and four at $13.99. The two sides match, so $11.99 is in the middle.

More questions like this

The original DOE question
Question 87 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 87. Open page 200 of the handbook (PDF)

Question 77 of 108

2023 Test B Math, question 32

Question

A number line with arrows at both ends and three points marked on it. From left to right, point P is at 1 1/3, point Q is at 3, and point R is at 5 1/6.

On the number line above, how many units longer is QR\overline{QR} than PQ\overline{PQ}?

  • A16\frac{1}{6}
  • B12\frac{1}{2}
  • C23\frac{2}{3}
  • D3
Show the answer

Correct answer: B

First, find the lengths of the given segments:

QR=516−3=216\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}

PQ=3−113=123\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}

Then, find the difference:

216−123=2\frac{1}{6}-1\frac{2}{3}=

136−53=\frac{13}{6}-\frac{5}{3}=

136−106=\frac{13}{6}-\frac{10}{6}=

36=12\frac{3}{6}=\frac{1}{2}

Question analysis

A number line shows three points labeled with mixed numbers and a whole number, and students find how many units longer one segment is than another. It tests subtracting mixed numbers with unlike denominators to find distances. Regrouping when the fraction being subtracted is the larger one is where most errors happen.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding distances between points on a number line is grade 7 work. Here the points are mixed numbers with different denominators, so finding two distances and their difference makes it a typical but careful grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is three subtractions, so most errors are in the fractions rather than the plan. Taking 1 1/3 from a whole number such as 3 tempts students to subtract only the whole parts and keep the 1/3, which reverses which segment looks longer. Rewriting every coordinate in sixths, 8/6, 18/6 and 31/6, turns the work into whole-number subtraction of numerators.

  1. Error-prone arithmetic moderate Working it out

    Both lengths subtract mixed numbers with different denominators, 5 1/6 − 3 and 3 − 1 1/3, and the second needs regrouping before the two results are compared in sixths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The three coordinates come from the labels under P, Q and R, and the segments QR and PQ must each be measured from Q at 3, the point they share. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find each length by subtracting the left endpoint from the right one, writing the mixed numbers in sixths, then subtract the shorter length from the longer.

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.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Length on a number line
length=right point−left point\text{length} = \text{right point} - \text{left point}
The length of a segment is the distance between its two end points. Subtract the number on the left from the number on the right. Subtracting tells how many units you move to get from one point to the other.
Mixed number to improper fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part. b over c is the fraction part, with b on top and c on the bottom. Each whole holds c parts, so a wholes hold a × c parts. Then add the b extra parts. An improper fraction has a top bigger than its bottom.

  1. Read the number line

    Point P is at 1131\frac{1}{3}. Point Q is at 3. Point R is at 5165\frac{1}{6}.

    The bar over QR means segment QR, the piece of the line from Q to R. You need how much longer QR is than PQ. So find both lengths, then subtract.

  2. Find the length of QR

    R is on the right, and Q is on the left. Use the length rule.

    QR=516−3=216\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}

    QR is 2162\frac{1}{6} units long.

  3. Find the length of PQ

    Q is on the right, and P is on the left.

    PQ=3−113\text{PQ} = 3 - 1\frac{1}{3}

    Take it in two parts. First, 3 − 1 = 2. Then take away the extra 13\frac{1}{3}. That gives 2−13=1232 - \frac{1}{3} = 1\frac{2}{3}. So PQ is 1231\frac{2}{3} units long.

    Watch out: Don’t just do 3 − 1 and keep the 13\frac{1}{3}. That gives 2132\frac{1}{3}, which is too long. It leads to choice A.

  4. Write both lengths in sixths

    The fractions have different bottoms, 6 and 3. Change both lengths to sixths so you can subtract. Use the mixed number rule.

    216=2×6+16=1362\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}

    123=1×3+23=53=1061\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}

  5. Subtract the lengths

    Take the shorter length away from the longer one.

    136−106=36=12\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}

    QR is 12\frac{1}{2} unit longer than PQ.

    So the answer is B, 12\frac{1}{2}.

Check your answer

Write every point in sixths. P is 86\frac{8}{6}, Q is 186\frac{18}{6} and R is 316\frac{31}{6}. QR is 31 − 18 = 13 sixths. PQ is 18 − 8 = 10 sixths. The difference is 3 sixths, which is one half.

More questions like this

The original DOE question
Question 89 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 89. Open page 200 of the handbook (PDF)

Question 78 of 108

2023 Test B Math, question 47

Question

Four slips of paper numbered 1, 2, 3, and 4 were placed in a bag:

  • One slip was selected from the bag, and the number recorded.
  • Without replacement, a second slip was chosen.
  • This process continued until no more slips remained in the bag.
  • One part of a tree diagram representing the sample space is shown.

A partial tree diagram. The top number, 1, branches down to 2, 3, and 4. Below 2 are 3 and 4, below 3 are 2 and 4, and below 4 are 2 and 3. The 4 under the 2 has one branch leading down to an empty box.

Which number should be placed in the box to complete this part of the tree diagram?

  • A1
  • B2
  • C3
  • D4
Show the answer

Correct answer: C

The “without replacement” statement means that the bag contained four slips for the first draw, three slips for the second draw, two slips for the third draw, and one slip for the last draw. Following the branch from the top and navigating down to the box, the 1 was chosen first, the 2 was chosen second, and the 4 was chosen third. The only remaining number is 3. The element would be 1243.

Question analysis

Four slips numbered 1 to 4 are drawn from a bag one at a time without replacement, and part of a tree diagram of the outcomes is shown. Students decide which number belongs in the empty box at the end of one branch. It tests reading a tree diagram and understanding what drawing without replacement means.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Tree diagrams for compound events are a grade 7 topic, but here the tree is already drawn and only one box is empty. Following a single branch to that box is shorter than typical grade 7 work with sample spaces.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 68 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

Only one branch matters, and once it is read as the slips drawn so far, 1, 2 and 4, the last slip left in the bag is the only one missing. Errors come from not counting the top 1 as a draw, from forgetting that a drawn slip stays out, or from copying a nearby number in the diagram instead of following the path.

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

    The box hangs below the branch 1, then 2, then 4, and students must read that path from the top as the order of the first three draws. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Follow the branch from the top down to the box, list the slips already drawn on that path, and name the one still left in the bag.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

  1. Sort out the question

    The bag holds four slips: 1, 2, 3 and 4. Slips are drawn one at a time. “Without replacement” means a drawn slip does not go back in the bag. So each number can be drawn only once.

  2. Read the tree diagram

    A tree diagram shows the orders the draws can happen in. Each row going down is one draw. The 1 at the top is the first slip drawn. The numbers below it are the choices for the second draw, and so on.

  3. Follow the path to the box

    Start at the top and go down the lines to the box.

    • First draw: 1
    • Second draw: 2
    • Third draw: 4
    • Fourth draw: the box

    Slips 1, 2 and 4 are already out of the bag.

  4. Find the slip that is left

    The bag started with 1, 2, 3 and 4. Only slip 3 has not been drawn. So it must be the fourth draw.

    Watch out: The 1 at the top counts as a draw, so 1 is not left. That trap is choice A. Slip 2 was already drawn, so choice B can’t be right. And 4 is just the number above the box, choice D.

    So the answer is C, 3.

Check your answer

Read the whole path: 1, 2, 4, 3. Each number from 1 to 4 shows up exactly once. That is just what should happen when no slip goes back in the bag.

More questions like this

The original DOE question
Question 104 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 104. Open page 204 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 79 of 108

2023 Test B Math, question 52

Question

An equilateral triangle drawn with dashed sides. On each side a semicircle bulges outward, with that side as its diameter. The three curved arcs together form the outline of the garden.

The garden design shown above is created in the shape of three semicircles whose diameters lie on the sides of an equilateral triangle. If each diameter is 2 meters long, what is the perimeter of the garden, in meters?

  • A6
  • B12
  • C3π3\pi
  • D6π6\pi
Show the answer

Correct answer: C

The perimeter (or circumference) of one of the three semicircles is 12(diameter)π=12(2π)=π\frac{1}{2}(\text{diameter})\pi=\frac{1}{2}(2\pi)=\pi. Since there are 3 semicircles that make up the perimeter, the perimeter is 3π3\pi.

Question analysis

A garden is outlined by three semicircles whose diameters, each 2 meters long, are the sides of an equilateral triangle. Students find the perimeter of the garden in meters. It tests the circumference of a circle applied to half circles, and the figure mixes straight dashed sides with curved ones, so deciding what counts as the outline matters.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying the circumference formula is a grade 7 standard. Here it is used for a semicircle and repeated three times, a short extension of routine classwork, with a dashed triangle inside the figure to look past.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 8, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The perimeter is just three half-circles, and each is half of π × 2. The figure’s dashed triangle invites measuring straight sides instead, and remembering the whole circle’s circumference without halving it gives a choice twice too large. Tracing the garden’s outer edge with a finger before computing shows that only the curves count.

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

    The dashed triangle lies inside the garden, so its three 2-meter sides are not part of the perimeter; only the three curved arcs bound the garden. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The circumference formula is not given, and each arc is half of a circle with a 2-meter diameter, so students must take half of π × 2. (Anderson & Krathwohl, 2001)

Strategy: Find the circumference of a full circle with a 2-meter diameter, take half of it for one semicircle, and multiply by the number of semicircles in the outline.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle
C=π×dC = \pi \times d
C is the circumference, the distance around a circle. d is the diameter, the straight line across the circle through its center. π is a number a little more than 3.14. The distance around any circle is π times as long as the distance across it.
Arc of a semicircle
arc=12×π×d\text{arc} = \frac{1}{2} \times \pi \times d
A semicircle is half of a circle. Its arc is the curved part. d is the diameter, the straight side. The arc is half of the full circle’s circumference, so take half of π times d.

  1. Sort out the figure

    The garden’s edge is made of three semicircles, or half circles. Each one sits on a side of the dashed triangle. Each side is a diameter, 2 meters long. The perimeter is the distance around the outside. Here that is only the three arcs, the curved parts. The dashed triangle is inside, so it does not count.

  2. Find a full circle’s circumference

    Picture the whole circle that one semicircle comes from. Its diameter is 2 meters. Use the circumference formula.

    C=π×2=2πC = \pi \times 2 = 2\pi

    The full circle would measure 2π2\pi meters around.

  3. Find the length of one arc

    Each semicircle’s arc is half of that circle. Use the arc formula.

    arc=12×π×2=π\text{arc} = \frac{1}{2} \times \pi \times 2 = \pi

    One arc is π\pi meters long.

    Watch out: Using the full circle’s 2π2\pi for each arc doubles the answer. That gives 6π6\pi, choice D.

  4. Add the three arcs

    The garden’s edge is three equal arcs.

    π+π+π=3π\pi + \pi + \pi = 3\pi

    The perimeter is 3π3\pi meters.

    Watch out: The dashed sides add to 2 + 2 + 2 = 6 meters, choice A. That is the triangle, not the curved edge.

    So the answer is C, 3π3\pi meters.

Check your answer

Estimate. 3π3\pi is about 3 × 3.14 = 9.42 meters. A curved path is longer than the straight line across it. So the edge should be longer than the triangle’s 6 meters, and 9.42 is.

More questions like this

The original DOE question
Question 109 as printed on page 206 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 206, question 109. Open page 206 of the handbook (PDF)

Question 80 of 108

2023 Test B Math, question 56

Question

A coordinate grid with the origin O, the x-axis marked 1, 2, and 3, and the y-axis marked 1 through 5. Line t passes through the point (0, 1) and line s passes through the point (1, 0); they intersect at P (3, 5), and dashed lines run from P across to 5 on the y-axis and down to 3 on the x-axis. The shaded region is the four-sided figure with vertices O, (0, 1), P (3, 5), and (1, 0).

Lines s and t intersect at point P (3, 5). What is the area, in square units, of the shaded region in the graph above?

  • A2
  • B3
  • C4
  • D7
Show the answer

Correct answer: C

First, find the area of the rectangle in the graph:

3×5=153\times 5=15

Then, find the areas of the two right triangles that are not shaded:

Upper triangle:

12×4×3=6\frac{1}{2}\times 4\times 3=6

Lower triangle:

12×2×5=5\frac{1}{2}\times 2\times 5=5

The area of the shaded region, in square units, is calculated by subtracting the total area of the two right triangles from the area of the rectangle:

15−(6+5)=15−11=415-(6+5)=15-11=4

Question analysis

Two lines meet at the point (3, 5) on a coordinate grid, and the shaded region between them and the axes is a four-sided shape. Students find its area in square units. The shape is not a standard figure, so no single area formula applies to it directly.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Areas of triangles and rectangles are grade 6 content, and grade 7 students often find composite areas. This region is an irregular four-sided shape with two slanted sides, so students need a plan, and the lengths come from coordinates rather than labels.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 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

The question rewards a plan more than computation: the region has no name, so students must see it inside the dashed rectangle from O to P, with two right triangles cut away. Reading each triangle’s legs from the axes is the second hurdle, since the 1-unit intercepts shorten one leg of each, to 4 instead of 5 and to 2 instead of 3. Only one wrong choice, from using one coordinate as both heights, has a clear working behind it, so most errors here are guesses after a failed plan.

  1. No obvious method moderate Working it out

    The shaded shape has four sides, two of them slanted, and no formula of its own; it must be boxed in the 3-by-5 rectangle with two triangles removed, or cut along OP. (Schoenfeld, 1985)

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

    No lengths are labeled; every one comes from the coordinates of P (3, 5), the y-intercept 1 of line s and the x-intercept 1 of line t. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    The rectangle, two triangle areas and a subtraction make four computations, 15, 6 and 5, then 15 − 11, each built from different coordinates. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Draw the rectangle from O to P, find the areas of the two unshaded right triangles inside it from the coordinates, and subtract both from the rectangle's area.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, in square units. l is the length and w is the width. Multiplying them counts the unit squares that fill the rectangle.
Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, one side. h is the height: the distance from the base straight up to the far corner. A right triangle has a square corner, and the two sides that meet there are the base and height. A triangle is half of a rectangle with the same base and height.

  1. Sort out the shaded region

    The shaded region has four corners: O at (0, 0), (0, 1) on the y-axis, P at (3, 5), and (1, 0) on the x-axis. Line t meets the y-axis at (0, 1), and line s meets the x-axis at (1, 0). There is no easy formula for this shape. But the dashed lines make a rectangle around it. So find the rectangle’s area and take away the white parts.

  2. Find the rectangle’s area

    The rectangle has corners O, (3, 0), P and (0, 5). It is 3 units wide and 5 units tall.

    A=3×5=15A = 3 \times 5 = 15

    The rectangle’s area is 15 square units.

  3. Find the upper triangle’s area

    The white triangle at the top left has corners (0, 1), (0, 5) and P. Its square corner, or right angle, is at (0, 5). One side runs up the y-axis from 1 to 5, so it is 4 units. The other runs across from 0 to 3, so it is 3 units.

    A=12×4×3=6A = \frac{1}{2} \times 4 \times 3 = 6

    Its area is 6 square units.

  4. Find the lower triangle’s area

    The white triangle at the bottom right has corners (1, 0), (3, 0) and P. Its square corner is at (3, 0). One side runs along the x-axis from 1 to 3, so it is 2 units. The other goes up from 0 to 5, so it is 5 units.

    A=12×2×5=5A = \frac{1}{2} \times 2 \times 5 = 5

    Its area is 5 square units.

    Watch out: The lines cross the axes at 1, not at O. So these sides are 4 and 2, not 5 and 3.

  5. Subtract the white triangles

    Take both triangles away from the rectangle.

    15−(6+5)=15−11=415 - (6 + 5) = 15 - 11 = 4

    The shaded area is 4 square units.

    So the answer is C, 4.

Check your answer

Cut the shaded region along a line from O to P. One piece has a base of 1 on the x-axis and a height of 5, since P is 5 units up. Its area is 2.5. The other has a base of 1 on the y-axis and a height of 3, since P is 3 units across. Its area is 1.5. Then 2.5 + 1.5 = 4. Using 3 for both heights gives 3, choice B.

More questions like this

The original DOE question
Question 113 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 113. Open page 208 of the handbook (PDF)

Question 81 of 108

2018 Test A English, question 25
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

The table contributes to the development of the topic of the passage mainly by

  • Aemphasizing that people have continued to swim across the channel and have significantly reduced the speed record.
  • Bsuggesting that Ederle inspired women to swim across the channel in an attempt to break the current speed record.
  • Crevealing that Ederle is not the only woman who has set a record time for swimming across the English Channel.
  • Dproviding a comparison between channel-swimming records of the early twentieth century and current records.
Show the answer

Correct answer: A

The answer is (A)

The question asks for an explanation of how the table at the end of the passage contributes to the development of the topic of the passage. The table provides information about the earliest speed records set for crossing the channel, including Ederle’s, and information about present-day records, which are much faster.

Option A is correct because the information in the table develops the topic of Ederle’s historic swim across the channel by indicating that people have continued to swim the channel and have found new ways to set speed records over time.

Option B is incorrect because, even though the table presents the time for the current female record holder, there is no indication in the passage or in the table that the record holder was inspired by Ederle.

Option C is incorrect because, although the table shows that another woman, Yvetta Hlavacova, set a record time, the passage itself does not discuss other female swimmers’ successful swims across the channel, so this detail is not the main way the table develops the topic.

Option D is incorrect because, although the table allows for the comparison of times, the topic of the passage is related to Ederle’s record in comparison with other swimmers of her time, not present-day swimmers.

Question analysis

This reading question asks how the table of English Channel swim records at the end of the Gertrude Ederle passage contributes to the topic. Students must read times from two eras, compare them, and decide what the table mainly adds, which is harder than it looks because several choices say something true about the table.

NYS standards
  • 7R5 structure and genre features
  • 7R1 textual evidence and inferences
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Explaining how a table adds to a passage goes beyond typical grade 7 reading questions: the student must compare times across eras, notice who set them, and choose among statements several of which are partly true of the table.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from the whole text
Reading level
Passage: about grade 9, 790 words; question and choices: about grade 11, 84 words.
Corrected from the DOE handbook
  • An inaccurate statement 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 says something about the table, and two are literally true of it, so the question hinges on the word “mainly”. Students lose the point by describing what the table holds, a comparison or a second woman, instead of what the comparison shows: the record has kept falling since Ederle's day. Reading the times, not just the headings, makes that drop visible.

  1. A very close wrong answer major Choosing the answer

    The right choice and the comparison one cover the same contrast between old and current records; the right one states its result, a much faster record, while the other only names the comparison and its “early twentieth century” leaves out 1875. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The table splits records into earliest and current sets, and the student must compare times across them, from 21 hours 45 minutes in 1875 down to 6 hours 55 minutes in 2012. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. How, not what moderate Working it out

    The question asks how the table develops the passage's topic, so the student must say what it adds beyond the text, not just what it lists. (Anderson & Krathwohl, 2001)

Strategy: Read the table's headings and compare the times, then ask what a reader learns from it that the passage alone does not say, and pick the choice that names that.

Traps

  • Wrong purpose or structure: Names a purpose, structure or effect that the part does not serve, or describes what a part says when the question asks what it does.
  • Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook rejects choice C because “the passage does not provide information about other female swimmers’ successful swims across the channel.” But the table lists a second woman, Yvetta Hlavacova (2006, 7 hours, 25 minutes), so C is true of the table. The explanation now says C is wrong because that one detail is not the table’s main contribution: the passage does not discuss other female swimmers, and the table mainly shows how far the record has fallen since Ederle’s time.

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 table adds to the passage. Don’t just say what is in the table. Say what you learn from it that the passage itself does not tell you.

  2. Read the key rows of the table

    The table has two parts, “Earliest Speed Records” and “Current Speed Records.” Here are three of its rows:

    YearSwimmerTime
    1875Matthew Webb21 hours, 45 minutes
    1926Gertrude Ederle14 hours, 31 minutes
    2012Trent Grimsey6 hours, 55 minutes

  3. Compare the times

    Ederle took about 14 and a half hours. Half of that is about 7 and a quarter hours. The 2012 record is under 7 hours. So today’s record is less than half of Ederle’s time. This shows two things. People kept swimming the channel long after Ederle. And they got much faster.

  4. See what the passage leaves out

    The passage ends with Ederle’s parade and the people she inspired. It never says what happened to the record later. The table adds that part of the story: the record kept falling, by a lot.

  5. Check each choice

    • A: people kept swimming the channel and “significantly reduced the speed record.” Significantly means by a lot. The table shows this.
    • B: nothing says later women were inspired by Ederle.
    • C: the table does show another woman, Yvetta Hlavacova, with a record. But that is one row, not the table’s main point.
    • D only names what is in the table, not what it shows. Also, the first record is from 1875, not the twentieth century.

    Watch out: D describes the table’s contents. The question asks what the table contributes, or adds.

  6. Pick the answer

    The table shows that the record has fallen a long way since Ederle’s time. So the answer is A.

Why this answer and not the closest wrong one

D is the closest wrong answer, because the table does set old and new records side by side. But D stops at naming the comparison. A says what the comparison shows. The record dropped from over 21 hours to under 7 hours. That is what the table adds to the passage.

More questions like this

The original DOE question
Question 25 as printed on page 60 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 60, question 25. Open page 60 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.
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 82 of 108

2018 Test A Math, question 1

Question

Parallelogram PQRS with side PS on a horizontal line that extends past P and S, and side QR at the top. At S, the angle between side SR and the line extended to the right of S is marked 72°. At Q, a vertical line rises above the figure at a right angle to side QR (right-angle mark), and angle x° is marked on the left of Q, between that vertical line and side QP.

In the figure above, PQRS is a parallelogram. What is the value of xx?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 162

First, find the measure of angle PQR. The measure of angle PQR is equal to the measure of angle PSR.

m∠PSR = 180 − 72 = 108.

So, the measure of angle PQR is also 108.

108 + 90 + x = 360

198 + x = 360

x = 162

Question analysis

A grid-in geometry question about angles around a parallelogram PQRS. A 72° angle sits outside corner S on the extended base, and at corner Q a right angle and the unknown angle x complete the turn around the point. Students chain three angle facts, and the angle asked for lies outside the shape, not inside it.

NYS standards
  • 7.G.5 angle relationships and equations
  • 8.G.5 angle sums, transversals, angle-angle similarity
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle work usually applies one or two facts, such as supplementary or vertical angles, to find an angle. This chains three, and needs the equal opposite angles of a parallelogram, which rest on grade 8 parallel-line facts.
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 7, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is two subtractions; the difficulty is seeing which angles the marks refer to. Students lose points by treating the 72° as the parallelogram's own angle at S, or by stopping at angle PQR, the most natural number to grid. The question rewards tracing a route from S to Q before computing anything, and checking at the end that x is the angle outside the shape that the arc marks.

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

    The 72° is marked outside the parallelogram, between side SR and the base extended past S, and x sits outside corner Q, between QP and a vertical ray that meets QR at a right-angle mark. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Three angle facts have to be chained: a straight line gives the angle inside S, opposite angles of a parallelogram carry it to Q, and the angles around Q make a full 360° turn. (Embretson, 1998; Daniel & Embretson, 2010)

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

    As a grid-in there are no choices to check against, so a student who stops at angle PQR, inside the parallelogram, gets no signal that the job is unfinished. (Enright, Morley & Sheehan, 2002)

Strategy: Use the straight line at S to find the parallelogram's angle there, carry it to the opposite corner Q, then take that angle and the right angle away from 360.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b=180a + b = 180
a and b are two angles side by side on a straight line. Together they make a half turn, so they always add up to 180 degrees. Use it when an angle you need shares a straight line with an angle you know.
Opposite angles of a parallelogram
∠PQR=∠PSR\angle PQR = \angle PSR
The symbol ∠ means angle. ∠PQR is the angle at corner Q, between sides QP and QR. ∠PSR is the angle at the opposite corner, S. In a parallelogram, opposite corners always have equal angles, because the shape leans the same way at both.
Angles around a point
a+b+c=360a + b + c = 360
a, b and c are angles that fit around one point with no gaps and no overlaps. Going all the way around is a full turn, which is 360 degrees. So the angles add up to 360.

  1. Sort out what the figure shows

    PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, which means they never meet. You need the value of x, in degrees.

    Look closely at the marks. The 72° angle is outside the shape, at S. It sits between side SR and the base line, which keeps going past S. At Q, a vertical line meets side QR. The small square means a right angle, 90°. The angle x is outside the shape too. It runs from that vertical line around to side QP.

  2. Find the angle inside corner S

    Angle PSR is the shape’s own angle at S. It sits right next to the 72° angle, on the same straight base line. So the two angles add up to 180°.

    a+b=180a + b = 180

    ∠PSR+72=180\angle PSR + 72 = 180

    ∠PSR=180−72=108\angle PSR = 180 - 72 = 108

    So angle PSR is 108°. Watch out: 72° is not the inside angle at S. Using 72 as the inside angle leads to the wrong answer 198.

  3. Carry the angle over to Q

    Corners Q and S are opposite each other. In a parallelogram, opposite angles are equal. So angle PQR matches angle PSR.

    ∠PQR=∠PSR=108\angle PQR = \angle PSR = 108

    So angle PQR is 108°. Watch out: this is not x. Angle PQR is inside the shape, and x is outside it. Stopping here and gridding 108 is the most common mistake.

  4. Add the angles around Q

    Three angles fill the space all the way around point Q. They are the right angle (90°), angle PQR (108°) and x. Together they make a full turn.

    a+b+c=360a + b + c = 360

    90+108+x=36090 + 108 + x = 360

    198+x=360198 + x = 360

    Watch out: don’t skip the right angle at Q. Leaving out the 90 gives 252.

  5. Solve for x

    Subtract 198 from 360 to find what is left for x.

    x=360−198=162x = 360 - 198 = 162

    So x is 162°. So you would grid 162.

Check your answer

Add the three angles at Q: 90+108+162=36090 + 108 + 162 = 360. That is a full turn. Also, the arc for x is wider than a right angle but not as wide as a straight line. So x should be between 90° and 180°, and 162° is.

More questions like this

The original DOE question
Question 58 as printed on page 81 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 58. Open page 81 of the handbook (PDF)

Question 83 of 108

2018 Test A Math, question 11

Question

PEOPLE PER VEHICLE AT CHECKPOINT

Number of People in Vehicle Percent of Vehicles
1 40%
2 35%
3 15%
4 7%
5 or more 3%

A researcher recorded the number of people in each vehicle that passed through a checkpoint. The table above shows the percent distribution for the 420 vehicles that passed through the checkpoint yesterday morning. How many of the 420 vehicles contained at least 3 people?

  • A42
  • B63
  • C105
  • D315
Show the answer

Correct answer: C

First, add the percentage of cars containing 3 people, 4 people, and 5 or more people:

15% + 7% + 3% = 25%

Thus, 25% of the cars contained at least 3 people, so use that to calculate the number of cars:

420 × 0.25 = 105 cars.

Question analysis

A data table question with percents: a table gives the percent of 420 vehicles carrying 1, 2, 3, 4, or 5 or more people. Students find how many vehicles held at least 3 people by adding the right rows and taking that percent of 420. Reading at least correctly decides which rows count.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Taking a percent of a number is grade 6 work, and grade 7 adds multistep percent problems. This two-step table problem sits at that level, with the phrase at least as the careful-reading point.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 7, 56 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is easy, since 25% of 420 is a quarter of it; the points are lost on the phrase "at least 3". Each common misreading, more than 3, exactly 3 or fewer than 3, gives a listed choice, so a wrong reading never leads to a dead end. Adding the percents before converting saves time and avoids three separate multiplications.

  1. Turning words into math moderate Understanding the question

    The bold "at least 3" has to become the rows for 3, 4 and "5 or more", taking in both the 3-person row and the open-ended last row. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table gives percents, not counts, so the rows have to be added as percents and then taken of the 420 vehicles named only in the text below the table. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add the percents for every row that meets the condition first, then take that single percent of 420 instead of working row by row.

Traps

  • 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).
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
x=p100×Wx = \frac{p}{100} \times W
W is the whole amount. p is the percent, and x is the part you want. Percent means out of 100, so p percent is the fraction p over 100. Multiplying the whole by that fraction gives the part.

  1. Sort out the question

    420 vehicles passed the checkpoint. The table gives the percent of vehicles with each number of people. You need how many vehicles had “at least 3” people. At least 3 means 3 or more. So count vehicles with 3 people, 4 people, and 5 or more.

  2. Pick the right rows

    Three rows fit: 3 people (15%), 4 people (7%), and 5 or more (3%). Watch out: at least 3 includes 3 itself. Leaving out the 3 row gives 42, choice A. Using only the 3 row gives 63, choice B.

  3. Add the percents first

    Adding the percents first saves work. Then you only multiply once.

    15%+7%+3%=25%15\% + 7\% + 3\% = 25\%

    So 25% of the vehicles had at least 3 people.

  4. Find 25% of 420

    Use the percent formula with p = 25 and W = 420.

    x=p100×Wx = \frac{p}{100} \times W

    x=25100×420=105x = \frac{25}{100} \times 420 = 105

    25% is one quarter, so this is the same as 420÷4420 \div 4. Watch out: 315 is the number of vehicles with fewer than 3 people, choice D. So the answer is C, 105.

Check your answer

The other rows, 1 and 2 people, make 40%+35%=75%40\% + 35\% = 75\%. 75% of 420 is 315. Then 105+315=420105 + 315 = 420, which is all the vehicles. The two groups fit together.

More questions like this

The original DOE question
Question 68 as printed on page 83 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 68. Open page 83 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 84 of 108

2018 Test A Math, question 12

Question

Square pyramid with apex R over the square base MNPQ; corner N is at the back, and its edges are dashed as hidden. Segment RS runs down face QRP from R to point S on base edge QP, with a right-angle mark at S.

In the pyramid above, each triangular face has the same area, and the base MNPQ is a square that measures 8 centimeters on each side. If the length of RS\overline{\mathrm{RS}} = 6 centimeters, what is the surface area of the pyramid excluding the base?

  • A48 sq cm
  • B96 sq cm
  • C128 sq cm
  • D160 sq cm
Show the answer

Correct answer: B

Line segment RS\overline{\mathrm{RS}} is the altitude, or height, of triangle QRP. The length of QP\overline{\mathrm{QP}} is 8 cm. Use that information to find the area of triangle QRP:

A=12bh=12(8)(6)=24A=\frac{1}{2}bh=\frac{1}{2}(8)(6)=24 sq cm.

There are 4 congruent triangles in the pyramid, so the surface area is

4 × 24 = 96 sq cm

Question analysis

A geometry question about the surface area of a square pyramid, leaving out its base. The base edges are 8 cm and segment RS, the height of one triangular face drawn from the top to a base edge, is 6 cm. Students find one face's area and multiply by the number of faces, reading excluding the base carefully.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.4 nets and surface area
  • 6.G.1 area of polygons
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Surface area of a right pyramid from its faces is named in the grade 7 standards, and this is a direct two-step case. It matches typical grade 7 work, though the drawing's slant height must be read correctly.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 49 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once RS is read as the height of a face, the computation is short: one triangle, then four of them. Points go to the figure and to one word. Students who count only the faces drawn in front, or who include the square base, reach listed choices, and the bold "excluding" is all that separates the lateral area from the total. The statement that every face has the same area is what makes multiplying by four valid.

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

    The drawing shows two lateral faces in front and two behind dashed edges, and RS, with its right-angle mark at S, is the height of face QRP, not of the pyramid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The triangle area formula, one half times base times height, and the fact that a square pyramid has four lateral faces are both assumed, not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of one triangular face with one half times base times height, using RS as the height, then multiply by the four congruent faces.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

  1. Sort out the pyramid

    The pyramid sits on a square base, MNPQ. Each side of the square is 8 cm. Four triangle faces rise from the base and meet at the top point, R. Every face has the same area. Segment RS is 6 cm. You need the area of the four faces, “excluding the base”. Excluding means leaving it out.

  2. See what RS measures

    RS runs from R down face QRP to point S on edge QP. The small square at S marks a right angle. So RS is the height of triangle QRP. The base of that triangle is QP, one side of the square, so it is 8 cm.

  3. Find the area of one face

    Use the triangle formula with base 8 cm and height 6 cm.

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×8×6=24A = \frac{1}{2} \times 8 \times 6 = 24

    One face has an area of 24 sq cm.

  4. Add the four faces

    All four faces are the same, so multiply by 4.

    4×24=964 \times 24 = 96

    Watch out: only two faces are drawn in front. The other two hide behind the dashed lines. Counting just two faces gives 48, choice A. Watch out: don’t add the base, 8×8=648 \times 8 = 64. That gives 160, choice D. So the answer is B, 96 sq cm.

Check your answer

Each face fits inside an 8 by 6 rectangle, with an area of 48 sq cm. The triangle is half of that, 24 sq cm, and four faces make 96 sq cm. That is less than 160, which would include the base.

More questions like this

The original DOE question
Question 69 as printed on page 83 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 69. Open page 83 of the handbook (PDF)

Question 85 of 108

2018 Test A Math, question 14

Question

Which number line below shows the solution to the inequality −4<x2<2-4< \frac{x}{2}< 2 ?

  • ANumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −2 and 1, with the segment between them shaded.
  • BNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −4 and 2, with the segment between them shaded.
  • CNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the line shaded to the left of −8 and to the right of 4.
  • DNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the segment between them shaded.
Show the answer

Correct answer: D

Multiply each term by 2 to eliminate the fraction, and isolate xx:

−4(2)<(x2)(2)<2(2)-4(2)< \left(\frac{x}{2}\right)(2)< 2(2)

−8<x<4-8< x< 4

Therefore, xx must be between −8 and 4.

Question analysis

An inequality question: choose the number line that shows the solution of a compound inequality in which x divided by 2 lies between −4 and 2. Students undo the division in every part and match the graph's endpoints and shading. The four number lines differ in their endpoints and in which part of the line is shaded.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve two-step inequalities and graph them on a number line. This compound inequality needs only one operation applied to all three parts at once, so it matches typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 11 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

Solving takes one step, multiplying through by 2, so most of the work is matching the result to a picture. The choices pair each common slip with a graph: the bounds left as written, the bounds halved, and the right bounds shaded on the wrong side. Reading the open circles against the labeled ticks at −5 and 5 settles it.

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

    The four number lines are labeled only at −10, −5, 0, 5 and 10, so open circles at −8 and 4, or at −4 and 2, have to be located by counting ticks. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic minor Working it out

    All three parts of −4 < x/2 < 2 must be multiplied by 2, and the negative bound −4 becomes −8; halving instead gives −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply all three parts of the inequality by 2 to isolate x, then pick the graph whose open circles and shading match the new bounds.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying an inequality by a positive number
if a<b, then a×2<b×2\text{if } a < b \text{, then } a \times 2 < b \times 2
a and b are numbers, and the less-than sign means the left side is smaller. If you multiply both sides by the same positive number, like 2, the smaller side stays smaller. So the sign keeps pointing the same way. A negative number would flip it, but 2 is positive.

  1. Sort out the inequality

    An inequality compares two amounts that may not be equal. The sign << means less than. Here x2\frac{x}{2} means x divided by 2. The inequality says x divided by 2 is between −4 and 2. You need the values of x itself. Then you pick the number line that shows them.

  2. Multiply all three parts by 2

    To undo dividing by 2, multiply by 2. Multiply all three parts, so the whole statement stays true.

    −4×2<x2×2<2×2-4 \times 2 < \frac{x}{2} \times 2 < 2 \times 2

    −8<x<4-8 < x < 4

    Watch out: multiply, don’t divide. Dividing gives −2 and 1, choice A. Leaving the ends at −4 and 2 is choice B.

  3. Read what the answer means

    x is greater than −8 and less than 4. So x is any number between −8 and 4. x cannot equal −8 or 4 exactly. On a number line, that means open circles at −8 and 4, with the part between them shaded.

  4. Find the matching number line

    The number lines are labeled only at −10, −5, 0, 5 and 10. Count the tick marks. −8 is 3 ticks left of −5, and 4 is 1 tick left of 5. Watch out: choice C has circles at −8 and 4 but shades the parts outside them. So the answer is D.

Check your answer

Test a number from the shaded part, like 0. 0÷2=00 \div 2 = 0, and 0 is between −4 and 2. Now test 6, which is outside. 6÷2=36 \div 2 = 3, and 3 is not less than 2. So only the part between −8 and 4 works.

More questions like this

The original DOE question
Question 71 as printed on page 84 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 71. Open page 84 of the handbook (PDF)

Question 86 of 108

2018 Test A Math, question 18

Question

Coordinate grid with the origin O, the x-axis and y-axis each marked 1 and 2, and grid lines at 1 and 2. A straight line starts at O and rises through the grid points (1, 1) and (2, 2). The triangle under that line with corners O, (1, 0) and (1, 1) is shaded.

What is the area of the shaded region in the graph above?

  • A0.25 square unit
  • B0.5 square unit
  • C1 square unit
  • D1.5 square units
Show the answer

Correct answer: B

The shaded region is a right triangle. Each leg is 1 unit in length.

So the area is

A=12bh=12(1)(1)=12A=\frac{1}{2}bh=\frac{1}{2}(1)(1)=\frac{1}{2} or 0.5 sq unit

Question analysis

A coordinate grid question: find the area of a shaded right triangle with corners at the origin, (1, 0) and (1, 1), under a diagonal line. Students read the base and height from the grid and apply the triangle area formula. The figure does most of the work; the task is recalling the formula correctly.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a triangle's area from a grid is grade 6 work, and this takes one step with whole-unit legs. It is easier than typical grade 7 area problems, which use composite figures or scale drawings.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 2, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a quick formula question with legs of 1. The main risk is the one-half in the triangle formula: leaving it out gives the area of the whole grid square, which is a listed choice. Seeing that the diagonal cuts the unit square exactly in half gives a check that needs no formula at all.

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

    The legs of the shaded triangle, from O to (1, 0) and up to (1, 1), are read off the grid lines; the line running on to (2, 2) is not part of the region. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The triangle area formula, one half times base times height, is needed and not given; the shaded triangle is half of one grid square. (Anderson & Krathwohl, 2001)

Strategy: Read the two legs straight off the grid, then apply the triangle area formula, base times height divided by two.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

  1. Sort out the shaded region

    The shaded region is a triangle. Its corners are O, the point (1, 0) and the point (1, 1). O is the origin, the point (0, 0) where the axes meet. The line keeps going up to (2, 2), but only the small triangle is shaded. You need its area in square units.

  2. Read the base and height

    The base lies along the x-axis, from O to 1. So the base is 1 unit. The height goes straight up from (1, 0) to (1, 1). So the height is 1 unit too. The base and height meet at a right angle, as the grid lines show.

  3. Use the triangle formula

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×1×1=12=0.5A = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} = 0.5

    The area is 0.5 square unit. Watch out: don’t forget the one half. 1×1=11 \times 1 = 1 is the whole grid square, choice C. So the answer is B, 0.5 square unit.

Check your answer

The grid square from 0 to 1 has an area of 1 square unit. The slanted line cuts it corner to corner into two equal triangles. The shaded one is half the square, so 0.5 square unit.

More questions like this

The original DOE question
Question 75 as printed on page 84 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 75. Open page 84 of the handbook (PDF)

Question 87 of 108

2018 Test A Math, question 23

Question

SCORES ON MATH QUIZ

Score Number of Students
85 4
75 4
65 2

What is the mean score of the 10 students in the table above?

  • A22.5
  • B75
  • C77
  • D85
Show the answer

Correct answer: C

For each row, multiply the number of students by the score. Then add those together and divide by the total number of students to find the mean (average) of the 10 students.

85(4)+75(4)+65(2)10=340+300+13010=77010=77\frac{85(4)+75(4)+65(2)}{10}=\frac{340+300+130}{10}=\frac{770}{10}=77

Question analysis

This statistics question gives a frequency table of quiz scores: three scores and how many of the 10 students earned each. Students find the mean score of all 10 students. Each score has to count as many times as students earned it, which is easy to overlook when only three scores are listed.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Computing a mean is grade 6 work, and this weighted version from a table takes two short steps. It is a little easier than typical grade 7 statistics, which compares whole data sets.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 2, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The procedure is routine once students see that the table lists counts, not separate students. The tempting shortcut is to average the three scores in the first column, which gives a listed choice close to the right one. Multiplying each score by its count, then dividing by 10, takes well under a minute.

  1. A very close wrong answer moderate Choosing the answer

    77 and 75 are only 2 points apart, so a rough sense of the middle score cannot choose between them; only weighting each score by its 4, 4 or 2 students does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The table pairs each score with a number of students, so the 10 students are 4, 4 and 2, and each score must be weighted by its count. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Multiply each score by its number of students, add the three products, and divide by the 10 students.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean (average)
M=SnM = \frac{S}{n}
M is the mean, also called the average. S is the sum, all the values added together. n is how many values there are. Dividing shares the total out equally, so the mean is what each value would be if they were all the same.

  1. Sort out the table

    The table lists 3 scores and how many students got each one. 4 students scored 85. 4 students scored 75. 2 students scored 65. That is 4+4+2=104 + 4 + 2 = 10 students. You need the mean, or average, of all 10 scores.

  2. Find the sum of all 10 scores

    Each score counts once for every student who got it. So multiply each score by its number of students.

    85×4=34085 \times 4 = 340

    75×4=30075 \times 4 = 300

    65×2=13065 \times 2 = 130

    Add the three results.

    340+300+130=770340 + 300 + 130 = 770

    All 10 scores add up to 770 points.

  3. Divide by the number of students

    M=SnM = \frac{S}{n}

    M=77010=77M = \frac{770}{10} = 77

    Watch out: averaging just the three scores gives (85+75+65)÷3=75(85 + 75 + 65) \div 3 = 75, choice B. That acts as if only one student got each score. So the answer is C, 77.

Check your answer

More students scored 85 than 65. So the mean should be a little above the middle score, 75. 77 fits. A mean must also be between the lowest and highest scores, 65 and 85. That rules out 22.5.

More questions like this

The original DOE question
Question 80 as printed on page 85 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 80. Open page 85 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 88 of 108

2018 Test A Math, question 24

Question

Circle graph titled "How People Get to Work in Center City": Drive Alone 49%, Walk 22%, Car Pool 15%, Bus 10%, Bicycle 4%. Below the graph: Total number of people working in Center City = 15,000.

How many more people in Center City walk to work than ride their bicycle to work?

  • A2,500
  • B2,700
  • C2,800
  • D3,000
Show the answer

Correct answer: B

According to the chart, 22% of people walk to work and 4% ride a bicycle. Subtract to find the percentage of how many more people walk than bicycle:

22% − 4% = 18%

To find the exact number of people, multiply 18% (0.18) by the number of people working in Center City (15,000):

(15000)(0.18)=2700(15000)(0.18)=2700

Question analysis

This circle graph question shows how the 15,000 people working in Center City get to work, with 22% walking and 4% riding a bicycle. Students find how many more people walk than bike by combining a percent difference with a percent of a total. The choices are close together, so estimating is risky.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Reading a circle graph and taking a percent of a total are grade 6 skills. With only two short steps and friendly numbers, this is easier than typical grade 7 multistep percent work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is short: subtract the percents first, or find both groups and subtract, and both reach the same number. What makes the question less safe than it looks is the tight spread of choices, all within 500 of each other, so rounding 18% or slipping on one product lands on a wrong answer. Subtracting the percents first leaves one multiplication to get right instead of two.

  1. A very close wrong answer moderate Choosing the answer

    2,700 sits only 100 from 2,800 and 300 from 3,000, so estimating 18% of 15,000 as about a fifth cannot pick it; only the exact product does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The Walk 22% and Bicycle 4% labels must be picked out of five sectors, and the 15,000 total sits in a caption below the circle graph. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Error-prone arithmetic minor Working it out

    The percent gap has to become a count, 0.18 × 15,000, a decimal times a five-digit number worked by hand without a calculator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the two percents first, then take that single percent of 15,000, instead of finding each group and subtracting.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
x=p100×Wx = \frac{p}{100} \times W
W is the whole amount. p is the percent, and x is the part you want. Percent means out of 100, so p percent is the fraction p over 100. Multiplying the whole by that fraction gives the part.

  1. Sort out the circle graph

    A circle graph shows how a whole splits into parts. This one shows how 15,000 workers get to work. Walk is 22%. Bicycle is 4%. You need how many more people walk than ride a bicycle.

  2. Subtract the percents first

    Both percents are parts of the same 15,000 people. So you can subtract the percents first. Then you only need one multiplication.

    22%−4%=18%22\% - 4\% = 18\%

    The difference is 18% of all the workers.

  3. Find 18% of 15,000

    x=p100×Wx = \frac{p}{100} \times W

    x=18100×15,000x = \frac{18}{100} \times 15{,}000

    First find 1% by dividing by 100. Then multiply by 18.

    15,000÷100=15015{,}000 \div 100 = 150

    150×18=2,700150 \times 18 = 2{,}700

    That is 2,700 people.

  4. Avoid rounding and answer

    Watch out: don’t round 18% up to 20% to make it easy. 20% of 15,000 is 3,000, choice D. The choices are close together, so you need the exact number. So the answer is B, 2,700.

Check your answer

Find each group, then subtract. Walkers: 22% of 15,000 is 3,300. Bike riders: 4% of 15,000 is 600. 3,300−600=2,7003{,}300 - 600 = 2{,}700. You get the same answer.

More questions like this

The original DOE question
Question 81 as printed on page 86 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 81. Open page 86 of the handbook (PDF)

Question 89 of 108

2018 Test A Math, question 36

Question

Number line with two points: L, labeled 4 and 5/16, and N to the right of L, with no value given.

On the number line above, LN = 18\frac{1}{8} . Point M (not shown) is located between point L and point N. Which value below is a possible value for M?

  • A4.26
  • B4.31
  • C4.35
  • D4.58
Show the answer

Correct answer: C

Since LN=18LN=\frac{1}{8}, point NN is located at

4516+18=47164\frac{5}{16}+\frac{1}{8}=4\frac{7}{16}.

So MM must be between point LL, 45164\frac{5}{16}, and point NN, 47164\frac{7}{16}. Point LL can also be written as 4.3125, and point NN can be written as 4.4375. The only option given that lies between those two points is 4.35.

Question analysis

This number line question says the distance from point L to point N is one eighth and asks which decimal could be the position of a point M between them. Students take L's position, a mixed number in sixteenths, find N, and compare with decimal choices. Converting sixteenths to decimals precisely is the tricky part.

NYS standards
  • 7.NS.2d fractions to decimals by long division
  • 7.NS.1b rational number addition
  • 6.NS.7a inequality as relative position
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students convert fractions to decimals and compare them. This item needs an exact conversion of sixteenths and a comparison to the thousandths, where rounding too early can mislead, but it stays within grade 7 content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The interval from L to N is only an eighth wide, and three of the four choices crowd near it, so precision decides the question. Students comfortable working in sixteenths can avoid decimals until the end, since 4.35 lies between 4 5/16 and 4 7/16. Those who convert loosely, or who misread a mixed number as a decimal, are pulled to a neighboring choice.

  1. Error-prone arithmetic moderate Working it out

    Converting 4 5/16 to 4.3125 and N to 4.4375 needs four decimal places, while the choices stop at hundredths, so early rounding blurs the left boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The choices 4.31 and 4.35 both lie within 0.04 of L, so estimating cannot separate them; only the exact 4.3125 shows 4.31 just outside the interval and 4.35 inside. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    L's value appears only on the number line, and the picture shows N to the right of L, which fixes which way the one-eighth gap runs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add one eighth, or two sixteenths, to L's position to locate N, convert both endpoints to four-place decimals, and keep only a choice strictly between them.

Traps

  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Answer in another form: The right answer is written differently from the way you worked it out.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top number (numerator) and b is the bottom number (denominator). Multiplying both by the same number k gives an equal fraction. It is like cutting each piece into k smaller pieces. Use it to give fractions the same bottom before you add or subtract.
Fraction to decimal
ab=a÷b\frac{a}{b} = a \div b
A fraction bar means divide. a is the top number and b is the bottom number. Dividing the top by the bottom turns the fraction into a decimal. Then you can compare it with decimal choices.

  1. Sort out what you know

    Point L is at 45164\frac{5}{16}. Point N is to the right of L. The distance from L to N, called LN, is 18\frac{1}{8}.

    Point M is somewhere between L and N. You need a choice that fits between them. The choices are decimals, so you will change the fractions into decimals.

  2. Find where N is

    N is 18\frac{1}{8} to the right of L, so add. First give 18\frac{1}{8} a bottom of 16, to match L:

    18=1×28×2=216\frac{1}{8} = \frac{1 \times 2}{8 \times 2} = \frac{2}{16}

    4516+216=47164\frac{5}{16} + \frac{2}{16} = 4\frac{7}{16}

    Watch out: N is to the right of L in the figure. Going left instead gives a gap that holds 4.26, choice A.

  3. Change both points to decimals

    Divide the top of each fraction by the bottom:

    516=5÷16=0.3125\frac{5}{16} = 5 \div 16 = 0.3125

    716=7÷16=0.4375\frac{7}{16} = 7 \div 16 = 0.4375

    So L is at 4.3125 and N is at 4.4375.

  4. Test each choice

    M must be bigger than 4.3125 and smaller than 4.4375.

    • 4.26 is smaller than 4.3125. It is left of L.
    • 4.31 is 4.3100, just smaller than 4.3125. It is a hair left of L.
    • 4.35 is between them. It fits.
    • 4.58 is bigger than 4.4375. It is right of N.

    Watch out: if you round L to 4.31, choice B seems to fit. Keep all four decimal places. So the answer is C, 4.35.

Check your answer

Subtract to be sure. 4.35 − 4.3125 = 0.0375, which is more than 0, so 4.35 is right of L. 4.4375 − 4.35 = 0.0875, which is also more than 0, so 4.35 is left of N. It sits between them.

More questions like this

The original DOE question
Question 93 as printed on page 88 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 93. Open page 88 of the handbook (PDF)

Question 90 of 108

2018 Test A Math, question 41

Question

Bar graph titled "Number of Schools in Five Cities". The vertical axis, Number of Schools, runs from 0 to 10 in steps of 1; the horizontal axis is City. The bars show M 8, N 5, P 9, Q 6 and R 4 schools.

The graph above shows the number of schools per city for five small cities. Cities M and N each have 500 students per school. City P has 400 students per school. Cities Q and R each have 700 students per school. Which of the five cities has the greatest number of students?

  • ACity M
  • BCity P
  • CCity Q
  • DCity R
Show the answer

Correct answer: C

Take each city’s number of schools and multiply by the number of students. It is not necessary to calculate all 5 of these.

Cities M and N have the same number of students per school, so just calculate the number of students in City M because it has more schools than City N.

The same goes for Q and R — only Q needs to be calculated because it has more schools than R.

M= 8 × 500 = 4,000

P = 9 × 400 = 3,600

Q = 6 × 700 = 4,200

City Q has the greatest number of students.

Question analysis

This graph question shows the number of schools in each of five small cities and gives the number of students per school for each city. Students multiply schools by students per school to compare the total number of students. Comparing only the bars, or only the rates, can mislead.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Reading a bar graph and multiplying is familiar from grade 6. Grade 7 data tasks often add percents or compare distributions; this one needs only a few whole-number products, so it is easier than typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 2, 57 words.
Corrected from the DOE handbook
  • A misstatement 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

Neither the tallest bar nor the highest rate wins, which is the point of the question: total students depend on both. The explanation's shortcut, comparing only M, P and Q because N and R each share a rate with a taller bar, cuts the work to three products. The top two totals are close, so the products must actually be computed rather than estimated.

  1. A very close wrong answer moderate Choosing the answer

    City M's 4,000 students and City Q's 4,200 differ by only 200, so judging from bar heights and rates cannot separate them; only computing 8 × 500 and 6 × 700 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Much to keep track of minor Working it out

    The rates come in three groups in the stem, 500 for M and N, 400 for P, 700 for Q and R, and each must be matched to its bar. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

    The number of schools comes only from the bar heights, read against gridlines from 0 to 10; all five bars end exactly on a gridline. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: For each city, multiply the number of schools read from the graph by its students per school and compare the products; cities that share a rate need only their school counts compared.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “Cities M and N have the same number of students”. They have the same number of students per school (500), not the same total: M has 8 × 500 = 4,000 and N has 5 × 500 = 2,500, which is why only City M, with more schools, needs to be calculated. The sentence now says “the same number of students per school”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total students in a city
S=n×sS = n \times s
S is the total number of students in a city. n is the number of schools, read from the bar. s is the number of students in each school. You multiply because every school in that city has the same number of students.

  1. Read the graph

    The bar graph shows how many schools each city has. Read the top of each bar against the numbers on the left. M has 8, N has 5, P has 9, Q has 6 and R has 4.

    The question gives the students per school. You need the city with the most students in all.

  2. Multiply for each city

    Use the formula for each city:

    CitySchoolsStudents per schoolTotal students
    M85004,000
    N55002,500
    P94003,600
    Q67004,200
    R47002,800

  3. Compare the totals

    The biggest total is 4,200, for City Q.

    Watch out: City P has the tallest bar, but that means the most schools, not the most students. That leads to choice B. City M’s 4,000 is close, so do not stop before you work out Q. That is choice A.

    So the answer is C, City Q.

Check your answer

A shortcut: M and N both have 500 per school, and M has more schools, so M beats N. Q and R both have 700, and Q beats R. So only M, P and Q matter: 4,000, 3,600 and 4,200. Q still wins.

More questions like this

The original DOE question
Question 98 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 98. Open page 89 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.

Question 91 of 108

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 92 of 108

2018 Test A Math, question 54

Question

Number line with points J, K, L and M marked from left to right. Only K has a value: 3/8.

On the number line above, JK=312JK=3\frac{1}{2} , JM=934JM=9\frac{3}{4} , and LM=118LM=1\frac{1}{8} . What is the position of point LL?

  • A5185\frac{1}{8}
  • B5145\frac{1}{4}
  • C5125\frac{1}{2}
  • D6146\frac{1}{4}
Show the answer

Correct answer: C

First, use JK=312JK=3\frac{1}{2} to find the location of JJ:

38−J=312\frac{3}{8}-J=3\frac{1}{2}

J=38−312=−318J=\frac{3}{8}-3\frac{1}{2}=-3\frac{1}{8}

Now, use JM=934JM=9\frac{3}{4} to find the location of MM:

M−(−318)=934M-\left(-3\frac{1}{8}\right)=9\frac{3}{4}

M+318=934M+3\frac{1}{8}=9\frac{3}{4}

M=934−318=658M=9\frac{3}{4}-3\frac{1}{8}=6\frac{5}{8}

Finally, use LM=118LM=1\frac{1}{8} to find the location of LL:

658−L=1186\frac{5}{8}-L=1\frac{1}{8}

L=658−118=548=512L=6\frac{5}{8}-1\frac{1}{8}=5\frac{4}{8}=5\frac{1}{2}

Question analysis

This number line question gives the lengths JK, JM and LM as mixed numbers and asks for the position of point L. Students must chain three distance steps, moving left from K to J, right to M and back to L, with fractions in eighths and a negative position. Losing track of where K sits is the main danger.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students add and subtract rational numbers and find distances on a number line one at a time. This item chains three such steps with mixed numbers and a negative position, more than typical grade 7 work demands.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
4 steps
Reading level
Question and choices: about grade 9, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every number is a mixed fraction, one position is negative, and the figure's even spacing hides how long KL really is. The distance route, KL = 9 3/4 − 3 1/2 − 1 1/8, avoids the negative entirely but ends one step short of a position unless K's 3/8 is added back. That last step is the one most students skip, which is why the question sits at the top of the scale.

  1. A very close wrong answer major Choosing the answer

    The choices 5 1/8, 5 1/4 and 5 1/2 lie within 3/8 of one another, so no estimate separates them; only exact eighths and adding K's 3/8 pick out the right one. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic moderate Working it out

    Every step mixes halves, quarters and eighths, and placing J means 3/8 − 3 1/2 = −3 1/8, a subtraction that crosses zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    Only K is labeled, and the points are drawn evenly spaced although KL is over four times LM, so the picture gives no help with sizes. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. Many steps minor Working it out

    Three linked moves, K to J, J to M and M back to L, each depend on the last, so an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Start from K's labeled position, subtract JK to place J, add JM to place M, then subtract LM, keeping every fraction in eighths.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
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.
Equal fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top number (numerator) and b is the bottom number (denominator). Multiplying both by the same number k gives an equal fraction. It is like cutting each piece into k smaller pieces. Use it to give fractions the same bottom before you add or subtract.

  1. Sort out what you know

    Only K has a value: 38\frac{3}{8}. JK means the distance from J to K. You know JK = 3123\frac{1}{2}, JM = 9349\frac{3}{4} and LM = 1181\frac{1}{8}.

    You need the position of L, the number where L sits. The points are drawn evenly spaced, but the distances are not equal. So use the numbers, not the picture.

  2. Write the fractions in eighths

    The fractions use halves, fourths and eighths. Give them all a bottom of 8:

    312=348934=9683\frac{1}{2} = 3\frac{4}{8} \qquad 9\frac{3}{4} = 9\frac{6}{8}

    Watch out: 34\frac{3}{4} is 68\frac{6}{8}, not 48\frac{4}{8}. That slip leads to choice B.

  3. Find the distance from K to M

    JM is JK and KM put together. So take JK away from JM:

    968−348=6289\frac{6}{8} - 3\frac{4}{8} = 6\frac{2}{8}

    KM is 6286\frac{2}{8}, or 6146\frac{1}{4}. Watch out: this is a distance, not where L sits. Choice D stops here.

  4. Find the distance from K to L

    KM is KL and LM put together. So take LM away:

    628−118=5186\frac{2}{8} - 1\frac{1}{8} = 5\frac{1}{8}

    KL is 5185\frac{1}{8}. Watch out: this is still a distance. Choice A stops here.

  5. Find the position of L

    L is 5185\frac{1}{8} to the right of K, and K is at 38\frac{3}{8}. Add:

    38+518=548=512\frac{3}{8} + 5\frac{1}{8} = 5\frac{4}{8} = 5\frac{1}{2}

    So the answer is C, 5125\frac{1}{2}.

Check your answer

Place every point. J is 3483\frac{4}{8} left of K, at −318-3\frac{1}{8}. M is 9689\frac{6}{8} right of J, at 6586\frac{5}{8}. L is 1181\frac{1}{8} left of M, at 5485\frac{4}{8}, which is 5125\frac{1}{2}. Same answer.

More questions like this

The original DOE question
Question 111 as printed on page 91 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 111. Open page 91 of the handbook (PDF)

Question 93 of 108

2018 Test A Math, question 56

Question

SERVINGS OF FRUITS AND VEGETABLES

Number of Servings of Fruits and Vegetables Number of Students
0 5
1 7
2 3
3 4
4 0
5 1

There are 20 students in a class. The frequency table above shows the number of students in this class who ate 0, 1, 2, 3, 4, or 5 servings of fruits and vegetables yesterday. What is the mean number of servings of fruits and vegetables eaten yesterday per student in this class?

  • A1121\frac{1}{2}
  • B33
  • C3133\frac{1}{3}
  • D44
Show the answer

Correct answer: A

First, determine the total number of servings of fruits and vegetables that the students ate by multiplying the number of servings by the number of students in each row of the table.

Then add that column to get the total number of servings:

Number of Servings of Fruits and Vegetables Number of Students Number of Servings × Number of Students
0 5 0
1 7 7
2 3 6
3 4 12
4 0 0
5 1 5

Total: 30

Calculate the mean by dividing the total number of servings of fruits and vegetables by the total number of students:

3020=112\frac{30}{20}=1\frac{1}{2}

Question analysis

This statistics question uses a frequency table showing how many of 20 students ate 0, 1, 2, 3, 4 or 5 servings of fruits and vegetables, and asks for the mean servings per student. Students must weight each serving amount by its number of students, add, and divide. Averaging the wrong column of the table is the common mistake.

NYS standards
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a mean is grade 6 content; doing it from a frequency table, where each value must be weighted by its count, is the multi-step version grade 7 students meet regularly.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 64 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The answer is smaller than every wrong choice, which surprises students who average the table's columns directly. The question rewards seeing that each row stands for several students, so the total servings come from weighting each value by its count. The stem's reminder that there are 20 students points to the right divisor.

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

    The frequency table has a value column and a count column; the mean needs each value multiplied by its count, so the two columns play different roles. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Six products, 0, 7, 6, 12, 0 and 5, must be found and added to 30 before dividing by the 20 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply each serving amount by its number of students, add those products to get the total servings, then divide by the 20 students.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean
m=Snm = \frac{S}{n}
m is the mean, or average. S is the sum of all the values, here the total servings eaten. n is the number of values, here the number of students. The mean shares the total equally, as if every student ate the same amount.

  1. Sort out the table

    This is a frequency table. Each row says how many students ate a certain number of servings. For example, the second row says 7 students ate 1 serving each.

    There are 20 students. You need the mean, or average, number of servings per student.

  2. Find the servings in each row

    In each row, multiply the servings by the number of students:

    ServingsStudentsServings × Students
    050
    177
    236
    3412
    400
    515

  3. Add up all the servings

    0+7+6+12+0+5=300 + 7 + 6 + 12 + 0 + 5 = 30

    The class ate 30 servings in all. The students add up to 20, as the question says: 5 + 7 + 3 + 4 + 0 + 1 = 20.

  4. Divide by the number of students

    m=3020=11020=112m = \frac{30}{20} = 1\frac{10}{20} = 1\frac{1}{2}

    Watch out: do not just average the serving numbers 1 to 5. That gives 3, choice B, and ignores how many students ate each amount. Dividing 20 students by 6 rows gives choice C.

    So the answer is A, 1121\frac{1}{2}.

Check your answer

12 of the 20 students ate 0 or 1 serving. Only 5 ate 3 or more. So the mean should be small, between 1 and 2. 1121\frac{1}{2} fits. Choices B, C and D are all 3 or more, which is too big.

More questions like this

The original DOE question
Question 113 as printed on page 92 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 92, question 113. Open page 92 of the handbook (PDF)

Question 94 of 108

2018 Test B English, question 53
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

How does the graph support the ideas in paragraph 8?

  • AIt indicates how welcome the improvement of long-distance communication was in the United States.
  • BIt provides evidence of the dramatic increase in the number of telegraph messages as Morse’s system expanded across the United States.
  • CIt reveals that by the twentieth century millions of people had used the telegraph despite earlier hesitations about the system.
  • DIt shows how improvements that allowed Morse code to be read in real time made relaying telegraph messages faster and increased the system’s usage.
Show the answer

Correct answer: B

The question asks how the graph supports the ideas in paragraph 8. The graph shows the number of messages transmitted from 1870 to 1920. In 1920 almost 160 million messages were transmitted, showing a “dramatic”—more than double—increase from 1910, which makes Option B the correct answer.

Option A is incorrect because, while it can be implied from the expansion shown on the graph that people possibly welcomed the improvement in long-distance communication, the graph does not state this explicitly.

Option C is incorrect because the graph does not show how many people used the telegraph, but rather it shows the number of messages sent. Additionally, paragraph 8 provides no details about hesitation on the part of individuals.

Option D is incorrect because, while improvements in recording and interpreting code were made, the graph does not show the effect of this advancement on the speed or number of messages being transmitted.

Question analysis

This question asks how a graph that accompanies “Invention of the Telegraph” supports the ideas in paragraph 8, the passage’s closing account of what followed the first telegraph line. Students must connect a data display to a paragraph of text, and the choices differ in how much they claim a graph of this kind could prove.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R5 structure and genre features
Grade level
Grade 7 reading and task
Compared with grade 7
At grade 7 Connecting a graph to a paragraph is standard grade 7 work with mixed formats. The demand is typical: read the paragraph, read the data, and check that a claim about the graph does not go beyond it.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Evidence
Answered from one paragraph or stanza
Reading level
Passage: about grade 11, 910 words; question and choices: about grade 11, 88 words.
Corrected from the DOE handbook
  • A misreading of the graph 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 graph is simple to read; the challenge is deciding how much a count of messages can prove. Wrong choices claim the graph shows public feeling, numbers of people, or the effect of one improvement, and the last of these is tempting because paragraph 8 mentions that improvement. Students who ask what the vertical axis measures can reject all three quickly.

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

    The bar graph counts messages in millions, rising from about 10 in 1870 to about 155 in 1920; it measures neither people, nor speed, nor causes, which is what rules out three choices. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer moderate Choosing the answer

    The key and the real-time-reading choice both tie the rise in messages to a change described in paragraph 8; they part on one detail, the claim that relaying became “faster,” which a count of messages cannot show. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Combines ideas minor Working it out

    Students must join the graph’s fifty years of growth to paragraph 8’s spread of overhead wires along the coast, westward and across the ocean. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Reread paragraph 8 and look at what the graph measures, then for each choice ask whether both the paragraph and the graph back it up.

Traps

  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
  • Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says 1920 showed an “almost double” increase from 1910. The graph shows about 75 million messages in 1910 and about 155 million in 1920, so the number more than doubled. The site now says “more than double.”

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This asks how the bar graph supports paragraph 8. To answer, you need to know two things. What does paragraph 8 say? And what does the graph actually measure?

  2. Read paragraph 8

    Paragraph 8 says that “overhead wires connected cities up and down the Atlantic coast.” Then “Telegraph lines soon extended westward, and within Morse’s own lifetime they connected the continents of Europe and America.” So the main idea is growth. The telegraph spread farther and farther.

  3. Read the bar graph

    The graph’s title is Telegraph Messages Transmitted, 1870–1920. Its side axis counts messages, in millions. The bars rise every ten years. There were about 10 million messages in 1870 and about 75 million in 1910. In 1920 there were about 155 million. That last jump is more than double. The graph counts messages. It does not count people, speed or feelings.

  4. Check each choice

    • A is wrong. A count of messages can’t show how welcome something was. That is a feeling.
    • B fits. The bars show a huge rise in messages as the lines spread.
    • C is wrong. The graph counts messages, not people. Nothing mentions “earlier hesitations” either.
    • D is wrong. Paragraph 8 does mention reading code “in real time,” or right away. But the graph shows totals only. It can’t show what caused the rise or how fast messages moved.

  5. Choose the answer

    The graph shows messages rising from about 10 million to about 155 million as the wires spread. So the answer is B, “It provides evidence of the dramatic increase in the number of telegraph messages as Morse’s system expanded across the United States.”

Why this answer and not the closest wrong one

D is the closest trap, since paragraph 8 really does mention reading code in real time. But ask what the side of the graph measures. It counts messages, nothing else. A count can show that use grew, as B says. It can’t show that messages got faster, or why.

More questions like this

The original DOE question
Question 53 as printed on page 150 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 150, question 53. Open page 150 of the handbook (PDF) 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 95 of 108

2018 Test B English, question 55

Uses the passage above: “Invention of the Telegraph”.

Question

How does the map provide additional support for a central idea of the passage?

  • Aby demonstrating that Morse’s telegraph system greatly exceeded the limitations of previous long-distance communication systems
  • Bby indicating that Morse was lacking foresight by seeking a patent only in North America
  • Cby proving that Gale’s advice helped Morse extend the telegraph’s range much farther than first thought possible
  • Dby providing a graphic that shows the surprisingly immense influence Morse’s telegraph had across the world
Show the answer

Correct answer: A

The question asks how the map provides additional support for a central idea of the passage. The map shows the telegraph line crossing continents in 1902, just forty-four years after the first transatlantic message was sent in 1858. This is a much greater distance than the first line Morse built between “the Supreme Court chamber of the Capitol building and the railroad station in Baltimore” (paragraph 7), making Option A correct.

Option B is incorrect because the map shows the distance the line covers; it does not indicate Morse’s lack of foresight in seeking patents only in North America.

Option C is incorrect because, while Gale’s advice was crucial to Morse’s success, the map shows the physical line stretching across continents many years later, after their initial work.

Option D is incorrect because, while it can be inferred that the telegraph had influence across the world, the amount of influence cannot be determined.

Question analysis

This question asks how a map that accompanies “Invention of the Telegraph” supports a central idea of the passage about Samuel Morse’s invention. Students must link the information in a map to the passage’s main point, and choices that sound alike differ in how much they claim, so careful reading of each one decides it.

NYS standards
  • 7R1 textual evidence and inferences
  • 7R2 theme or central idea
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Connecting a map to a central idea is grade 7 work, but this one requires tying a visual to the whole passage and weighing choices that differ by a single word or claim, which pushes it above typical demand.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from the whole text
Reading level
Passage: about grade 11, 910 words; question and choices: about grade 11, 77 words.
Corrected from the DOE handbook
  • An inaccurate detail 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

Two choices describe the map’s reach in nearly the same terms, and the difference is in what each claims the map proves. A map of lines can show distance covered, which outruns the hilltop towers of paragraph 1, but it cannot measure influence or reveal surprise. Students lose time because the right choice depends on a paragraph read many minutes earlier.

  1. A very close wrong answer major Choosing the answer

    The choice about the telegraph’s “surprisingly immense influence” sounds like a fair reading of a map spanning five continents and differs from the key mainly in claiming influence rather than reach. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Evidence in several places moderate Working it out

    The previous systems are in paragraph 1, towers on hills relaying flag and light signals, far from the map at the end of the passage, and the two must be compared. (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)

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

    The map shows a 1902 British line from London across the Atlantic to Canada, south to Bermuda, Barbados and Cape Town, and on to India, Australia and New Zealand, with Washington left off it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: First state the passage’s central idea in a sentence, then ask which choice describes something a map of telegraph lines could actually show that also fits that idea.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says the 1902 line came “forty-four years after the first message was sent in 1858.” The first telegraph message was Morse’s in 1844 (paragraph 1); 1858 is the year of the first transatlantic message, as the map’s caption says. The site now says “the first transatlantic message.”

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This asks how the map supports a central idea, or main message, of the passage. You need two things: what the map actually shows, and which big idea of the passage it backs up.

  2. Read the map and its caption

    The map shows the British All-Red Telegraph Line in 1902. A thick black line stands for telegraph wire. It runs from London across the Atlantic Ocean to Canada and on to Vancouver. It also goes to Bermuda, Barbados and Cape Town. Then it crosses the Indian Ocean to India, Australia and New Zealand. The caption says that by 1902 the line “connected most of the world.”

  3. Compare it with earlier systems

    Now look back at paragraph 1. Earlier systems used flags or lights. An observer had to “decipher a signal from a high tower on a distant hill and then send it on to the next station.” Decipher means figure out. Those signals only worked if you could see the next tower. The line on the map crosses whole oceans. That shows how far the new system went past the old limits.

  4. Check each choice

    • A fits. Lines across oceans go far past towers on hills.
    • B is wrong. A map of wires shows nothing about patents. Paragraph 5 even says Morse tried to get patents overseas.
    • C is wrong. The map is from 1902, long after Gale helped. It can’t prove what his advice did.
    • D is wrong. The map shows how far the lines reached, not how much they changed lives. It also can’t show that anything was surprising.

  5. Choose the answer

    The map shows wires circling the globe, far beyond hilltop flags and lights. So the answer is A, “by demonstrating that Morse’s telegraph system greatly exceeded the limitations of previous long-distance communication systems.”

Why this answer and not the closest wrong one

D is the closest trap, because the map does cover the world. But a map of lines shows reach, meaning how far something goes. Influence means how much it changed people’s lives, and a map can’t measure that. A claims only what the map can show: distance far past the old towers.

More questions like this

The original DOE question
Question 55 as printed on page 151 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 151, question 55. Open page 151 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 96 of 108

2018 Test B Math, question 5

Question

Four straight lines cross at point P: one horizontal, one vertical and two slanted. Above the horizontal line, from right to left: a 60° angle between the horizontal line and the slanted line rising to the right, a 30° angle between that line and the vertical line, an unlabeled angle between the vertical line and the slanted line rising to the left, and the angle y° between that line and the horizontal line. Below the horizontal line, a 25° angle lies between the vertical line and the slanted line falling to the right.

Four straight lines intersect at point P as shown above. What is the value of yy ?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 65

Call the missing angle in the top half of the figure xx.

The sum of the four angles on the top of the figure is equal to 180°.

x+y+30+60=180x+y+30+60=180

Since xx is a vertical angle with the 25° angle, then xx is also 25°. Use that to solve for yy.

25+y+30+60=18025+y+30+60=180

y+115=180y+115=180

y=65y=65

Question analysis

This grid-in angle question shows four straight lines crossing at a single point P and asks for the size of an unknown angle y. Students combine vertical angles with the fact that angles along a straight line add to 180 degrees. One angle they need is unlabeled and must be found from its vertical partner.

NYS standards
  • 7.G.5 angle relationships and equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using vertical angles and straight angles to write an equation for an unknown angle is a core grade 7 geometry task. Four lines create more angles than usual, but the reasoning stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is one subtraction; the work is in seeing the figure correctly. Students have to notice that a fourth angle sits between the 30° angle and y, recognize it as the vertical partner of the 25° angle, and then use the straight line. The drawing is not to scale: y is drawn close to the 60° angle across the vertical line, which rewards anyone who computes rather than eyeballs.

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

    Four lines make eight angles around P but only three are labeled; above the horizontal line sit 60°, 30°, an unlabeled angle and y°, and that unlabeled angle is easy to overlook. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    The solution joins two facts: the unlabeled angle is vertical to the 25° angle below P, and the four angles above the horizontal line make a straight angle of 180°. (Embretson, 1998; Daniel & Embretson, 2010)

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

    As a grid-in, nothing confirms the result, and the drawing shows y at about 60°, so a reading of the picture produces a number that looks right. (Enright, Morley & Sheehan, 2002)

Strategy: Pick a straight line through P and list the angles on one side of it; replace the unlabeled one with the labeled angle vertical to it, then subtract from 180.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b+c+d=180a + b + c + d = 180
a, b, c and d are angles that sit side by side along one side of a straight line, all meeting at one point. Together they make a half turn, which is 180 degrees. This works for any number of angles, not just four.
Vertical angles
∠1=∠2\angle 1 = \angle 2
When two lines cross, they make an X. Angle 1 and angle 2 are opposite each other across the crossing point. Angles like these are called vertical angles. They are always equal.

  1. Look closely at the figure

    Four lines cross at point P. Look above the flat, or horizontal, line. From right to left there are four angles: 60°, 30°, an angle with no label, and y°.

    Below the horizontal line there is a 25° angle. You need to find y.

  2. Find the unlabeled angle

    The unlabeled angle sits between the up-and-down, or vertical, line and the slanted line rising to the left.

    That slanted line keeps going through P and down to the right. The 25° angle sits between that same line and the vertical line, below P.

    So the two angles are opposite each other across P. They are vertical angles, so they are equal. The unlabeled angle is 25°.

  3. Add the angles above the line

    The four angles above the horizontal line fill one side of a straight line. So they add up to 180°.

    a+b+c+d=180a + b + c + d = 180

    y+25+30+60=180y + 25 + 30 + 60 = 180

    y+115=180y + 115 = 180

  4. Solve for y

    Take 115 away from 180 to get y alone.

    y=180−115=65y = 180 - 115 = 65

    Watch out: Do not skip the unlabeled angle. Using only 30 and 60 gives 90. Do not stop at the sum, 115, either. And do not guess from the picture. Angle y looks like 60°, but the drawing is not exact.

    So you would grid 65.

Check your answer

Add all four angles above the horizontal line: 65+25+30+60=18065 + 25 + 30 + 60 = 180. They fill a straight line exactly, so 65 fits.

More questions like this

The original DOE question
Question 62 as printed on page 153 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 62. Open page 153 of the handbook (PDF)

Question 97 of 108

2018 Test B Math, question 7

Question

Parallelogram PQRS with Q and R along the top and P and S along the bottom. Point T is on the bottom side PS, and segment QT cuts off triangle PQT. Angle PQT, at Q, is marked 50°; angle PTQ, at T, is marked 70°; angle QRS, at R, is marked with an arc and no measure.

In the figure above, PQRS is a parallelogram. The measure of ∠PQT is 50°, and the measure of ∠PTQ is 70°. What is the measure of ∠QRS?

  • A60°
  • B70°
  • C80°
  • D120°
Show the answer

Correct answer: A

Find the missing angle, angle QPT, of triangle PQT:

180° − 70° − 50° = 60°

In parallelogram PQRS, angle QPT is congruent to angle QRS, so the measure of angle QRS is also 60°.

Question analysis

This geometry question uses parallelogram PQRS and a triangle PQT inside it, giving two of the triangle's angles and asking for angle QRS. Students need the triangle angle sum and the fact that opposite angles of a parallelogram are equal. The challenge is connecting the angles found in the triangle to the angles of the parallelogram.

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 work uses vertical, adjacent and supplementary angles. This also needs the triangle angle sum and the opposite angles of a parallelogram, ideas New York places in grade 8, though the arithmetic is short.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 4, 25 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

Few steps are needed, but the figure misleads. It draws triangle PQT as isosceles, so the angle at P looks the same as the marked 70°, and the angle at R looks the same again; a student who trusts the drawing never computes. Those who do compute the angle at P must still choose the opposite angle of the parallelogram, not the angle next to it, which is where 120° comes from.

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

    The figure carries facts the stem leaves out: T lies on side PS, the 50° is only part of the angle at Q, and the 70° is at T, not at a corner of the parallelogram. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Two separate facts link up: the angles of triangle PQT sum to 180°, and the angle at P, once known, equals the opposite angle at R of parallelogram PQRS. (Embretson, 1998; Daniel & Embretson, 2010)

  3. A very close wrong answer moderate Choosing the answer

    60° and 70° are close enough that judging the angle at R by eye cannot separate them, and the drawing even favors 70°; only computing the triangle's third angle does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Something you must already know minor Working it out

    Neither the triangle angle sum nor the rule that opposite angles of a parallelogram are equal is given; students must bring both to the figure. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of triangle PQT from the angle sum, work out which angle of the parallelogram it matches, then use the parallelogram's angle facts to reach angle QRS.

Traps

  • Trusting the picture: The figure suggests something it does not state; work from the given facts.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles in a triangle
a+b+c=180a + b + c = 180
a, b and c are the three angles inside any triangle. They always add up to 180 degrees. If you know two of the angles, subtract them from 180 to find the third.
Opposite angles of a parallelogram
∠P=∠R\angle P = \angle R
A parallelogram is a four-sided shape whose opposite sides are parallel, meaning they never meet. In parallelogram PQRS, the corners at P and R are opposite each other. Opposite angles of a parallelogram are always equal.

  1. Sort out the figure

    PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, meaning they never meet.

    Point T is on the bottom side PS. Segment QT cuts off triangle PQT on the left. In that triangle, angle PQT is 50° and angle PTQ is 70°.

    You need angle QRS, the corner at R.

  2. Find the third angle of the triangle

    Triangle PQT has angles at Q, T and P. You know two of them. The angle at P is angle QPT. All three must add to 180°.

    a+b+c=180a + b + c = 180

    ∠QPT+50+70=180\angle QPT + 50 + 70 = 180

    ∠QPT=180−50−70=60\angle QPT = 180 - 50 - 70 = 60

    So angle QPT is 60°.

  3. Match angle P to angle R

    T lies on side PS. So angle QPT is also the corner of the parallelogram at P. Angle P and angle R are opposite corners. Opposite angles are equal, so angle QRS is also 60°.

    Watch out: Angle Q is next to angle P, not opposite it. Angle Q is 120°, choice D, but the question asks about angle R. Also, do not trust the drawing. Angle R may look like 70°, choice B.

    So the answer is A, 60°.

Check your answer

The four corners of the parallelogram are 60°, 120°, 60° and 120°. They add to 360°, as the angles of any four-sided shape must. Also, the 50° angle at Q fits inside the 120° corner at Q.

More questions like this

The original DOE question
Question 64 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 64. Open page 154 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 98 of 108

2018 Test B Math, question 13

Question

DESSERT CHOICES

Dessert Number of Times Ordered
Cookies 42
Pie 23
Cake 47
Ice Cream 48

The table above shows the number of times that different desserts were ordered at a restaurant. Based on this information, what is the probability of a customer ordering ice cream as a dessert?

  • A25%
  • B30%
  • C40%
  • D48%
Show the answer

Correct answer: B

The total number of desserts ordered is

42 + 23 + 47 + 48 = 160

The probability that ice cream was chosen is

48160=310=30%\frac{48}{160}=\frac{3}{10}=30\%.

Question analysis

This probability question gives a table of how many times four desserts were ordered at a restaurant and asks how likely a customer is to order ice cream, as a percent. Students total the orders and divide. The trap is using a raw count, or the wrong total, in place of the full number of desserts ordered.

NYS standards
  • 6.SP.8b probability models from observed frequencies
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a probability from observed frequencies is grade 6 content in New York. Here it takes one sum, one division and a conversion to a percent with friendly numbers, lighter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
3 steps
Reading level
Question and choices: about grade 5, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a direct relative-frequency question with friendly numbers, and most students finish it quickly. The risk is using a number from the table as if it were already a percent, or treating the four desserts as equally likely. Remembering that a probability compares one count with the total of all counts prevents both mistakes.

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

    The four counts sit in a table, and the probability needs all of them: the ice cream row gives the numerator and the sum of the column gives the denominator. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic minor Working it out

    Adding 42 + 23 + 47 + 48 by hand and then simplifying 48/160 to 3/10 are small steps where a careless total spoils the percent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four counts in the table to get all the orders, divide the ice cream count by that total, and convert the fraction to a percent.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}
P is the probability, the chance that something happens. The top counts the outcomes you want, here ice cream orders. The bottom counts every outcome, here all dessert orders. It works because each order counts the same.
Fraction to percent
percent=fraction×100\text{percent} = \text{fraction} \times 100
Percent means out of 100. Multiply a fraction or a decimal by 100 to write it as a percent. For example, 3/10 times 100 is 30, so 3/10 is 30%.

  1. Sort out the table

    The table shows how many times each dessert was ordered. Ice cream was ordered 48 times.

    You need the chance that a customer orders ice cream. The orders in the table show how often each dessert is picked.

  2. Find the total orders

    Add all four rows of the table.

    42+23+47+48=16042 + 23 + 47 + 48 = 160

    There were 160 dessert orders in all. This is the bottom of the fraction, because it counts every order.

  3. Write the probability

    Put the counts into the probability formula. The top is the 48 ice cream orders. The bottom is all 160 orders.

    P=outcomes you wantall outcomes=48160P = \frac{\text{outcomes you want}}{\text{all outcomes}} = \frac{48}{160}

    Simplify the fraction. Both numbers can be divided by 16.

    48÷16160÷16=310\frac{48 \div 16}{160 \div 16} = \frac{3}{10}

  4. Change to a percent

    310×100=30\frac{3}{10} \times 100 = 30

    The probability is 30%.

    Watch out: 48 orders is not 48%, choice D. You must divide by the total. Also, the four desserts are not equally popular. So the chance is not 1 out of 4, or 25%, choice A.

    So the answer is B, 30%.

Check your answer

Work backward: 0.30×160=480.30 \times 160 = 48. That matches the ice cream row. Also, 48 is a bit less than a third of 160, and 30% is a bit less than a third.

More questions like this

The original DOE question
Question 70 as printed on page 155 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 70. Open page 155 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 99 of 108

2018 Test B Math, question 18

Question

PRICES FOR NEWSPAPER ADVERTISING

Page Space Price
14\frac{1}{4} page $200
12\frac{1}{2} page $350
full page $600

The table above shows prices for newspaper advertising. A store purchased 14\frac{1}{4} pages, 12\frac{1}{2} pages, and full pages of page space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?

  • A1341\frac{3}{4} pages
  • B10 pages
  • C161216\frac{1}{2} pages
  • D171217\frac{1}{2} pages
Show the answer

Correct answer: D

The question says that an equal number (xx) of each type of page space was purchased. To find the number of each type of page space that was purchased, multiply the price per type by xx and set it equal to the total amount spent, then solve for xx:

200x+350x+600x=11,500200x+350x+600x=11,500

1,150x=11,5001,150x=11,500

x=10x=10

Thus, the store purchased 10 units of each type of page space. To find the total amount of page space purchased, multiply each type of page space by 10, and add:

(10×14 page)+(10×12 page)+(10×1 page)\left(10\times\frac{1}{4}\text{ page}\right)+\left(10\times\frac{1}{2}\text{ page}\right)+(10\times 1\text{ page})

=1712 pages=17\frac{1}{2}\text{ pages}

Question analysis

This word problem gives a table of newspaper advertising prices for quarter-page, half-page and full-page ads, and says a store bought equal numbers of each for a set total. Students write and solve an equation for how many ads of each size were bought. The question wants total page space, which takes a second calculation with fractions.

NYS standards
  • 7.EE.4a two-step equations
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students write and solve equations like px = r from word problems. This adds a table, combines three prices into one coefficient and needs a second step with fractions after solving, a longer chain than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 5, 59 words.
Corrected from the DOE handbook
  • An arithmetic 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

Solving for the number of ads is the obvious goal, and finding it feels like the finish, but the question asks for page space, which needs one more pass through the table. The awkward wording about buying pages in equal numbers slows students before any arithmetic starts. The wrong choices mark each place to stop: one set of ads, the number of each ad, or a slip in adding the fractional pages.

  1. Many steps major Working it out

    The work has two phases: turn the money into a count, 1,150x = 11,500 so x = 10, then turn the count into space by multiplying 10 by 1/4, 1/2 and 1 and adding. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math moderate Understanding the question

    The phrase “purchased 1/4 pages, 1/2 pages, and full pages of page space in equal numbers” has to be read as the same number of ads of each size, so the three prices add to one coefficient. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    Both columns of the table are needed: the price column builds the equation, and the page-space column turns the count of ads into the final total. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Let x be the number of each ad size, add the three prices to write one equation for the total, solve it, then multiply x by the page space in one set of ads.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook adds 200x + 350x + 600x and writes 1,500x = 11,500. The sum is 1,150x, and the handbook’s next line, x = 10, only follows from 1,150x = 11,500. The site shows 1,150x = 11,500.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal sets
number of sets=total spentcost of one set\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}}
A set here means one ad of each size. The cost of one set is what one of each costs together. Dividing the total money by it tells how many sets were bought, because every set costs the same.

  1. Sort out the question

    The store bought the same number of each ad size: quarter page, half page and full page. It spent $11,500 in all.

    You need the total page space. That is not the number of ads, and not the money.

  2. Find the cost of one set

    Think of one quarter page, one half page and one full page as a set. The store bought equal numbers of each, so it bought whole sets.

    200+350+600=1,150200 + 350 + 600 = 1{,}150

    One set costs $1,150.

  3. Find how many sets

    number of sets=total spentcost of one set=11,5001,150=10\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}} = \frac{11{,}500}{1{,}150} = 10

    The store bought 10 sets. That means 10 ads of each size.

    Watch out: 10 is the number of each ad, not the page space. Stopping here gives choice B.

  4. Find the page space in one set

    Add the sizes in one set. Write them in quarters so they have the same bottom number.

    14+12+1=14+24+44=74\frac{1}{4} + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}

    One set is 74\frac{7}{4}, or 1341\frac{3}{4}, pages.

    Watch out: 1341\frac{3}{4} pages is only one set. That is choice A.

  5. Multiply by the number of sets

    There are 10 sets, so multiply the space in one set by 10.

    10×74=704=171210 \times \frac{7}{4} = \frac{70}{4} = 17\frac{1}{2}

    So the answer is D, 171217\frac{1}{2} pages.

Check your answer

Add the sizes another way. 10 quarter pages make 2.5 pages, not 1.5. 10 half pages make 5 pages. 10 full pages make 10 pages. Then 2.5+5+10=17.52.5 + 5 + 10 = 17.5, the same answer.

More questions like this

The original DOE question
Question 75 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 75. Open page 156 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 100 of 108

2018 Test B Math, question 23

Question

A number line with point M on the left and point N on the right. N is labeled 1 1/3; no other coordinates are marked.

On the number line above, MN=556MN=5\frac{5}{6}. What is the position of point MM?

  • A−716-7\frac{1}{6}
  • B−412-4\frac{1}{2}
  • C4124\frac{1}{2}
  • D7167\frac{1}{6}
Show the answer

Correct answer: B

To find MM, subtract N−MN-M and set it equal to the length:

113−M=5561\frac{1}{3}-M=5\frac{5}{6}

−M=556−113-M=5\frac{5}{6}-1\frac{1}{3}

−M=556−126-M=5\frac{5}{6}-1\frac{2}{6}

−M=436-M=4\frac{3}{6}

M=−412M=-4\frac{1}{2}

Question analysis

This number line question gives the distance between points M and N as a mixed number and asks for the position of M. Students subtract fractions with unlike denominators, moving past zero into negative numbers. Direction along the line and the sign of the result are where it gets tricky.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students find distances on a number line as the absolute value of a difference. Doing it with mixed numbers that have unlike denominators, and crossing zero, is a typical and slightly demanding grade 7 task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The distance is given and one endpoint is known, so the question is really about direction and sign. Because the drawing shows no zero, students must reason that moving 5 5/6 to the left from 1 1/3 carries M past zero. The choices pair each size with each sign, so getting the size right is only half the work; the sign decides between two choices.

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

    The number line labels only N, at 1 1/3, and shows no zero, so students must see from the picture that M lies left of N and work out whether it passes zero. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic moderate Working it out

    The subtraction 1 1/3 − 5 5/6 mixes unlike denominators and crosses zero: in sixths it is 8/6 − 35/6 = −27/6, which then reduces to a mixed number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer moderate Choosing the answer

    4 1/2 and −4 1/2 are written the same except for the sign; only noticing that M lies left of N, so below 1 1/3, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Read N's coordinate from the number line, convert everything to sixths, and move the given distance from N toward M's side of the line.

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.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
On a number line, numbers grow as you move right. The distance between two points is the larger number minus the smaller one. That is the right point minus the left point, so a distance is never negative.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.

  1. Read the number line

    Point N is at 1131\frac{1}{3}. Point M is to the left of N. MN means the distance from M to N, and it is 5565\frac{5}{6}.

    You need the number at point M.

  2. Decide which way to move

    On a number line, numbers get smaller as you move left. M is left of N, so N is the right point.

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    556=113−M5\frac{5}{6} = 1\frac{1}{3} - M

    So M is what you get when you take the distance away from N.

    M=113−556M = 1\frac{1}{3} - 5\frac{5}{6}

    The distance is bigger than 1131\frac{1}{3}, so you will pass 0. M will be negative.

  3. Change both numbers to sixths

    To subtract, the fractions need the same bottom number. Use sixths, since 3 goes into 6. Change each mixed number to an improper fraction, a fraction whose top is bigger than its bottom.

    113=1×3+13=43=861\frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{4}{3} = \frac{8}{6}

    556=5×6+56=3565\frac{5}{6} = \frac{5 \times 6 + 5}{6} = \frac{35}{6}

  4. Subtract and simplify

    M=86−356=−276M = \frac{8}{6} - \frac{35}{6} = -\frac{27}{6}

    27 sixths is 4 wholes (24 sixths) and 3 sixths more. And 36=12\frac{3}{6} = \frac{1}{2}. So M is −412-4\frac{1}{2}.

    Watch out: 4124\frac{1}{2} without the minus sign is choice C. That point would be to the right of N, not the left.

    So the answer is B, −412-4\frac{1}{2}.

Check your answer

Find the distance from your M to N. From −412-4\frac{1}{2} to 0 is 4124\frac{1}{2}, or 276\frac{27}{6}. From 0 to N is 86\frac{8}{6}. Together, 276+86=356=556\frac{27}{6} + \frac{8}{6} = \frac{35}{6} = 5\frac{5}{6}. That matches MN.

More questions like this

The original DOE question
Question 80 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 80. Open page 157 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 101 of 108

2018 Test B Math, question 32

Question

PET SURVEY

Number of Pets Number of Students
0 12
1 16
2 7
3 or more 5

Amy surveyed students at her school about the number of pets they have. What is the probability that a student who participated in the survey has at least 2 pets?

  • A740\frac{7}{40}
  • B112\frac{1}{12}
  • C18\frac{1}{8}
  • D310\frac{3}{10}
Show the answer

Correct answer: D

First, add the number of students for each category to find out how many total students were in the survey:

12 + 16 + 7 + 5 = 40

The number of students who had at least 2 pets are the ones who have 2 pets (7) plus the ones who have 3 or more (5). The total number of students with at least 2 pets is

7 + 5 = 12

The probability of a student in the survey having at least two pets is:

1240=310\frac{12}{40}=\frac{3}{10}

Question analysis

Amy's pet survey table shows how many pets each student has, and the question asks for the probability that a randomly chosen student has at least 2 pets. It tests finding a probability from survey data. The tricky part is reading "at least 2" correctly and using the right total as the denominator.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Simple probability from a frequency table is grade 6 work in New York. The only extra demand is interpreting "at least" and adding a column, two short steps a grade 7 student should handle easily.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light, so the question is decided by the phrase "at least 2" and by remembering that the denominator is everyone surveyed. Students who include only one of the two qualifying rows reach an answer with the right denominator, which makes the mistake hard to notice. Rereading the condition before choosing is the main safeguard.

  1. Turning words into math minor Understanding the question

    "At least 2 pets" has to be turned into two rows of the table, the 2-pet row and the 3-or-more row, not one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The denominator is not printed anywhere; it has to be built by adding all four rows of the table, 12 + 16 + 7 + 5 = 40. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Total the students in every row to get the denominator, then add the counts for each row that satisfies "at least 2" and simplify the fraction.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out the question

    The table shows how many students have each number of pets. You pick one student who took the survey. You need the chance that the student has “at least 2 pets”.

    At least 2 means 2 or more. So two rows count: the 2 pets row and the 3 or more row.

  2. Find the total number of students

    The bottom of the fraction is everyone in the survey. Add all four rows.

    12+16+7+5=4012 + 16 + 7 + 5 = 40

    There are 40 students in all.

  3. Count the students you want

    Add the 2 pets row and the 3 or more row.

    7+5=127 + 5 = 12

    So 12 students have at least 2 pets.

    Watch out: using only the 7 students with exactly 2 pets gives 740\frac{7}{40}, choice A. Using only the 5 with 3 or more gives 540=18\frac{5}{40} = \frac{1}{8}, choice C.

  4. Write and simplify the probability

    Use the probability formula.

    P=1240P = \frac{12}{40}

    Divide the top and bottom by 4, since 4 goes into both.

    1240=12÷440÷4=310\frac{12}{40} = \frac{12 \div 4}{40 \div 4} = \frac{3}{10}

    So the answer is D, 310\frac{3}{10}.

Check your answer

The students with fewer than 2 pets are 12+16=2812 + 16 = 28. Then 28+12=4028 + 12 = 40, so every student is counted once. Also, 12 out of 40 is a bit less than a third. So is 310\frac{3}{10}.

More questions like this

The original DOE question
Question 89 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 89. Open page 159 of the handbook (PDF)

Question 102 of 108

2018 Test B Math, question 35

Question

Square MNPQ with N at the top left, P at the top right, Q at the bottom right and M at the bottom left; side PQ is labeled 6 cm. S is on the top side NP, T is on the bottom side MQ, and R is on the left side MN. Segments ST, RS and RT are drawn, and the two triangles RNS (top left corner) and TMR (bottom left corner) are shaded.

R, S, and T are midpoints of the sides of square MNPQ, as shown above. What is the sum of the areas of the shaded triangles?

  • A9 sq cm
  • B12 sq cm
  • C18 sq cm
  • D36 sq cm
Show the answer

Correct answer: A

The length of one side of the square is 6 cm. Since RR, SS, and TT are midpoints, then TM\overline{TM}, MR\overline{MR}, RN\overline{RN}, and NS\overline{NS} are all equal to 3 cm. Triangles TMRTMR and RNSRNS are both right triangles, so the area of one of the triangles is

12×3×3=92\frac{1}{2}\times 3\times 3=\frac{9}{2}

The triangles are congruent, so the sum of the areas is

92+92=9 sq cm\frac{9}{2}+\frac{9}{2}=9\text{ sq cm}

Question analysis

Points R, S and T are midpoints of the sides of square MNPQ, whose sides are 6 cm long, and students find the total area of the two shaded triangles in its corners. It tests the area of a triangle. The common slips are forgetting the one-half or measuring the wrong region.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Triangle area is grade 6 content, and the corner triangles here are right triangles with easy legs. Typical grade 7 area problems are similar or harder; the only extra step is using midpoints to get the lengths.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 1, 34 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The shaded triangles are right triangles tucked into the corners, so their legs lie along the square's sides and are easy to measure once the midpoints are used. Most lost points come from the one half in the area formula, or from reporting a larger region of the figure, such as the left half or the whole square. Labeling each 3 cm leg on the figure before computing pays off here.

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

    Only one length is labeled, the 6 cm side PQ; the legs NR, NS, MR and MT must each be read as half a side, 3 cm, from the word midpoints. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The area of a triangle, one half times base times height, is not given, and leaving out the one half doubles each 4.5 sq cm triangle. (Anderson & Krathwohl, 2001)

Strategy: Use the midpoints to get each triangle's legs as half the square's side, apply one-half base times height, and remember there are two shaded triangles.

Traps

  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.

  1. Read the figure

    MNPQ is a square, and side PQ is 6 cm. All sides of a square are equal, so every side is 6 cm. A midpoint is the point exactly halfway along a side. R is the midpoint of side NM. S is the midpoint of NP, and T is the midpoint of MQ.

    The shaded triangles are RNS at the top and RMT at the bottom.

  2. Find the short sides

    Each midpoint cuts a 6 cm side into two equal parts.

    6÷2=36 \div 2 = 3

    So NR, NS, RM and MT are each 3 cm long.

  3. Find the area of one triangle

    Triangle RNS has a square corner at N. A square corner is a right angle. So NS can be the base and NR the height.

    A=12×3×3=4.5A = \frac{1}{2} \times 3 \times 3 = 4.5

    One triangle is 4.5 sq cm. Triangle RMT has the same sides, so it is also 4.5 sq cm.

    Watch out: do not forget the one half. Without it, each triangle is 9 sq cm and the total is 18. That is choice C.

  4. Add the two triangles

    There are two shaded triangles, so add their areas to get the total.

    4.5+4.5=94.5 + 4.5 = 9

    So the answer is A, 9 sq cm.

Check your answer

The left half of the square is rectangle NSTM, 3 cm by 6 cm. Its area is 3×6=183 \times 6 = 18 sq cm. The white triangle RST fills half of it. So the shaded part is the other half, 9 sq cm. Choice D, 36, is the whole square.

More questions like this

The original DOE question
Question 92 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 92. Open page 159 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 103 of 108

2018 Test B Math, question 38

Question

SURVEY OF CATS PER FAMILY

Number of Cats Number of Families
0 15
1 42
2 35
3 or more 8

The table above shows the number of cats per family in 100 households in the Blaine neighborhood. By what percentage is the number of families with 1 cat greater than the number of families with 2 cats?

  • A7%
  • B10%
  • C17%
  • D20%
Show the answer

Correct answer: D

To find by what percent the number of families with 1 cat is greater than the number of families with 2 cats, calculate the difference between the two numbers and divide by the number of families with 2 cats:

42−3535=735=15=0.20 or 20%\frac{42-35}{35}=\frac{7}{35}=\frac{1}{5}=0.20\text{ or }20\%

Question analysis

A table lists how many cats live in each of 100 households in the Blaine neighborhood, and the question asks by what percent the number of one-cat families exceeds the number of two-cat families. It tests percent difference. The tricky part is choosing the right base for the comparison when the table's total is a convenient 100.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent increase and decrease is grade 7 content, and this is one standard comparison. What makes it harder than a routine exercise is choosing the base and ignoring the tempting 100-family total.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 6, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is a single subtraction and division, so the question is really about choosing the base of a percent comparison. The 100-household total makes the count difference look like a percent already, and dividing by the larger count is the other classic slip. Students who say "greater than the 2-cat count" to themselves pick 35 as the base.

  1. Turning words into math moderate Understanding the question

    "Greater than the number of families with 2 cats" fixes the base of the percent as 35, the count being compared against, not 42 or the 100 households. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    17% and the right 20% both come from dividing the 7-family gap by a count, 42 or 35; only choosing 35, the count being compared against, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Only two rows of the four-row table matter, the 42 families with 1 cat and the 35 with 2; the 0-cat and 3-or-more rows go unused. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the difference between the two counts, decide which count the comparison is measured against, divide by it, and convert to a percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent greater than
percent=differenceamount you compare to×100\text{percent} = \frac{\text{difference}}{\text{amount you compare to}} \times 100
The difference is how much bigger one amount is than the other. The amount you compare to is the one after the word than. Dividing shows what share of that amount the difference is. Times 100 turns that share into a percent.

  1. Sort out the question

    From the table, 42 families have 1 cat and 35 families have 2 cats. You need the percent by which 42 is “greater than the number of families with 2 cats”. The rows for 0 cats and 3 or more cats do not matter here.

  2. Find the difference

    Subtract the two counts.

    42−35=742 - 35 = 7

    There are 7 more families with 1 cat.

    Watch out: 7 is a number of families, not a percent. There are 100 households, so 7% looks tempting. It is choice A, and it is wrong.

  3. Divide by the number you compare to

    The question compares with the families that have 2 cats. So divide by 35, not 42.

    735=15\frac{7}{35} = \frac{1}{5}

    Watch out: dividing by 42 gives about 17%. That is choice C.

  4. Change the fraction to a percent

    Use the percent formula. One fifth of 100 is 20.

    percent=735×100=15×100=20\text{percent} = \frac{7}{35} \times 100 = \frac{1}{5} \times 100 = 20

    So the answer is D, 20%.

Check your answer

Find 20% of 35. That is 0.20×35=70.20 \times 35 = 7. Add it on: 35+7=4235 + 7 = 42. That is the number of families with 1 cat, so 20% is right.

More questions like this

The original DOE question
Question 95 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 95. Open page 160 of the handbook (PDF)

Question 104 of 108

2018 Test B Math, question 43

Question

A quarter circle whose center, marked with a dot, is at the bottom left corner. A segment labeled x + 1 runs from the center to a point on the arc, and from that point a vertical segment drops to the bottom edge, meeting it at a right angle. Along the bottom edge, the part from the center to the foot of that segment is labeled y, and the part from the foot to the end of the arc is labeled 2.

In the quarter circle above, what is yy in terms of xx ?

  • Ax−1x-1
  • Bx+1x+1
  • Cx+12\frac{x+1}{2}
  • D(x+1)22\sqrt{\frac{\left ( x+1 \right )^{2}}{2}}
Show the answer

Correct answer: A

Both x+1x+1 and y+2y+2 are radii.

So, set them equal to each other and solve for yy.

y+2=x+1y+2=x+1

y=x−1y=x-1

Question analysis

In a quarter circle, one segment is labeled x + 1 and the bottom edge is split into lengths y and 2, and students must write y in terms of x. It tests basic circle facts together with solving a simple equation. The challenge is seeing which marked lengths must be equal.

NYS standards
  • 7.G.4 circle area and circumference
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the definition of a radius and a one-step equation, but spotting the relationship in a figure is less routine than plugging into a formula.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 9, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure looks like a right-triangle problem, and that is the trap: the right angle and the slanted radius pull students toward the Pythagorean theorem, which cannot work without the vertical leg. The solution needs only the fact that all radii of a circle are equal, applied to the bottom edge. Students who mark every segment that starts at the center see it at once.

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

    The labels sit on two different radii: x + 1 on the slanted one, and y + 2 along the bottom edge, which runs from the center all the way to the arc. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Variables instead of numbers minor Understanding the question

    The answer is an expression in x, so there is no number to check; trying x = 5, which makes the radius 6 and y = 4, is the fallback. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    x + 1 differs from the right x − 1 by a single sign; only seeing that y plus the 2-unit piece makes a radius, so the 2 is subtracted, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Look for two lengths in the figure that are both radii of the quarter circle, set their expressions equal, and solve for y.

Traps

  • Trusting the picture: The figure suggests something it does not state; work from the given facts.
  • 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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radii of a circle are equal
r1=r2r_1 = r_2
A radius is a segment from the center of a circle to its curved edge. r₁ and r₂ stand for any two radii of the same circle. They are always equal, because every point on the curve is the same distance from the center.

  1. Read the figure

    The shape is a quarter circle, one fourth of a whole circle. Its center is the corner with the dot. A radius is a segment from the center to the curved edge. The slanted segment, labeled x+1x+1, goes from the center to the curve. The bottom edge is split into two parts, yy and 2.

  2. Spot the two radii

    The slanted segment is a radius, so the radius is x+1x+1. The whole bottom edge also runs from the center to the curve. Its length is y+2y+2, so it is a radius too.

    Watch out: the slanted radius only looks like it is at 45 degrees. Do not assume the triangle’s two short sides are equal. That idea leads to choice D.

  3. Set the radii equal

    Use the rule that all radii of a circle are equal.

    y+2=x+1y + 2 = x + 1

    Now you have one equation with xx and yy.

  4. Solve for y

    Get yy alone by taking 2 from both sides.

    y=x+1−2=x−1y = x + 1 - 2 = x - 1

    Watch out: yy is only part of the bottom radius. Setting yy equal to x+1x+1 gives choice B. Dividing by 2 instead of subtracting 2 gives choice C.

  5. Give the answer

    yy is 2 less than the radius, and the radius is x+1x+1. So the answer is A, x−1x-1.

Check your answer

Pick a number. If x=5x = 5, the radius is 6. Then y=5−1=4y = 5 - 1 = 4, and 4+2=64 + 2 = 6, which matches the radius. Also, yy is a short side of the right triangle. It should be shorter than the slanted side, and 4 is less than 6.

More questions like this

The original DOE question
Question 100 as printed on page 161 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 100. Open page 161 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 105 of 108

2018 Test B Math, question 44

Question

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

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

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

Correct answer: C

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

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

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

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

R is 3 sections away from M, so add:

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

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

Question analysis

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

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students place fractions and negatives on number lines, but here they must subtract a negative fraction, divide by five, and then add back. Four fraction steps with unlike denominators make it harder than typical classwork.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 6, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic major Working it out

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

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

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

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

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Read the number line

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

  2. Find the distance from M to T

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

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

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

  3. Find the length of one space

    The 5 spaces are equal, so divide by 5.

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

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

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

  4. Step from M to R

    R is 3 spaces to the right of M.

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

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

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

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

  5. Give the answer

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

Check your answer

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

More questions like this

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

Question 106 of 108

2018 Test B Math, question 48

Question

Coordinate axes with origin O. A shaded triangle stands on the x-axis: its base runs from the point marked m to the point marked n, both to the right of O, and its top vertex touches a dashed horizontal line drawn at 4 on the y-axis.

What is the area of the shaded triangle shown above?

  • Am+nm+n
  • Bn−mn-m
  • C2(n−m)2\left ( n-m \right )
  • D4(n−m)4\left ( n-m \right )
Show the answer

Correct answer: C

The height of the triangle is 4 units. The length of the base is n−mn-m. So the area is

A=12(n−m)(4)=2(n−m)A=\frac{1}{2}(n-m)(4)=2(n-m)

Question analysis

Students find the area of a shaded triangle drawn on coordinate axes, with its base between the points m and n on the x-axis and its top vertex at a height of 4. It tests the triangle area formula with an algebraic side length. Writing the base as an expression in m and n is the step that needs care.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Triangle area is grade 6 content, but a base written with two variables is less familiar. With only two steps, the demand matches typical grade 7 work that mixes geometry with algebraic expressions.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 7, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is trivial, but students must treat a length as a difference of coordinates and keep a formula's one half when the answer is a letter expression. Those who forget the half, or who stop at the base length, find their result among the choices. Trying m = 1 and n = 4 gives a triangle of area 6 that checks the expressions quickly.

  1. Variables instead of numbers moderate Understanding the question

    The base runs from m to n on the x-axis, so its length is the expression n − m, and the area stays in letters with no number to check. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The height, 4, is read from the dashed line through the apex rather than labeled on the triangle, and the base endpoints appear only as tick labels. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. A very close wrong answer moderate Choosing the answer

    4(n − m) is the right 2(n − m) doubled, written the same way except for the coefficient; only remembering the one half in a triangle's area separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the base as the difference between its endpoint values, read the height from the figure, and apply one-half times base times height.

Traps

  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.
Length along the x-axis
length=right end−left end\text{length} = \text{right end} - \text{left end}
The right end and the left end are the numbers where a segment starts and stops on the x-axis. Subtracting the smaller from the larger tells how long the segment is.

  1. Read the figure

    The shaded triangle sits on the x-axis, the flat number line. Its base runs from mm to nn. Its top corner touches the dashed line at height 4. You need the area, written with the letters mm and nn.

  2. Find the length of the base

    The base starts at mm and ends at nn. Use the length rule: right end minus left end.

    base=n−m\text{base} = n - m

    Watch out: adding the ends, m+nm+n, does not give a length. That is choice A.

  3. Find the height

    The height is the straight-up distance from the base to the top corner. The top corner is on the dashed line, which is at 4 on the y-axis. So the height is 4.

  4. Find the area

    Use the triangle area formula.

    A=12×(n−m)×4=2(n−m)A = \frac{1}{2} \times (n - m) \times 4 = 2(n - m)

    Half of 4 is 2, so the area is 2(n−m)2(n-m).

    Watch out: forgetting the one half gives 4(n−m)4(n-m), choice D. Stopping at the base gives n−mn-m, choice B.

  5. Give the answer

    The area is 2 times the length of the base. So the answer is C, 2(n−m)2(n-m).

Check your answer

Try numbers. Say m=1m = 1 and n=4n = 4. The base is 3 and the height is 4, so the area is 12×3×4=6\frac{1}{2} \times 3 \times 4 = 6. Choice C gives 2(4−1)=62(4-1) = 6. It matches. Choice D would give 12.

More questions like this

The original DOE question
Question 105 as printed on page 162 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 105. Open page 162 of the handbook (PDF)

Question 107 of 108

2018 Test B Math, question 49

Question

ANIMAL CARDS

Number of Cards Picture on Card
8 cat
6 dog
5 bird
4 fish
1 horse

The cards in the table above are mixed in a box. Which animal pictured on a card has exactly a 1 in 4 chance of being picked at random from the box?

  • Acat
  • Bdog
  • Cfish
  • Dhorse
Show the answer

Correct answer: B

The total number of cards in the box is

8 + 6 + 5 + 4 + 1 = 24

Set up a proportion to figure out which card has exactly a 1 in 4 chance of being picked at random.

x24=14\frac{x}{24}=\frac{1}{4}

x=6x=6

The dog card has a 1 in 4 chance of being randomly selected.

Question analysis

A table lists picture cards for five animals, from 8 cat cards down to 1 horse card, all mixed in one box. Students decide which animal has exactly a 1 in 4 chance of being drawn at random. It tests simple probability as a fraction of the total, and the numbers in the table invite matching the wrong count.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Simple probability from a list of counts is grade 6 work. This takes one sum and one division with small whole numbers, a shorter and more routine task than typical grade 7 probability problems.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 1, 47 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

Once the 24 cards are totaled, one fourth of them is 6 and the answer is a lookup. The wrong choices mostly attract students who match the numbers 1 and 4 in the phrase to the horse and fish rows, or who assume the most common card is the one asked about. Little time is needed, so the main risk is skipping the total.

  1. Turning words into math minor Understanding the question

    "Exactly a 1 in 4 chance" has to be turned into a count, one fourth of the 24 cards, rather than matched to a 1 or a 4 in the table. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The counts sit in the table's first column with the animals in the second, and the total of 24 has to be added up from all five rows. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add the cards to find the total in the box, find one-fourth of that total, and look for the animal with exactly that many cards.

Traps

  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out the question

    The table shows 8 cat cards, 6 dog, 5 bird, 4 fish and 1 horse. All the cards are mixed in a box. You need the animal with “exactly a 1 in 4 chance” of being picked. A 1 in 4 chance means a probability of 14\frac{1}{4}.

  2. Find the total number of cards

    Add every row of the table. This total is the bottom of the probability fraction.

    8+6+5+4+1=248 + 6 + 5 + 4 + 1 = 24

    There are 24 cards in the box.

  3. Find one fourth of the cards

    The animal you want must have one fourth of all the cards.

    24÷4=624 \div 4 = 6

    So you need the animal with exactly 6 cards. The dog has 6 cards.

    P=624=14P = \frac{6}{24} = \frac{1}{4}

  4. Rule out the other choices

    Watch out: do not match the numbers 1 and 4 to the table. Fish has 4 cards, but its chance is 424=16\frac{4}{24} = \frac{1}{6}. So C is wrong. Horse has 1 card, a chance of 124\frac{1}{24}, so D is wrong. Cat has 824=13\frac{8}{24} = \frac{1}{3}, so A is wrong.

  5. Give the answer

    Only the dog has exactly one fourth of the cards. So the answer is B, dog.

Check your answer

Four groups of 6 cards make 24, the whole box. So 6 cards is one of four equal shares. That is a 1 in 4 chance. The dog cards are exactly one of those shares.

More questions like this

The original DOE question
Question 106 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 106. Open page 163 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 108 of 108

2018 Test B Math, question 50

Question

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

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

Correct answer: C

Separate the compound inequality into two pieces:

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

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

2x−2≤12x-2\leq1

2x≤32x\leq 3

x≤32x\leq\frac{3}{2}

1≤4x+101\leq 4x+10

−9≤4x-9\leq4x

−94≤x-\frac{9}{4}\leq x

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

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph one two-step inequality. This compound inequality in two variables requires substituting, solving two inequalities and combining them into one interval, a step toward algebra 1.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 9, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps moderate Working it out

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

  2. Error-prone arithmetic moderate Working it out

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

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

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

  4. A very close wrong answer minor Choosing the answer

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

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

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Split it into two parts

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

    2x−2≤12x - 2 \leq 1

    1≤4x+101 \leq 4x + 10

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

  3. Solve the first part

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

    2x≤32x \leq 3

    x≤32x \leq \frac{3}{2}

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

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

  4. Solve the second part

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

    −9≤4x-9 \leq 4x

    −94≤x-\frac{9}{4} \leq x

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

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

  5. Put the two parts together

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

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

  6. Give the answer

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

Check your answer

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

More questions like this

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