Practice Test

2022 Handbook

2022 Test B Math

Question 1 of 57

2022 Test B Math, question 1

Question

$A cube whose right edge is labeled 18 mm. A hole shaped like a square prism runs straight down through the middle of the cube from the top face to the bottom face; its square opening on the top face is shaded gray, and one side of the opening is labeled 6 mm. The hidden edges of the cube and of the hole are dashed.$

Beth makes a bead in the shape of a cube with side length 18 millimeters. She cuts a hole in the shape of a square prism out of the center of the cube so that the bead will fit on a string. The base of the square prism has a side length of 6 millimeters. After the hole is cut out, what is the volume of the bead in cubic millimeters?

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

Show the answer

Correct answer: 5184

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

The length, width, and height of the cube are equal (18 mm).

$V_{c} = 18 \times 18 \times 18 = 5{,}832\text{ cubic mm}$

Next, find the volume of the cut-out square prism, with equal length and width (6 mm), and height of 18 mm.

$V_{p} = 6 \times 6 \times 18 = 648\text{ cubic mm}$

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

$5{,}832 - 648 = 5{,}184\text{ cubic mm}$

Question analysis

Beth makes a bead from a cube 18 millimeters on each side and cuts a square prism-shaped hole, 6 millimeters across, through its center for a string. Students find the volume of the bead that is left, in cubic millimeters. It tests the volume of rectangular prisms, and the hole's height is not given in words, so it has to be read from the 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 by removing one rectangular prism from another is grade 7 work. This case is direct, but the hole's height is not stated in words, and cubing 18 by hand is heavier arithmetic, typed with no choices.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Error-prone arithmetic minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 69 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, cube minus hole, but the hole’s height is never stated and has to be read from the drawing. Students who picture the hole as a 6 mm cube take away only 216 cubic millimeters, and those who stop at either volume grid a number that looks finished. Seeing the hole as a square shaft, 6 by 6 by 18, and checking that the bead holds less than the cube catches all three slips.

Reading a figure, table or graph moderate Understanding the question
The stem gives only the 6 mm side of the hole’s square base; that the hole runs the full 18 mm, from the top face to the bottom, shows only in the figure’s dashed edges. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Neither volume formula is given: the cube needs edge × edge × edge, and the square hole needs its 36 square millimeter base times its 18 mm length. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
Cubing 18 by hand, 324 × 18 = 5,832, and then 6 × 6 × 18 = 648 are multi-digit products, and a carrying slip in either changes the final difference. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The volume is gridded with no choices to compare, so entering the hole’s 648 or the whole cube’s 5,832 brings no warning that a step is missing. (Enright, Morley & Sheehan, 2002)

Strategy: Find the volume of the whole cube, then the volume of the square hole, which runs from the top face to the bottom face, 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.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

648 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the volume of the hole, 6 × 6 × 18 = 648 cubic millimeters, and enters the material removed instead of the bead that is left.
5616 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Takes the hole to be a 6 mm cube, 6 × 6 × 6 = 216, though the figure shows it running the full 18 mm, so 5,832 − 216 = 5,616.
5832 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds the volume of the whole cube, 18 × 18 × 18 = 5,832 cubic millimeters, and stops before taking away the hole cut through its center.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a rectangular prism: $V = l \times w \times h$; V is the volume, the space inside a solid, in cubic units. l is the length, w is the width and h is the height. A cube is a prism whose length, width and height are all equal. Multiplying counts the unit cubes that fill the box, layer by layer.

Sort out the question
The bead starts as a cube, a box whose edges are all 18 millimeters long. A hole is cut through its middle. The hole is a square prism, a box with a square base. Each side of that square is 6 millimeters.
You need the volume of the bead that is left, in cubic millimeters. Volume is the space a solid takes up.
Read the hole’s height from the figure
The words don’t say how tall the hole is. Look at the figure. The gray square sits on the top face. Its dashed edges run all the way down to the bottom face.
So the hole is as tall as the cube, 18 millimeters. The hole measures 6 by 6 by 18 millimeters.
Find the volume of the cube
Use the volume formula. All three measures of the cube are 18.
$V = l \times w \times h = 18 \times 18 \times 18 = 5{,}832$
Work it in two parts: $18 \times 18 = 324$ , then $324 \times 18 = 5{,}832$ . The whole cube is 5,832 cubic millimeters.
Find the volume of the hole
Use the same formula for the hole: 6 by 6 by 18.
$V = l \times w \times h = 6 \times 6 \times 18 = 648$
The hole is 648 cubic millimeters.
Watch out: the hole is not a small 6 mm cube. Using $6 \times 6 \times 6 = 216$ would give 5,616, which is wrong.
Subtract the hole from the cube
The bead is the cube with the hole taken out. So subtract.
$5{,}832 - 648 = 5{,}184$
The bead’s volume is 5,184 cubic millimeters.
Watch out: 648 is the part cut out, and 5,832 is the cube before the cut. Neither one is the bead. So you would grid 5184.

Check your answer

Picture the cube cut into tall columns, each the size of the hole. It is 3 columns wide and 3 deep, so it holds $3 \times 3 = 9$ columns. The hole takes 1, leaving 8. And $8 \times 648 = 5{,}184$ . It matches.

More questions like this

The original DOE question

Question 58 as printed on page 197 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 197, question 58. Open page 197 of the handbook (PDF)

Question 2 of 57

2022 Test B Math, question 2

Question

The table shows the proportional relationship between x cups of oatmeal and y ounces of raisins in a bread recipe.

BREAD RECIPE

Cups of Oatmeal (x)	Ounces of Raisins (y)
2	1.0
3	1.5
6	3.0

What is the constant of proportionality of the number of ounces of raisins to the number of cups of oatmeal?

Enter your answer in the space.

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

Show the answer

Correct answer: 0.5

To find p, the constant of proportionality of ounces of raisins, y, to cups of oatmeal, x, use the equation $y = px$ .

The table shows that there is 1 ounce of raisins for 2 cups of oatmeal.

$1 = 2p$

$\frac{1}{2} = \frac{2p}{2}$

$0.5 = p$

The constant of proportionality is 0.5.

Question analysis

A table for a bread recipe lists cups of oatmeal and the ounces of raisins that go with them, in a proportional relationship. Students type the constant of proportionality of raisins to oatmeal. It tests reading a unit rate from a table, and the order in which the question names the two quantities matters.

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from a table is a grade 7 standard, and here a single division in one row gives it. The demand is lighter than typical grade 7 work; the order of the ratio in the wording needs the most care.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 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

Every row gives the same ratio, so the division itself is easy; the question tests which quantity goes on top. Because oatmeal comes first in the table and its numbers are the larger ones, dividing cups by ounces feels natural and produces a clean whole number. Writing the relationship as y = kx, raisins equal k times oatmeal, and checking it on a second row such as 3 cups and 1.5 ounces fixes the order.

Turning words into math moderate Understanding the question
The phrase “of the number of ounces of raisins to the number of cups of oatmeal” fixes the order y ÷ x, while the table lists oatmeal first, inviting 2 ÷ 1.0. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed in with no choices, so an upside-down ratio of 2, a whole number, looks as finished as the decimal the table actually gives. (Enright, Morley & Sheehan, 2002)

Strategy: Divide an ounces-of-raisins value by the cups of oatmeal in the same row, in the order the question names them, and check that every row gives the same value.

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).
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

2 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides cups of oatmeal by ounces of raisins, 2 ÷ 1.0, turning the ratio upside down: cups per ounce instead of ounces per cup.
1 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads the first raisin amount, 1.0 ounce, as the constant without dividing it by the 2 cups of oatmeal in the same row.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $y = kx \quad \text{so} \quad k = \frac{y}{x}$; x and y are two amounts that grow together at the same rate. k is the constant of proportionality, the number you always multiply x by to get y. Here y is ounces of raisins and x is cups of oatmeal, so k is ounces of raisins per cup of oatmeal.

Sort out the question
The table shows cups of oatmeal, x, and ounces of raisins, y. Proportional means the raisins are always the same multiple of the oatmeal. That multiple is the constant of proportionality. The question asks for raisins to oatmeal. So you want ounces of raisins for each cup of oatmeal: y ÷ x.
Divide y by x in one row
Use the first row of the table: 2 cups of oatmeal go with 1.0 ounce of raisins.
$k = \frac{y}{x} = \frac{1.0}{2} = 0.5$
So there is 0.5 ounce of raisins for each cup of oatmeal.
Check the other rows
A constant must be the same in every row:
$\frac{1.5}{3} = 0.5 \qquad \frac{3.0}{6} = 0.5$
Every row gives 0.5, so the rule is $y = 0.5x$ .
Watch out: the table lists oatmeal first, which tempts you to divide 2 by 1.0. That gives 2, which is cups per ounce, the ratio upside down. And 1.0 alone is just one raisin amount, not the constant.
Enter the answer
Type the constant in the box as a number, with no units. On this site, 0.5, .5 and 1/2 all count.
So you would enter 0.5.

Check your answer

Multiply each oatmeal amount by 0.5: 2 × 0.5 = 1.0, 3 × 0.5 = 1.5 and 6 × 0.5 = 3.0. Each result matches the raisins in its row. Also, every raisin number is smaller than its oatmeal number, so the constant must be less than 1.

More questions like this

The original DOE question

Question 16 as shown on the DOE’s digital test, page 89 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 89, question 16.

Question 3 of 57

2022 Test B Math, question 3

Question

Kelly had 7.3 grams of sugar. She used 2.7 grams in her coffee. How many grams of sugar does she have left?

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

Show the answer

Correct answer: 4.6

Subtract the number of grams of sugar Kelly used in her coffee from the original amount she had.

$7.3 - 2.7 = 4.6\text{ grams left}$

Question analysis

Kelly has 7.3 grams of sugar and uses 2.7 grams of it in her coffee. This short grid-in word problem asks how many grams she has left. It tests subtracting decimals, and because the tenths digit being taken away is the larger one, the subtraction needs regrouping.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Subtracting decimals is grade 6 fluency work, and this needs one subtraction in a short, familiar setting. Grade 7 problems usually bring in negative numbers or several steps; the only care here is regrouping in the tenths.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
An answer in an unexpected form minor
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

One subtraction in a two-sentence story, so the only real risk is the regrouping in the tenths column. Students who skip the borrow, or who borrow and forget to reduce the ones, still produce a tidy decimal, and nothing on the page flags it. Adding the result back to the 2.7 grams used should return the 7.3 grams Kelly started with.

Error-prone arithmetic minor Working it out
The 7 tenths being taken away is more than the 3 tenths in 7.3, so one of the 7 wholes must be regrouped as ten tenths before 13 − 7 = 6. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The difference is gridded with no choices to compare, so a regrouping slip that gives 5.4 or 5.6 looks just as finished as the right amount. (Enright, Morley & Sheehan, 2002)

Strategy: Line up the decimal points and subtract the grams used from the grams Kelly started with, regrouping one whole as ten tenths when the tenths column needs it.

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.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
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

5.4 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the smaller digit from the larger in each column, 7 − 3 = 4 tenths and 7 − 2 = 5 ones, instead of regrouping a whole as ten tenths.
5.6 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Regroups to get 13 − 7 = 6 tenths but forgets to reduce the ones after borrowing, so the ones digit stays 7 − 2 = 5.
10 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the two amounts, 7.3 + 2.7 = 10, instead of taking the sugar used in the coffee away from the 7.3 grams Kelly had.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Amount left: $\text{left} = \text{start} - \text{used}$; Start is the amount you had at first. Used is the amount taken away. Left is what remains. Taking part of an amount away is subtraction, so you subtract the amount used from the start.

Sort out the question
Kelly started with 7.3 grams of sugar. She used 2.7 grams in her coffee. You need the grams she has left.
Left means what remains after some is taken away. So subtract the amount used from the amount she started with.
$\text{left} = 7.3 - 2.7$
Regroup to subtract the tenths
Line up the decimal points, so tenths sit over tenths. A tenth is one of ten equal parts of a whole. Start with the tenths. You can’t take 7 tenths from 3 tenths. So break 1 of the 7 wholes into 10 tenths. Now you have 6 wholes and 13 tenths.
$13 - 7 = 6$
That leaves 6 tenths.
Watch out: taking the smaller digit from the bigger one, 7 − 3, skips the regrouping. It gives 5.4, which is wrong.
Subtract the whole grams
After regrouping, the wholes column has 6, not 7.
$6 - 2 = 4$
That leaves 4 wholes. Put the decimal point back: 4 wholes and 6 tenths is 4.6 grams.
Watch out: forgetting to change the 7 to a 6 gives 5.6. Adding instead of subtracting gives 10. So you would grid 4.6.

Check your answer

Add back the sugar she used: $4.6 + 2.7 = 7.3$ . That is the amount Kelly started with, so 4.6 grams is right. An estimate agrees too: about 7 − 3 = 4 grams.

More questions like this

The original DOE question

Question 60 as printed on page 197 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 197, question 60. Open page 197 of the handbook (PDF)

Question 4 of 57

2022 Test B Math, question 4

Question

The total cost of a book is given by the expression $1.08x$ , where x is the price of the book before tax. The tax rate is m%. What is the value of m?

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

Show the answer

Correct answer: 8

The price of a book before tax is x and after tax is $1.08x$ . Subtract to determine the tax rate.

$1.08x - x = 0.08x$ , so the tax rate is 0.08.

The tax rate m% can be written as $\frac{m}{100}$ .

Solve for m:

$\frac{m}{100} = 0.08$

$100\left(\frac{m}{100}\right) = 100(0.08)$

$m = 8$

Question analysis

The total cost of a book, tax included, is given by the expression 1.08x, where x is the price before tax, and the tax rate is m%. Students find the value of m and type it in. It tests reading a percent increase from an algebraic expression, and the rate has to be entered as a number of percent, not as a decimal.

NYS standards: 7.EE.2 rewriting expressions to show relationships
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Reading a percent increase from an expression such as 1.08x is a named grade 7 idea. It takes one short step, but students must turn a decimal into a percent and type the value with no choices to check.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: An answer in an unexpected form moderate
Something you must already know minor
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 33 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 expression already holds the answer, since 1.08x is the price, 1x, plus a tax of 0.08x. The work lies in pulling the tax apart from the price and then writing it in the form the stem asks for, the number in front of a percent sign. Students who stop at the decimal, or who read the whole 1.08 as 108 percent, grid a value that looks reasonable. Trying a price of $100, which costs $108 with tax, makes the 8 visible.

An answer in an unexpected form moderate Choosing the answer
The rate is asked for as the m in m%, so the 0.08 found from the expression must become 8, and the grid takes 0.08 or 108 just as readily. (Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Nothing in the stem says that multiplying a price by 1.08 adds 8% of it; students must split 1.08x into 1x for the price and 0.08x for the tax. (Anderson & Krathwohl, 2001)
Variables instead of numbers minor Understanding the question
The price is the letter x, never a dollar amount, so the tax has to be read from the coefficient 1.08 rather than worked out from a number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Split 1.08x into the price before tax plus the tax added to it, then write the tax part's coefficient as a percent.

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.
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

0.08 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the tax as 0.08x and enters the decimal rate 0.08, one step short of multiplying by 100 to give m as a number of percent.
108 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads 1.08 as 108% and enters it, giving the whole cost as a percent of the price instead of only the tax added to it.
1.08 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Copies the coefficient 1.08 from the expression as the rate, without separating the original price, 1x, from the tax added on top.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction: $m\% = \frac{m}{100}$; Percent means out of 100. So m% is m hundredths, the fraction m over 100. To change a decimal to a percent, multiply it by 100. To change a percent to a decimal, divide it by 100.

Sort out the question
x is the price of the book before tax. The total cost with tax is 1.08x, which means 1.08 times x. The tax rate is m%. You need the number m.
The tax rate is the percent of the price that is added on as tax.
Split the total into price and tax
The total is the price plus the tax. Split 1.08 into 1 and 0.08.
$1.08x = 1x + 0.08x = x + 0.08x$
The x part is the price you started with. So the extra 0.08x is the tax. The tax is 0.08 times the price.
Change the decimal to a percent
The rate is 0.08 as a decimal. Use the formula and set it equal to 0.08.
$\frac{m}{100} = 0.08$
To get m alone, multiply both sides by 100.
$m = 0.08 \times 100 = 8$
So the tax rate is 8%.
Give the answer
The question asks for m, the number in front of the % sign. That is 8.
Watch out: 0.08 is the rate as a decimal, one step short. And 1.08, or 108%, counts the price as well as the tax. So you would grid 8.

Check your answer

Try a price of $100. The total is $1.08 \times 100 = 108$ dollars. The tax is 108 − 100 = 8 dollars. And $8 out of $100 is 8%. So m = 8.

More questions like this

The original DOE question

Question 61 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 2023 Admissions, page 198, question 61. Open page 198 of the handbook (PDF)

Question 5 of 57

2022 Test B Math, question 5

Question

A row of seats in a theater contains 20 seats numbered from 101 to 120. The probability that a randomly chosen seat in this row will be numbered 104 is x%. What is the value of x?

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

Show the answer

Correct answer: 5

The probability that any 1 seat will be selected at random from 20 seats is 1 out of 20. To find the probability as a percentage, convert the fraction to a decimal.

$\frac{1}{20} = 0.05$

Then convert the decimal to a percentage by multiplying by 100.

$0.05 \times 100 = 5$

The probability is 5%.

Question analysis

A theater row has 20 seats numbered 101 to 120, and one seat is chosen at random. Students find the probability that it is one particular seat, written as x%, and type the value of x. It tests simple probability and changing a fraction to a percent; the seat numbers themselves play no part in the answer.

NYS standards: 6.SP.8a uniform probability models
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The probability of one equally likely outcome is grade 6 content, and turning it into a percent is a short grade 6 step. Grade 7 probability usually involves compound events, which this question does not.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: An answer in an unexpected form moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 33 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 seat numbers, 101 to 120, and the chosen seat 104 play no part: the row holds 20 equally likely seats, so the chance is 1 in 20. What the item tests is the change of form, from a fraction to the number in front of a percent sign. Knowing that 1/20 is half of 1/10, which is 10%, gives the value at once and rules out both 0.05 and 50.

An answer in an unexpected form moderate Choosing the answer
The probability is asked for as the x in x%, so 1 out of 20 must become the number 5, not the fraction 1/20 or the decimal 0.05 it is first found as. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Dividing 1 by 20 gives 0.05, with a zero in the tenths place, and losing that zero makes the percent ten times too large. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the seats, write the chance of one particular seat as a fraction, then change it to a percent and enter only the number in front of the percent sign.

Traps

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

0.05 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the probability, 1 ÷ 20 = 0.05, and enters the decimal without multiplying by 100 to give x as a number of percent.
50 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Loses the zero after the decimal point when dividing, taking 1 ÷ 20 as 0.5, as 1 ÷ 2 would be, and so enters 50.
20 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reads 1 chance in 20 as 20 percent, putting the number of seats in place of the percent instead of dividing 1 by 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance that something happens, from 0 to 1. The top counts the outcomes you want. The bottom counts all the outcomes that could happen. Use it when every outcome is equally likely, like one seat picked from a row.
Fraction to percent: $\text{percent} = \text{fraction} \times 100$; A percent is a number out of 100. Multiplying a fraction or decimal by 100 tells how many out of 100 it is. So 1/2 times 100 is 50, which means 50%.

Sort out the question
The row has 20 seats, numbered 101 to 120. One seat is picked at random. At random means every seat has the same chance. You need the chance that it is seat 104, written as x%. Then you grid x.
The seat numbers don’t change the math. Only the number of seats matters.
Write the chance as a fraction
Seat 104 is 1 seat. There are 20 seats in all.
$P = \frac{\text{outcomes you want}}{\text{all possible outcomes}} = \frac{1}{20}$
So the chance is 1 out of 20.
Change the fraction to a percent
Use the formula: multiply the fraction by 100.
$\text{percent} = \frac{1}{20} \times 100 = \frac{100}{20} = 5$
So the chance is 5%. As a decimal, it is 0.05.
Give the answer
The question asks for x, the number in front of the % sign. That is 5.
Watch out: 0.05 is the decimal, not the percent. Writing 1 ÷ 20 as 0.5 gives 50. And 20 is the number of seats, not the percent. So you would grid 5.

Check your answer

All 20 seats share 100% equally. Each seat gets $100 \div 20 = 5$ percent. That includes seat 104. Also, 1/20 is half of 1/10, and 1/10 is 10%. Half of 10% is 5%.

More questions like this

The original DOE question

Question 62 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 2023 Admissions, page 198, question 62. Open page 198 of the handbook (PDF)

Question 6 of 57

2022 Test B Math, question 6

Question

What is the prime factorization of 3,575?

A $5 \cdot 11 \cdot 13$
B $5^{2} \cdot 11 \cdot 13$
C $5 \cdot 715$
D $5^{2} \cdot 143$

Show the answer

Correct answer: B

By observation, 3,575 ends in 75; therefore, 3,575 is divisible by 25.

$3{,}575 \div 25 = 143$

Test 143 for factors and eliminate 3, 5, and 7, but 11 is a factor.

$143 \div 11 = 13$ ; both 11 and 13 are prime numbers.

The first factor, 25, is not prime but is the product of primes, $5 \times 5 = 5^{2}$ .

The prime factorization of 3,575 is $5^{2} \times 11 \times 13$ .

Question analysis

This number question asks for the prime factorization of 3,575, chosen from four products of factors. Students divide by primes, using divisibility tests, until only primes remain. The four products look much alike, so the work must be carried all the way through, keeping count of every prime divided out along the way.

NYS standards: 6.NS.4 common factors and multiples
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Breaking a number into primes is grade 6 number work that relies on divisibility tests. The number here is larger than in most class examples, but the work is still a few divisions, lighter than typical grade 7 problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 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

Three of the four choices multiply back to 3,575, so checking a product rules out only the one that equals 715. The question turns on whether 143 is prime: it survives the quick tests for 2, 3 and 5, and the factor 11 is easy to miss. Students who keep dividing until every factor is prime, and who write a prime as often as it divides, finish in under a minute.

A very close wrong answer moderate Choosing the answer
5² · 143 matches the right factorization in its first factor and multiplies back to 3,575 just as well, so a product check cannot separate them; only splitting 143 into 11 × 13 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Nothing on the page defines a prime factorization or gives a test for 11, and 143 passes the familiar quick tests for 2, 3 and 5, so it looks prime. (Anderson & Krathwohl, 2001)

Strategy: Divide by the smallest prime that goes in evenly, again and again, writing down every prime you use until the quotient reaches 1, then compare that list with each choice.

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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Lists each prime that divides 3,575 only once, 5, 11 and 13, as if a factorization named which primes divide it, though their product is only 715.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Divides out one 5 and stops at 5 · 715, a true factorization but not a prime one, though 715 ends in 5 and splits further.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Divides by 25 and stops at 143, which passes the tests for 2, 3 and 5 and looks prime, though 143 = 11 × 13.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisibility by 25: $\text{last two digits } 00, 25, 50 \text{ or } 75 \rightarrow \text{divisible by } 25$; A number is divisible by 25 when 25 goes into it with no remainder. You only need to look at the last two digits. If they are 00, 25, 50 or 75, the number is divisible by 25. This works because 100 is 4 × 25, so the hundreds never leave a remainder.
Writing a repeated factor with an exponent: $a^{2} = a \times a$; a is any number. The small raised 2 is an exponent. It tells you how many times a is used as a factor, so 5² means 5 × 5 = 25. Use an exponent in a prime factorization when the same prime shows up more than once.

Sort out the question
A prime number is a whole number greater than 1 whose only factors are 1 and itself. Examples are 2, 3, 5, 7, 11 and 13. The prime factorization of 3,575 writes it as a product (the answer to a multiplication) of prime numbers only. So you must break 3,575 down until every factor is prime.
Divide by 25 first
3,575 ends in 75. So it is divisible by 25:
$3{,}575 \div 25 = 143$
That means 3,575 = 25 × 143. But 25 is not prime. It is 5 × 5, which you can write as 5².
Test 143 for prime factors
Now see if 143 splits. Try the primes in order. 143 is odd, so 2 does not work. Its digits add to 1 + 4 + 3 = 8, which is not a multiple of 3, so 3 does not work. It does not end in 0 or 5, so 5 does not work. Since 7 × 20 = 140, 143 ÷ 7 leaves a remainder of 3, so 7 does not work. Now try 11:
$143 \div 11 = 13$
So 143 = 11 × 13, and both 11 and 13 are prime.
Watch out: 143 looks prime, but it is not. Stopping at 5² · 143 leads to choice D.
Put all the primes together
Collect every prime you found. The 5 appears twice, then 11 and 13 once each. The dot · means times, just like ×.
$3{,}575 = 5 \times 5 \times 11 \times 13 = 5^{2} \cdot 11 \cdot 13$
Watch out: write the 5 twice, since it divides 3,575 twice. Writing it once gives 5 · 11 · 13, choice A, which equals only 715. Choice C, 5 · 715, stops too early, because 715 ends in 5 and splits further.
So the answer is B, $5^{2} \cdot 11 \cdot 13$ .

Check your answer

Multiply the primes back together: 25 × 11 = 275, and 275 × 13 = 3,575. That is the number you started with. Also, 5, 11 and 13 are all prime, so nothing can be split further.

More questions like this

The original DOE question

Question 1 as shown on the DOE’s digital test, page 74 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 74, question 1.

Question 7 of 57

2022 Test B Math, question 7

Question

If $\frac{x + 2y}{5} = 3y$ , what is the value of x in terms of y?

A $\frac{7y}{5}$
B $\frac{13y}{5}$
C13
D $13y$

Show the answer

Correct answer: D

The goal is to isolate the variable x by using inverse operations. First, multiply both sides by 5 and then subtract $2y$ from both sides.

$\frac{x + 2y}{5} = 3y$

$x + 2y = 15y$

$x = 13y$

Another approach is to make the right side of the equation a fraction and then use cross multiplication for the first step.

$\frac{x + 2y}{5} = \frac{3y}{1}$

$x + 2y = 15y$

$x = 13y$

Question analysis

An equation says that x plus 2y, all divided by 5, equals 3y, and students choose the expression for x in terms of y. This multiple-choice algebra question tests solving a literal equation with inverse operations. Because the 5 divides the whole left side, clearing the fraction first and keeping track of the y terms is the tricky part.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve two-step equations with numbers, such as (n + 4)/3 = 6. Solving for one letter in terms of another, with a fraction to clear first, is grade 8 work and a step beyond.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

Two inverse operations solve it, but the fraction bar spans the whole sum x + 2y, so multiplying both sides by 5 has to come first. Students who split the fraction and stop at x/5 find a matching choice, and those who divide 3y by 5 instead of multiplying end up with fifths of y as well. Substituting y = 1, which makes x + 2 equal 15, checks any choice in seconds.

Variables instead of numbers moderate Understanding the question
Both sides hold y, so the result must stay a multiple of y: after clearing the 5, the like terms 15y and 2y are combined, and there is no number to check against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Multiplying the whole of 3y by 5 gives 15y, but dividing it by 5 or splitting the fraction into x/5 + 2y/5 leaves fifths of y in the working. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply both sides by 5 to clear the fraction, then subtract 2y from both sides; check by putting y = 1 into both the equation and the chosen expression.

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.
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

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 3y by 5 instead of multiplying, so x + 2y = 3y/5 and x = −7y/5, then takes 7y/5, the choice that matches its size.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Splits the fraction and subtracts 2y/5 from 3y, reaching x/5 = 13y/5, then stops without multiplying both sides by 5.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reaches x = 15y − 2y but drops the letter, as if subtracting like terms cancelled the y, and gives the number 13 instead of 13y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Do the same to both sides: $\text{if } a = b, \text{ then } a \times c = b \times c$; a and b are two equal amounts, and c is any number. If you do the same thing to both sides of an equation, they stay equal, like a balanced scale. Use it to undo each step until the letter you want is alone.

Sort out the question
The equation says: take x plus 2y, divide the whole sum by 5, and you get 3y. You need x “in terms of y.” That means x alone on one side, with an expression using y on the other. An expression is a math phrase with no equal sign.
Clear the fraction first
The fraction bar divides the whole top, x + 2y, by 5. To undo dividing by 5, multiply both sides by 5.
$5 \times \frac{x + 2y}{5} = 5 \times 3y$
$x + 2y = 15y$
Watch out: multiply 3y by 5. Don’t divide it by 5. Dividing leaves fifths of y, as in choice A.
Get x alone
Now 2y is added to x. To undo adding, subtract 2y from both sides.
$x + 2y - 2y = 15y - 2y$
$x = 13y$
15y − 2y means 15 y’s take away 2 y’s. That leaves 13 y’s, so the y stays.
Give the answer
Watch out: you might split the fraction into x/5 + 2y/5 instead. Then you reach x/5 = 13y/5. Stopping there gives choice B, but you still have to multiply by 5. Dropping the y gives 13, choice C. So the answer is D, $13y$ .

Check your answer

Pick y = 1. Then x = 13 × 1 = 13. Put both into the equation: $\frac{13 + 2 \times 1}{5} = \frac{15}{5} = 3$ . And 3y = 3 × 1 = 3. Both sides match, so x = 13y works.

More questions like this

The original DOE question

Question 64 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 2023 Admissions, page 199, question 64. 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 8 of 57

2022 Test B Math, question 8

Question

What is the value of z in $\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}$ ?

A1
B5
C $y$
D $2y - x$

Show the answer

Correct answer: D

First, multiply both sides of the equation by 10 to eliminate the fractions. In other words, multiply each term by 10.

$\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}$

$10\left(\frac{x}{10} + \frac{y - x}{5}\right) = 10\left(\frac{z}{10}\right)$

$\frac{10x}{10} + \frac{10(y - x)}{5} = \frac{10z}{10}$

$x + 2(y - x) = z$

Then, distribute and combine like terms to find the value for z.

$x + 2(y - x) = z$

$x + 2y - 2x = z$

$2y - x = z$

Question analysis

This algebra question gives an equation with three letters and fractions, x over 10 plus (y − x) over 5 equals z over 10, and asks for the value of z. Students clear the denominators, distribute and combine like terms. The two fractions on the left have different denominators, which is where the usual slip happens.

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add and expand linear expressions with fraction coefficients, usually in one letter. Here three letters appear, one numerator is a difference, and the result is an expression rather than a number, which adds abstraction but no new skill.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 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

The equation has three fractions and three letters, and the work is three routine moves: multiply through by 10, distribute, collect the x terms. Most errors come from the middle term, whose denominator 5 differs from the other two; treating it as tenths makes z simply y. Students who test values should avoid x = y, since that pair makes y and 2y − x agree, and x = y = 1 even makes the number 1 fit.

Variables instead of numbers moderate Understanding the question
The answer is z in terms of x and y, not a number, yet two choices are plain numbers, 1 and 5, which tempt students who expect the equation to give z a single value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Clearing the denominators turns (y − x)/5 into 2(y − x), and the −2x from distributing must then combine with the lone x to leave −x, two places where a factor or a sign slips. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply every term on both sides by 10 so no fractions remain, distribute across the parentheses, and combine the x terms; substituting small numbers for x and y checks the result.

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

A Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: weak
Tests x = 1 and y = 1, gets 1/10 + 0 = z/10, so z = 1, and takes the choice that reads 1; any pair with x ≠ y rules it out.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Expects z to be a number and takes the 5 from the middle denominator, a number copied from the equation rather than the result of any step.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the numerators x and y − x over 10 as if (y − x)/5 were already in tenths, so the left side becomes y/10 and z = y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } 10 \times a = 10 \times b$; a and b are the two sides of an equation. Multiplying both sides by the same number keeps them equal, like adding the same weight to both pans of a balance. Here 10 is a good choice, because it clears away every fraction.
The distributive property: $a(b - c) = a \times b - a \times c$; a is the number outside the parentheses, and b and c are the terms inside. Multiply a by each term inside, and keep the minus sign between them. It works because a groups of (b − c) is a groups of b minus a groups of c.

Sort out the question
The equation has three letters, x, y and z. You need z written using x and y. The answer will not be one number, because x and y can be anything. The fractions make the work messy, so clear them first.
Multiply every term by 10
The bottoms are 10, 5 and 10, and each one divides into 10. So multiply every term by 10:
$10 \times \frac{x}{10} + 10 \times \frac{y - x}{5} = 10 \times \frac{z}{10}$
10 times x/10 is x. Since 10 ÷ 5 = 2, 10 times (y − x)/5 is 2(y − x). And 10 times z/10 is z.
$x + 2(y - x) = z$
Distribute the 2
Use the distributive property. Multiply the 2 by both y and x:
$2(y - x) = 2y - 2x$
So the equation becomes:
$x + 2y - 2x = z$
Combine like terms
Like terms have the same letter. Here x and −2x are like terms. Since 1 − 2 = −1, they combine to −x:
$x - 2x = -x$
That leaves:
$2y - x = z$
Watch out: do not add the tops of x/10 and (y − x)/5 straight away. Their bottoms are different. Doing it anyway gives z = y, choice C. Choices A and B are plain numbers, but z changes when x and y do.
So the answer is D, $2y - x$ .

Check your answer

Pick numbers. Let x = 2 and y = 7. The left side is 2/10 + 5/5 = 0.2 + 1 = 1.2. Choice D gives z = 2 × 7 − 2 = 12, and 12/10 = 1.2. Both sides match.

More questions like this

The original DOE question

Question 2 as shown on the DOE’s digital test, page 75 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 75, question 2.

Question 9 of 57

2022 Test B Math, question 9

Question

Two sets, R and S, are described below. The sum of the elements in set R equals the sum of the elements in set S.

$\mathrm{R} = \{5, x, 3, 8\}$

$\mathrm{S} = \{6, y, 4, 1\}$

What is the value of $x - y$ ?

A−7
B−5
C5
D7

Show the answer

Correct answer: B

First, find the sum for each set by adding the terms in each set separately.

The sum for set R is $5 + x + 3 + 8$ , which is $x + 16$ .

The sum for set S is $6 + y + 4 + 1$ , which is $y + 11$ .

The question states that these two sums are equal to each other, so write the equation and then rearrange using inverse operations.

$x + 16 = y + 11$

$x = y - 5$

$x - y = -5$

If the rearranging to obtain the target value is unclear, use the equation $x + 16 = y + 11$ and find a few possible values for x and y.

For example, if $x = 1$ , then $y = 6$ because $1 + 16 = 6 + 11$ . If $x = 2$ , then $y = 7$ because $2 + 16 = 7 + 11$ . Then find the value of $x - y$ . No matter which values are chosen, $x - y$ will always be $-5$ .

Question analysis

Two sets of four numbers, R and S, each contain one unknown, x and y, and the sums of the two sets are equal. Students find the value of x − y without being able to find x or y alone. It tests writing and rearranging an equation, and deciding which way round the difference goes is the tricky part.

NYS standards: 7.EE.4 equations and inequalities from problems
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students write and solve equations with one unknown. This one has two unknowns that are never found separately; students must rearrange to get their difference, a less familiar goal, though the arithmetic is light.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 33 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 sums take seconds; the point is lost on the sign. Because set R's known numbers add to more than set S's, x has to be the smaller unknown, so x − y is negative, but subtracting the two known sums in the order they appear gives a positive 5. Choosing one pair of values that balances the sets, such as x = 1 and y = 6, settles the sign at once.

A very close wrong answer moderate Choosing the answer
The choices pair −5 with 5 and −7 with 7, so the size 5 is easy to reach and only the order of subtraction, x − y rather than y − x, separates the right choice from its opposite. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
Neither x nor y can be found, only the gap between them, so students who look for a value of each letter stall; the question rewards rearranging x + 16 = y + 11. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Add the known numbers in each set, set the two sums equal, then move both letters to one side so it reads x − y; testing one pair of values confirms the sign.

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

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives −7; it needs a slip in a sum, such as 18 for 5 + 3 + 8, so it mostly draws guesses near −5.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Subtracts the known sums in the order the sets appear, 16 − 11 = 5, which is y − x, missing that R's larger known sum makes x the smaller unknown.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two unrelated errors, a slip in a sum and the subtraction reversed, so no single line of working reaches 7 and it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } a - c = b - c$; a and b are two equal amounts, and c is any number or letter. Taking the same thing away from both sides keeps them equal. You use it to move terms around until the equation shows what you want.

Sort out the question
Set R holds 5, x, 3 and 8. Set S holds 6, y, 4 and 1. The elements (the members) of R add up to the same total as the elements of S. You cannot find x or y alone. But you can find $x - y$ , the gap between them.
Add up each set
Add the known numbers in each set, and keep the letter:
$5 + x + 3 + 8 = x + 16$
$6 + y + 4 + 1 = y + 11$
The two sums are equal, so:
$x + 16 = y + 11$
Rearrange to get x − y
You want $x - y$ on one side. Subtract y from both sides:
$x - y + 16 = 11$
Then subtract 16 from both sides:
$x - y = 11 - 16 = -5$
Make sense of the sign
Set R already has 16 from its numbers, but set S has only 11. So y must be 5 more than x to make the sums match. That makes $x - y$ negative.
Watch out: 16 − 11 = 5 is $y - x$ , not $x - y$ . Using it leads to choice C.
So the answer is B, −5.

Check your answer

Try real values. If x = 1, then R adds to 17. So y + 11 = 17, and y = 6. Then $x - y = 1 - 6 = -5$ . Try x = 10: then y = 15, and 10 − 15 = −5 again.

More questions like this

The original DOE question

Question 3 as shown on the DOE’s digital test, page 76 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 76, question 3.

Question 10 of 57

2022 Test B Math, question 10

Question

If $x = \frac{1}{4}$ , what is the value of $\frac{2}{2 - x}$ ?

A $\frac{3}{4}$
B $\frac{8}{9}$
C $1\frac{1}{7}$
D $3\frac{1}{2}$

Show the answer

Correct answer: C

The first step is to substitute the value for x into the expression to get $\frac{2}{2 - \frac{1}{4}}$ .

Then, evaluate the denominator and finally divide the numerator by the denominator.

$\frac{2}{\left(\frac{7}{4}\right)} = 2 \div \left(\frac{7}{4}\right) = 2\left(\frac{4}{7}\right) = \frac{8}{7} = 1\frac{1}{7}$

Question analysis

This substitution question sets x equal to one fourth and asks for the value of 2 divided by (2 − x). Students substitute, subtract a fraction from a whole number, and divide by the result. The expression is a stacked fraction, so treating the whole denominator as one quantity and dividing rather than multiplying are where students slip.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Substituting a value and dividing by a fraction are grade 6 skills, and here they take three short steps with friendly numbers. The stacked fraction asks for care, but the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 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

Substitution is one step, so the question is about handling a fraction inside a fraction. The bar over 2 − x groups the whole denominator, and the division by 7/4 must flip it; both errors lead to listed choices. A size check helps: 2 − 1/4 is a little less than 2, so the value must be a little more than 1, which rules out every wrong choice.

Error-prone arithmetic moderate Working it out
The denominator 2 − 1/4 has to become 7/4 before anything is divided, and dividing 2 by 7/4 means multiplying by 4/7; multiplying by 7/4 instead is the usual slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The work ends at the improper fraction 8/7, but the choices show it only as a mixed number, listed among 3/4, 8/9 and 3 1/2. (Enright, Morley & Sheehan, 2002)

Strategy: Simplify the denominator to a single fraction first, then divide 2 by that fraction by multiplying by its reciprocal; treating the fraction bar as a grouping symbol avoids the common slip.

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).
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
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

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Lets the fraction bar cover only the first 2, working 2/2 − 1/4 = 1 − 1/4 = 3/4 instead of dividing 2 by the whole denominator.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Adds 1/4 to 2 instead of subtracting it, so the denominator becomes 9/4, and 2 ÷ 9/4 = 8/9.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds the denominator 7/4 correctly but multiplies 2 by it instead of dividing, 2 × 7/4 = 14/4 = 3 1/2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b/c is a fraction. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing asks how many copies of b/c fit into a.
Changing an improper fraction to a mixed number: $\frac{a}{b} = q\frac{r}{b}$; Use it when the top, a, is bigger than the bottom, b. q is how many whole times b fits into a. r is the remainder, what is left over. For 8/7, the 7 fits once with 1 left, so 8/7 = 1 1/7.

Sort out the question
You know $x = \frac{1}{4}$ . You need the value of 2 divided by (2 − x). A fraction bar means divide. It also groups, so you work out the whole bottom, 2 − x, before you divide.
Work out the bottom
Put in 1/4 for x. To subtract, write 2 as fourths, since 2 = 8/4:
$2 - \frac{1}{4} = \frac{8}{4} - \frac{1}{4} = \frac{7}{4}$
The bottom is 7/4.
Watch out: the sign is minus. Adding 1/4 gives 9/4 and then 8/9, choice B.
Divide 2 by 7/4
Now the problem is 2 ÷ 7/4. Use the rule for dividing by a fraction. Flip 7/4 to get 4/7, then multiply:
$2 \div \frac{7}{4} = 2 \times \frac{4}{7} = \frac{8}{7}$
Watch out: multiplying by 7/4 instead of 4/7 gives 14/4 = 3 1/2, choice D.
Write it as a mixed number
The choices use mixed numbers. 7 fits into 8 once, with 1 left over:
$\frac{8}{7} = 1\frac{1}{7}$
Watch out: the fraction bar covers all of 2 − 1/4. Working 2/2 = 1 first and then taking away 1/4 gives 3/4, choice A.
So the answer is C, $1\frac{1}{7}$ .

Check your answer

Estimate. The bottom, 1 3/4, is a bit less than 2. So 2 divided by it should be a bit more than 1, and 1 1/7 fits. Also check by multiplying: 7/4 × 8/7 = 56/28 = 2. That is the top, so the division is right.

More questions like this

The original DOE question

Question 4 as shown on the DOE’s digital test, page 77 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 77, question 4.

Question 11 of 57

2022 Test B Math, question 11

Question

The cost of shipping a package is $12, plus $2 for each additional pound over 3 pounds. Which expression represents the total cost, in dollars, of shipping a package that weighs p pounds? Assume that $p \geq 3$ .

A $12 + 2(p + 3)$
B $12 + 2(p - 3)$
C $14p + 3$
D $14p - 3$

Show the answer

Correct answer: B

Since the additional cost per pound is not added until the package weighs more than 3 pounds, the total cost for 1, 2, or 3 pounds is $12.

1 pound is $12
2 pounds is $12
3 pounds is $12

Since p is greater than or equal to 3 pounds, the total cost for:

4 pounds is

$12 + (4 - 3)(2) = 14$

5 pounds is

$12 + (5 - 3)(2) = 16$

p pounds is

$12 + (p - 3)(2) = 12 + 2(p - 3)$

You may try to write the expression without generating the pattern by recognizing that subtracting 3 from the number of pounds for the package before multiplying by $2 and then adding $12 results in the total cost.

Question analysis

Shipping a package costs $12, plus $2 for each additional pound over 3 pounds, and students choose the expression for the total cost of a package that weighs p pounds. It tests writing an algebraic expression from a word problem. The per-pound charge starts only after the first 3 pounds, and turning that phrase into algebra is the tricky part.

NYS standards: 6.EE.6 variables representing unknown numbers
6.EE.2a writing algebraic expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Writing an expression from a cost description is grade 6 algebra. Because the per-pound charge starts only after the first 3 pounds, it is less routine than a fee plus a simple rate, in line with typical grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards modeling the fee in two parts: a flat $12 that covers the first 3 pounds, and $2 for each pound beyond them. The two choices that start with 12 look equally reasonable until a weight is tried, and the two that begin with 14p come from merging the fee with the per-pound charge. Testing p = 3, which must cost exactly $12, rules out three choices in one step.

Turning words into math moderate Understanding the question
“$2 for each additional pound over 3 pounds” must become 2(p − 3), a charge on the pounds beyond the first 3, and the word “over” invites writing p + 3 instead. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
12 + 2(p + 3) and 12 + 2(p − 3) differ only in the sign inside the parentheses, so their shape cannot separate them; only a test weight or a careful reading can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
The weight is the letter p, so there is no cost to work out and compare; choosing a weight such as 5 pounds, which costs $16, turns every choice into a number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Test a weight such as 5 pounds: work out its cost from the description, then substitute 5 for p in each choice and keep the one that gives the same cost.

Traps

Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
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

A Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Reads “over 3 pounds” as adding 3, so the $2 charge applies to p + 3 pounds instead of only the pounds above the first 3.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Merges the $12 and the $2 into $14 for every pound and tacks on the 3 from the stem, building an expression from its numbers without a model.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Charges $14 for every pound by merging the fee with the $2 rate, then subtracts 3 for the first 3 pounds, mixing dollars with pounds.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost with a base fee and a rate: $\text{total} = \text{base fee} + \text{rate} \times \text{extra pounds}$; The base fee is the set price, $12, which covers the first 3 pounds. The rate is the price for each pound after that, $2. The extra pounds are the pounds over 3, which is p − 3. Use it when a price has a fixed part plus a part per unit.

Sort out the question
Shipping costs $12 to start. Each pound over 3 pounds adds $2. The package weighs p pounds, and p is at least 3. You need an expression for the total cost in dollars.
Count the extra pounds
Only the pounds above 3 cost extra. To find them, take 3 away from the weight:
$\text{extra pounds} = p - 3$
For example, a 5-pound package has 5 − 3 = 2 extra pounds.
Watch out: over 3 pounds means more than 3. It does not tell you to add 3. Writing p + 3 leads to choice A.
Build the expression
Each extra pound costs $2, so the extra pounds cost 2(p − 3) dollars. Add the $12 base fee:
$\text{total} = 12 + 2(p - 3)$
Watch out: the $12 is paid once, not for every pound. Choices C and D charge $14 for every pound, which mixes up the fee and the rate.
Test each choice with 5 pounds
A 5-pound package costs $12 plus two $2 charges, which is $16. Put p = 5 into each choice:
Choice Value when p = 5
A 12 + 2(8) = 28
B 12 + 2(2) = 16
C 14(5) + 3 = 73
D 14(5) − 3 = 67
Only B gives $16. So the answer is B, $12 + 2(p - 3)$ .

Choice	Value when p = 5
A	12 + 2(8) = 28
B	12 + 2(2) = 16
C	14(5) + 3 = 73
D	14(5) − 3 = 67

Check your answer

Try p = 3, the smallest weight allowed. There are no extra pounds, so it should cost just $12. Choice B gives 12 + 2(0) = 12. Try p = 4: 12 + 2(1) = 14, which is $12 plus one $2 charge.

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 D Math, page 79, question 6.

Question 12 of 57

2022 Test B Math, question 12

Question

Julie is at a clothing store. She wants to buy a bag for $18.75 and some T-shirts for $9.25 each. She can spend no more than $50.00. Which inequality can be used to find the number of T-shirts, x, Julie can purchase?

A $9.25 + 18.75x \leq 50.00$
B $9.25 + 18.75x \geq 50.00$
C $18.75 + 9.25x \leq 50.00$
D $18.75 + 9.25x \geq 50.00$

Show the answer

Correct answer: C

The question uses the phrase “no more than,” which translates to “less than or equal to” (≤).

Since T-shirts cost $9.25 each and Julie bought x T-shirts, the total cost just for the T-shirts is found by multiplying $9.25 by x, or $9.25x$ .

Combining the cost for one bag ($18.75) plus the cost for x T-shirts, the inequality is $18.75 + 9.25x \leq 50.00$ .

Question analysis

Julie wants to buy one bag and some T-shirts at a clothing store, with a price for each, and can spend no more than $50.00. Students choose the inequality for the number of T-shirts, x, she can buy. It tests writing an inequality from a word problem: the choices swap the two prices and the direction of the sign.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an inequality for a budget with one fixed cost and a cost per item is standard grade 7 work. With choices given, the task is to match each price to its role and the phrase to its symbol.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 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 choices form a grid of two decisions: which price goes with x, and which way the symbol points. The prices appear in the order they are used, bag first, so the expression is the easier half and most lost points come from the symbol. Pricing the bag with 2 T-shirts, $37.25, and with 3, $46.50, shows that 9.25 is the amount that repeats with each shirt.

Turning words into math moderate Understanding the question
The $18.75 bag is bought once and the $9.25 is paid for each of x T-shirts, and “no more than $50.00” has to become ≤, which allows spending exactly $50.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choices 18.75 + 9.25x ≤ 50.00 and 18.75 + 9.25x ≥ 50.00 match term for term and differ only in the direction of the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Price one bag and 2 T-shirts by hand to see which price is multiplied by x, then translate “no more than” into an inequality symbol.

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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Keeps the right symbol but multiplies the number of T-shirts by $18.75 and adds $9.25 once, putting the bag’s price where the price per shirt belongs.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Puts x on the $18.75 bag and also turns “no more than $50.00” into ≥, so both the expression and the symbol are wrong.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Builds the total, 18.75 + 9.25x, correctly but reads “no more than $50.00” as at least, so ≥ would make Julie spend $50.00 or more.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost: $\text{total} = \text{one-time cost} + \text{price each} \times \text{number bought}$; The one-time cost is paid once, like one bag. The price each is paid again for every item, so it is multiplied by the number bought. Add them to get the total. Use it when one cost happens once and another repeats.
Spending limit: $\text{total} \leq \text{limit}$; The total is what Julie spends, and the limit is $50.00. The sign ≤ means less than or equal to. So the total can be under the limit or exactly equal to it, but never over. That is what the words “no more than” mean.

Sort out the question
Julie buys one bag for $18.75. She also buys some T-shirts for $9.25 each. x is the number of T-shirts. She can spend no more than $50.00. You need the right inequality. An inequality is a math sentence that uses a sign like ≤ instead of =.
Write the cost of the T-shirts
Each T-shirt costs $9.25. For x T-shirts, she pays $9.25 x times.
$9.25 \times x = 9.25x$
The bag is bought only once. So its $18.75 is not multiplied by x.
Watch out: putting x on 18.75 prices x bags and one T-shirt. That gives choices A and B.
Add to get the total
Use the total cost formula. The bag is the one-time cost.
$\text{total} = 18.75 + 9.25x$
Test it with 2 T-shirts: $18.75 + 9.25 \times 2 = 18.75 + 18.50 = 37.25$ dollars. That is one bag and two shirts, as it should be.
Choose the sign
“She can spend no more than $50.00” means she can spend less than $50.00, or exactly $50.00. That is less than or equal to, the sign ≤.
$18.75 + 9.25x \leq 50.00$
Watch out: ≥ means greater than or equal to. Choice D says she must spend $50.00 or more, the opposite. So the answer is C, $18.75 + 9.25x \leq 50.00$ .

Check your answer

Test whole numbers of T-shirts. With 3, the total is 18.75 + 27.75 = 46.50 dollars. That is not over $50.00. With 4, it is 18.75 + 37.00 = 55.75 dollars, too much. Choice C allows 3 T-shirts and rules out 4, which fits.

More questions like this

The original DOE question

Question 69 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 2023 Admissions, page 200, question 69. Open page 200 of the handbook (PDF)

Question 13 of 57

2022 Test B Math, question 13

Question

$2x + 11 > 3x + 9$

For what values of x is the above inequality true?

A $x < 2$
B $x > 2$
C $x < 20$
D $x > 20$

Show the answer

Correct answer: A

Use inverse operations to isolate the variable x by first getting the x variable on one side.

$2x + 11 > 3x + 9$

$2x - 3x + 11 > 3x - 3x + 9$

$-x + 11 > 9$

$-x + 11 - 11 > 9 - 11$

$-x > -2$

Now, when dividing by $-1$ , the direction of the inequality sign changes.

$-x > -2$

$x < 2$

Another way to solve is to keep the coefficient of x positive and then recognize that $2 > x$ is the same as $x < 2$ .

$2x + 11 > 3x + 9$

$11 > x + 9$

$2 > x$

$x < 2$

Question analysis

This algebra question asks which values of x make the inequality 2x + 11 > 3x + 9 true, choosing from four solution sets. It tests solving a linear inequality with the variable on both sides. Collecting the x terms can leave a negative coefficient, and handling the inequality sign at that point is where most mistakes happen.

NYS standards: 7.EE.4b two-step inequalities
8.EE.7b linear equations with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 Two-step inequalities are grade 7 work, and having x on both sides, a grade 8 idea, takes only one extra subtraction here. The demand matches a careful grade 7 problem, with one sign decision that is easy to get wrong.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 10, 15 words.
Corrected from the DOE handbook: A grammatical error in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The boundary 2 comes out of two routine moves; the direction is where points are lost. Whichever side the x terms are collected on, the last step either divides by a negative or leaves x on the right, and both moments invite a reversed answer. A second risk is moving the 9 without changing its sign, which yields 20. Substituting x = 0, which makes 11 > 9 true, confirms the direction.

Error-prone arithmetic moderate Working it out
Subtracting 3x from both sides leaves −x > −2, and dividing by −1 must reverse the sign; subtracting 2x instead leaves 2 > x, which still has to be read from the x side. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
x < 2 and x > 2 share the boundary and differ only in direction, so finding the 2 settles nothing; only the sign rule or a test value such as x = 0 decides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract 2x from both sides so the x term stays positive, undo the + 9, then rewrite the result with x first; testing a value in the original inequality checks the direction.

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.)
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

B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches −x > −2 but divides by −1 without reversing the inequality, or reads 2 > x as x > 2, so the direction comes out backwards.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Moves the 9 across the inequality without changing its sign, 11 + 9 = 20 instead of 11 − 9 = 2, and keeps the direction right.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves the 9 without changing its sign to get 20 > x, then also rewrites that statement backwards as x > 20, two errors at once.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “Another way to solve is to keep the coefficient of x positive and then recognizing that 2 > x is the same as x < 2.” The two verbs joined by “and then” should have the same form, so the site now says “and then recognize that”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides of an inequality: $\text{if } a > b \text{, then } a - c > b - c$; a and b are the two sides, and c is any number or term. Subtracting (or adding) the same amount on both sides keeps the inequality true and its sign the same. That is how you move terms from one side to the other.
Dividing by a negative number: $\text{if } a > b \text{, then } -a < -b$; a and b are any numbers. Multiplying or dividing both sides by a negative number flips the sign, so > becomes <. For example, 3 > 2, but −3 < −2. Use it whenever you divide by a negative number.

Sort out the question
The sign > means is greater than. You need every value of x that makes 2x + 11 greater than 3x + 9. The answer is a range of values, not one number. Solve it like an equation, but keep an eye on the direction of the sign.
Gather the x terms on one side
The right side has more x’s, 3x, than the left side, 2x. So subtract 2x from both sides. That keeps the x term positive:
$2x + 11 - 2x > 3x + 9 - 2x$
$11 > x + 9$
Get x by itself
Subtract 9 from both sides:
$11 - 9 > x + 9 - 9$
$2 > x$
Watch out: to move the 9, subtract it from both sides. Adding it instead gives 20 > x, which leads to choice C or D.
Read it from the x side
$2 > x$ says 2 is greater than x. That is the same as saying x is less than 2:
$x < 2$
Watch out: when you swap the sides, the sign must point the other way. Writing $x > 2$ gives choice B. If you subtract 3x instead, you get $-x > -2$ . Dividing by −1 then flips the sign, and you get $x < 2$ again.
So the answer is A, $x < 2$ .

Check your answer

Test x = 0, which is less than 2. The left side is 11 and the right side is 9, and 11 > 9 is true. Test x = 5, which is more than 2. The left side is 21 and the right side is 24, and 21 > 24 is false. So only values below 2 work.

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 D Math, page 83, question 10. This is the DOE’s version, before the corrections listed in the analysis.

Question 14 of 57

2022 Test B Math, question 14

Question

In a scale drawing of a triangular park, the sides measure 5 inches, 7 inches, and 8 inches. The perimeter of the actual park is 25 miles. What is the actual length of the shortest side of the park?

A6.25 mi
B8.75 mi
C10 mi
D15.625 mi

Show the answer

Correct answer: A

To find the perimeter of a triangle, add the measures of the three sides. The perimeter of the scale drawing is $5 + 7 + 8 = 20$ inches.

Now, use the perimeters to find the scale factor for the scale drawing to the actual perimeter.

The scale factor is 20 inches to 25 miles, which reduces to a scale factor of 4 inches to 5 miles. Note that the scale for the perimeter is the same as the scale for the side lengths.

Since the smallest side on the scale drawing is 5 inches, use the scale factor to find the actual side length of the smallest side:

$5\text{ inches}\left(\frac{5\text{ miles}}{4\text{ inches}}\right) = 6.25\text{ miles}$

Question analysis

A scale drawing of a triangular park has sides of 5, 7 and 8 inches, and the real park's perimeter is 25 miles. Students find the actual length of the park's shortest side. It tests scale drawings and proportional reasoning; the scale is not given directly, so it must be worked out from the two perimeters first.

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are a grade 7 topic, and this is a standard scale problem. The one twist is that the scale is not stated: it must come from comparing the two perimeters before any side can be converted.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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

A scale drawing multiplies every length by one factor, and the perimeter shares that factor, so the work is short once the two perimeters are paired. The common loss is reasoning with differences: the perimeter grows by 5, so a side seems to grow by 5 as well. Converting to miles per inch first, then checking that the three real sides add back to 25 miles, guards against that error and against scaling the wrong side.

Turning words into math minor Understanding the question
The 25 miles is the park's perimeter, not a side, so it must be matched with the drawing's perimeter, 5 + 7 + 8 = 20 inches, before any side can be scaled. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three operations in a row, adding the sides, dividing 25 by 20 for 1.25 miles per inch, and multiplying the 5-inch side, each produce a number the next step needs. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the three drawn sides to get the drawing's perimeter, divide the real perimeter by it to find the miles per inch, then multiply the shortest drawn side by that scale.

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

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the scale of 1.25 miles per inch but applies it to the 7-inch side, the middle one, giving 8.75 miles instead of the shortest side.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Keeps a constant difference instead of a constant ratio: the perimeter grows by 5, from 20 to 25, so it adds 5 to the 5-inch side.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Pairs the 25 miles with the longest side, 8 inches, instead of the 20-inch perimeter, so each inch seems to be 3.125 miles and 5 inches 15.625.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a triangle: $P = a + b + c$; P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. You add them because going once around the triangle covers each side once.
Using a scale: $\text{actual length} = \text{drawing length} \times \text{miles per inch}$; The drawing length is measured on the drawing, in inches. Miles per inch is how many real miles each inch stands for. In a scale drawing, every length is multiplied by the same number, so the perimeter and each side use the same scale.

Sort out the question
A scale drawing is a smaller copy, with every length shrunk by the same amount. The drawing’s sides are 5, 7 and 8 inches. The real park’s perimeter, the distance all the way around, is 25 miles. You need the real length of the shortest side.
Find the drawing’s perimeter
Add the three sides of the drawing:
$P = 5 + 7 + 8 = 20$
The drawing’s perimeter is 20 inches. It stands for the real perimeter of 25 miles.
Watch out: 25 miles is the whole way around, not one side. Matching it with the 8-inch side gives 15.625 miles, choice D.
Find the scale
20 inches stand for 25 miles. Divide to find how many miles 1 inch stands for:
$25 \div 20 = 1.25$
So each inch on the drawing is 1.25 miles in the park.
Scale the shortest side
The shortest side on the drawing is 5 inches. Multiply it by the scale:
$5 \times 1.25 = 6.25$
The real shortest side is 6.25 miles.
Watch out: the perimeter grew by 5, from 20 to 25, but you cannot just add 5 to a side. That gives 10 miles, choice C. Scaling multiplies. Also use the 5-inch side, not the 7-inch side, which gives 8.75 miles, choice B.
So the answer is A, 6.25 mi.

Check your answer

Scale all three sides: 5 × 1.25 = 6.25, 7 × 1.25 = 8.75 and 8 × 1.25 = 10. Add them: 6.25 + 8.75 + 10 = 25 miles. That is the park’s perimeter, so the scale is right.

More questions like this

The original DOE question

Question 11 as shown on the DOE’s digital test, page 84 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 84, question 11.

Question 15 of 57

2022 Test B Math, question 15

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?

Show the answer

Correct answer: D

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

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

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

$x + 5 = 24$

$x = 19$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
No obvious method minor
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.

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)
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

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.
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

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads x as the radius RP, as if its arrow ended at the center P, though it runs past P to the foot of the vertical segment.
B Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a surface strategy Expected pull: weak
Judges x from the drawing, where it looks about three times as long as the piece labeled 5, instead of working from the radius of 12.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Adds the two labeled numbers, 12 + 5 = 17, without placing the 12 on the diameter, as if x were a radius plus the 5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius and diameter: $d = 2r$; 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.

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.
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.
Find the diameter RT
RT passes through the center, so it is a diameter. Use the formula with r = 12:
$\mathrm{RT} = 2r = 2 \times 12 = 24$
RT is 24 long. It is the radius RP plus the radius PT.
Solve for x
RT is also split into x and 5, so:
$x + 5 = 24$
Subtract 5 from both sides:
$x = 24 - 5 = 19$
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.
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 16 of 57

2022 Test B Math, question 16

Question

$A right triangle with the right angle marked at its bottom left vertex. The angle at the top vertex is labeled (2x + 9)°, and the angle at the bottom right vertex is labeled (3x − 14)°.$

In the right triangle shown above, what is the measure of the smallest angle?

A43°
B44°
C46°
D47°

Show the answer

Correct answer: A

Since the three angles of a triangle have a sum of 180°, and since one angle is a right angle, the other two angles have a sum of 90°. By writing an equation and using inverse operations, the value of x can be found.

$(2x + 9) + (3x - 14) = 90$

$5x - 5 = 90$

$5x = 95$

$x = 19$

Now, use substitution to find the measure of the two angles to determine the smallest angle of the triangle.

$2x + 9 = 2(19) + 9 = 47$

$3x - 14 = 3(19) - 14 = 43$

The smallest angle is 43°.

Question analysis

A right triangle has its two other angles labeled (2x + 9)° and (3x − 14)°, and students find the measure of the smallest angle. It tests the angle sum of a triangle and solving a linear equation. Finding x is not the end: both angles must be worked out and compared to see which one is smallest.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students write equations for unknown angles using complementary and supplementary angles. This also needs the angle sum of a triangle, which New York places in grade 8, plus substituting back and comparing, though each step is short.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Many steps minor
Something you must already know minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 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

Setting the sum of the two labeled angles to 90° is the decision; the algebra after it is routine. The trap waits at the end: x = 19 is not an angle, and the expression at the top vertex, 2x + 9, gives the larger acute angle, 47°. Evaluating both expressions and checking that they add to 90° takes seconds and shows which one is smaller.

Combines ideas moderate Working it out
The angle facts and the algebra meet in one equation: the right angle means the two labeled angles add to 90°, and only that sum makes (2x + 9) + (3x − 14) solvable. (Embretson, 1998; Daniel & Embretson, 2010)
Many steps minor Working it out
After solving 5x − 5 = 90, both angle expressions must be evaluated and compared, because the question wants an angle, not x, and not the first angle found. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
The 180° angle sum of a triangle is not printed, and only the small square at the bottom left shows that one angle is 90°. (Anderson & Krathwohl, 2001)

Strategy: Use the right angle to see that the other two angles add to 90°, solve that equation for x, then substitute into both expressions and compare them.

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.

The wrong answers, classified

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 44°; it sits one degree from the right measure and fits no consistent value of x, so it mostly draws estimates and guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 46°, which would need x = 20 and two angles adding to 95°, so it mostly catches guesses near 45°.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Solves x = 19 correctly but answers with 2x + 9 = 47°, the larger acute angle, instead of evaluating both expressions to find the smaller.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles in a triangle: $a + b + c = 180$; a, b and c are the three angles of any triangle, in degrees. They always add up to 180 degrees. If you tear off the three corners and set them side by side, they make a straight line.
The two other angles of a right triangle: $a + b = 90$; a and b are the two angles that are not the right angle. A right angle is 90 degrees, so the other two share what is left of the 180: 180 − 90 = 90.

Sort out the question
The small square at the bottom left corner marks a right angle, which is 90°. The top angle is (2x + 9)°, and the bottom right angle is (3x − 14)°. You need the smallest of the three angles. First find x, then find each angle.
Write an equation
The three angles add to 180°. The right angle uses 90° of that, so the other two angles add to 90°:
$(2x + 9) + (3x - 14) = 90$
Solve for x
Combine the x terms, 2x + 3x = 5x. Combine the numbers, 9 − 14 = −5:
$5x - 5 = 90$
Add 5 to both sides, then divide both sides by 5:
$5x = 95$
$x = 95 \div 5 = 19$
Find both angles
Put x = 19 into each expression:
$2x + 9 = 2 \times 19 + 9 = 38 + 9 = 47$
$3x - 14 = 3 \times 19 - 14 = 57 - 14 = 43$
The three angles are 47°, 43° and 90°.
Watch out: x = 19 is not an angle. And the first angle you find, 47°, is not the smallest one. Stopping there gives choice D.
Pick the smallest angle
Compare 47°, 43° and 90°. The smallest is 43°. The figure is not drawn to scale, so do not judge the angles by how they look. Use the numbers.
So the answer is A, 43°.

Check your answer

Add the three angles: 47 + 43 + 90 = 180. That is the right total for a triangle. Also, 47 + 43 = 90, so the two smaller angles fill exactly what the right angle leaves.

More questions like this

The original DOE question

Question 13 as shown on the DOE’s digital test, page 86 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 86, question 13.

Question 17 of 57

2022 Test B Math, question 17

Question

James must read 84 pages of a book for a class assignment. If he reads $\frac{1}{4}$ of the pages in the morning and $\frac{1}{3}$ of the remaining pages in the afternoon, how many pages will he still need to read?

Show the answer

Correct answer: B

Pages read in the morning: $\left(\frac{1}{4}\right)(84) = 21$

Pages remaining after the morning: $84 - 21 = 63$

Pages read in the afternoon: $\left(\frac{1}{3}\right)(63) = 21$

Pages he still needs to read: $63 - 21 = 42$

Question analysis

James must read 84 pages for a class assignment: 1/4 of them in the morning, then 1/3 of the remaining pages in the afternoon. Students find how many pages he still has left to read. It tests finding a fraction of a quantity in stages, and the bold word remaining decides which amount the second fraction is taken of.

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking fractions of a quantity in stages is typical grade 7 work. The numbers divide evenly and the bold word flags the trap, so the four short steps match routine grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 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

Each step is simple, but the second fraction has a different whole from the first. Students who take 1/3 of the full 84 pages get a tidy 28 and a listed answer, despite the bold word that warns against it. Here the pages read and the pages left both come to 42, so confusing those two quantities costs nothing, and checking that 21 + 21 + 42 = 84 confirms the work.

Many steps moderate Working it out
Four operations in order, a quarter of 84, a subtraction, a third of the 63 left, and a second subtraction, each feed the next, so a slip early carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The bold “remaining” moves the base of the second fraction from all 84 pages to the 63 left after the morning, and “still need to read” asks for what is left, not what was read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work in order: find the morning's pages, subtract to get what remains, take 1/3 of that remainder, and subtract again; a quick bar sketch of the 84 pages helps.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Takes 1/3 of all 84 pages for the afternoon, 28, instead of 1/3 of the 63 remaining, so 84 − 21 − 28 = 35.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 45; it is not a fraction or a difference of any page count in the problem, so it mostly draws guesses near 42.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Takes the afternoon's third from all 84 pages and subtracts only that, 84 − 28 = 56, overlooking the 21 pages already read in the morning.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of a number: $\frac{a}{b} \text{ of } n = \frac{a}{b} \times n$; n is the whole amount, and a/b is the fraction you take. The word of means multiply. When the top, a, is 1, this is just n divided by b. For example, 1/4 of 84 is 84 ÷ 4.

Sort out the question
James has 84 pages to read. In the morning he reads 1/4 of all the pages. In the afternoon he reads 1/3 of the remaining pages, the ones still left after the morning. You need the number of pages he still has to read after that.
Find the morning pages
Take 1/4 of 84:
$\frac{1}{4} \times 84 = 84 \div 4 = 21$
He reads 21 pages in the morning. That leaves:
$84 - 21 = 63$
63 pages are left after the morning.
Find the afternoon pages
The afternoon fraction is 1/3 of the 63 pages left, not of all 84:
$\frac{1}{3} \times 63 = 63 \div 3 = 21$
He reads 21 pages in the afternoon.
Watch out: taking 1/3 of all 84 pages gives 28. Then 84 − 21 − 28 = 35, choice A. The bold word remaining tells you to use 63.
Find the pages still to read
Take the afternoon pages away from the 63 pages left after the morning:
$63 - 21 = 42$
He still needs to read 42 pages.
Watch out: subtract both readings, morning and afternoon. Taking only 28 from 84 gives 56, choice D.
So the answer is B, 42.

Check your answer

Add up all the parts: 21 pages in the morning, 21 in the afternoon and 42 still to read. 21 + 21 + 42 = 84, the whole assignment. Another way: after he reads 1/3 of the 63 pages, 2/3 of them are left, and 2/3 of 63 is 42.

More questions like this

The original DOE question

Question 14 as shown on the DOE’s digital test, page 87 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 87, question 14.

Question 18 of 57

2022 Test B Math, question 18

Question

$\frac{1}{11} + \frac{1}{22} + \frac{1}{33}$

What is the value of the expression shown above?

A $\frac{1}{22}$
B $\frac{1}{11}$
C $\frac{1}{6}$
D $\frac{3}{22}$

Show the answer

Correct answer: C

To add fractions, find a common denominator and rename the fractions using the common denominator. The least common denominator for 11, 22, and 33 is 66. After adding the fractions, the final fraction can be reduced to lowest terms.

$\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}$

$\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}$

$\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}$

$\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66} = \frac{1}{6}$

Question analysis

This fractions question asks for the value of 1/11 + 1/22 + 1/33. Students find a common denominator for three fractions, rename each one, add and simplify. The denominators are all multiples of 11, and adding the tops and bottoms straight across, or forgetting to rename a numerator, leads to tempting wrong choices.

NYS standards: 7.NS.1d properties for adding and subtracting
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding fractions with unlike denominators is taught before grade 7. Finding a common denominator for three related numbers and simplifying is short, routine work, lighter than typical grade 7 problems with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 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 question checks whether students rename fractions before adding, and each wrong choice comes from a shortcut that skips that step or stops early. Seeing that 22 and 33 are multiples of 11 points to 66 as the common denominator. A size check helps too: the sum must be more than 1/11, its largest term, which rules out two of the wrong choices at once.

Error-prone arithmetic moderate Working it out
Three unlike denominators must be renamed over 66, not over the product of all three, and the numerators 6, 3 and 2 have to be found one by one before adding. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The sum comes out as 11/66, which must be reduced to match a choice; its simplest form has a denominator, 6, unlike any in the expression. (Enright, Morley & Sheehan, 2002)

Strategy: Notice that 22 and 33 are multiples of 11, rename all three fractions over their least common multiple, add the numerators, and simplify the result.

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the numerators and the denominators straight across, 3 over 11 + 22 + 33 = 66, and simplifies 3/66 to 1/22.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Puts the three 1s over the largest denominator, 33, without renaming the other two fractions, so 3/33 simplifies to 1/11.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds the first two fractions correctly, 2/22 + 1/22 = 3/22, and stops before adding the third fraction, 1/33.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction. It works because n/n is 1, and multiplying by 1 changes nothing. Use it to give fractions the same bottom.
Adding fractions with the same denominator: $\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}$; d is the denominator, the bottom number, which tells the size of each piece. a, b and c are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

Sort out the question
You need to add three fractions: 1/11, 1/22 and 1/33. Their denominators (bottom numbers) are all different. You can only add fractions when the pieces are the same size. So first give all three the same denominator.
Find a common denominator
A common denominator is a number that 11, 22 and 33 all divide into. Notice that 22 = 2 × 11 and 33 = 3 × 11. So the smallest number that all three divide into is:
$2 \times 3 \times 11 = 66$
Rename each fraction
Multiply the top and bottom of each fraction by the number that makes the bottom 66:
$\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}$
$\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}$
$\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}$
Add the tops
Now every piece is a sixty-sixth. Add the tops and keep the bottom:
$\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66}$
Watch out: never add the bottoms. Adding straight across gives 3/66 = 1/22, choice A. Putting the three 1s over 33 gives 3/33 = 1/11, choice B.
Simplify the fraction
To simplify, divide the top and bottom by a number that goes into both. Here 11 goes into 11 and into 66:
$\frac{11}{66} = \frac{11 \div 11}{66 \div 11} = \frac{1}{6}$
Watch out: add all three fractions. Stopping after the first two gives 3/22, choice D.
So the answer is C, $\frac{1}{6}$ .

Check your answer

Estimate with decimals. 1/11 is about 0.091, 1/22 is about 0.045 and 1/33 is about 0.030. They add to about 0.166. And 1/6 is 0.1666…, a match. The other choices are all 0.136 or less.

More questions like this

The original DOE question

Question 15 as shown on the DOE’s digital test, page 88 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 88, question 15.

Question 19 of 57

2022 Test B Math, question 19

Question

At Midville High School, 64 students are members of either the orchestra or the choir, or both. Of these students, 38 are members of the orchestra, and 46 are members of the choir. What fraction of the total is the number of students who are members of both the orchestra and the choir?

A $\frac{16}{21}$
B $\frac{10}{19}$
C $\frac{10}{23}$
D $\frac{5}{16}$

Show the answer

Correct answer: D

First, add the number of students that are members of the orchestra and the number of students that are members of the choir.

$38 + 46 = 84$

Then, subtract to find the number of students that are members of both orchestra and choir.

$84 - 64 = 20$

Then, find the fraction of the 64 students that are members of both orchestra and choir.

$\frac{20}{64} = \frac{5}{16}$

Question analysis

At Midville High School, 64 students belong to the orchestra, the choir or both; 38 are in the orchestra and 46 in the choir. Students find what fraction of the 64 belong to both groups. It is a Venn diagram counting question with a fraction, and the students in both groups are included in each membership number.

NYS standards: 7.SP.8b sample spaces for compound events
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Overlapping groups are seldom taught directly in grade 7, though the arithmetic is whole-number addition and subtraction. Students must first find the overlap and then choose the right whole for the fraction, two decisions where errors are common.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 54 words.
Corrected from the DOE handbook: A misstatement in the explanation’s first step 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 question hides its key number: the students in both groups appear only as the amount by which 38 + 46 overshoots 64. A sketch of two overlapping circles, with 20 in the middle, 18 in orchestra only and 26 in choir only, shows the whole group at once. After that, the remaining risk is the denominator, since the question asks for a fraction of all 64 students, not of one group.

Turning words into math moderate Understanding the question
“Either the orchestra or the choir, or both” means the 64 counts each student once, while 38 + 46 counts the students in both groups twice; the overlap itself is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method minor Working it out
Nothing names the step that finds the overlap: the excess of 38 + 46 = 84 over 64 is the number in both groups, an idea students must bring themselves. (Schoenfeld, 1985)
An answer in an unexpected form minor Choosing the answer
The fraction 20/64 appears only in lowest terms, so it must be simplified by 4 before it can be matched with a choice. (Enright, Morley & Sheehan, 2002)

Strategy: Add the orchestra and choir counts, subtract the 64 students to find how many were counted twice, then put that number over the total and simplify.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

A Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: strong
Takes 38 + 46 = 84 as the number of students, counting those in both groups twice, and puts the stated 64 over it to get 16/21.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the 20 students in both groups but compares them with the 38 orchestra members instead of all 64 students, so 20/38 = 10/19.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the 20 students in both groups but divides by the 46 choir members, using one group as the whole instead of all 64, so 20/46 = 10/23.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation begins “First, add the number of students that are members of both orchestra and choir. 38 + 46 = 84.” But 38 is the number of orchestra members and 46 the number of choir members; the number of students in both is what the next step finds (84 − 64 = 20). The site now says “First, add the number of students that are members of the orchestra and the number of students that are members of the choir.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap: $\text{total} = O + C - \text{both}$; O is the number in the orchestra, and C is the number in the choir. Students in both groups are counted in O and again in C. Subtracting them once makes each student count one time. Here every student is in at least one group, so this gives everyone.
Fraction of a total: $\text{fraction} = \frac{\text{part}}{\text{whole}}$; The part is the group you are asked about. The whole is the full group it comes from. Here the part is the students in both groups, and the whole is all 64 students. Simplify by dividing the top and bottom by the same number.

Sort out the question
There are 64 students in all. Each one is in the orchestra, the choir or both. 38 are in the orchestra, and 46 are in the choir. Some students are in both, but the question does not say how many. You need that number as a fraction of all 64 students.
Add the two groups
Add the number in the orchestra and the number in the choir:
$38 + 46 = 84$
That is more than the 64 students there are. Why? A student in both groups is counted once with the orchestra and once more with the choir.
Find the students in both
The extra amount over 64 is the students who were counted twice:
$84 - 64 = 20$
So 20 students are in both groups. This is the overlap rule turned around: both = O + C − total.
Watch out: 84 is not the number of students. Putting 64 over 84 gives 16/21, choice A.
Write the fraction and simplify
The part is 20 students, and the whole is all 64 students. Both numbers divide by 4:
$\frac{20}{64} = \frac{20 \div 4}{64 \div 4} = \frac{5}{16}$
Watch out: the whole is all 64 students, not one group. Using 38 gives 10/19, choice B. Using 46 gives 10/23, choice C.
So the answer is D, $\frac{5}{16}$ .

Check your answer

Split the 64 students into three groups that do not overlap. Only orchestra: 38 − 20 = 18. Only choir: 46 − 20 = 26. Both: 20. Add them: 18 + 26 + 20 = 64. That matches the total, so 20 is right.

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 D Math, page 90, question 17. This is the DOE’s version, before the corrections listed in the analysis.

Question 20 of 57

2022 Test B Math, question 20

Question

Which of the following decimals is equivalent to $1\frac{2}{3} + 2\frac{4}{9}$ ?

A $3.\overline{1}$
B $3.\overline{6}$
C $4.\overline{1}$
D $4.\overline{9}$

Show the answer

Correct answer: C

First, add the mixed numbers. Then, convert the fraction $\frac{1}{9}$ to a decimal.

$1\frac{2}{3} + 2\frac{4}{9}$

$1\frac{6}{9} + 2\frac{4}{9}$

$3 + \frac{10}{9}$

$3 + 1\frac{1}{9}$

$4\frac{1}{9} = 4.111\ldots = 4.\overline{1}$

Question analysis

This question asks which repeating decimal equals the sum of the mixed numbers 1 2/3 and 2 4/9. Students rename the fractions with a common denominator, add, and convert the fraction part to a decimal with a bar over its repeating digit. The choices differ in both the whole number and the repeating digit, so each step has to be right.

NYS standards: 7.NS.2d fractions to decimals by long division
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding mixed numbers and writing a fraction as a repeating decimal are standard grade 7 skills. This combines them in three short steps; renaming a fraction and reading the bar notation are the only places that need care.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 13, 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

Two fraction moves and one conversion stand between the sum and the choices. The carry is the main hazard: 10/9 has to become 1 1/9, and students who keep only the 1/9 land on a choice with the wrong whole number. Estimating first, about 1.7 + 2.4 = 4.1, rules out both choices that begin with 3 and the one that equals 5.

Error-prone arithmetic moderate Working it out
Adding 2/3 and 4/9 needs 2/3 renamed as 6/9, and the sum 10/9 is more than 1, so a whole must be carried into the 3 before converting. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The choices are repeating decimals with bars, so a mixed number in ninths has to be recognized as a digit repeating forever; knowing that ninths repeat their numerator makes the match quick. (Enright, Morley & Sheehan, 2002)

Strategy: Rename 2/3 as ninths, add the whole numbers and the fractions separately, regroup any fraction of 9/9 or more as a whole, then divide the leftover numerator by 9.

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.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Adds the fractions to 10/9 but keeps only the 1/9 left over, forgetting to carry the whole 1 into the 3, so the sum becomes 3 1/9.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators 2 and 4 over 9 without renaming 2/3 as 6/9, so gets 3 6/9, which is 3.666…
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Reaches 4 1/9 but repeats the denominator's 9 instead of the numerator's 1, as if one ninth were 0.999… rather than 0.111…

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction, because n/n is 1. Use it to give two fractions the same bottom before you add them.
Ninths as repeating decimals: $\frac{1}{9} = 0.111\ldots = 0.\overline{1}$; 1 divided by 9 is 0.111…, with 1s that go on forever. The bar over a digit means that digit repeats forever. Ninths from 1/9 to 8/9 repeat their top digit, so 4/9 is 0.444… and 2/9 is 0.222…

Sort out the question
You need to add two mixed numbers, 1 2/3 and 2 4/9. A mixed number is a whole number plus a fraction. Then write the sum as a decimal. The choices are repeating decimals, where the bar over a digit means it repeats forever.
Rename two thirds as ninths
The fractions 2/3 and 4/9 have different bottoms. Since 3 × 3 = 9, multiply the top and bottom of 2/3 by 3:
$\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}$
So 1 2/3 is the same as 1 6/9.
Watch out: adding the tops 2 and 4 over 9 without renaming gives 3 6/9, choice B.
Add the mixed numbers
Add the whole numbers, then add the fractions:
$1 + 2 = 3 \qquad \frac{6}{9} + \frac{4}{9} = \frac{10}{9}$
10/9 is more than 1. It is 9/9 plus 1/9, or 1 1/9. Carry the 1 into the whole number:
$3 + 1\frac{1}{9} = 4\frac{1}{9}$
Watch out: forgetting to carry the 1 gives 3 1/9, choice A.
Change it to a decimal
One ninth is 0.111…, with the 1 repeating forever. So:
$4\frac{1}{9} = 4.111\ldots = 4.\overline{1}$
Watch out: ninths repeat the top digit, not the 9. Choice D, 4.999…, does not match 4 1/9.
So the answer is C, $4.\overline{1}$ .

Check your answer

Use rough decimals. 1 2/3 is about 1.667, and 2 4/9 is about 2.444. Their sum is about 4.111. That matches 4.111… and rules out both choices that start with 3. And 4.999… is almost 5, which is too big.

More questions like this

The original DOE question

Question 18 as shown on the DOE’s digital test, page 91 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 91, question 18.

Question 21 of 57

2022 Test B Math, question 21

Question

The decimal 0.4 can be written as the fraction $\frac{x}{25}$ . What is the value of x?

A0.016
B0.16
C10
D16

Show the answer

Correct answer: C

First, convert the decimal 0.4, which is read as four-tenths, to the fraction $\frac{4}{10}$ , and then set up a proportion to solve for x using cross products.

$\frac{4}{10} = \frac{x}{25}$

$10x = 100$

$x = 10$

Or, reduce the fraction to $\frac{2}{5}$ before setting up the proportion and then use proportional reasoning.

$\frac{4}{10} = \frac{2}{5}$

$\frac{2 \times 5}{5 \times 5} = \frac{10}{25}$

Therefore, $x = 10$ .

Question analysis

This question asks students to write the decimal 0.4 as an equivalent fraction whose denominator is 25 and to choose the missing numerator. It tests converting between decimals and fractions. Tenths do not turn into twenty-fifths in one obvious step, so the numerator has to be scaled with care, in the same way as the denominator.

NYS standards: 6.NS.3 multi-digit decimal operations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Rewriting a decimal as an equivalent fraction is upper elementary and grade 6 work. It is one conversion with no context, and the friendly denominator keeps it well below the demand of typical grade 7 problems.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

A single conversion: 0.4 is 4/10, or 2/5, and 2/5 renamed over 25 is 10/25. The only risk is the direction of the step, since x/25 = 0.4 means multiplying 0.4 by 25, and dividing instead leads to the small decimals among the choices. Any choice can be checked by dividing it by 25, and only one gives back 0.4.

Error-prone arithmetic minor Working it out
Four tenths has to be renamed in twenty-fifths, and 10 does not divide 25, so the numerator is reached through 2/5 = 10/25 or by multiplying 0.4 by 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the decimal as a fraction in tenths, simplify it, then multiply the numerator and denominator by the same number until the denominator is 25.

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).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 0.4 by 25 instead of multiplying, undoing the division by 25 with another division, which gives 0.016.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides by 25 instead of multiplying and also drops the decimal point of 0.4, so 4 ÷ 25 = 0.16, ten times 0.016.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single mistake gives 16; it is 4 ÷ 25 = 0.16 with the decimal point moved two places, a chain with no reason behind it, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a is the top number of the fraction, and b is the bottom number. n is any number except 0. Multiplying the top and the bottom by the same number gives an equal fraction. It is like cutting every piece into n smaller pieces.

Sort out the question
You need a fraction with 25 on the bottom that equals 0.4. Its top is x. So you need the x that makes $\frac{x}{25}$ the same amount as 0.4.
Write the decimal as a fraction
0.4 is read four tenths, so it is 4 over 10. Simplify it by dividing the top and the bottom by 2.
$0.4 = \frac{4}{10} = \frac{2}{5}$
Rename the fraction over 25
To turn the bottom 5 into 25, multiply it by 5. Do the same to the top, so the fraction keeps its value.
$\frac{2}{5} = \frac{2 \times 5}{5 \times 5} = \frac{10}{25}$
So x = 10.
Give the answer
Another way: x/25 = 0.4 means x divided by 25 is 0.4. Undo the division by multiplying: $x = 0.4 \times 25 = 10$ .
Watch out: dividing 0.4 by 25 instead gives 0.016, choice A. Dropping the decimal point as well gives 0.16, choice B. So the answer is C, 10.

Check your answer

Divide the answer by 25 and see if you get 0.4. $10 \div 25 = 0.4$ , because 25 × 0.4 = 10. It works. Choice D fails: 16 ÷ 25 = 0.64, not 0.4.

More questions like this

The original DOE question

Question 78 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 202, question 78. Open page 202 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 22 of 57

2022 Test B Math, question 22

Question

What is the value of $\frac{-9(-4)^{2} + 36(-4) + 304}{-4}$ ?

A−76
B−4
C4
D148

Show the answer

Correct answer: B

Use the order of operations to evaluate the expression. It is important to recognize that the fraction bar is a grouping symbol and so the entire numerator should be evaluated first. Exponents are evaluated first.

$\frac{-9(-4)^{2} + 36(-4) + 304}{-4}$

$\frac{-9(16) + 36(-4) + 304}{-4}$

Then multiplication, recognizing that the product of a negative factor and a positive factor is negative.

$\frac{-9(16) + 36(-4) + 304}{-4}$

$\frac{-144 - 144 + 304}{-4}$

Then addition and subtraction in order from left to right, recognizing the rules for adding and subtracting signed numbers.

$\frac{-144 - 144 + 304}{-4}$

$\frac{-288 + 304}{-4} = \frac{16}{-4}$

And finally the fraction can be evaluated by division, recognizing that the quotient of a positive and a negative number is negative.

$\frac{16}{-4} = -4$

Question analysis

This arithmetic question asks for the value of a fraction whose numerator, −9(−4)² + 36(−4) + 304, is divided by −4. Students follow the order of operations with negative numbers: a square, two products, a sum and a division. Every step carries a sign, and one dropped negative changes the result completely.

NYS standards: 7.NS.2a rational number multiplication
7.NS.2b rational number division
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Order of operations with negative numbers is standard grade 7 practice. This expression stacks a squared negative, two products and a division by a negative into one fraction, so it takes more steps than usual, each with a sign to track.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 6 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 here is conceptually new; the question measures care with signs under the order of operations. The parentheses in (−4)² decide whether −9 multiplies 16 or −16, and the two products of −144 then nearly cancel the 304, leaving a small numerator whose sign is easy to lose. Writing each intermediate value with its sign, line by line, prevents most errors.

Error-prone arithmetic moderate Working it out
Five sign decisions sit in one expression: squaring −4, two products with a negative factor, a sum that climbs from −288 past zero, and a division by −4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The numerator alone takes an exponent, two multiplications and two additions before the single division, and all of it must be finished before dividing. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The right value and the choice 4 differ only in sign, so a student with the right size must still check the sign of a positive numerator over −4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Treat the fraction bar as grouping: square −4 first, do the two multiplications, combine the numerator from left to right, and only then divide by −4, writing each sign as you go.

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.)
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−4)² as −16, so −9 × (−16) = 144, the numerator becomes 304, and 304 ÷ (−4) = −76.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Works the numerator out correctly as 16 but divides by −4 as if the signs did not matter, giving a positive quotient.
D Negatives and direction (Brown & Burton, 1978)
Reveals a surface strategy Expected pull: weak
Drops every negative sign, adding 144 + 144 + 304 = 592 and dividing by 4, so 148 keeps none of the signs in the problem.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{grouping} \rightarrow \text{exponents} \rightarrow \times, \div \rightarrow +, -$; This is the order for working out an expression. Grouping means parentheses and the fraction bar. Exponents are small raised numbers, like the 2 in (−4)². Next, multiply and divide from left to right. Last, add and subtract from left to right.
Squaring a negative number: $(-a)^{2} = (-a) \times (-a) = a^{2}$; a is a positive number, so −a is negative. Squaring means multiplying a number by itself. A negative times a negative is positive, so a negative number squared is positive. For example, (−4)² = 16.
Signs when you multiply or divide: $\frac{\text{positive}}{\text{negative}} = \text{negative}$; When two numbers have different signs, their product or quotient is negative. When they have the same sign, it is positive. A quotient is the answer to a division.

Sort out the question
You need the value of one big fraction. The fraction bar is a grouping symbol, so finish the whole top first. Then divide by −4. Follow the order of operations, and track every minus sign.
Do the exponent first
The only exponent is (−4)². Squaring means multiplying a number by itself:
$(-4)^{2} = (-4) \times (-4) = 16$
Watch out: (−4)² is 16, not −16. Using −16 makes the top 304, which gives −76, choice A.
Do the multiplication
Now multiply. A negative times a positive is negative:
$-9 \times 16 = -144$
$36 \times (-4) = -144$
So the top becomes:
$-144 + (-144) + 304$
Add from left to right
Add the two negatives first:
$-144 + (-144) = -288$
Then add 304. Since 304 is more than 288, the result is positive:
$-288 + 304 = 16$
The whole top is 16.
Divide by −4
Now divide the top by the bottom. A positive divided by a negative is negative:
$\frac{16}{-4} = -4$
Watch out: positive 4, choice C, ignores the negative bottom. Dropping every minus sign gives 592 ÷ 4 = 148, choice D.
So the answer is B, −4.

Check your answer

Multiply back. If the answer is −4, then −4 times the bottom, −4, should give the top. It does: (−4) × (−4) = 16. Also, the top is positive and the bottom is negative, so the answer must be negative.

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 D Math, page 93, question 20.

Question 23 of 57

2022 Test B Math, question 23

Question

Maxie borrowed $7.75 from her mother, $11.00 from her father, and $4.50 from her brother to purchase a video game. Her grandmother gave her $25.00 as a gift. How much money will Maxie have left or still owe if she uses the money her grandmother gave her to pay back the money she borrowed?

AMaxie will still owe $2.25.
BMaxie will still owe $1.75.
CMaxie will have $1.75 left.
DMaxie will have $2.25 left.

Show the answer

Correct answer: C

First, determine the total amount Maxie borrowed, in dollars:

$7.75 + 11.00 + 4.50 = 23.25$

Then use subtraction to determine the difference between the amount her grandmother gave her and the amount she borrowed:

$25.00 - 23.25 = 1.75$

The difference is positive, which means that Maxie will still have $1.75 left from the amount her grandmother gave her.

Question analysis

Maxie borrowed money from her mother, father and brother to buy a video game, then received $25.00 from her grandmother to pay them back. Students decide how much money she will have left over or still owe. It tests adding and subtracting decimals with money, and the choices also test which way the difference goes.

NYS standards: 6.NS.3 multi-digit decimal operations
6.NS.5 positive and negative numbers
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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

Few students miss this one, since it needs only one addition and one subtraction with money. The choices come in pairs, so the right amount must also be matched with the right outcome. Lining up the decimal points, and asking whether $25.00 is more or less than the $23.25 borrowed, settles both the amount and whether Maxie owes or keeps money.

Error-prone arithmetic minor Working it out
The loans add to $23.25, and subtracting that from $25.00 means borrowing across two zeros, a classic place for a digit to go wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Each amount appears twice among the choices, once as money left and once as money owed, so the result must also be read for direction by asking which is larger, the gift or the loans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the three amounts borrowed, compare the total with the $25.00 gift, and decide from which is larger whether Maxie ends up with money left or still owes some.

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.)
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

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two errors at once, a wrong difference such as $2.25 and the reversed outcome, so no single line of working reaches it and it mostly draws guesses.
B Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Finds the $1.75 difference correctly but reverses its meaning, as if the $25.00 gift fell short of the $23.25 borrowed instead of covering it.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts $23.25 from $25.00 column by column, taking the smaller digit from the larger instead of borrowing, so 5 − 0 and 2 − 0 give $2.25.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Money left over: $\text{left} = \text{money she has} - \text{money she owes}$; Money she has is the $25.00 gift. Money she owes is the total she borrowed. If the result is positive, she has that much left. If it were negative, she would still owe that much. Comparing the two amounts first tells you which.

Sort out the question
Maxie borrowed from three people: $7.75, $11.00 and $4.50. Her grandmother gave her $25.00. She uses the $25.00 to pay everyone back. You need to know if money is left over or if she still owes some, and how much.
Add up what she borrowed
Line up the decimal points and add:
$7.75 + 11.00 + 4.50 = 23.25$
She owes $23.25 in all. That is the amount she must pay back.
Compare the gift with the debt
The gift, $25.00, is more than the $23.25 she owes. So she can pay everyone back and still have money left. She will not owe anything.
Watch out: once you find $1.75, keep its meaning straight. Saying she still owes $1.75, choice B, gets the direction backwards.
Subtract to find what is left
Take the debt away from the gift:
$25.00 - 23.25 = 1.75$
You need to regroup (borrow) across the zeros. Think of $25.00 as 24 dollars and 100 cents. Then 100 − 25 = 75 cents, and 24 − 23 = 1 dollar.
Watch out: taking the smaller digit from the larger one in each column gives $2.25, choice D.
So the answer is C: Maxie will have $1.75 left.

Check your answer

Add back. The $23.25 she pays plus the $1.75 she keeps is 23.25 + 1.75 = 25.00. That is the whole gift, so the subtraction is right.

More questions like this

The original DOE question

Question 21 as shown on the DOE’s digital test, page 94 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 94, question 21.

Question 24 of 57

2022 Test B Math, question 24

Question

Seth grows strawberries in his garden.

He picked $2\frac{1}{2}$ cups of strawberries on Monday.
He picked 4 cups of strawberries on Tuesday.
He ate $\frac{3}{4}$ cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.

How many cups of strawberries did Seth have left?

A $3\frac{1}{2}$
B $4\frac{1}{4}$
C6
D9

Show the answer

Correct answer: B

When Seth picks strawberries, the value represents a positive number, and when he eats the strawberries, the value represents a negative number. Write an expression and then evaluate the expression, where the negative value is multiplied by 3 because he ate the same amount on 3 different days.

$2\frac{1}{2} + 4 + (3)\left(-\frac{3}{4}\right)$

$2\frac{1}{2} + 4 - 2\frac{1}{4}$

$6\frac{1}{2} - 2\frac{1}{4}$

$6\frac{2}{4} - 2\frac{1}{4}$

$4\frac{1}{4}$

If the multiplication is confusing, evaluate the expression below which represents the sum for the 5 days with each day listed separately.

$2\frac{1}{2} + 4 + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right)$

$4\frac{1}{4}$

Question analysis

Seth picked strawberries on Monday and Tuesday, in amounts given as a mixed number and a whole number, then ate the same fraction of a cup at breakfast on three days. Students find how many cups he has left. It tests adding, multiplying and subtracting fractions in a short word problem, and the eaten amount repeats each day.

NYS standards: 7.NS.3 four operations with rational numbers
7.EE.3 multi-step rational number problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding mixed numbers and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 3 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 bulleted list lays out the story day by day, so setting up the problem is easy; the points are lost in the fractions. The number of breakfast days, 3, sits right beside the 3/4 cup, and subtracting 3 cups instead of three quarters of a cup each day is the tempting shortcut. Writing every amount in fourths, 26/4 picked and 9/4 eaten, makes the last step a whole-number subtraction.

Error-prone arithmetic moderate Working it out
Three breakfasts of 3/4 cup make 9/4 = 2 1/4 cups, and taking that from 6 1/2 means renaming the half as 2/4, two fraction steps where a slip is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Picking adds and eating subtracts, and “each day on Wednesday, Thursday, and Friday” means the 3/4 cup counts three times, not once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the two amounts picked, multiply the breakfast amount by the number of days, and subtract that from the total; writing every amount in fourths keeps the arithmetic simple.

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.
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

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Subtracts 3 cups, the number of breakfast days, instead of 3 × 3/4 = 2 1/4 cups, so 6 1/2 − 3 = 3 1/2.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives exactly 6; it is near the 6 1/2 cups picked and mostly catches students who drop the fractions or guess.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the 2 1/4 cups eaten to the 6 1/2 cups picked instead of subtracting, 8 3/4 cups, and takes the nearest choice, 9.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total of a repeated amount: $\text{total} = \text{number of times} \times \text{amount each time}$; Use it when the same amount happens again and again. Here Seth eats 3/4 cup on each of 3 days, so the number of times is 3 and the amount each time is 3/4 cup.
A whole number times a fraction: $n \times \frac{a}{b} = \frac{n \times a}{b}$; n is a whole number, and a/b is a fraction. Multiply the whole number by the top only, and keep the bottom. It works because n groups of a pieces make n × a pieces, and each piece is still size 1/b.

Sort out the question
Seth picked 2 1/2 cups on Monday and 4 cups on Tuesday. Then he ate 3/4 cup each day on Wednesday, Thursday and Friday. That is 3 days of eating. Picking adds strawberries, and eating takes them away. You need the number of cups he has left.
Add what he picked
Add the two days of picking:
$2\frac{1}{2} + 4 = 6\frac{1}{2}$
He picked 6 1/2 cups in all, before he ate any.
Find how much he ate
He ate 3/4 cup on each of 3 days. Multiply:
$3 \times \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4}$
To change 9/4 to a mixed number, note that 4 fits into 9 twice, with 1 left over. He ate 2 1/4 cups.
Watch out: he ate 3/4 cup three times, not 3 cups. Subtracting 3 gives 3 1/2, choice A.
Take away what he ate
Write the half as fourths, so both fractions have the same bottom: 1/2 = 2/4. Then subtract the wholes and the fourths:
$6\frac{1}{2} - 2\frac{1}{4} = 6\frac{2}{4} - 2\frac{1}{4} = 4\frac{1}{4}$
Watch out: eating takes away. Adding the 2 1/4 cups gives 8 3/4, and the nearest choice is 9, choice D.
So the answer is B, $4\frac{1}{4}$ cups.

Check your answer

Use decimals. 2.5 + 4 = 6.5 cups picked. 3 × 0.75 = 2.25 cups eaten. Then 6.5 − 2.25 = 4.25 cups, which is 4 1/4.

More questions like this

The original DOE question

Question 22 as shown on the DOE’s digital test, page 95 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 95, question 22.

Question 25 of 57

2022 Test B Math, question 25

Question

Which situation involves quantities that combine to equal zero?

Areceiving $5 as a gift and then giving $5 to a friend
Bbuying a book for $10 and then buying lunch for $10
Cgetting on an elevator at the ground floor, rising 3 floors, and then rising 3 more floors
Dstarting in a cave 20 feet below the ground, climbing up 20 feet, and then climbing up 20 feet more

Show the answer

Correct answer: A

When looking for a situation where quantities combine to make zero, find two values with the same absolute value where one value is positive and one value is negative.

In the situation in option A, receiving a gift of $5 is represented by $+5$ , and giving a friend $5 is represented by $-5$ . These values combine to equal zero because $+5 - 5 = 0$ .

Question analysis

This question gives four everyday situations involving money, an elevator and a cave, and asks which one involves quantities that combine to make zero. It tests representing gains and losses, or ups and downs, with positive and negative numbers. Equal numbers alone do not make zero, so the direction of each amount must be read carefully.

NYS standards: 7.NS.1a opposite quantities making 0
6.NS.5 positive and negative numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Describing situations where opposite quantities make zero is named in the grade 7 standards, and this is its most direct form: one reading of four everyday situations, easier than typical grade 7 work with signed numbers.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Turning words into math minor
Every choice must be checked minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 60 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 idea that a quantity and its opposite add to zero is familiar, and the right situation states it plainly. The wrong choices are built from equal amounts: two of them repeat a number but move in one direction, and the cave climb reaches zero partway before going 20 feet past it. Writing each situation as a sum of signed numbers takes a few seconds and settles it.

Turning words into math minor Understanding the question
Each situation must be turned into signed numbers, receiving or rising as plus and giving or buying as minus, before its total can be judged. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
Every choice repeats one amount, $5, $10, 3 floors or 20 feet, so equal numbers alone do not identify the answer; the direction of each move has to be checked. (Embretson & Wetzel, 1987)

Strategy: Write each situation as signed numbers, using + for gains or upward moves and − for losses or downward moves, add them, and see which total is zero.

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

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Treats the two equal $10 amounts as opposites that cancel, though buying a book and buying lunch both take money away, for a total of −$20.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Pairs the two 3s as amounts that cancel, missing that rising and then rising again both go up, ending 6 floors above the ground floor.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Sees that starting 20 feet below ground and climbing 20 feet reaches zero, then overlooks the last 20 feet climbed, which ends 20 feet above ground.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Opposites add to zero: $a + (-a) = 0$; a is any number, and −a is its opposite: the same size, but the other sign. A number and its opposite always add to 0. On a number line, they sit the same distance from 0 on opposite sides. Equal moves in opposite directions cancel out.

Sort out the question
You need the situation whose amounts add up to zero. Give each amount a sign. Use + for getting money or going up. Use − for spending money or going down. Then add the amounts in each choice. The direction matters, not just the numbers.
Write each choice with signs
Turn each situation into signed numbers and add them up:
Choice Amounts Total
A +5, then −5 0
B −10, then −10 −20
C +3, then +3 +6
D start at −20, then +20, then +20 +20
Look at the tricky choices
In B, buying a book and buying lunch both take money away. Equal amounts cancel only when they go in opposite directions. In C, rising and rising again both go up, so you end 6 floors up.
Watch out: in D, starting at −20 and climbing 20 feet does reach 0, ground level. But then you climb 20 feet more. You end at +20, 20 feet above the ground, so D is wrong.
Choose the zero total
Only choice A has two opposite amounts of the same size. Getting $5 is +5, and giving $5 away is −5:
$5 + (-5) = 0$
So the answer is A: receiving $5 as a gift and then giving $5 to a friend.

Choice	Amounts	Total
A	+5, then −5	0
B	−10, then −10	−20
C	+3, then +3	+6
D	start at −20, then +20, then +20	+20

Check your answer

Think about where you end up. In A, you end with exactly the money you started with, so the total change is zero. In every other choice, you end with less money, higher up, or above the ground.

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 D Math, page 96, question 23.

Question 26 of 57

2022 Test B Math, question 26

Question

$\frac{\left(\frac{2}{3} - \frac{4}{5} \times \frac{1}{3}\right)}{\left(\frac{5}{3} + \frac{1}{4} \div \frac{3}{4}\right)} =$

A $-\frac{2}{115}$
B $-\frac{1}{45}$
C $\frac{2}{9}$
D $\frac{1}{5}$

Show the answer

Correct answer: D

First, simplify the numerator and the denominator separately.

To simplify the numerator, multiply the last two fractions and then subtract from the first fraction.

$\frac{2}{3} - \left(\frac{4}{5} \times \frac{1}{3}\right)$

$\frac{2}{3} - \frac{4}{15}$

$\frac{10}{15} - \frac{4}{15}$

$\frac{6}{15} = \frac{2}{5}$

To simplify the denominator, divide the last two fractions and then add to the first fraction.

$\frac{5}{3} + \left(\frac{1}{4} \div \frac{3}{4}\right)$

$\frac{5}{3} + \left(\frac{1}{4} \times \frac{4}{3}\right)$

$\frac{5}{3} + \frac{1}{3}$

$\frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$

Finally, divide the numerator by the denominator.

$\frac{\frac{2}{5}}{2} = \frac{2}{5} \div 2 = \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}$

Question analysis

This fraction arithmetic question gives a complex fraction: a numerator with a subtraction and a multiplication over a denominator with an addition and a division, all with fractions. Students choose its value. It tests order of operations and the four operations with fractions, and working left to right instead of following the order of operations leads to a wrong result.

NYS standards: 7.NS.3 four operations with rational numbers
6.NS.1 dividing fractions by fractions
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students compute with fractions, usually two or three operations at a time. This expression needs five, in an order set by the rules rather than read left to right, so it is longer and more trap-prone than typical work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Something you must already know minor
Solution: 5 steps
Reading level: Question and choices: about grade 16+, 5 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 here is beyond grade 7, but the expression stacks five fraction operations, and the top and the bottom each hide an order-of-operations decision. Working left to right in the numerator alone gives a negative value, and doing so in both parts gives an even stranger fraction; both are listed. The sign is a quick check: 4/15 is less than 2/3, so the numerator is positive, and so is the value.

Many steps moderate Working it out
The numerator and the denominator are each simplified on their own, to 2/5 and 2, before the final division, so five operations stand between the expression and its value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Every step is fraction work: 2/3 − 4/15 needs fifteenths, 1/4 ÷ 3/4 needs the divisor flipped, and the last step, 2/5 ÷ 2, is easily turned into 2/5 × 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
No parentheses inside the numerator or the denominator show the order, so students must know to multiply before subtracting and to divide before adding. (Anderson & Krathwohl, 2001)

Strategy: Simplify the numerator and the denominator separately, doing the multiplication and the division before the subtraction and the addition, then divide the two results.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Works left to right in both parts, (2/3 − 4/5) × 1/3 = −2/45 over (5/3 + 1/4) ÷ 3/4 = 23/9, and −2/45 ÷ 23/9 = −2/115.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Subtracts before multiplying in the numerator, (2/3 − 4/5) × 1/3 = −2/45, but handles the denominator correctly, so −2/45 ÷ 2 = −1/45.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/9: it is positive like the answer, but no single order or fraction slip reaches it, so it mostly draws guesses between the two positive choices.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\times \text{ and } \div \text{ before } + \text{ and } -$; When no parentheses show the order, multiply and divide first. Then add and subtract. Everyone follows the same order, so an expression has only one value.
Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the top numbers, and b and d are the bottom numbers. Multiply top times top and bottom times bottom.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. The reciprocal of c/d is d/c. It works because c/d times d/c is 1. So multiplying by d/c undoes c/d, just as dividing by c/d does.

Sort out the question
The big fraction bar means divide: the top part divided by the bottom part. Work out the top and the bottom on their own. Then divide.
Inside each part, no parentheses show the order. So follow the order of operations.
Multiply first in the top
The top is $\frac{2}{3} - \frac{4}{5} \times \frac{1}{3}$ . Multiply before you subtract.
$\frac{4}{5} \times \frac{1}{3} = \frac{4 \times 1}{5 \times 3} = \frac{4}{15}$
Watch out: subtracting first, 2/3 − 4/5, makes the top negative. That leads to −1/45, choice B. Working left to right in the bottom as well gives −2/115, choice A.
Subtract to finish the top
To subtract, the bottoms must match. Change 2/3 to fifteenths by multiplying its top and bottom by 5.
$\frac{2}{3} - \frac{4}{15} = \frac{10}{15} - \frac{4}{15} = \frac{6}{15} = \frac{2}{5}$
To simplify 6/15, divide its top and bottom by 3. The top part is 2/5.
Divide first in the bottom
The bottom is $\frac{5}{3} + \frac{1}{4} \div \frac{3}{4}$ . Divide before you add. Flip 3/4 and multiply.
$\frac{1}{4} \div \frac{3}{4} = \frac{1}{4} \times \frac{4}{3} = \frac{4}{12} = \frac{1}{3}$
To simplify 4/12, divide its top and bottom by 4.
Add to finish the bottom
The bottoms already match, so add the tops and keep the bottom.
$\frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$
Six thirds make 2 wholes. The bottom part is 2.
Divide the top by the bottom
Now divide 2/5 by 2. Flip 2, which is 2/1, to get 1/2.
$\frac{2}{5} \div 2 = \frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = \frac{1}{5}$
Dividing by 2 is the same as taking half, and half of 2/5 is 1/5. So the answer is D, $\frac{1}{5}$ .

Check your answer

Look at the signs. 4/15 is less than 2/3, so the top is positive. The bottom is positive too. So the answer must be positive, which rules out A and B. Then multiply back: $\frac{1}{5} \times 2 = \frac{2}{5}$ , the top. It matches.

More questions like this

The original DOE question

Question 83 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 2023 Admissions, page 204, question 83. Open page 204 of the handbook (PDF)

Question 27 of 57

2022 Test B Math, question 27

Question

An athlete runs $\frac{1}{8}$ kilometer in $\frac{3}{4}$ minute. At this rate, how many kilometers would the athlete run in 1 minute?

A $\frac{3}{32}$
B $\frac{1}{6}$
C $1\frac{1}{2}$
D6

Show the answer

Correct answer: B

To determine the number of kilometers in 1 minute, find the unit rate in kilometers per minute: $\frac{\frac{1}{8}\text{ kilometer}}{\frac{3}{4}\text{ minute}}$ .

To evaluate the fraction, use the reciprocal when dividing to get $\frac{1}{6}$ kilometer per 1 minute.

$\frac{1}{8} \div \frac{3}{4} = \frac{1}{8} \times \frac{4}{3} = \frac{4}{24} = \frac{1}{6}$

Question analysis

An athlete runs 1/8 kilometer in 3/4 minute, and students find how many kilometers the athlete would run in 1 minute at the same rate. It tests unit rates with fractions. Both the distance and the time are fractions, so deciding which one to divide by which is the main challenge.

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 Computing a unit rate from a ratio of fractions is the grade 7 standard itself, and this is a standard example: one fraction divided by another in two short steps, with the order of division the only real decision.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 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 is a unit rate with fractions, and its wrong choices are the other ways to combine 1/8 and 3/4. Since 1 minute is only a third longer than 3/4 minute, the distance must be only a third more than 1/8 kilometer, which rules out the product and both large choices at once. Dividing with the units written in, kilometers over minutes, keeps the order right.

Error-prone arithmetic moderate Working it out
The rate is a fraction divided by a fraction, 1/8 ÷ 3/4, and only the 3/4 may be flipped; multiplying straight across or dividing the other way round both give listed choices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Kilometers in 1 minute means kilometers per minute, so the 1/8 goes on top even though the time, 3/4, is the larger number. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide the distance by the time, kilometers over minutes, by multiplying 1/8 by the reciprocal of 3/4, and check that the units of the result are kilometers per minute.

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).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Multiplies 1/8 by 3/4 instead of dividing, getting 3/32, which is less than the 1/8 kilometer already run in under a minute.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: weak
Splits the 3/4 minute into quarters but multiplies 1/8 by 3 instead of dividing, then multiplies by 4 for a full minute, giving 12/8 = 1 1/2.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides 3/4 by 1/8, minutes by kilometers, which gives 6 minutes per kilometer, the rate turned upside down.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{rate} = \text{distance} \div \text{time}$; A unit rate tells how much happens in 1 unit of time. Here the distance is in kilometers and the time is in minutes, so the rate is kilometers per minute. You divide because you are sharing the distance equally over the minutes.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). Flip only the second fraction, the one you divide by.

Sort out the question
The athlete runs 1/8 kilometer in 3/4 minute, which is less than a full minute. You need how far the athlete runs in 1 whole minute at the same speed. That is the unit rate, in kilometers per minute.
Set up the division
Kilometers per minute means kilometers divided by minutes. So the distance comes first and the time second:
$\text{rate} = \frac{1}{8} \div \frac{3}{4}$
Watch out: dividing the other way, 3/4 ÷ 1/8, gives 6. That is minutes per kilometer, the rate turned upside down, choice D.
Divide the fractions
Flip 3/4 to get 4/3, then multiply. Multiply the tops and multiply the bottoms:
$\frac{1}{8} \times \frac{4}{3} = \frac{4}{24}$
Simplify by dividing the top and bottom by 4:
$\frac{4}{24} = \frac{1}{6}$
Watch out: multiplying 1/8 × 3/4 without flipping gives 3/32, choice A.
Make sense of the answer
In 1 minute, the athlete runs 1/6 kilometer. That is a bit more than 1/8 kilometer, which makes sense. A full minute is a bit longer than 3/4 minute, so the athlete goes a bit farther. Choice C, 1 1/2 kilometers, would be 12 times as far in barely more time.
So the answer is B, $\frac{1}{6}$ kilometer.

Check your answer

Work in quarter minutes. In 3 quarter minutes, the athlete runs 1/8 km. So in 1 quarter minute, the athlete runs 1/8 ÷ 3 = 1/24 km. A full minute has 4 quarters: 4 × 1/24 = 4/24 = 1/6 km.

More questions like this

The original DOE question

Question 24 as shown on the DOE’s digital test, page 97 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 97, question 24.

Question 28 of 57

2022 Test B Math, question 28

Question

A bus trip takes 9 hours if the mean speed is 50 miles per hour. How many hours would this trip take if the mean speed was 45 miles per hour?

Show the answer

Correct answer: B

First, find the length of the bus trip in miles by multiplying the time, 9 hours, by the average (mean) speed, 50 miles per hour, to get 450 miles.

Then, divide the total distance, 450 miles, by the average (mean) speed, 45 miles per hour, to get the total time:

$450 \div 45 = 10\text{ hours}$

Question analysis

A bus trip takes 9 hours at a mean speed of 50 miles per hour, and students find how long the same trip takes at 45 miles per hour. It tests speed, distance and time. Working out the fixed distance first is the key, since guessing how the time should change leads many students astray.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 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

Two short calculations solve it, but only after noticing that speed and time move in opposite directions: a slower bus takes longer. Students who scale the 9 hours by 45/50, as in a direct proportion, get a little over 8 hours, a result that should feel wrong for a slower trip. Finding the distance first turns the question into one division with no direction to decide.

No obvious method moderate Working it out
The question never mentions the trip's length, yet the 450-mile distance is what stays fixed; setting up a proportion between hours and speeds, the usual reflex, points the wrong way. (Schoenfeld, 1985)
Something you must already know minor Working it out
Students must bring distance = speed × time themselves, to turn 9 hours at 50 miles per hour into miles before dividing by the new speed. (Anderson & Krathwohl, 2001)

Strategy: Find the trip's length in miles from the first speed and time, since that length stays the same, then divide it by the new mean speed.

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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Treats time as proportional to speed, scaling 9 hours by 45/50 to 8.1, about 8, though a slower bus needs more time, not less.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12 hours; it needs a slip such as 540 miles for 9 × 50, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the 5-mile-per-hour drop in speed to the 9 hours, 9 + 5 = 14, combining numbers from the question without a model of the trip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, speed and time: $d = r \times t$; d is the distance traveled. r is the rate, or speed, such as miles per hour. t is the time. It works because at 50 miles each hour, 9 hours covers 9 groups of 50 miles.
Finding the time: $t = d \div r$; This is the same rule turned around. d is the distance and r is the speed. Divide the distance by the speed to find t, the number of hours. Use it when you know how far and how fast.

Sort out the question
A bus trip takes 9 hours at a mean speed of 50 miles per hour. The mean speed is the average speed over the whole trip. You need the time for the same trip at 45 miles per hour. The length of the trip does not change, so find it first.
Find the length of the trip
Use distance equals speed times time, with a speed of 50 miles per hour and a time of 9 hours:
$d = 50 \times 9 = 450$
The trip is 450 miles long.
Find the new time
Now the bus goes 45 miles per hour over the same 450 miles. Divide the distance by the speed:
$t = 450 \div 45 = 10$
The trip takes 10 hours.
Check that it makes sense
A slower bus needs more time for the same trip. 10 hours is more than 9 hours, so it fits.
Watch out: time does not shrink with speed. Scaling 9 hours by 45/50 gives about 8 hours, choice A, which would make the slower bus arrive sooner. Adding the drop of 5 miles per hour to 9 hours gives 14, choice D, which has no reason behind it.
So the answer is B, 10 hours.

Check your answer

Check the distance both ways. At 50 miles per hour for 9 hours, the bus goes 50 × 9 = 450 miles. At 45 miles per hour for 10 hours, it goes 45 × 10 = 450 miles. Same trip, same distance.

More questions like this

The original DOE question

Question 25 as shown on the DOE’s digital test, page 98 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 98, question 25.

Question 29 of 57

2022 Test B Math, question 29

Question

A birdseed mixture is $\frac{2}{5}$ sunflower seeds. The cost of the sunflower seeds is $0.10 per ounce. At this rate, what is the cost of the sunflower seeds in 25 pounds of this birdseed? (Note: 1 lb = 16 oz.)

A$10.00
B$16.00
C$40.00
D$160.00

Show the answer

Correct answer: B

First, calculate the amount of sunflower seeds in 25 pounds of this birdseed by using ratio reasoning to find that there are 10 pounds of sunflower seeds in the birdseed.

$\frac{2}{5} = \frac{10}{25}$

Then, use ratio reasoning to find the cost of the sunflower seeds per pound.

$\left(\frac{\$0.10}{1 \text{ ounce}}\right)\left(\frac{16 \text{ ounces}}{1 \text{ pound}}\right) = \$1.60$

The cost of the sunflower seeds is $1.60 per pound.

Finally, find the total cost using ratios by multiplying the unit cost per pound by 10 pounds.

$\frac{\$1.60}{1 \text{ pound}} \times \frac{10 \text{ pounds}}{1} = \$16.00$

The first two steps above could be done in the other order; that is, find $1.60 per pound first and 10 pounds of sunflower seeds second.

Question analysis

A birdseed mixture is two-fifths sunflower seeds, which cost 10 cents an ounce, and students find the cost of the sunflower seeds in 25 pounds of the mix. It combines a fraction of a whole, a pounds-to-ounces conversion and a unit price. Each step's result looks like a possible answer, so keeping track of which quantity is in hand matters.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3d converting measurement units
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multistep ratio problems with a unit conversion are typical grade 7 work. This one chains three short steps, a fraction of the mixture, pounds to ounces and a price per ounce, and the numbers are chosen to come out evenly.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 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

Each step is routine, but the steps can be done in more than one order, and every partial result looks like a price once a dollar sign is put in front of it. Each wrong choice leaves something out: the 2/5, the price per ounce, or the conversion and the price together. Labelling each number with its unit, pounds, ounces or dollars, as it is found keeps the chain straight and shows when it is finished.

Many steps moderate Working it out
Three operations are chained: 2/5 of 25 pounds gives 10 pounds of sunflower seeds, the note turns them into 160 ounces, and $0.10 an ounce prices them; stopping at 10 pounds or 160 ounces, or skipping the 2/5, gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Combines ideas minor Working it out
A fraction of a mixture, a pounds-to-ounces conversion and a price per ounce meet in one question, and the price is per ounce while the amount is given in pounds. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the weight of the sunflower seeds first, change it to ounces with the note, then multiply by the price per ounce; or turn the price into dollars per pound first.

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.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds that 10 of the 25 pounds are sunflower seeds and stops, giving that weight as $10.00 without converting it to ounces or pricing it.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Prices all 25 pounds of birdseed, 400 ounces at $0.10 an ounce, never taking the 2/5 of the mixture that is sunflower seeds.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Converts the 10 pounds of sunflower seeds to 160 ounces and stops, reading the number of ounces as the cost without multiplying by $0.10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of an amount: $\text{part} = \text{fraction} \times \text{whole}$; The whole is the full amount, here 25 pounds of birdseed. The fraction tells how much of it you want. To find 2/5 of something, split it into 5 equal parts and take 2 of them. Multiplying does both at once.
Changing pounds to ounces: $\text{ounces} = \text{pounds} \times 16$; The note says 1 pound is 16 ounces. So each pound turns into 16 ounces. An ounce is smaller than a pound, so you get a bigger number.
Total cost: $\text{cost} = \text{number of units} \times \text{price of one unit}$; The units here are ounces. The price of one unit is $0.10 for one ounce. Every ounce costs the same, so you multiply the number of ounces by the price of one.

Sort out the question
You know:
- The birdseed is $\frac{2}{5}$ sunflower seeds.
- There are 25 pounds of birdseed.
- Sunflower seeds cost $0.10 per ounce.
- 1 pound = 16 ounces.
You need the cost, in dollars, of just the sunflower seeds. The price is per ounce, but the weight is in pounds. So you will need to change pounds to ounces.
Find the weight of sunflower seeds
Only 2/5 of the birdseed is sunflower seeds. Find 2/5 of 25 pounds:
$\frac{2}{5} \times 25 = \frac{50}{5} = 10$
Another way: 25 ÷ 5 = 5, and 2 fifths is 2 × 5 = 10. So there are 10 pounds of sunflower seeds.
Watch out: 10 is a number of pounds, not dollars. Stopping here and calling it $10.00 leads to choice A.
Change pounds to ounces
The price is per ounce, so turn the 10 pounds into ounces. Each pound is 16 ounces:
$10 \times 16 = 160$
There are 160 ounces of sunflower seeds.
Watch out: 160 is a number of ounces, not dollars. Calling it $160.00 leads to choice D.
Find the cost
Each ounce costs $0.10. Multiply:
$160 \times 0.10 = 16.00$
The sunflower seeds cost $16.00.
Watch out: skipping the 2/5 prices all 25 pounds. That is 400 ounces, and $400 \times 0.10 = 40.00$ , choice C.
So the answer is B, $16.00.

Check your answer

Do it in another order. One pound of sunflower seeds costs $16 \times 0.10 = 1.60$ dollars. Ten pounds cost $10 \times 1.60 = 16.00$ dollars. You get $16.00 again.

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 D Math, page 100, question 27.

Question 30 of 57

2022 Test B Math, question 30

Question

A robot moves forward at the rate of 9 steps every 6 seconds. If each step is x feet long, what must x be in order for the robot to travel exactly 270 feet per minute?

Show the answer

Correct answer: A

First, convert steps per second to steps per minute.

$\frac{9\text{ steps}}{6\text{ seconds}}=\frac{90\text{ steps}}{60\text{ seconds}}=\frac{90\text{ steps}}{1\text{ minute}}$

Then, use ratio reasoning to determine that each step must be 3 feet since 90 times 3 is 270 (or 270 divided by 90 is 3).

$\frac{90\text{ steps}}{1\text{ minute}}\times\frac{3\text{ feet}}{1\text{ step}}=\frac{270\text{ feet}}{1\text{ minute}}$

$\frac{270\text{ feet}}{1\text{ minute}}\div\frac{90\text{ steps}}{1\text{ minute}}=$

$\frac{270\text{ feet}}{1\text{ minute}}\times\frac{1\text{ minute}}{90\text{ steps}}=$

$\frac{3\text{ feet}}{1\text{ step}}$

Question analysis

A robot moves forward 9 steps every 6 seconds, and each step is x feet long. Students find the step length that makes the robot travel exactly 270 feet per minute. It tests rates and unit conversion: the rate is given per 6 seconds but the target is per minute, so the units must be matched before solving.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 rate problems often give one rate and ask for a total. This one links steps per time, feet per step and feet per minute, with seconds to convert, a typical but full multistep ratio problem.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 33 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 numbers are friendly, but the rate is given per 6 seconds and the target per minute, so the units have to meet first. A minute holds ten 6-second stretches, so the robot takes 90 steps in it, and 270 feet over 90 steps settles the length. Students who divide 270 by the 9 steps forget that those steps take only 6 seconds, and those who scale 270 down to 6 seconds stop at 27 feet, a distance rather than a step length.

Much to keep track of moderate Working it out
Three rates are in play, 9 steps per 6 seconds, x feet per step and 270 feet per minute, and their units must be lined up before any two are combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Many steps minor Working it out
The rate is first scaled from 6 seconds to 60, giving 90 steps a minute, and only then compared with the 270 feet covered in that minute. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change the rate to steps per minute first, since 6 seconds goes into a minute evenly, then ask what step length makes that many steps cover 270 feet.

Traps

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

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 9 steps by the 6 seconds, 54, and divides 270 by it, combining the stem’s numbers without a model of the rate.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Scales 270 feet per minute down to 27 feet per 6 seconds but stops before dividing by the 9 steps taken in that time.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides 270 feet by 9 steps, as if the robot took 9 steps a minute, never converting the 6-second rate to a minute.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate: $\text{rate} = \frac{\text{amount}}{\text{time}}$; A rate compares an amount with the time it takes. Here the robot’s rate is 9 steps for every 6 seconds. To compare two rates, first write them for the same length of time, such as one minute, which is 60 seconds.
Distance from steps: $\text{distance} = \text{number of steps} \times \text{length of each step}$; Every step covers the same length. So the distance is the number of steps times the length of one step. Here the distance and the step length are both in feet.

Sort out the question
The robot takes 9 steps every 6 seconds. Each step is x feet long. You need the x that makes the robot go exactly 270 feet per minute.
The two rates use different times: 6 seconds and 1 minute. Match the times before you compare them.
Find the steps in one minute
One minute is 60 seconds. Find how many 6-second stretches fit in a minute.
$60 \div 6 = 10$
In each stretch, the robot takes 9 steps. So in one minute it takes 10 times as many.
$9 \times 10 = 90$
The robot takes 90 steps per minute.
Find the length of each step
In one minute, those 90 steps must cover 270 feet. Use the distance formula.
$\text{distance} = \text{number of steps} \times \text{length of each step}$
$270 = 90 \times x$
To get x alone, divide both sides by 90.
$x = 270 \div 90 = 3$
Each step must be 3 feet long.
Give the answer
Watch out: dividing 270 by 9 gives 30, choice D. That treats 9 steps as a whole minute’s steps, but they take only 6 seconds. Cutting 270 feet per minute down to 27 feet per 6 seconds and stopping gives 27, choice C. That is a distance, not a step length. So the answer is A, 3.

Check your answer

Check the units. 90 steps per minute times 3 feet per step gives $90 \times 3 = 270$ feet per minute. That is the goal. Or think in 6 seconds: 9 steps of 3 feet is 27 feet, and 10 of those stretches make 270 feet.

More questions like this

The original DOE question

Question 87 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 205, question 87. Open page 205 of the handbook (PDF)

Question 31 of 57

2022 Test B Math, question 31

Question

There were 36 people at Emily’s party. If each person at the party drank 2 cups of juice, how many gallons of juice did they drink altogether?

Note: 1 gallon = 4 quarts; 1 quart = 4 cups.

A $2\frac{1}{4}$
B $4\frac{1}{2}$
C9
D72

Show the answer

Correct answer: B

Since 36 people each drank 2 cups of juice, the total amount of juice they drank was $36 \times 2 = 72$ cups.

Convert 72 cups to gallons:

$72 \text{ cups} \times \frac{1 \text{ quart}}{4 \text{ cups}} \times \frac{1 \text{ gallon}}{4 \text{ quarts}} =$

$\frac{72}{16} \text{ gallons} =$

$\frac{9}{2} \text{ gallons} =$

$4\frac{1}{2} \text{ gallons}$

Question analysis

At Emily's party 36 people each drank 2 cups of juice, and this unit conversion question asks how many gallons they drank in all, given 4 quarts to a gallon and 4 cups to a quart. Students multiply to get the total in cups, then convert through quarts to gallons, and the bold word gallons sets the unit of the answer.

NYS standards: 6.RP.3d converting measurement units
6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting units with given factors is grade 6 work. This adds a first step, the total number of cups, and a two-stage conversion through quarts, but the numbers are friendly and the path is routine for grade 7 students.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 32 words.
Corrected from the DOE handbook: A stray word at the end of the explanation was removed.
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

With both conversion facts printed, the question is mainly about keeping the units straight. Students who skip the word each convert 36 cups instead of 72, and those who stop at cups find a total that is right in the wrong unit. Working out that a gallon is 4 × 4 = 16 cups before dividing turns two conversions into one and avoids adding the 4s.

Many steps minor Working it out
A multiplication, 36 × 2, comes before two divisions by 4, cups to quarts and quarts to gallons, and stopping after the first step leaves a believable 72. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The word “each” turns 36 people into 72 cups, and the bold word gallons sets the unit of the answer, a change the note supplies in two steps. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the total number of cups first, then divide by 4 to get quarts and by 4 again to get gallons, or work out how many cups make a gallon and divide once.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Converts 36 cups to gallons, 36 ÷ 16 = 2 1/4, missing the word each, so every person’s second cup of juice is left out.
C Units or scale (King et al., 2004)
Reveals a misconception Expected pull: weak
Uses 8 cups to a gallon, adding the two 4s in the note instead of multiplying them, so 72 cups become 72 ÷ 8 = 9 gallons.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds the 72 cups of juice and stops, giving the amount in cups though the bold word gallons asks for it in gallons.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation ends with a line that reads only “lineExtend”, a formatting code left in the document by mistake. The site leaves it out.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total amount: $\text{total} = \text{number of people} \times \text{amount each}$; When every person has the same amount, multiply the number of people by the amount each one has. The word each tells you to multiply.
Cups in a gallon: $\text{cups in 1 gallon} = 4 \times 4 = 16$; A gallon holds 4 quarts, and each quart holds 4 cups. So a gallon holds 4 groups of 4 cups. You multiply, not add, because each of the 4 quarts has its own 4 cups.
Changing cups to gallons: $\text{gallons} = \text{cups} \div 16$; It takes 16 cups to make 1 gallon. Dividing by 16 finds how many groups of 16 cups there are. A gallon is bigger than a cup, so the number of gallons is smaller.

Sort out the question
You know:
- 36 people were at the party.
- Each person drank 2 cups of juice.
- 1 gallon = 4 quarts, and 1 quart = 4 cups.
You need the total amount of juice in gallons. The bold word tells you the unit of the answer.
Find the total in cups
Each of the 36 people drank 2 cups:
$36 \times 2 = 72$
They drank 72 cups in all.
Watch out: skipping the word each uses only 36 cups. Then $36 \div 16 = 2\frac{1}{4}$ gallons, choice A.
Find the cups in a gallon
A gallon is 4 quarts, and each quart is 4 cups:
$4 \times 4 = 16$
So 1 gallon holds 16 cups.
Watch out: adding the two 4s gives 8 cups in a gallon. Then $72 \div 8 = 9$ , choice C.
Change cups to gallons
Divide the cups by 16. Simplify the fraction by dividing the top and bottom by 8:
$72 \div 16 = \frac{72}{16} = \frac{9}{2} = 4\frac{1}{2}$
They drank $4\frac{1}{2}$ gallons of juice.
Watch out: 72 is the amount in cups. The question asks for gallons, so 72, choice D, is in the wrong unit.
So the answer is B, $4\frac{1}{2}$ .

Check your answer

Work backward. $4\frac{1}{2}$ gallons is $4\frac{1}{2} \times 4 = 18$ quarts. 18 quarts is $18 \times 4 = 72$ cups. Shared among 36 people, that is $72 \div 36 = 2$ cups each. It 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 D Math, page 105, question 32. This is the DOE’s version, before the corrections listed in the analysis.

Question 32 of 57

2022 Test B Math, question 32

Question

A student reads $\frac{1}{12}$ of a book in $\frac{1}{3}$ hour. What fraction of the book can the student read in 1 hour?

A $\frac{1}{36}$
B $\frac{1}{4}$
C $\frac{5}{12}$
D $\frac{12}{3}$

Show the answer

Correct answer: B

To find the fraction of the book the student can read in 1 hour, find the unit rate per hour. First, set up the ratio.

$\frac{\frac{1}{12} \text{ book}}{\frac{1}{3} \text{ hour}}$

Then use division to find the unit rate.

$\frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}$

Another way to solve the problem is to set up the ratio and then multiply both the numerator and denominator by 3, since $\frac{1}{3}$ hour times 3 equals 1 hour.

$\frac{3 \times \frac{1}{12} \text{ book}}{3 \times \frac{1}{3} \text{ hour}} = \frac{\frac{1}{4} \text{ book}}{1 \text{ hour}}$

Question analysis

A student reads 1/12 of a book in 1/3 of an hour, and this rate question asks what fraction of the book the student can read in a whole hour. It tests finding a unit rate when both amounts are fractions. Deciding whether to multiply or divide, and which fraction goes on top, decides the result.

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 Computing a unit rate from a ratio of fractions is the grade 7 standard itself. This is a short example with unit fractions, where scaling the time up to a whole hour is as quick as dividing.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 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

Thinking in thirds of an hour makes this quick: an hour holds three of them, so the student reads three twelfths of the book. Students who reach for a fraction rule instead must choose between multiplying and dividing, and between two orders of division. A sense check rules out two wrong choices at once: a whole hour must cover more than 1/12 of the book, and no more than the whole book.

Error-prone arithmetic minor Working it out
Both amounts are unit fractions, and 1/12 ÷ 1/3 means multiplying by 3, which gives 3/12 that still has to be simplified to match a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “fraction of the book … in 1 hour” asks for book per hour, so the fraction of the book is divided by the time, not the other way round. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Ask how many thirds of an hour make 1 hour, then scale the fraction of the book by the same factor; or divide the fraction of the book by the time.

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).
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

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 1/12 by 1/3 instead of dividing, getting 1/36, which would mean reading less of the book in an hour than in a third of one.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two fractions, 1/12 + 4/12 = 5/12, combining an amount of the book with an amount of time as if they were alike.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Divides the time by the amount read, 1/3 ÷ 1/12 = 12/3, which is hours per book, the rate turned upside down.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{part of the book}}{\text{hours}}$; A unit rate tells how much happens in 1 unit, here 1 hour. Put the part of the book on top and the time on the bottom. Dividing finds the part of the book for one whole hour.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by.

Sort out the question
The student reads $\frac{1}{12}$ of a book in $\frac{1}{3}$ of an hour. You need the part of the book read in 1 whole hour. That is a unit rate.
Scale up to one hour
One hour holds three thirds of an hour. So in 1 hour the student reads 3 times as much:
$3 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$
To simplify 3/12, divide the top and bottom by 3. The student reads 1/4 of the book in 1 hour.
Check with the unit rate
Divide the part of the book by the time. Multiply by the reciprocal of 1/3, which is 3/1:
$\frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}$
Both ways give 1/4.
Avoid the wrong operations
Watch out: multiplying gives $\frac{1}{12} \times \frac{1}{3} = \frac{1}{36}$ , choice A. That is less than 1/12. But in a full hour, the student must read more than in a third of an hour. Dividing the time by the book gives 12/3, choice D. That is 4 whole books.
So the answer is B, $\frac{1}{4}$ .

Check your answer

If the student reads 1/4 of the book each hour, the whole book takes 4 hours. In 1/3 of an hour, the student reads $\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$ of the book. That matches the question.

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 D Math, page 115, question 42.

Question 33 of 57

2022 Test B Math, question 33

Question

The perimeter of square W is twice the perimeter of equilateral triangle X. What is the ratio of the length of one side of square W to the length of one side of equilateral triangle X?

A3:8
B2:3
C3:2
D2:1

Show the answer

Correct answer: C

First, set up the ratio of the perimeter of the square to the triangle as 2:1 because the question states that the perimeter of the square is twice the perimeter of the triangle.

Recognizing that the square has 4 equal sides and that an equilateral triangle has 3 sides, use division to find the ratio of the sides.

Perimeter: $2 : 1 = \frac{2}{1}$

Sides:

$\frac{\frac{2}{4}}{\frac{1}{3}} = \frac{2}{4} \div \frac{1}{3} =$

$\frac{2}{4} \times \frac{3}{1} = \frac{6}{4} =$

$\frac{3}{2} = 3 : 2$

Question analysis

The perimeter of square W is twice the perimeter of equilateral triangle X, and students find the ratio of a side of the square to a side of the triangle. This ratio question uses perimeters of regular shapes with no lengths given. Turning a comparison of perimeters into one of sides, and keeping the ratio in the order asked, are the tricky parts.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4 equations and inequalities from problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 ratio problems usually supply numbers to work with. This one gives only a relationship between two perimeters, so students must bring in the number of sides of each shape themselves and keep the ratio in the order asked.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 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 turns a statement about perimeters into a ratio of sides, and the natural guess, that the sides keep the perimeters’ 2 to 1, ignores that the square shares its perimeter among 4 sides and the triangle among 3. Choosing a triangle perimeter such as 12 makes the comparison concrete. The order of the ratio still has to be checked at the end, because its reverse is listed too.

Turning words into math moderate Understanding the question
“The perimeter of square W is twice the perimeter of equilateral triangle X” must become 4s = 2 × 3t, with the 2 on the triangle’s side; written backward, it gives the triangle the larger perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers moderate Understanding the question
No side length is given, so students must either work with letters for the two sides or choose a convenient perimeter themselves, such as 12 for the triangle. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The ratios 3:2 and 2:3 use the same two numbers, so only keeping the square’s side first, as the question orders it, separates the answer from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Pick an easy side length for the triangle, find its perimeter, double that for the square, divide by the square's number of sides, and write square to triangle.

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 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).
Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

A Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: weak
Writes the comparison backward, 2 × 4s = 3t, giving the triangle the larger perimeter, so the ratio of the sides comes out 3:8.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Finds the right relationship but turns the ratio upside down, giving the triangle’s side to the square’s instead of the square’s side first.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Assumes the sides are in the same ratio as the perimeters, 2:1, ignoring that the square has 4 sides and the triangle only 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a square: $P = 4s$; P is the perimeter, the distance around the shape. s is the length of one side. A square has 4 equal sides, so its perimeter is 4 times one side.
Perimeter of an equilateral triangle: $P = 3t$; An equilateral triangle has 3 equal sides. t is the length of one side. So its perimeter is 3 times one side.
Ratio: $a : b$; A ratio compares two amounts in order. a : b means a compared with b. The order matters, so write the amounts in the order the question names them.

Sort out the question
Square W has a perimeter twice as long as the perimeter of triangle X. Equilateral means all 3 sides of X are equal.
You need the ratio of one side of the square to one side of the triangle, in that order. No lengths are given, so you can pick a size to try.
Pick a size for the triangle
Choose an easy side length for the triangle, such as 2 units. All 3 sides are 2, so its perimeter is:
$P = 3 \times 2 = 6$
Find the square’s side
The square’s perimeter is twice the triangle’s:
$2 \times 6 = 12$
The square has 4 equal sides, so divide by 4:
$s = 12 \div 4 = 3$
Each side of the square is 3 units.
Write the ratio in order
The square comes first, then the triangle. Their sides are 3 and 2, so the ratio is 3 : 2.
Watch out: 2 : 1, choice D, compares the perimeters, not the sides. The square splits its perimeter into 4 sides, the triangle into only 3. Choice B, 2 : 3, puts the triangle first. Choice A, 3 : 8, comes from giving the triangle the bigger perimeter.
So the answer is C, 3:2.

Check your answer

Try a different size. Give the triangle sides of 4. Its perimeter is 12, so the square’s is 24. Each side of the square is $24 \div 4 = 6$ . The ratio 6 : 4 simplifies to 3 : 2. The size does not change the ratio.

More questions like this

The original DOE question

Question 48 as shown on the DOE’s digital test, page 121 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 121, question 48.

Question 34 of 57

2022 Test B Math, question 34

Question

A recipe uses $\frac{2}{3}$ cup of sugar for every $1\frac{1}{4}$ cups of flour. A cook wants to increase the recipe. How many cups of sugar are needed for each cup of flour?

A $\frac{8}{15}$
B $\frac{5}{6}$
C $1\frac{7}{8}$
D $3\frac{2}{3}$

Show the answer

Correct answer: A

To find the cups of sugar for each cup of flour, find the unit rate using this ratio:

$\frac{\frac{2}{3} \text{ cup of sugar}}{1\frac{1}{4} \text{ cups of flour}}$

Then, convert the denominator to an improper fraction and divide to find the cups of sugar per cup of flour.

$\frac{\frac{2}{3}}{\frac{5}{4}} = \frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$

Question analysis

A recipe uses 2/3 cup of sugar for every 1 1/4 cups of flour, and this unit rate question asks how many cups of sugar go with each cup of flour. Students divide one fraction by a mixed number. Deciding which amount to divide by which, and converting the mixed number first, are where mistakes creep in.

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 Computing a unit rate from a ratio of fractions is exactly the grade 7 standard. This one divides by a mixed number, and the cook's plan to enlarge the recipe adds wording without adding any numbers to use.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
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

The work is a single fraction division, but it involves a mixed number and an awkward result. Students unsure of the rule multiply the two amounts or divide them the other way round, and each gives a tidy listed fraction. A size check settles it: 1 1/4 cups of flour take only 2/3 cup of sugar, so one cup of flour needs less than 2/3 cup, which leaves a single choice.

Error-prone arithmetic moderate Working it out
The divisor is the mixed number 1 1/4, which must become 5/4 before dividing; then 2/3 × 4/5 gives 8/15, a fraction that does not simplify and looks unfinished. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “for each cup of flour” puts flour in the denominator, sugar ÷ flour, and the sentence about increasing the recipe adds nothing to the calculation. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write the sugar over the flour, change 1 1/4 to an improper fraction, and divide by multiplying by its reciprocal, keeping sugar on top because the question asks for sugar per cup of flour.

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).
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

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies 2/3 by 5/4 instead of dividing, getting 10/12 = 5/6, more sugar for one cup of flour than the recipe uses for 1 1/4 cups.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides the flour by the sugar, 5/4 ÷ 2/3 = 15/8 = 1 7/8, which is cups of flour for each cup of sugar, the rate upside down.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Splits the mixed number apart, dividing 2/3 by only the 1/4 to get 2 2/3, then adding the whole 1 to make 3 2/3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{cups of sugar}}{\text{cups of flour}}$; A unit rate tells how much of one thing goes with 1 of another. Here it is the cups of sugar for 1 cup of flour. The amount you want goes on top. The amount you want to make 1 goes on the bottom.
Mixed number to fraction: $w\frac{n}{d} = \frac{w \times d + n}{d}$; w is the whole number, and n/d is the fraction part. Multiply the whole number by the bottom, add the top, and keep the bottom. It works because each whole holds d pieces of size 1/d.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by.

Sort out the question
The recipe uses $\frac{2}{3}$ cup of sugar for every $1\frac{1}{4}$ cups of flour. You need the cups of sugar for each 1 cup of flour. The part about making more of the recipe gives no numbers, so you can set it aside.
Set up the unit rate
You want sugar for each cup of flour. So sugar goes on top and flour goes on the bottom:
$\text{unit rate} = \frac{2}{3} \div 1\frac{1}{4}$
Watch out: dividing the other way, flour by sugar, gives $\frac{15}{8} = 1\frac{7}{8}$ . That is cups of flour for each cup of sugar, choice C.
Change the mixed number
You cannot divide by a mixed number directly. Change it to a fraction first. Each whole has 4 fourths:
$1\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}$
Divide the fractions
Multiply by the reciprocal of 5/4, which is 4/5. Multiply the tops and multiply the bottoms:
$\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$
The fraction 8/15 cannot be simplified, since 8 and 15 share no factor but 1.
Watch out: multiplying without flipping gives $\frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$ , choice B.
So the answer is A, $\frac{8}{15}$ .

Check your answer

More than 1 cup of flour takes only 2/3 cup of sugar. So 1 cup of flour needs a bit less than 2/3. Since $\frac{2}{3} = \frac{10}{15}$ , the answer 8/15 is a bit less. Also, $\frac{8}{15} \times \frac{5}{4} = \frac{40}{60} = \frac{2}{3}$ , which brings back the recipe.

More questions like this

The original DOE question

Question 30 as shown on the DOE’s digital test, page 103 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 103, question 30.

Question 35 of 57

2022 Test B Math, question 35

Question

READING ASSIGNMENTS

Assignment	Pages Assigned
I	3–17, inclusive
II	25–38, inclusive
III	45–60, inclusive

Over a two-week period, students are assigned a total of 250 pages to read from a science textbook. The table above shows the first three reading assignments within the larger assignment. What percentage of the 250 pages is the total number of pages in these first three assignments?

A16.8%
B18.0%
C22.8%
D24.0%

Show the answer

Correct answer: B

First determine the pages read for each assignment. Since the pages per assignment are inclusive, subtract the first page from the last page assigned and then add 1 to determine the pages read for each assignment.

Assignment I: $(17-3)+1=15$

Assignment II: $(38-25)+1=14$

Assignment III: $(60-45)+1=16$

The student was assigned a total of 45 pages, since $15+14+16=45$ .

To determine the percent, express 45 pages out of 250 pages as a fraction.

$\frac{45}{250}=\frac{45\div 5}{250\div 5}=\frac{9}{50}$

$\frac{9}{50}=\frac{18}{100}=18\%$

Question analysis

A table lists the first three reading assignments from a science textbook as page ranges marked inclusive, out of 250 pages assigned over two weeks. Students find what percentage of the 250 pages the three assignments make up in total. It tests percents, but the harder part comes first: counting the pages in a range when both end pages are included.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Expressing one number as a percent of another is familiar from grades 6 and 7. Here that number must first be counted from three inclusive page ranges, a step most grade 7 percent problems leave out.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Many steps moderate
Reading a figure, table or graph minor
Solution: 5 steps
Reading level: Question and choices: about grade 6, 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

Counting pages in a range is the real test; the percent at the end is routine once the three counts are right. Subtracting first page from last feels natural, and the 42 pages it gives lead straight to a listed choice. The table also invites shortcuts, the span from page 3 to page 60 or the last page alone, and each lands on a choice too. Checking the rule on a tiny range, pages 1 to 3, which hold 3 pages and not 2, fixes the count.

Something you must already know moderate Working it out
Each row says inclusive, but nothing says that pages 3 to 17 hold 17 − 3 + 1 = 15 pages; subtracting alone loses one page from each of the three assignments. (Anderson & Krathwohl, 2001)
Many steps moderate Working it out
Three page counts, their sum, a fraction of the 250 pages and a change to a percent come in sequence, and a miscount in the first step carries through to the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
The table’s ranges leave gaps, pages 18 to 24 and 39 to 44, so the first and last pages listed, 3 and 60, do not bound the pages actually assigned. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Work out how many pages each range holds when both end pages count, total the three, then write that total as a fraction of 250 and change it to a percent.

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.
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

A Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Subtracts the first page from the last in each row, 14 + 13 + 15 = 42 pages, leaving out one end of each inclusive range, and 42/250 = 16.8%.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Treats pages 3 to 60 as one stretch, 60 − 3 = 57, counting the unassigned pages between assignments, and 57/250 = 22.8%.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the last assigned page, 60, as the number of pages read, as if the assignments covered pages 1 to 60, and 60/250 = 24%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range: $\text{pages} = \text{last page} - \text{first page} + 1$; Inclusive means the first and last pages are both read. Subtracting counts the steps from the first page to the last. That misses one page, so add 1. Test it: pages 1 to 3 are 3 pages, and 3 − 1 + 1 = 3.
Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you have, here the pages in the three assignments. The whole is the amount you compare with, here 250 pages. Dividing gives a fraction or decimal. Multiplying by 100 turns it into a percent, a number out of 100.

Sort out the question
The whole assignment is 250 pages. The table gives three page ranges:
- I: pages 3 to 17
- II: pages 25 to 38
- III: pages 45 to 60
Each range is inclusive. That means its first and last pages are read too. You need the percent of 250 pages these three make up.
Count the pages in each range
Use the rule: last page minus first page, plus 1.
$17 - 3 + 1 = 15$
$38 - 25 + 1 = 14$
$60 - 45 + 1 = 16$
Watch out: subtracting alone gives 14 + 13 + 15 = 42 pages. That leads to 16.8%, choice A.
Add the three ranges
$15 + 14 + 16 = 45$
The three assignments hold 45 pages in all.
Watch out: don’t count pages 3 to 60 as one stretch. Pages 18 to 24 and 39 to 44 are not assigned. That mistake gives 57 pages and 22.8%, choice C. Using 60 pages gives 24.0%, choice D.
Change to a percent
$\text{percent} = \frac{45}{250} \times 100$
Simplify first. Divide the top and bottom by 5, then multiply them both by 2.
$\frac{45}{250} = \frac{9}{50} = \frac{18}{100}$
That is 18 out of 100, or 18%. So the answer is B, 18.0%.

Check your answer

Work backward. 10% of 250 pages is 25 pages, so 1% is 2.5 pages. Then 18% is $18 \times 2.5 = 45$ pages. That matches the count of 45.

More questions like this

The original DOE question

Question 92 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 2023 Admissions, page 206, question 92. 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 36 of 57

2022 Test B Math, question 36

Question

A box contains an unknown number of green marbles. An experimenter adds 100 red marbles to the box, mixes the marbles thoroughly, and then draws out 10 marbles at random. Assume that this sample is representative of the proportion in the box. If there are 2 red marbles and 8 green marbles in this sample, approximately how many green marbles are in the box?

A80
B300
C400
D500

Show the answer

Correct answer: C

In this context, the phrase “this sample is representative” means that the ratio of 2 red marbles to 8 green marbles is the same as the ratio of red to green marbles in the entire box. Use this ratio of 2 to 8 to set up a proportion and solve, where x represents the number of green marbles in the box.

$\frac{2}{8} = \frac{100}{x}$

$2x = 800$

$x = 400$

Question analysis

An experimenter adds 100 red marbles to a box of green marbles, mixes them, and draws a sample of 10 with 2 red and 8 green. This sampling question asks students to estimate how many green marbles the box holds. It applies a ratio from a representative sample, worked from a count known for the box rather than for the sample.

NYS standards: 6.SP.1b samples and populations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a representative sample to reason about a population is grade 6 and 7 work. The twist is that the known count, the 100 red marbles added, belongs to the box rather than the sample, so the usual scaling runs in an unfamiliar direction.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 60 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 the logic of estimating a population by tagging, and the twist is that the known count belongs to the marbles that were added, not to the ones being estimated. Scaling the sample up by 50, so that its 2 red become the box’s 100, turns its 8 green into the estimate in one step. Students who scale by way of the whole box often stop at the box’s total, and those who treat 100 as the whole box scale the wrong amount.

Turning words into math moderate Understanding the question
The 100 red marbles are counted in the box, while the 2 red and 8 green are counted in a sample of 10, so “representative of the proportion” must become 2 : 8 = 100 : x. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Three ratios can be read from the sample, red to green, red to all and green to all, and only the first two connect to the known 100 red marbles. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Use the sample's ratio of red to green marbles, then scale it up until the red part matches the 100 red marbles known to be in the box.

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.
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

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Scales the sample’s 8 green in 10 up to 100 marbles, getting 80, as if the 100 red marbles added were every marble in the box.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reads the sample’s 2 to 8 as red to all marbles, so 2/8 = 100/400 makes 400 marbles in the box, then subtracts the 100 red ones.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Uses 2 red in 10 marbles to find 500 marbles in the whole box, then stops without taking away the 100 red ones to leave the green.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios (a proportion): $\frac{\text{red in sample}}{\text{green in sample}} = \frac{\text{red in box}}{\text{green in box}}$; A ratio compares two amounts. A proportion says two ratios are equal. A sample that is representative has the same mix as the whole box. So the box is the sample scaled up.

Sort out the question
You know:
- The box has an unknown number of green marbles.
- 100 red marbles are added to the box.
- A sample of 10 marbles has 2 red and 8 green.
Representative means the sample has the same mix as the box. You need the number of green marbles in the box. Notice that the 100 red marbles are counted in the box, not in the sample.
Set up the proportion
In the sample, red to green is 2 to 8. In the box, red to green is 100 to x, where x is the number of green marbles. The two ratios are equal:
$\frac{2}{8} = \frac{100}{x}$
Scale up the sample
How do you get from 2 red to 100 red? Multiply by 50, since $2 \times 50 = 100$ . The green marbles grow by the same factor:
$8 \times 50 = 400$
So the box has about 400 green marbles.
Avoid the tempting wrong totals
Watch out: there is another way that can trip you up. In the sample, 2 of 10 marbles are red. So the 100 red marbles are 2/10 of the box, and the box holds 500 marbles in all. But 100 of those are red. Stopping at 500 gives choice D.
Taking away the red ones gives $500 - 100 = 400$ green marbles, the same answer. Treating 100 as the whole box gives 80 green, choice A. Reading 2 to 8 as red to all marbles gives 300, choice B.
So the answer is C, 400.

Check your answer

Picture the finished box: 400 green and 100 red, so 500 marbles. The red part is $\frac{100}{500} = \frac{1}{5}$ . The sample had $\frac{2}{10} = \frac{1}{5}$ red too. The mix matches, so 400 makes sense.

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 D Math, page 104, question 31.

Question 37 of 57

2022 Test B Math, question 37

Question

A car costing $24,000 is subject to a sales tax of 8%. If Bindu made a $10,000 down payment on this car, what is the total amount she has left to pay?

A$14,000
B$15,620
C$15,920
D$25,920

Show the answer

Correct answer: C

First, find the sales tax. Then, add the sales tax to the cost of the car. Finally, subtract the down payment to find out how much she has left to pay.

$(0.08)(24{,}000) = 1{,}920$

$\$24{,}000 + \$1{,}920 = \$25{,}920$

$\$25{,}920 - \$10{,}000 = \$15{,}920$

Question analysis

Bindu buys a $24,000 car with an 8% sales tax and makes a $10,000 down payment, and students find how much she still has to pay. This percent word problem combines finding a tax, adding it to the price and subtracting a payment. Every step must be done in order, since leaving one out gives an amount that looks just as reasonable.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding sales tax to a price is a standard grade 7 percent task. This one also subtracts a down payment at the end, three routine steps with round numbers, much like a typical classroom problem.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 29 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 the standard one for tax on a purchase, so the question checks order and completeness. Forgetting the tax or forgetting the down payment each produces an amount as reasonable-looking as the answer, and a careless 8 × 24 produces one just $300 below it. Multiplying the price by 1.08 in one step, which gives $25,920, is a second route that catches a slip in the tax.

Many steps minor Working it out
Three operations come in a fixed order, 8% of $24,000, then adding the tax, then subtracting the $10,000, and leaving out the first or the last gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Finding 8% of $24,000 comes down to 8 × 24 = 192 with two zeros added, and a lost carry in that product turns $1,920 into $1,620. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find 8% of the price and add it, or multiply the price by 1.08, to get the full cost with tax, then subtract what Bindu has already paid.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Subtracts the $10,000 down payment from the $24,000 price and never adds the 8% sales tax, so the tax is simply left out.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Drops the carried 3 when multiplying 8 × 24, so the tax comes out $1,620 instead of $1,920 and the balance $300 too low.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Adds the $1,920 tax to the price correctly but stops at $25,920, the full cost, forgetting to subtract the $10,000 Bindu already paid.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; Percent means out of 100. So 8% is 8/100, or 0.08. To find a percent of a number, change the percent to a decimal and multiply it by the whole.
Amount left to pay: $\text{amount left} = \text{price} + \text{tax} - \text{down payment}$; The price plus the tax is the full cost. A down payment is money paid at the start. Subtract it, because that part is already paid.

Sort out the question
You know:
- The car costs $24,000.
- The sales tax is 8%. Sales tax is extra money added to a price.
- Bindu paid $10,000 at the start. That is the down payment.
You need the amount Bindu still has to pay. Work in order: tax, then full cost, then subtract.
Find the sales tax
Change 8% to the decimal 0.08. Multiply it by the price:
$0.08 \times 24{,}000 = 1{,}920$
A quick way: 1% of 24,000 is 240, so 8% is $8 \times 240 = 1{,}920$ . The tax is $1,920.
Watch out: in 8 × 24 you carry a 3 from 8 × 4 = 32. Dropping it gives 162, so the tax looks like $1,620. That leads to $15,620, choice B.
Find the full cost
Add the tax to the price:
$24{,}000 + 1{,}920 = 25{,}920$
The car costs $25,920 with tax.
Watch out: this is not the answer yet. Stopping here gives $25,920, choice D. Bindu already paid part of it.
Subtract the down payment
Bindu already paid $10,000. Take it away:
$25{,}920 - 10{,}000 = 15{,}920$
She has $15,920 left to pay.
Watch out: leaving out the tax gives $24{,}000 - 10{,}000 = 14{,}000$ , choice A.
So the answer is C, $15,920.

Check your answer

Use a shortcut for the full cost. The price plus 8% more is 108% of the price, or 1.08 times it: $24{,}000 \times 1.08 = 25{,}920$ . Then $25{,}920 - 10{,}000 = 15{,}920$ . Same answer.

More questions like this

The original DOE question

Question 33 as shown on the DOE’s digital test, page 106 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 106, question 33.

Question 38 of 57

2022 Test B Math, question 38

Question

WINTER TEMPERATURES IN CHICAGO

Day	1	2	3	4	5	6	7
Temperature (°F)	17	7	2	4	−3	−5	1

The table above shows temperatures for a one-week period during the winter in Chicago. What is the range of temperatures shown in the table?

Show the answer

Correct answer: C

Range is the difference between the largest and the smallest values in a given set of data. In this case, the highest temperature is 17 and the lowest temperature is −5. Therefore, the range is $17-(-5)=17+5=22$ .

Question analysis

A table gives Chicago's temperatures in degrees Fahrenheit for seven winter days, some above zero and some below. Students find the range of the temperatures. It tests the meaning of range and subtracting a negative number, and treating a below-zero reading as if it were positive gives a wrong spread.

NYS standards: 7.NS.1c rational number subtraction
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Range is a grade 6 idea, and subtracting a negative number is grade 7. Here they meet in a single subtraction from a short table, typical grade 7 practice with integers rather than anything longer.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Reading a figure, table or graph minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 31 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 a range is routine, but the lowest reading is below zero, so the subtraction crosses zero: from −5 up to 0 is 5 degrees, and from 0 up to 17 is 17 more. Students who treat −5 as 5 find a smaller gap that is listed. Picturing a thermometer, or a number line, shows the full spread at a glance.

Error-prone arithmetic moderate Working it out
The lowest reading is −5°F, so the range is 17 − (−5), and subtracting a negative number means adding 5, the step where a gap of 12 appears instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph minor Understanding the question
The seven readings run across one table row in day order, 17, 7, 2, 4, −3, −5, 1, so the extremes have to be picked out, including the lower of the two below zero. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The word range is not defined in the question; students must know it means the highest value minus the lowest, not the highest value alone. (Anderson & Krathwohl, 2001)

Strategy: Pick out the highest and the lowest temperature in the table, then subtract the lowest from the highest, writing the subtraction of a negative as an addition.

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.)
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

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts 5 from 17 as if −5 were 5, so the range comes out 12, missing that a reading below zero lies further from 17, not closer.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: weak
Gives the highest temperature, 17°F, as the range, taking the top of the readings for the spread between the highest and the lowest.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 27: it is 5 more than the true spread and matches no sign or subtraction mistake with these readings, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range: $\text{range} = \text{highest} - \text{lowest}$; The range tells how spread out the data are. Find the highest value and the lowest value, then subtract. It is the distance between the two ends of the data on a number line.
Subtracting a negative: $a - (-b) = a + b$; a is any number, and −b is a negative number. Taking away a negative is the same as adding the positive number b. On a number line, the gap from −5 up to 17 is 5 to reach 0, then 17 more.

Sort out the question
The table gives seven temperatures, in degrees Fahrenheit (°F): 17, 7, 2, 4, −3, −5 and 1. A minus sign means below zero. You need the range, the highest temperature minus the lowest.
Find the highest and lowest
The highest is 17°F, on day 1. For the lowest, look at the numbers below zero: −3 and −5. −5 is colder, since it is further below zero. So the lowest is −5°F, on day 6.
Subtract the lowest from the highest
$\text{range} = 17 - (-5) = 17 + 5 = 22$
Subtracting a negative is the same as adding. The range is 22 degrees Fahrenheit.
Watch out: −5 is not 5. Doing 17 − 5 = 12 gives choice A. And 17 alone is just the highest temperature, choice B. So the answer is C, 22.

Check your answer

Picture a thermometer. From −5°F up to 0°F is 5 degrees. From 0°F up to 17°F is 17 more. In all, $5 + 17 = 22$ degrees. That matches.

More questions like this

The original DOE question

Question 95 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 2023 Admissions, page 207, question 95. Open page 207 of the handbook (PDF)

Question 39 of 57

2022 Test B Math, question 39

Question

$A graph titled ERRORS ON JOB TEST. The horizontal x-axis, labeled Number of Errors, is marked 1 through 7, and the vertical y-axis, labeled Number of Applicants, is marked 1 through 7, with the origin O. Points are plotted at (1, 4), (2, 7), (3, 5), (4, 6), (5, 3) and (6, 1).$

A total of 26 job applicants took a computer-competency test, and the number of errors (x) was recorded for each applicant. The graph above shows how many applicants made x errors, for each value of x. How many applicants made fewer than 3 errors?

Show the answer

Correct answer: B

Each (x, y) ordered pair represents y, the number of applicants that made x errors.

First, determine the points with x-values less than 3.

The point (1, 4) means that 4 applicants made 1 error.

The point (2, 7) means that 7 applicants made 2 errors.

Then, add the y-values to find the total applicants who made fewer than 3 errors.

$4+7=11$

Question analysis

A graph titled Errors on Job Test shows, for each number of errors, how many of 26 job applicants made that many. Students find how many applicants made fewer than 3 errors. It tests reading a data display in which each point's height is a count of people, and the words fewer than decide which points to use.

NYS standards: 6.SP.4 dot plots and histograms
6.SP.5a number of observations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading counts from a data display and adding them is grade 6 statistics. It needs only a few readings from the graph and one addition, less than typical grade 7 data work, which compares distributions or uses box plots.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 42 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 reading question more than a computing one: two heights from the graph and one addition. The point at 3 errors is the one to leave out, since fewer than 3 stops at 2. Checking that all six heights add up to the 26 applicants named in the stem confirms the readings before they are used.

Reading a figure, table or graph minor Understanding the question
Each point’s height is a number of applicants, so the counts for 1 and 2 errors, 4 and 7, are read on the vertical axis above those error values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Turning words into math minor Understanding the question
The words “fewer than 3 errors” mean the points at 1 and 2 errors only, leaving out the 5 applicants who made exactly 3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Read the number of applicants at each error count below 3 from the heights of the points, then add those numbers.

Traps

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

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads the axes the wrong way round, so the one point with a height below 3, (6, 1), seems to show 6 applicants making 1 error.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single misreading gives 12; working from the total, 26 − 15 = 11, reaches it only if a point is skipped, so it mostly draws guesses.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the 26 applicants stated in the stem, everyone who took the test, instead of only those who made fewer than 3 errors.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reading a point: $(x, y) = (\text{errors}, \text{applicants})$; Each point on the graph is an ordered pair. The first number, x, is read on the bottom axis: the number of errors. The second number, y, is read on the side axis: how many applicants made that many errors.

Sort out the question
In all, 26 applicants took a test. The graph shows how many made each number of errors. You need how many made “fewer than 3 errors.”
Fewer than 3 means 0, 1 or 2 errors. It does not include 3.
Read the points you need
Find the error count on the bottom axis. Then go up to the point and read across to the side axis.
- At 1 error, the point is at height 4. So 4 applicants made 1 error.
- At 2 errors, the point is at height 7. So 7 applicants made 2 errors.
There is no point at 0 errors.
Add the two counts
$4 + 7 = 11$
So 11 applicants made fewer than 3 errors.
Watch out: leave out the point at 3 errors, (3, 5). Those 5 applicants made exactly 3 errors, not fewer. Reading the axes the wrong way round can give 6, choice A. And 26 is everyone, choice D. So the answer is B, 11.

Check your answer

Add all six heights: $4 + 7 + 5 + 6 + 3 + 1 = 26$ . That is the 26 applicants in the question, so the readings are right. It also shows nobody made 0 errors. The other $26 - 11 = 15$ made 3 or more.

More questions like this

The original DOE question

Question 96 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 2023 Admissions, page 208, question 96. Open page 208 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 40 of 57

2022 Test B Math, question 40

Question

Last week, Emily rode the bus to school on 3 of the 5 mornings and rode the bus home on 4 of the 5 afternoons. Based on last week’s events, what is the probability that Emily will ride the bus in the morning and the afternoon next Monday?

A $\frac{1}{5}$
B $\frac{7}{25}$
C $\frac{12}{25}$
D $\frac{7}{10}$

Show the answer

Correct answer: C

Using the joint probability formula, find the product of the two independent events. The probability that Emily will ride the bus to school on any given morning is $\frac{3}{5}$ . The probability that Emily will ride the bus home on any given afternoon is $\frac{4}{5}$ . Therefore, the probability that Emily will ride the bus in the morning and the afternoon is $\frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$ .

Question analysis

Last week Emily rode the bus to school on 3 of 5 mornings and home on 4 of 5 afternoons. This probability question asks how likely she is to ride the bus both ways next Monday, based on those frequencies. Students turn each count into a probability and combine the two, deciding how chances join when both events must happen.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.8b probability models from observed frequencies
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Compound probability is grade 7 work, often done with a table of outcomes. Each chance here comes from one week's frequencies, and joining the morning and afternoon events in one step is about the usual grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 48 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

Once the two chances are written, the question is how they combine. Pooling the week into 7 bus rides out of 10 trips, or adding fractions in some way, gives listed choices that look reasonable, since each lies between 0 and 1. Picturing the 25 pairings of a morning with an afternoon, 3 × 4 = 12 of which have both rides, shows why the chances multiply.

Something you must already know moderate Working it out
The rule that two independent events both happen with the product of their chances, 3/5 times 4/5, is not given, and “the morning and the afternoon” is the only signal for it. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The counts “3 of the 5 mornings” and “4 of the 5 afternoons” have to become two separate probabilities based on last week, not one pool of 10 trips. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Turn each count into a probability, then multiply them, since both rides must happen; picturing a table of 5 mornings by 5 afternoons shows why multiplying works.

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.
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.
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

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Subtracts the morning chance from the afternoon chance, 4/5 − 3/5 = 1/5, an operation with no meaning for two rides that must both happen.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators, 3 + 4, but multiplies the denominators, 5 × 5, mixing the rule for adding fractions with the rule for multiplying them.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Pools the week’s trips into 7 bus rides out of 10, a rate for a single trip, as if riding both ways next Monday were just as likely.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “The probability that Emily will ride the bus to school on any given afternoon is 4/5.” The question says she rode the bus home on 4 of the 5 afternoons; her rides to school are the morning ones. The site now says “ride the bus home on any given afternoon”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability from past results: $P = \frac{\text{times it happened}}{\text{times it could have happened}}$; P is the probability, the chance something happens. You use what happened before to predict the chance it happens again. Emily rode in 3 of 5 mornings, so the chance for one morning is 3/5.
Both of two independent events: $P(A \text{ and } B) = P(A) \times P(B)$; A and B are two events. Independent means one does not change the chance of the other. To find the chance that both happen, multiply their chances. The result is smaller than either one, since both must happen.

Sort out the question
Last week Emily:
- rode the bus to school on 3 of the 5 mornings;
- rode the bus home on 4 of the 5 afternoons.
You need the chance that she rides the bus in the morning and in the afternoon next Monday. The word and means both rides must happen.
Write each chance
Use last week to find each chance:
- Bus to school on a morning: $\frac{3}{5}$
- Bus home on an afternoon: $\frac{4}{5}$
Keep them apart. The morning rides go to school, and the afternoon rides go home.
Multiply the two chances
Both rides must happen, so multiply:
$\frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$
Why multiply? Pair each of the 5 mornings with each of the 5 afternoons. That makes 25 pairs. In $3 \times 4 = 12$ of them, she rides both times. So the chance is 12 out of 25.
Avoid the tempting wrong choices
Watch out: pooling the week gives 7 rides out of 10 trips, or 7/10, choice D. That is the chance for one trip, not for two trips on one day.
Adding the tops but multiplying the bottoms gives 7/25, choice B. Subtracting the chances gives 1/5, choice A.
So the answer is C, $\frac{12}{25}$ .

Check your answer

The chance of both rides must be less than the chance of each ride alone. As decimals, 12/25 = 0.48. That is less than 3/5 = 0.6 and 4/5 = 0.8. Choice D, 7/10 = 0.7, is more than 3/5, so it cannot be right.

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 D Math, page 107, question 34. This is the DOE’s version, before the corrections listed in the analysis.

Question 41 of 57

2022 Test B Math, question 41

Question

XYZ Corporation hired 6 people as summer workers. Of these 6 people, 3 will be hired as permanent employees in the fall. If 2 of the 6 people are male, how many of the possible groups of 3 people hired in the fall will include only 1 male?

Show the answer

Correct answer: C

An organized list can be used to determine the number of groups of 3 people that include only 1 male.

Let the letters A, B, C, and D represent the females, and let the numbers 1 and 2 represent the males.

Then, create a list of all possible combinations of 3 people with only 1 male.

AB1, AB2, AC1, AC2, AD1, AD2

BC1, BC2, BD1, BD2

CD1, CD2

There are 12 possible groups with only 1 male.

Question analysis

A company hired 6 summer workers, 2 of them male, and will keep 3 as permanent employees in the fall. This counting question asks how many possible groups of 3 include only 1 male. Students count combinations where order does not matter, and the bold word only, with two kinds of workers to combine, makes careful listing essential.

NYS standards: 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 students list outcomes with organized lists and tree diagrams, usually ordered ones. Counting unordered groups of three under an "only one" condition, built from two kinds of people, goes beyond that typical work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 41 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 a two-stage count in a hiring story. Students tend to get one stage right and forget the other: they count the pairs of women and stop, or they count ordered picks and so find every group twice. Naming the women A to D and the men 1 and 2, then listing the pairs of women alphabetically with each man attached, makes every group visible and checkable.

Combines ideas moderate Working it out
Two counts are joined: the 6 ways to pick 2 of the 4 women, then the 2 ways to pick the man, and the total is their product. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Students must already know that a group is the same whichever order its members are named, so picking one woman and then another repeats the reverse pick. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The bold word only means exactly one man, so the other two places in each group of 3 must go to two of the 4 women, a step the stem leaves unsaid. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Name the women A to D and the men 1 and 2, list every pair of women without repeats, then attach each man to each pair and count the groups.

Traps

Stops one step early: A choice is a number from partway through the solution.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 2 men by the 2 places left for women, 2 × 2 = 4, using numbers from the question without listing a single group.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Counts the 6 pairs of women correctly but stops before pairing each one with either of the 2 men, which doubles the count.
D Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Counts ordered picks of the two women, 4 × 3 = 12 instead of 6, so each group appears twice, then doubles for the men to reach 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting with two choices: $\text{number of groups} = \text{ways to pick the man} \times \text{ways to pick the women}$; Each way of picking the man can go with each way of picking the two women. So you multiply the two counts. It is like outfits: 2 shirts with 6 pairs of pants make 12 outfits.

Sort out the question
Six people work for the summer. Two are male, so the other 4 are female. Three will be hired in the fall.
You need the number of groups of 3 with only 1 male. That means exactly 1 man and 2 women. In a group, order does not matter. Ann, Bea and Carl is the same group as Carl, Ann and Bea.
Give each person a name
Call the women A, B, C and D. Call the men 1 and 2. Every group you want has 2 letters and 1 number, like AB1.
List the pairs of women
List each pair once. Start with A, then pair B only with letters after it, and so on:
- AB, AC, AD
- BC, BD
- CD
That is 6 pairs. BA is not new, since it is the same pair as AB.
Watch out: counting the order gives $4 \times 3 = 12$ pairs, each one twice. Then doubling for the men gives 24, choice D.
Add a man to each pair
Each pair of women can go with man 1 or man 2. Use the counting rule:
$6 \times 2 = 12$
Watch out: stopping at the 6 pairs of women leaves out the men. That gives 6, choice B.
So the answer is C, 12.

Check your answer

Write out every group. With man 1: AB1, AC1, AD1, BC1, BD1, CD1. With man 2: AB2, AC2, AD2, BC2, BD2, CD2. Each list has 6 groups, and $6 + 6 = 12$ . No group repeats, and each has just one man.

More questions like this

The original DOE question

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

Question 42 of 57

2022 Test B Math, question 42

Question

At any given time, one person out of every 20 people has the common cold. Assume that the common cold can be caused by any one of 200 different viruses, each of which is equally likely to cause the cold. One of these 200 viruses is cold virus V144. What is the probability that a randomly selected person is suffering from a cold caused by virus V144?

A $\frac{1}{10}$
B $\frac{1}{20}$
C $\frac{1}{200}$
D $\frac{1}{4{,}000}$

Show the answer

Correct answer: D

To calculate the joint probability, you must first determine the probability of each independent event. The probability that a randomly selected person has a common cold is $\frac{1}{20}$ . The probability that a cold is caused by V144 is $\frac{1}{200}$ . The probability that a person has a cold and the cold is caused by the V144 virus is the product of those two probabilities, $\frac{1}{20}\times\frac{1}{200}=\frac{1}{4{,}000}$ .

Question analysis

One person in 20 has a common cold at any given time, and each cold comes from one of 200 equally likely viruses. Students find the probability that a randomly selected person has a cold caused by one particular virus, V144. It tests the probability of a compound event, which is hidden here inside a single phrase.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.8a uniform probability models
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students find compound probabilities, usually from organized lists or tree diagrams. Here the two stages are hidden in one phrase, a cold caused by one particular virus, and there are far too many outcomes to list.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know moderate
Solution: 2 steps
Reading level: Question and choices: about grade 6, 68 words.
Corrected from the DOE handbook: A probability in the explanation that was stated for the wrong event 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 one multiplication; the difficulty is seeing that the event has two stages. Each stated chance, 1 in 20 for a cold and 1 in 200 for the virus behind it, is listed as a choice on its own, and each answers only half the question. A chance that both stages happen must be smaller than either one alone, which leaves a single choice.

Turning words into math moderate Understanding the question
The single phrase “a cold caused by virus V144” holds two events, having a cold at all and that cold coming from V144, and they must be separated before any rule applies. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know moderate Working it out
The stem never says that the chance of both stages is the product of their chances, so students must bring that rule to the 1 in 20 and the 1 in 200 themselves. (Anderson & Krathwohl, 2001)

Strategy: Split the event into two stages, having a cold and that cold coming from V144, write each stage's chance as a fraction, and combine them with the rule for events that must both happen.

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).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides one chance by the other, 1/200 ÷ 1/20 = 20/200 = 1/10, instead of multiplying them, giving a chance larger than either stage.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Gives the chance that a person has a cold of any kind, 1 in 20, and stops before multiplying by the 1/200 chance that V144 caused it.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Gives the chance that a cold is caused by V144, one virus out of 200, forgetting that only 1 person in 20 has a cold at all.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “The probability that a person has a cold caused by V144 is 1/200.” A person having a cold caused by V144 is exactly what the question asks about, and the same paragraph goes on to give its probability as 1/4,000; as printed, the sentence gives that event the value of choice C. The 1/200 is the probability that a cold is caused by V144 (one of 200 equally likely viruses), so the site reads “The probability that a cold is caused by V144 is 1/200.” The working and the answer are unchanged: 1/20 × 1/200 = 1/4,000.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance that something happens, from 0 to 1. The top counts the outcomes you want. The bottom counts all the outcomes that could happen. Use it when every outcome is equally likely, like one virus out of 200.
Chance that two events both happen: $P(A \text{ and } B) = P(A) \times P(B \text{ given } A)$; A is the first event and B is the second. P(A) is the chance of A. P(B given A) is the chance of B among the times A has happened. Multiply them to get the chance that both happen. The result is smaller than either chance, since both things must happen.

Sort out the question
One person in 20 has a cold. Every cold comes from one of 200 viruses, all equally likely. V144 is one of them. You need the chance that a random person has a cold caused by V144.
That phrase holds two events. First, the person has a cold. Second, that cold comes from V144.
Find the chance of a cold
The question says “one person out of every 20 people has the common cold.” So 1 person is the part with a cold, and 20 is the whole group.
$P(\text{cold}) = \frac{1}{20}$
Find the chance it is V144
Now think only about people who have a cold. Each cold comes from one of 200 viruses, all equally likely. V144 is 1 of the 200.
$P(\text{V144 given a cold}) = \frac{1}{200}$
This chance is only about people who already have a cold. So the second event depends on the first.
Multiply the two chances
Use the rule for two events that must both happen. Event A is a cold, and event B is V144 as its cause.
$P(A \text{ and } B) = P(A) \times P(B \text{ given } A)$
$P(\text{cold and V144}) = \frac{1}{20} \times \frac{1}{200} = \frac{1}{4{,}000}$
Multiply the tops: 1 × 1 = 1. Multiply the bottoms: 20 × 200 = 4,000.
Give the answer
Watch out: 1/20 is the chance of any cold, choice B. And 1/200 is the chance that a cold comes from V144, choice C. Each one answers only half the question. Dividing one by the other gives 1/10, choice A. So the answer is D, $\frac{1}{4{,}000}$ .

Check your answer

Picture 4,000 people. One in 20 has a cold, so $4{,}000 \div 20 = 200$ people have colds. Of those 200 colds, 1 in 200 comes from V144. That is 1 person out of 4,000. It is also smaller than both 1/20 and 1/200, as it should be.

More questions like this

The original DOE question

Question 99 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 209, question 99. Open page 209 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 43 of 57

2022 Test B Math, question 43

Question

Karl has one red spinner and one blue spinner. Each spinner is divided into 4 equal sections, numbered 1 through 4. He spins each spinner once and writes down the number that each lands on. What is the probability that the two numbers, when multiplied together, will have 4 as a product?

A $\frac{1}{16}$
B $\frac{3}{16}$
C $\frac{4}{7}$
D $\frac{7}{16}$

Show the answer

Correct answer: B

There are four possible outcomes for each spin of the spinner:

1, 2, 3, 4

The possible outcomes of two spins are:

$\begin{array}{cccc} 11 & 12 & 13 & \boxed{14} \\ 21 & \boxed{22} & 23 & 24 \\ 31 & 32 & 33 & 34 \\ \boxed{41} & 42 & 43 & 44 \end{array}$

Of the 16 possible outcomes, only 3 pairings have a product of 4: 14, 22, and 41.

Therefore, the probability that the two numbers resulting from two consecutive spins have a product of 4 is $\frac{3}{16}$ .

Question analysis

Karl spins a red spinner and a blue spinner, each with four equal sections numbered 1 to 4, and multiplies the two results. This compound probability question asks for the chance that the product is 4. Students build the sample space of outcomes, and listing every pair rather than the ones that come to mind is the safe route.

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: At grade 7 Finding a compound probability from a list or table of equally likely outcomes is standard grade 7 work. Two four-section spinners give a small sample space that is quick to write out, and the event needs one careful scan of the products.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 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 is decided by the count of favorable outcomes, not by any calculation. Students who think of 4 only as 2 × 2 find one outcome, and those who think in factor pairs find two, which is not a listed choice and should send them back to the table. A 4-by-4 grid of products, red down the side and blue across, shows all three cells at once.

Something you must already know moderate Working it out
Nothing in the stem says that red 1 with blue 4 and red 4 with blue 1 are separate outcomes, or that two four-part spinners give 16 equally likely pairs. (Anderson & Krathwohl, 2001)
Much to keep track of minor Working it out
All 16 pairs must be scanned for a product of 4, and the two pairs that use 1 and 4 are easy to miss beside the obvious 2 times 2. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Make a 4-by-4 table with the red spinner's numbers down the side and the blue spinner's across the top, write each product in its cell, and count the cells that match.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds only the double, 2 and 2, and stops, missing the two outcomes in which a 1 on one spinner meets a 4 on the other.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No count of the 16 outcomes gives a denominator of 7, so 4/7 has no working behind it and mostly attracts guesses.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Counts the 7 outcomes in which either spinner shows a 4, reading “4 as a product” as a pair that simply contains a 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of outcomes you want}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. An outcome is one possible result. This works when every outcome is equally likely, like landing on equal sections of a spinner.
Counting outcomes of two actions: $\text{total outcomes} = a \times b$; a is the number of results for the first action, and b is the number for the second. Each result of the first can go with each result of the second, so you multiply.

Sort out the question
Karl spins two spinners, one red and one blue. Each has 4 equal sections, numbered 1 to 4. He multiplies the two numbers. You need the probability that the product, the answer to a multiplication, is 4.
Count all the outcomes
The red spinner has 4 results, and so does the blue one. Each red result can go with each blue result:
$4 \times 4 = 16$
So there are 16 outcomes, all equally likely. Red 1 with blue 4 is not the same as red 4 with blue 1. They come from different spinners.
Make a table of products
Write each product in a table. Red goes down the side and blue goes across the top.
Red × Blue 1 2 3 4
1 1 2 3 4
2 2 4 6 8
3 3 6 9 12
4 4 8 12 16
Three cells show 4: red 1 with blue 4, red 2 with blue 2, and red 4 with blue 1.
Watch out: thinking of 4 only as 2 × 2 finds one outcome. That gives 1/16, choice A.
Find the probability
Use the probability rule. Three outcomes give 4, out of 16 in all:
$P = \frac{3}{16}$
Watch out: 7 outcomes have a 4 on at least one spinner. But that counts pairs that show a 4, not pairs whose product is 4. That mistake gives 7/16, choice D.
So the answer is B, $\frac{3}{16}$ .

Red × Blue	1	2	3	4
1	1	2	3	4
2	2	4	6	8
3	3	6	9	12
4	4	8	12	16

Check your answer

List the ways to multiply two numbers from 1 to 4 and get 4: $1 \times 4$ , $2 \times 2$ and $4 \times 1$ . That is 3 ways. The table has 16 cells, and 13 of them are not 4. So a small chance like 3/16 makes sense.

More questions like this

The original DOE question

Question 29 as shown on the DOE’s digital test, page 102 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 102, question 29.

Question 44 of 57

2022 Test B Math, question 44

Question

Melissa has a bag of marbles that are all the same size. The bag contains three red (R) marbles, three green (G) marbles, and two white (W) marbles. For an experiment, she will take two marbles out of the bag without looking. Which list shows the sample space for Melissa’s experiment?

AR, G, W
BRG, RW, GW
CR, R, R, G, G, G, W, W
DRR, RG, RW, GG, GW, WW

Show the answer

Correct answer: D

The sample space of a random experiment is the collection of all possible outcomes. For this experiment, R represents a red marble, G represents a green marble, and W represents a white marble.

Since two marbles are being drawn, the possible outcomes for the first marble drawn are R, G, W.

List the possible results for each initial outcome combined with the second choice of marble. For example, if the first marble and second marble are both red, the sample space will include RR (red, red).

RR, RG, RW

GR, GG, GW

WR, WG, WW

Since order does not matter in this selection, remove any pairing that presents a duplicate choice, such as RG and GR—only one of the two pairings will be included in the sample space. Those are removed and leave a sample space of:

RR, RG, RW, GG, GW, WW

Question analysis

A bag holds three red, three green and two white marbles, and Melissa will take out two without looking. This probability question asks which list shows the sample space for her experiment. Students must know what a sample space is and what one outcome of drawing two marbles looks like, then decide which color pairs are possible.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Representing the sample space of a compound event is a named grade 7 skill. Nothing is computed here; the demand is knowing what one outcome of drawing two marbles looks like and which pairs the bag allows, much as in grade 7 classwork.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 3, 71 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 tests a definition more than a skill, and three choices describe something real: the colors of one marble, the contents of the bag, and the pairs of different colors. The strongest pull comes from treating the colors as three single objects, which leaves out the pairs of one color although the bag holds three reds, three greens and two whites. Asking what Melissa could be holding after the draw settles it.

Something you must already know moderate Working it out
The term sample space is not defined in the question: students must know it means every possible result of the whole experiment, here a pair of colors drawn together. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
The three mixed pairs, RG, RW and GW, form a complete-looking list that is the right list minus RR, GG and WW, so only asking whether two marbles of one color can come out together separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Describe one possible result of taking out two marbles, then build the list systematically, starting from each color and pairing it with every color that could come out with it.

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

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Lists the colors a single marble can be, the sample space for drawing one marble, though Melissa takes out two.
B Without replacement (Fischbein & Schnarch, 1997)
Reveals a misconception Expected pull: strong
Leaves out pairs of one color, as if drawing a red marble left no red behind, though the bag holds three reds, three greens and two whites.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Lists the eight marbles in the bag, describing what the bag holds rather than the possible results of taking two out.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sample space: $\text{sample space} = \text{the list of every possible outcome}$; An outcome is one possible result of an experiment. The sample space lists every outcome once. For this experiment, one outcome is the pair of colors Melissa holds after she takes out two marbles.

Sort out the question
The bag holds 3 red (R), 3 green (G) and 2 white (W) marbles. Melissa takes out 2 marbles without looking.
You need the sample space: every possible result of taking out 2 marbles. So each outcome must name two colors, like RG for one red and one green.
List pairs by the first color
Start with each color the first marble could be. Then pair it with each color the second could be:
- First red: RR, RG, RW
- First green: GR, GG, GW
- First white: WR, WG, WW
That gives 9 pairs.
Remove pairs that repeat
Melissa ends up holding two marbles. One red and one green is the same result whichever came out first. So GR is the same as RG. Cross out GR, WR and WG. Six pairs are left:
RR, RG, RW, GG, GW, WW
Check the same-color pairs
Can both marbles be the same color? Yes. The bag has 3 reds, 3 greens and 2 whites, so RR, GG and WW can all happen.
Watch out: choice B leaves out RR, GG and WW. That would only be right if the bag had one marble of each color. Choice A is the colors of one marble. Choice C lists the marbles in the bag, not results.
So the answer is D, RR, RG, RW, GG, GW, WW.

Check your answer

Ask what Melissa could be holding. Two reds? Yes, there are 3. Two whites? Yes, there are exactly 2. A green and a white? Yes. Every pair in D can happen, and no possible pair is missing. Choice B misses three pairs that can happen.

More questions like this

The original DOE question

Question 38 as shown on the DOE’s digital test, page 111 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 111, question 38.

Question 45 of 57

2022 Test B Math, question 45

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Reading a figure, table or graph minor
A very close wrong answer minor
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.

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)
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

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Pairs the numbers with the quantities in the order the stem names them, gallons first, so reads (3, 24) as 3 gallons in 24 hours.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes the first coordinate for a rate and pins it to gallons, mixing up one point on the graph with the constant rate, which is 8 gallons an hour.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 24 gallons as the amount added each hour, taking a point’s y-value for the rate, though 24 is the total after 3 hours.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

An ordered pair: $(x, y)$; 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: $\text{rate} = \frac{y}{x}$; 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.

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.
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.
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 $24 \div 3 = 8$ 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, $3 \times 8 = 24$ 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 46 of 57

2022 Test B Math, question 46

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:

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

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

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

$0.21 + 0.09 + 0.04 = 0.34.$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
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.

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)
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

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Adds only the rows for 4 and 5 goals, 0.09 + 0.04 = 0.13, reading three or more as more than three and leaving out the end row.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads only the row for exactly 3 goals, 0.21, overlooking the words “or more” and the two rows below it.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Counts 3 of the 6 rows, for 3, 4 and 5 goals, as half the outcomes, ignoring that the table gives them different probabilities.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding probabilities: $P(A \text{ or } B) = P(A) + P(B)$; 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: $P(\text{not } A) = 1 - P(A)$; 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.

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.
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.
Add the chances
A game can end with only one number of goals. So add the three chances:
$0.21 + 0.09 + 0.04 = 0.34$
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: $0.16 + 0.21 + 0.29 = 0.66$ . Then the chance of 3 or more is $1 - 0.66 = 0.34$ . 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 47 of 57

2022 Test B Math, question 47

Question

This graph shows points that represent a proportional relationship between x and y.

$A coordinate grid with the origin O, the x-axis marked 1 through 10 and the y-axis marked 1 through 10. Points are plotted at (0, 0), (1, 3) and (3, 9).$

What is the constant of proportionality for this relationship?

Show the answer

Correct answer: C

A constant of proportionality, k, is the constant value of the ratio between two proportional values. Two varying values are said to be in a relation of proportionality when their ratio yields a constant. In the graph shown, the proportional relationship between x and y is represented by:

$k=\frac{y}{x}$

Therefore, the constant of proportionality for the relationship on the graph is:

$k=\frac{9}{3}=\frac{3}{1}=3$

Question analysis

A coordinate graph shows three points that represent a proportional relationship between x and y, one of them at the origin. Students choose the constant of proportionality. It tests finding the unit rate of a proportional relationship from a graph, and the choices include coordinates read straight off the points.

NYS standards: 7.RP.2b constant of proportionality
7.RP.2d points on proportional graphs
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Identifying the constant of proportionality on a graph is a named grade 7 skill. With whole-number points on a clear grid, one division settles it, which makes this shorter than most grade 7 proportional reasoning.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 10, 22 words.
Corrected from the DOE handbook: A wrong statement of what makes two quantities proportional was corrected 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

A single division settles it once a point other than the origin is chosen. The point at x = 1 shows the constant directly, and the higher point confirms it. Every wrong choice is a number that can be read straight off the graph, the origin’s 0, a coordinate of 1, or the top point’s 9, so the item rewards dividing rather than reading.

Reading a figure, table or graph minor Understanding the question
The plotted points, (0, 0), (1, 3) and (3, 9), carry no labels, so their coordinates are read off the grid before y is divided by x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The question names the constant of proportionality but not how to find it: k is y ÷ x for any point other than the origin, here 3 ÷ 1 or 9 ÷ 3. (Anderson & Krathwohl, 2001)

Strategy: Pick a plotted point other than the origin, divide its y-coordinate by its x-coordinate, and confirm that another point gives the same quotient.

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes the value at the origin, (0, 0), as the constant, confusing the constant of proportionality with the y-intercept, which is 0 for every proportional graph.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads the x-coordinate of the unit-rate point (1, 3) instead of its y-coordinate, taking the wrong axis’s value as the constant.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Reads the y-value of the point (3, 9) and stops before dividing it by that point’s x-value, 3, to get the ratio.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “Two varying values are said to be in a relation of proportionality when either their ratio or their product yields a constant.” Some general definitions use “proportionality” for both kinds, but a constant product (y = k/x) is inverse proportion, a different relationship whose graph never passes through (0, 0). The proportional relationship this question asks about has a constant ratio y/x, which is how the explanation’s own first sentence defines the constant of proportionality. The site reads “…when their ratio yields a constant.” The answer is unchanged: k = 9/3 = 3/1 = 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; x and y are the two amounts, read from any point (x, y) except (0, 0). k is the constant of proportionality. In a proportional relationship, y divided by x gives the same number at every point. So y = kx, and y is always k times x.

Sort out the question
The graph shows a proportional relationship between x and y. In one, the ratio y ÷ x is the same at every point. A ratio compares two numbers by division. That steady ratio is the constant of proportionality, k. You need k.
Read the points
For each point, read the bottom axis for x and the side axis for y. The points are (0, 0), (1, 3) and (3, 9).
The point (0, 0) is the origin, where the axes meet. You can’t divide by 0, so use the other two points.
Divide y by x
Use the formula with each point.
$k = \frac{y}{x} = \frac{3}{1} = 3$
$k = \frac{y}{x} = \frac{9}{3} = 3$
Both points give 3. So y is always 3 times x.
Give the answer
Watch out: 9 is the y-value of (3, 9) before you divide, choice D. And 1 is the x-value of (1, 3), choice B. And 0, choice A, comes from the origin. Every proportional graph passes through (0, 0), so the origin can’t tell you k. So the answer is C, 3.

Check your answer

Test each point with y = 3x. When x = 1, $y = 3 \times 1 = 3$ . When x = 3, $y = 3 \times 3 = 9$ . When x = 0, y = 0. All three points fit, so k = 3.

More questions like this

The original DOE question

Question 104 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 211, question 104. Open page 211 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 48 of 57

2022 Test B Math, question 48

Question

The distance an eagle flies is proportional to the time the eagle spends flying. An eagle flies 90 miles in 3 hours. What is the constant of proportionality for this relationship?

Show the answer

Correct answer: B

The constant of proportionality is the ratio of y (90) to x (3). $90 \div 3 = 30$ . So the constant of proportionality is 30.

Question analysis

An eagle flies 90 miles in 3 hours, and the distance it flies is proportional to its flying time. This proportional relationship question asks for the constant of proportionality. Students need to know that the constant is the unit rate, here miles per hour, and find it from the two values given.

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students find constants of proportionality in tables, graphs and equations. This one is a single division of miles by hours given in a sentence, the same unit rate work students do in grade 6.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 29 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 division answers the question, so it is among the easiest on the test. The wrong choices are the two numbers in the stem and one unrelated value, so a student who knows what the constant means is not tempted. Checking that the constant times 3 hours gives back the 90 miles confirms it.

Something you must already know minor Working it out
Students must know that the constant of proportionality is the unit rate, distance over time, here the miles flown in a single hour; the stem gives only the term. (Anderson & Krathwohl, 2001)

Strategy: Divide the distance by the time, because the constant of proportionality is the distance flown in one hour, then check by multiplying it by the 3 hours.

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

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the 3 hours printed in the stem as the constant, instead of the miles flown in each hour.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 90 miles and 3 hours leads to 60, which may echo the 60 minutes in an hour, so it mostly attracts guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives the 90 miles, the distance for the whole 3-hour flight, as if the eagle covered it in a single hour.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx \text{, so } k = \frac{y}{x}$; x is one amount, here the time in hours, and y is the other, here the distance in miles. k is the constant of proportionality. It is the same for every pair, and it equals y divided by x. Here it is the miles flown in 1 hour.

Sort out the question
The eagle flies 90 miles in 3 hours. Proportional means the distance is always the same number times the time. That number is the constant of proportionality. You need to find it.
Divide the distance by the time
Use the formula. The distance is y and the time is x:
$k = \frac{90}{3} = 30$
The constant is 30. It means the eagle flies 30 miles in each hour.
Avoid the numbers in the question
Watch out: 3 is the number of hours, choice A. And 90 is the distance for all 3 hours, choice D. Neither is the distance for 1 hour.
So the answer is B, 30.

Check your answer

Multiply back. At 30 miles each hour, the eagle flies $30 \times 3 = 90$ miles in 3 hours. That matches the question. Hour by hour, it goes 30, 60, then 90 miles.

More questions like this

The original DOE question

Question 41 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 D Math, page 114, question 41.

Question 49 of 57

2022 Test B Math, question 49

Question

A machine part is supposed to be 0.38 millimeters thick. The manufacturer allows a variation of −2.5% to +2.5%. Which interval shows the range of acceptable thicknesses, in millimeters, for the part?

A0.285–0.475
B0.355–0.405
C0.3705–0.3895
D0.3724–0.3876

Show the answer

Correct answer: C

A variation of −2.5% to +2.5% means that the thickness of the machine part can be from 2.5% less than the standard of 0.38 mm to 2.5% greater than 0.38 mm.

The least allowed measure is

$0.38-0.025(0.38)=0.3705.$

The greatest allowed measure is

$0.38+0.025(0.38)=0.3895.$

Therefore, the allowable range of thicknesses is 0.3705 mm to 0.3895 mm.

Question analysis

A machine part should be 0.38 millimeters thick, and the manufacturer accepts a variation from −2.5% to +2.5%. Students choose the interval of acceptable thicknesses. It tests finding a percent of a decimal and applying it as a decrease and an increase; the percent must be turned into millimeters first, and the four-decimal-place choices need exact work.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking a percent of a number and adding or subtracting it is grade 7 percent work. The decimal percent and results to four decimal places make the arithmetic fussier than usual, so careful work matters more than the idea.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 29 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 turning the percent into millimeters before anything else. Students who take ±2.5% as ±0.025 mm find a tidy interval, 0.355 to 0.405, and the other listed intervals come from a misplaced decimal point or from 2% in place of 2.5%. Because the right interval and its near twin differ only from the third decimal place, the product 0.38 × 0.025 has to be exact; multiplying 0.38 by 0.975 and by 1.025 is a quick double-check.

Error-prone arithmetic moderate Working it out
Two and a half percent of 0.38 is 0.025 × 0.38 = 0.0095, a product of two decimals with four decimal places, and the limits 0.3705 and 0.3895 need the same care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
“A variation of −2.5% to +2.5%” has to become 2.5% of 0.38 millimeters, a length, before it is subtracted for the least thickness and added for the greatest. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The intervals 0.3705–0.3895 and 0.3724–0.3876 both sit about 0.01 mm either side of 0.38, so an estimate cannot separate them and only the exact 0.0095 decides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write 2.5% as a decimal, multiply it by 0.38 to find the allowed change in millimeters, then subtract it for the least thickness and add it for the greatest.

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.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Writes 2.5% as 0.25 instead of 0.025, so the change is 0.38 × 0.25 = 0.095 mm and the interval, 0.285 to 0.475, is ten times too wide.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reads ±2.5% as ±0.025 mm, a fixed amount, adding it to and subtracting it from 0.38 instead of first taking 2.5% of 0.38.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Uses 2% instead of 2.5%, dropping the half percent, so the change is 0.38 × 0.02 = 0.0076 and the interval, 0.3724 to 0.3876, is too narrow.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get. Dividing the percent by 100 turns it into a decimal, so you can multiply.

Sort out the question
The part should be 0.38 millimeters thick. It may be off by up to 2.5%, thinner or thicker. You need the thinnest and thickest sizes allowed, in millimeters.
The 2.5% is a percent of 0.38 mm. So first find how many millimeters 2.5% is.
Write 2.5% as a decimal
Percent means out of 100. So divide by 100, which moves the decimal point two places left.
$2.5\% = \frac{2.5}{100} = 0.025$
Watch out: 2.5% is 0.025, not 0.25. Using 0.25 makes the interval ten times too wide, choice A.
Find 2.5% of 0.38
$\text{part} = 0.025 \times 0.38 = 0.0095$
First, 25 × 38 = 950. The two numbers have 3 + 2 = 5 decimal places in all. So the answer has 5 decimal places: 0.00950, or 0.0095. The size may change by 0.0095 millimeters.
Watch out: adding and subtracting 0.025 mm itself gives 0.355 to 0.405, choice B.
Find the two limits
Subtract the change for the thinnest size. Add it for the thickest.
$0.38 - 0.0095 = 0.3705$
$0.38 + 0.0095 = 0.3895$
The part may be from 0.3705 to 0.3895 millimeters thick.
Watch out: using 2% instead of 2.5% gives 0.3724 to 0.3876, choice D. So the answer is C, 0.3705–0.3895.

Check your answer

A part 2.5% thinner is 97.5% of 0.38: $0.975 \times 0.38 = 0.3705$ . A part 2.5% thicker is 102.5% of it: $1.025 \times 0.38 = 0.3895$ . Both match. Also, 0.38 sits exactly in the middle of the interval.

More questions like this

The original DOE question

Question 106 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 211, question 106. Open page 211 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 57

2022 Test B Math, question 50

Question

A bakery charges $1.25 for each cupcake and $2.50 for a serving tray. The equation describes p, the price of c cupcakes and a serving tray, not including tax.

$p = 2.50 + 1.25c$

If Samantha bought some cupcakes and a serving tray for a total of $22.50, not including tax, how many cupcakes did she buy?

Show the answer

Correct answer: C

Using the specified equation $p = 2.50 + 1.25c$ , determine the number of cupcakes Samantha bought by substituting the total paid ($22.50) for p, and solve the equation for the value of c, the number of cupcakes bought.

$22.50 = 2.50 + 1.25c$

$22.50 - 2.50 = 1.25c$

$20.00 \div 1.25 = c$

$16 = c$

Question analysis

A bakery's price for c cupcakes and a serving tray is given by the equation p = 2.50 + 1.25c, and Samantha paid $22.50 before tax. This equation question asks how many cupcakes she bought. Students substitute the total and solve a two-step equation with decimal money amounts, undoing the steps in the right order.

NYS standards: 7.EE.4a two-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step equation of the form px + q = r in a money context is core grade 7 algebra. The equation is given, so the work is substituting the total and dividing by a decimal, a routine example.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 50 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 equation is supplied, so the question is a two-step solve with money. The tray’s $2.50 has to come off the total before dividing, and students who divide the whole $22.50 by $1.25 get a whole number that matches a choice, so nothing warns them. Substituting the chosen number back into p = 2.50 + 1.25c is a quick, decisive check.

Error-prone arithmetic minor Working it out
After subtracting the tray, $20 must be divided by $1.25, a decimal divisor that is easiest handled as 2,000 ÷ 125 or by noting that 4 cupcakes cost $5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute the total for p, subtract the tray's cost from both sides, then divide by the price of one cupcake; check by putting your number back into the equation.

Traps

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

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Notes that 4 cupcakes cost $5.00 and that the $20 left after the tray holds four such amounts, then gives the 4 groups instead of 16 cupcakes.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the $22.50 total by the $2.50 tray price, combining two numbers from the question without using the equation.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole $22.50 by $1.25 a cupcake, forgetting that $2.50 of the total paid for the serving tray.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } a - n = b - n \text{ and } \frac{a}{n} = \frac{b}{n}$; a and b are the two sides of an equation, and n is any number (not zero when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it level. That is how you get c, the number of cupcakes, by itself.

Sort out the question
The equation $p = 2.50 + 1.25c$ gives the price. p is the price in dollars, and c is the number of cupcakes. The 2.50 is the tray, and 1.25 is the price of one cupcake. Samantha paid $22.50. You need c.
Put in the total
Samantha paid $22.50 in all, so put 22.50 in place of p. Now the equation has only one unknown, c:
$22.50 = 2.50 + 1.25c$
Subtract the tray
Take 2.50 away from both sides. That removes the cost of the tray:
$22.50 - 2.50 = 1.25c$
$20.00 = 1.25c$
So $20.00 went to cupcakes alone.
Watch out: dividing all of $22.50 by $1.25 gives 18, choice D. That forgets the tray.
Divide by the price of one cupcake
Divide both sides by 1.25:
$c = 20.00 \div 1.25 = 16$
An easy way: 4 cupcakes cost $4 \times 1.25 = 5.00$ dollars. $20 is 4 groups of $5. So Samantha bought $4 \times 4 = 16$ cupcakes.
Watch out: 4 is the number of $5 groups, not cupcakes. That is choice A.
So the answer is C, 16.

Check your answer

Put 16 back into the equation. $1.25 \times 16 = 20.00$ , and $2.50 + 20.00 = 22.50$ . That is what Samantha paid, so 16 is right.

More questions like this

The original DOE question

Question 44 as shown on the DOE’s digital test, page 117 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 117, question 44.

Question 51 of 57

2022 Test B Math, question 51

Question

A clothing store buys shirts for a cost of m dollars each. To determine the selling price of the shirts, the manager increases the purchase cost by 55%. Which expression represents the selling price of each shirt?

A $0.45m$
B $0.55m$
C $1.55m$
D $m+0.55$

Show the answer

Correct answer: C

It is given that m represents the purchase cost.

$1m$ represents 100%. The amount the purchase cost was increased is represented by 55%.

Therefore, the selling price will be $(100\%+55\%)m$ .

$155\%=\frac{155}{100}=1.55$

The selling price can be represented by the expression $1.55m$ .

Question analysis

A clothing store buys shirts for m dollars each, and the manager sets the selling price by increasing that purchase cost by 55%. Students choose the expression for the selling price of each shirt. It tests writing a percent increase as an algebraic expression, where a percent of an unknown cost has to be handled in letters.

NYS standards: 7.EE.2 rewriting expressions to show relationships
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing a percent markup as an expression is a named grade 7 idea and a common classroom task. This is a direct case with one percent and a letter for the cost, with no further steps such as tax.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers minor
Solution: 2 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

The markup is an increase by 55%, so the selling price is the whole cost plus 55% of it. The commonest mistake writes only the 55% part, which is the markup and not the price. Others subtract the percent as a discount would, or add 55 cents. Choosing a cost of $100 and finding a $155 price picks out the right expression at once.

Turning words into math moderate Understanding the question
“Increases the purchase cost by 55%” means the cost plus 55% of it, 100% + 55% = 155% of m, not 55% of m on its own. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
The cost is the letter m, so the markup has to be written as 0.55m and added to m, with no dollar amount to test the expression against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Pick an easy cost such as m = 100 dollars, work out the selling price after a 55% increase, and see which expression gives that price.

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.
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

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: weak
Subtracts the 55% from 100% instead of adding it, writing 0.45m, which is the price after a 55% discount rather than a 55% increase.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Writes only the amount of the markup, 55% of m, and stops before adding it to the cost the store paid for the shirt.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds 0.55 to m as if the 55% markup were 55 cents, a fixed amount, instead of 55% of whatever the shirt cost.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new amount} = \text{old amount} \times \left(1 + \frac{\text{percent}}{100}\right)$; The old amount is what you start with, here the cost m. The 1 keeps the whole old amount. The percent over 100 adds the increase on top. Use it whenever an amount goes up by a percent.

Sort out the question
Each shirt costs the store m dollars. The manager raises that cost by 55% to get the selling price. You need an expression for the selling price. An expression is a math phrase with numbers, letters and signs, but no equal sign.
Find the amount of the increase
The increase is 55% of m. Write 55% as a decimal by dividing by 100.
$55\% = \frac{55}{100} = 0.55$
So the increase is 0.55m dollars, which means 0.55 times m.
Watch out: 0.55m is only the increase. Stopping here gives choice B.
Add the increase to the cost
The selling price is the cost plus the increase. Since m is the same as 1m, add the numbers in front.
$m + 0.55m = 1m + 0.55m = 1.55m$
The percent increase formula gives the same thing:
$\text{new amount} = m \times \left(1 + \frac{55}{100}\right) = m \times 1.55 = 1.55m$
Give the answer
Watch out: 0.45m is a 55% decrease, choice A. And m + 0.55 adds only 55 cents, choice D. The increase must grow when the cost grows. So the answer is C, $1.55m$ .

Check your answer

Pick a cost of $100, so m = 100. A 55% increase adds $55, for a price of $155. Now try choice C: $1.55 \times 100 = 155$ . It matches. Choice B gives only 55, and choice D gives 100.55.

More questions like this

The original DOE question

Question 108 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 2023 Admissions, page 212, question 108. Open page 212 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 52 of 57

2022 Test B Math, question 52

Question

A farmer has 100 apples and will eat 2 apples per day. Is there a proportional relationship between the number of apples remaining and the number of days that have passed?

Ano, because the relationship does not include 99 as a number of apples
Bno, because the ordered pair (0, 0) is not part of the relationship
Cyes, because all the values in the relationship are even
Dyes, because the number of apples is decreasing at a constant rate

Show the answer

Correct answer: B

Proportional relationships are relationships between two variables where their ratios are equivalent. The graph of a proportional relationship will always go through the point (0, 0).

In the text of the problem, it is given that the farmer begins with 100 apples. Therefore, the ordered pair (0, 100) would be the point at which the graph of the relationship crosses the y-axis, not (0, 0). Additionally, in a proportional relationship, each pair of values represents the same ratio of $\frac{y}{x}$ .

If the relationship were proportional, the ratio for the first two days (1, 98) and (2, 96) would be equivalent.

$\frac{98\text{ apples}}{1\text{ day}}=\frac{96\text{ apples}}{2\text{ days}}$

$2(98)\neq 96$

Question analysis

A farmer starts with 100 apples and eats 2 apples per day. Students decide whether the number of apples remaining and the number of days that have passed are in a proportional relationship, choosing a yes or no answer with its reason. It tests the definition of a proportional relationship, and every reason offered has to be judged, not just the verdict.

NYS standards: 7.RP.2a testing for proportionality
7.RP.2d points on proportional graphs
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Deciding whether two quantities are proportional is core grade 7 work. Here there is no table or graph to check, only a description, so students must produce pairs of values themselves before testing them.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Something you must already know moderate
Every choice must be checked moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 74 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The situation is clearly linear, a steady drop of 2 apples a day, and that is the trap: a constant rate of change sounds like proportionality. The relationship starts at 100 apples on day 0, not at 0, so the ratio of apples to days falls from 98 to 48 between days 1 and 2. Half the choices say no, and only one of those gives a reason that decides the question.

Something you must already know moderate Working it out
The tests for a proportional relationship are not given: apples over days must be one constant ratio, and the graph must pass through (0, 0), which day 0 with 100 apples does not. (Anderson & Krathwohl, 2001)
Every choice must be checked moderate Choosing the answer
Each choice pairs a verdict with a reason, so a right “no” with a wrong reason, the missing 99, has to be ruled out as firmly as a wrong “yes”. (Embretson & Wetzel, 1987)
Turning words into math minor Understanding the question
No table or graph is given, so the pairs have to be built from the words: 100 apples on day 0, 98 on day 1, 96 on day 2. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: List the apples left after 0, 1 and 2 days, recall the two tests for a proportional relationship, and keep only a choice whose reason is both true and decisive.

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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals partial knowledge Expected pull: weak
Has the right verdict, no, but rests it on a true detail that decides nothing: the count skips 99, yet missing values have no bearing on whether ratios are equal.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a surface strategy Expected pull: weak
Treats the even counts, 100, 98, 96 and so on, as a sign of proportionality, a pattern in the numbers that has nothing to do with a constant ratio.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Takes a constant rate of change for proportionality: the apples do fall by 2 each day, but they start from 100, not 0, so apples ÷ days is not constant.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx$; x and y are two amounts that change together. k is a fixed number, the constant of proportionality. In a proportional relationship, y ÷ x always equals k. When x is 0, y is 0 too, so the graph passes through (0, 0).

Sort out the question
The farmer starts with 100 apples and eats 2 apples each day. You must decide if apples left and days passed are proportional. Each choice gives a yes or a no and a reason. Both parts must be right.
Make a small table
Each day, take away 2 apples.
Days passed 0 1 2
Apples left 100 98 96
As pairs (days, apples), these are (0, 100), (1, 98) and (2, 96).
Test the ratios
If the relationship were proportional, apples ÷ days would be the same every day. Are $\frac{98}{1}$ and $\frac{96}{2}$ equal? Work out each one.
$\frac{98}{1} = 98$
$\frac{96}{2} = 48$
98 is not equal to 48. The ratio changes, so the relationship is not proportional.
Check the point (0, 0)
A proportional graph always passes through (0, 0). Here, on day 0 the farmer has 100 apples, not 0. So that pair is (0, 100). When the apples run out, 50 days have passed: (50, 0). The pair (0, 0) never happens. That is the reason in choice B.
Rule out the other choices
Watch out: choice D is tempting. The apples do drop at a constant rate, 2 each day. But they start at 100, not 0. A steady change alone is not enough.
Choice A is true, since 99 never appears, but it decides nothing. Choice C is about even numbers, which has nothing to do with a steady ratio. So the answer is B.

Days passed	0	1	2
Apples left	100	98	96

Check your answer

Try one more day. After 10 days the farmer has $100 - 2 \times 10 = 80$ apples. Then apples ÷ days is 80 ÷ 10 = 8, not 98 or 48. The ratio keeps changing, so the answer is no, and (0, 0) is missing.

More questions like this

The original DOE question

Question 109 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 2023 Admissions, page 212, question 109. Open page 212 of the handbook (PDF)

Question 53 of 57

2022 Test B Math, question 53

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
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.

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)
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

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Expects a slice to cut off a corner and make a triangle, as a cut through a pyramid might, not a cut parallel to this square base.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Describes a vertical slice, which runs down the 6-inch height and makes a 2-by-6 rectangle, mixing up horizontal and vertical.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Builds a triangle from the labels in the figure, one 2-inch and two 6-inch sides, a shape no horizontal cut of this prism can make.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A slice parallel to the 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.

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.
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.
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.
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 54 of 57

2022 Test B Math, question 54

Question

Anise has a piece of cloth measuring 45 inches by 75 inches. She wants to cut it into squares of equal size, without any cloth left over. What is the greatest possible side length, in inches, of one of those squares?

Show the answer

Correct answer: C

Given that each piece will be a square, the length and width will be the same. The greatest possible length can be determined by finding the greatest common factor of the length and width of the original piece of cloth.

Begin by finding the prime factorization of the length (75 in.) and width (45 in.).

$45 = 3 \times 3 \times 5$

$75 = 3 \times 5 \times 5$

The common factors can be seen as $3 \times 5$ .

The greatest possible side length, in inches, will be $3 \times 5 = 15$ .

Question analysis

Anise wants to cut a 45-inch by 75-inch piece of cloth into equal squares with nothing left over. This factors question asks for the greatest possible side length of each square, in inches. Students must see that the side has to divide both lengths evenly, then find the largest number that does.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Greatest common factors of two numbers up to 100 are grade 6 content. The cloth setting asks students to recognize that this is what is needed, but the work is one short factor search, below typical grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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

Once the side is seen as a common factor, the work is short, since 45 and 75 share both 3 and 5. Students who spot one shared factor and stop reach a listed choice, and those who think only of the largest square that fits use the full 45-inch width. Checking that 45 and 75 both divide by the chosen side, and that no larger number does, closes it.

Turning words into math minor Understanding the question
The words “squares of equal size, without any cloth left over” mean the side must divide both 45 and 75 exactly; nothing names the greatest common factor. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding the greatest common factor of 45 and 75 calls for prime factorization or a list of factors, a grade 6 skill the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Write each length as a product of primes, multiply the primes they share, and check that both 45 and 75 divide evenly by the result.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds a factor that 45 and 75 share, 3, and stops without combining it with the 5 they also share.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Spots the common factor 5, since both lengths end in 5, and stops before multiplying it by the 3 they also share.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Uses the cloth’s full width, 45 inches, as the side, the largest square that fits, but 45 does not divide 75, leaving a 45-by-30 strip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor: $\text{GCF} = \text{product of the prime factors both numbers share}$; A factor divides a number evenly. The greatest common factor, or GCF, is the biggest number that divides both. Write each number as a product of primes, numbers whose only factors are 1 and themselves. Then multiply the primes they share.

Sort out the question
The cloth is 45 inches by 75 inches. It is cut into equal squares, with no cloth left over. You need the greatest side the squares can have.
No cloth left over means the side must fit evenly into 45 and into 75. So the side is a common factor of both. The greatest one is the GCF.
Break each number into primes
Write each length as a product of prime numbers:
$45 = 3 \times 3 \times 5$
$75 = 3 \times 5 \times 5$
The two lists share one 3 and one 5.
Multiply the shared primes
Multiply the primes the two numbers share:
$\text{GCF} = 3 \times 5 = 15$
The squares can be 15 inches on a side.
Watch out: 3 and 5 are each common factors, choices A and B. But they are not the greatest. Use every prime the numbers share.
Rule out the full width
Watch out: a 45-inch square, choice D, fits the width. But 75 ÷ 45 = 1 with 30 left over. That leaves a strip of cloth 45 inches by 30 inches.
So the answer is C, 15.

Check your answer

Divide each length by 15: $45 \div 15 = 3$ and $75 \div 15 = 5$ . So the cloth cuts into 3 rows of 5 squares, with nothing left. Both divisions come out even.

More questions like this

The original DOE question

Question 49 as shown on the DOE’s digital test, page 122 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 122, question 49.

Question 55 of 57

2022 Test B Math, question 55

Question

This table shows the number of birds that came to a bird feeder on Monday and Tuesday.

BIRDS AT BIRD FEEDER

Day	Number of Birds
Monday	30
Tuesday	24

What was the percent decrease in the number of birds from Monday to Tuesday?

A6%
B20%
C24%
D25%

Show the answer

Correct answer: B

To determine the percent decrease from Monday to Tuesday, first calculate the decrease in the number of birds at the feeder.

$30 - 24 = 6$

The difference in the number of birds at the feeder divided by the number of birds on Monday will yield the percent decrease from Monday to Tuesday.

$\frac{6}{30} = 0.20$

0.20 represents 20%.

There was a 20% decrease from Monday to Tuesday.

Question analysis

A table shows that 30 birds came to a bird feeder on Monday and 24 on Tuesday. This percent change question asks for the percent decrease from Monday to Tuesday. Students find the change and compare it with one day's count, and choosing the right day as the base decides the result.

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 decrease is core grade 7 work, and this is a plain example: one subtraction and one division with friendly numbers. The only real demand is measuring the change against the right day's count.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 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 subtraction is trivial, so the question is about the base. Both counts sit in the table, and dividing by the later one gives an equally neat percent that is easy to accept. Saying “a decrease from Monday” and putting Monday’s 30 under the change avoids it, and checking that 20% of 30 is 6 confirms the result.

A very close wrong answer minor Choosing the answer
Dividing the drop of 6 by Monday’s 30 gives 20% and by Tuesday’s 24 gives 25%, two round, listed percents only 5 points apart, so only the choice of base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that a percent decrease is the change divided by the starting amount, here Monday’s count, a rule the question does not state. (Anderson & Krathwohl, 2001)

Strategy: Subtract to find how many fewer birds came on Tuesday, divide by Monday's count, the starting amount, and change the fraction to a percent.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the decrease of 6 birds and reads it as 6 percent, as if the count had dropped from 100 rather than from 30.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Reads Tuesday’s 24 birds as a percent, taking a number straight from the table without working out the change at all.
D Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the decrease of 6 by Tuesday’s 24 birds instead of Monday’s 30, measuring the change from the later day: 6/24 = 25%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent decrease: $\text{percent decrease} = \frac{\text{old} - \text{new}}{\text{old}} \times 100$; Old is the starting amount, here Monday’s count. New is the later amount, Tuesday’s count. The top is how much it dropped. You divide by the starting amount because a change is measured from where you started.

Sort out the question
The table shows 30 birds on Monday and 24 birds on Tuesday. You need the percent decrease from Monday to Tuesday. From Monday means Monday is the starting amount.
Find the drop
Subtract Tuesday’s count from Monday’s:
$30 - 24 = 6$
There were 6 fewer birds.
Watch out: 6 is a number of birds, not a percent. Calling it 6% leads to choice A.
Compare the drop with Monday
Use the formula. Divide the drop by Monday’s count, then multiply by 100:
$\frac{6}{30} \times 100 = 0.20 \times 100 = 20$
The number of birds went down by 20%.
Watch out: dividing by Tuesday’s 24 gives $\frac{6}{24} = 0.25$ , or 25%, choice D. Measure the change from the starting day. Choice C, 24%, just copies a number from the table.
So the answer is B, 20%.

Check your answer

Take 20% of Monday’s count: $0.20 \times 30 = 6$ . Then $30 - 6 = 24$ , which is Tuesday’s count. The answer checks out.

More questions like this

The original DOE question

Question 50 as shown on the DOE’s digital test, page 123 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 123, question 50.

Question 56 of 57

2022 Test B Math, question 56

Question

A team is participating in a contest to collect canned goods for a food drive. The team has 53 points and must have a total of at least 100 points to win a prize. The team earns 5 points for each can of food a member collects. Which inequality can be used to find x, the number of cans of food the team needs to collect to win a prize?

A $5 + 53x \leq 100$
B $5 + 53x \geq 100$
C $53 + 5x \leq 100$
D $53 + 5x \geq 100$

Show the answer

Correct answer: D

Since each can is worth 5 points, the total number of points earned from all the cans, x, is 5x. The team started with 53 points. That is a constant value and doesn’t change. So the total number of points the team will earn is $53 + 5x$ . The team wants to earn at least (≥) 100 points, so they need $53 + 5x \geq 100$ total points.

Question analysis

A team has 53 points in a canned-food drive, earns 5 points for each can, and needs at least 100 points to win a prize. Students choose the inequality for x, the number of cans the team must collect. It tests writing an inequality from words, where the starting points and the points per can must each take the right place.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an inequality for a word problem with a starting amount and a rate is grade 7 work. The four choices pair two expressions with two symbols, so students make one decision about the terms and one about the phrase "at least".
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 71 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 choices form a two-by-two grid: which number goes with x, and which way the symbol points. Each decision can be made separately, so a student who settles both gets the item, while one mistake lands on a listed choice. Testing two numbers of cans, 5, which leaves the team at 78 points, and 10, which lifts it to 103, rules out every wrong inequality.

Turning words into math moderate Understanding the question
The 53 points the team “has” are a fixed start, the “5 points for each can” go with x, and “at least 100” calls for ≥, though “least” can sound like less. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The four choices are two expressions, 5 + 53x and 53 + 5x, each with ≤ and with ≥, so the right one and the strongest wrong one differ only in the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the team's total points after 2 cans and then after x cans, translate "at least 100" into a symbol, and compare the result with each choice.

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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Attaches x to the 53 points the team already has and also reverses “at least”, combining the mistakes behind the other two wrong choices.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Keeps the right symbol but multiplies the number of cans by 53 and adds 5, putting the starting points where the points per can belong.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Writes the points correctly as 53 + 5x but reads “at least 100” as at most 100, reversing the inequality.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total with a start and a rate: $\text{total} = \text{start} + \text{rate} \times x$; The start is the amount you have now, here 53 points. The rate is what each item adds, here 5 points for each can. x is the number of items. The start is added once, and the rate is multiplied by x.
At least: $\text{total} \geq \text{goal}$; The sign ≥ means is greater than or equal to. At least 100 means 100 or more. So the total must be greater than or equal to 100.

Sort out the question
You know:
- The team already has 53 points.
- It earns 5 points for each can.
- It needs at least 100 points to win.
x is the number of cans. You need the inequality that shows this.
Write the total points
Try 2 cans first. The team gets 5 points for each, so $53 + 5 \times 2 = 63$ points. For x cans, the total is:
$53 + 5x$
Watch out: the 53 points do not repeat for each can. $5 + 53x$ gives 53 points for each can, as in choices A and B.
Choose the sign
At least 100 means 100 or more, so use ≥:
$53 + 5x \geq 100$
Watch out: “least” sounds like “less,” but at least means the total can be 100 or higher. Using ≤ gives choice C.
So the answer is D, $53 + 5x \geq 100$ .

Check your answer

Test two numbers of cans. With 10 cans, $53 + 5 \times 10 = 103$ , and 103 ≥ 100 is true. The team wins. With 9 cans, $53 + 5 \times 9 = 98$ , and 98 ≥ 100 is false. The team is short. Choice D gets both right.

More questions like this

The original DOE question

Question 40 as shown on the DOE’s digital test, page 113 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 113, question 40.

Question 57 of 57

2022 Test B Math, question 57

Question

A store manager purchases a piece of furniture for $456.00. To determine the selling price, the manager increases the purchase cost by 125%. A customer buys the furniture and pays an additional 8% sales tax. How much does the customer pay for the furniture?

A$615.60
B$627.48
C$1,108.08
D$1,846.80

Show the answer

Correct answer: C

Given a purchase price of $456.00, the 125% increase in price is calculated as:

$456 \times 1.25 = 570$

The selling price equals the purchase price plus the increase, or:

$456 + 570 = 1{,}026$

The sales tax on the selling price is:

$1{,}026(0.08) = 82.08$

The customer pays the selling price plus the sales tax:

$1{,}026 + 82.08 = 1{,}108.08$

Question analysis

A store manager buys a piece of furniture for $456.00 and raises the price by 125% to set the selling price, and the customer also pays 8% sales tax. Students find the total the customer pays. It tests percent increase and sales tax in sequence, and reading an increase of more than 100% correctly is the main challenge.

NYS standards: 7.RP.3 multistep ratio and percent problems
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 find markups and sales tax, usually with increases under 100%. An increase of more than 100% is unusual and easy to misread, and two percent changes must be applied in order, which goes beyond routine grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps moderate
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 41 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 percent above 100 is unusual in grade 7 work, and it is the center of the question: the markup more than doubles the price. Students who treat “by 125%” as “125% of” still find a listed answer after adding tax, and so do students who add $125. A rough check helps: the selling price must be more than twice $456, about $1,000, and the tax adds less than a tenth of that.

Turning words into math moderate Understanding the question
“Increases the purchase cost by 125%” adds 125% of $456 to the $456 itself, a multiplier of 2.25; reading it as 125% of the cost leaves the original price out. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Four steps in a row, the increase, the selling price, the tax and the total, each depend on the one before, and two of them are easy to merge or skip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Taking 8% of $1,026 means multiplying by 0.08, and a decimal point one place off, 0.8, makes the tax ten times too large. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn each percent change into a multiplier, remembering that an increase by a percent is added to the original amount, apply the markup and then the tax, and estimate to check the size.

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.
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

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the increase, 125% of $456 = $570, and uses it as the selling price without adding back the $456 cost, then adds 8% tax.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 125% increase as $125, a fixed amount, so the selling price is $581, and then adds 8% tax correctly to reach $627.48.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds the $1,026 selling price correctly but writes 8% as 0.8, adding $820.80 in tax instead of $82.08.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $p\% \text{ of } n = \frac{p}{100} \times n$; n is the starting amount, and p is the percent. Percent means out of 100, so 125% is 125/100 = 1.25 and 8% is 8/100 = 0.08. Multiply n by that decimal to find the part.
Price after an increase: $\text{new price} = \text{old price} + \text{increase}$; The old price is what you start with. The increase is the amount added on. To increase by a percent, find that percent of the old price, then add it to the old price. Sales tax is added the same way.

Sort out the question
The manager pays $456.00 for the furniture. The selling price is that cost increased by 125%. The customer then pays 8% sales tax on top. Sales tax is an extra charge, a percent of the price. You need the total the customer pays.
Find the increase
Find 125% of $456. As a decimal, 125% is 1.25, which is 1 and a quarter. A quarter of 456 is 114:
$456 \times 1.25 = 456 + 114 = 570$
The increase is $570.
Find the selling price
Increases by means the increase is added to the cost:
$456 + 570 = 1{,}026$
The selling price is $1,026.
Watch out: $570 is only the increase. Using it as the price gives $615.60, choice A. And 125% is a percent, not $125. Adding $125 leads to $627.48, choice B.
Find the sales tax
As a decimal, 8% is 0.08. Find 8% of the selling price:
$1{,}026 \times 0.08 = 82.08$
The tax is $82.08.
Watch out: 8% is 0.08, not 0.8. Using 0.8 makes the tax ten times too big and gives $1,846.80, choice D.
Add the tax to the price
The customer pays the selling price plus the tax:
$1{,}026 + 82.08 = 1{,}108.08$
So the answer is C, $1,108.08.

Check your answer

Estimate. The price more than doubles: $456 plus $570 is about $1,000. Tax of 8% adds about $80. So the total should be a bit over $1,080, and only $1,108.08 is close. Or multiply in one go: 456 × 2.25 × 1.08 = 1,108.08.

More questions like this

The original DOE question

Question 19 as shown on the DOE’s digital test, page 92 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 92, question 19.