Practice Test

2022 Handbook

2022 Test A Math

Question 1 of 57

2022 Test A Math, question 1

Question

The figure below shows a circle with a hole in the middle. The outer diameter of the circle is $1\frac{1}{4}$ inches. The diameter of the hole is $\frac{1}{4}$ inch.

$A large circle shaded gray, with a smaller white circle, the hole, at its center. A double-headed arrow across the hole is labeled 1/4 in., and a double-headed arrow below the figure, spanning the full width of the large circle, is labeled 1 1/4 in. The shaded region is the part of the large circle outside the hole.$

What is the area of the shaded region, to the nearest hundredth of a square inch?

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

Show the answer

Correct answer: 1.18

To determine the area of a circle, begin by finding the radius of the circle, which is half of the diameter.

The radius of the entire circle is $1\frac{1}{4}\text{ in.} \div 2 = \frac{5}{8}\text{ in.}$ , so the area of the entire circle is $\pi\left(\frac{5}{8}\right)^{2}\text{ in.}^{2} = \frac{25}{64}\pi\text{ in.}^{2}$

The radius of the hole in the circle is $\frac{1}{4}\text{ in.} \div 2 = \frac{1}{8}\text{ in.}$ , so the area of the hole in the circle = $\pi\left(\frac{1}{8}\right)^{2}\text{ in.}^{2} = \frac{1}{64}\pi\text{ in.}^{2}$

To determine the area of the shaded region, subtract the area of the hole in the circle from the area of the entire circle:

$\left(\frac{25}{64}\right)\pi\text{ in.}^{2} - \left(\frac{1}{64}\right)\pi\text{ in.}^{2} =$

$\left(\frac{24}{64}\right)\pi\text{ in.}^{2} =$

$\left(\frac{3}{8}\right)\pi\text{ in.}^{2} =$

$\left(\frac{3}{8}\right)\text{ in.}^{2}(3.14) =$

$1.1775\text{ in.}^{2}$

1.1775 square inches rounded to the nearest hundredth of a square inch = 1.18 square inches

Question analysis

A gray ring is a circle 1 1/4 inches across with a hole 1/4 inch across at its center, and this grid-in asks for the area of the shaded part to the nearest hundredth of a square inch. Students use the area of a circle twice. Turning diameters given as fractions into radii, and squaring them, is where mistakes happen.

NYS standards: 7.G.4 circle area and circumference
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students find circle areas and often subtract a hole from a shape. This ring gives its diameters as a mixed number and a fraction, so halving, squaring fractions and rounding add work, though no step goes beyond grade 7.
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
Error-prone arithmetic moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 45 words.
Corrected from the DOE handbook: A missing exponent in the explanation’s area formula was restored.
An arithmetic slip in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is the familiar one for a ring, the big circle's area minus the hole's, but the labels are diameters given as fractions, so each one has to be halved and then squared before π enters. Students lose points by skipping the halving, by stopping at the full disk, or by subtracting the radii before squaring. Keeping both areas as fractions of π, 25/64π − 1/64π = 3/8π, leaves one multiplication and one rounding at the very end.

Something you must already know moderate Working it out
Neither area formula is printed, and both circles are labeled by diameter, so students must recall A = πr² and halve 1 1/4 inches and 1/4 inch to radii of 5/8 and 1/8 first. (Anderson & Krathwohl, 2001)
Error-prone arithmetic moderate Working it out
Every value is a fraction: 5/4 ÷ 2 = 5/8, squares of 25/64 and 1/64, then 3/8 × 3.14 worked by hand, so a slip can enter at each stage. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Combines ideas minor Working it out
The gray ring has no formula of its own; it is the large circle's area minus the hole's, so the area formula is used twice and the results subtracted. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
The grid-in wants the nearest hundredth, so 1.1775 must be rounded to 1.18, and a whole-disk area such as 1.23 gets no warning from a list of choices. (Enright, Morley & Sheehan, 2002)

Strategy: Write 1 1/4 as 5/4 and halve both diameters to get the radii, subtract the hole's area from the area of the full circle, and round to the nearest hundredth only at the very end.

Traps

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

The wrong answers, classified

4.71 Wrong formula (King et al., 2004)
Reveals a slip Expected pull: strong
Uses the labeled diameters as radii, π(5/4)² − π(1/4)² = 24/16π = 1.5π ≈ 4.71, forgetting to halve them first, which makes the ring four times too big.
1.23 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the area of the whole circle, π(5/8)² = 25/64π ≈ 1.23 square inches, and stops before subtracting the hole's 1/64π.
0.79 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts the radii first, 5/8 − 1/8 = 1/2, then squares, π(1/2)² ≈ 0.79, as if the difference of two squares were the square of the difference.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes the area of the whole circle as “π(5/8) in.² = 25/64π in.²” and the area of the hole as “π(1/8) in.² = 1/64π in.²” (p. 128, question 58), squaring the unit but not the radius. The area of a circle is πr², and π(5/8) is only 5/8π. The site now reads π(5/8)² in.² = 25/64π in.² and π(1/8)² in.² = 1/64π in.². The areas that follow and the answer, 1.18, are unchanged.
Explanation: The handbook computes (3/8) in.²(3.14) as “1.175 in.²” and then rounds “1.175 square inches” to 1.18 (p. 128, question 58). But 3/8 × 3.14 = 1.1775, so the site now reads 1.1775 in both places. Rounded to the nearest hundredth it is still 1.18 square inches, the answer (with π itself, 3/8 × π = 1.178…, which also rounds to 1.18).

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius from diameter: $r = \frac{d}{2}$; d is the diameter, the distance straight across a circle through its center. r is the radius, the distance from the center to the edge. The diameter is two radii laid end to end, so the radius is half of it.
Area of a circle: $A = \pi \times r^{2}$; A is the area, the space inside the circle. π (pi) is a number a little more than 3, about 3.14. r is the radius. r² means r × r. Square the radius first, then multiply by π.

Sort out the question
The gray part is a ring. It is the big circle with the hole cut out. The big circle is $1\frac{1}{4}$ inches across. The hole is $\frac{1}{4}$ inch across. These are diameters. You need the area of the gray ring, in square inches. Round it to the nearest hundredth, two places after the decimal point.
Find both radii
The area formula uses the radius, so halve each diameter first. Write the mixed number as a fraction: $1\frac{1}{4} = \frac{5}{4}$ .
$r = \frac{d}{2}$
$r = \frac{5}{4} \div 2 = \frac{5}{4} \times \frac{1}{2} = \frac{5}{8}$
$r = \frac{1}{4} \div 2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
The big circle’s radius is 5/8 inch. The hole’s radius is 1/8 inch. Watch out: don’t use the diameters as radii. That makes the ring four times too big, and you would grid 4.71.
Find the area of the big circle
$A = \pi \times r^{2}$
$A = \pi \times \left(\frac{5}{8}\right)^{2} = \pi \times \frac{5 \times 5}{8 \times 8} = \frac{25}{64}\pi$
To square a fraction, square the top and square the bottom. The big circle’s area is 25/64 π square inches. Keep π for now. You will use 3.14 at the very end.
Find the area of the hole
$A = \pi \times \left(\frac{1}{8}\right)^{2} = \pi \times \frac{1 \times 1}{8 \times 8} = \frac{1}{64}\pi$
The hole’s area is 1/64 π square inches. Watch out: square each radius on its own. Don’t subtract the radii first. That gives $\frac{5}{8} - \frac{1}{8} = \frac{1}{2}$ , and $\pi \times \left(\frac{1}{2}\right)^{2}$ leads to 0.79.
Subtract the hole
The ring is the big circle minus the hole. Both fractions have 64 on the bottom, so subtract the tops.
$\frac{25}{64}\pi - \frac{1}{64}\pi = \frac{24}{64}\pi = \frac{3}{8}\pi$
To simplify 24/64, divide the top and bottom by 8. Watch out: don’t stop at the big circle. Its area alone is about 1.23 square inches.
Multiply by 3.14 and round
$\frac{3}{8} \times 3.14 = \frac{9.42}{8} = 1.1775$
The ring’s area is 1.1775 square inches. To round to the hundredths place, look at the next digit. It is 7, which is 5 or more, so round up. 1.1775 becomes 1.18. Using the π key instead gives 1.178…, which also rounds to 1.18. So you would grid 1.18.

Check your answer

Estimate. The big circle is about 1.23 square inches. The hole is tiny: $\frac{1}{64} \times 3.14$ is about 0.05 square inch. $1.23 - 0.05 = 1.18$ . That matches. The ring must also be smaller than the whole circle, and it is.

More questions like this

The original DOE question

Question 58 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 58. Open page 79 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 2 of 57

2022 Test A Math, question 2

Question

A homeowner has a budget of $135 for the installation of a light fixture. An electrician charges $60 per hour plus a one-time trip charge of $45. What is the greatest amount of time, in hours, the electrician can work on the installation while staying within the homeowner’s budget? Write your answer as a decimal.

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

Show the answer

Correct answer: 1.5

Let x represent the number of hours the electrician can work.

The cost for the number of hours worked is $60 times the number of hours worked, x. The sum of the one-time trip charge ($45) and the cost of the number of hours worked ( $60x$ ) must be less than or equal to $135. Express this relationship as an inequality:

$45 + 60x \leq 135$

Solve the inequality for x:

$45 + 60x \leq 135$

$60x \leq 90$

$x \leq \frac{90}{60} = \frac{9}{6} = \frac{3}{2}$

Written as a decimal, $\frac{3}{2} = 1.5$ .

Question analysis

A homeowner has $135 to spend, and an electrician charges $60 an hour plus a one-time $45 trip charge. This grid-in asks for the greatest number of hours the electrician can work within the budget, written as a decimal. Students write and solve a two-step inequality, and turning a length of time into decimal hours is the last trap.

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 Solving a word problem that leads to an inequality of the form px + q ≤ r is the grade 7 standard itself. The numbers are friendly; the only twists are treating the budget as a limit and writing the time in decimal hours.
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
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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 arithmetic is light, $135 − $45 = $90 and then $90 ÷ $60, so the question is decided by the setup and by the form of the entry. The trip charge is paid once, not every hour, and has to come off the budget before dividing. The time works out to an hour and a half, and students who think in hours and minutes may write it the way a clock shows it rather than as a decimal.

Turning words into math moderate Understanding the question
The stem never says “at most”; the limit comes from “a budget of $135”, and the one-time $45 trip charge is added once while the $60 multiplies the hours, giving 45 + 60x ≤ 135. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The grid-in demands a decimal, so an hour and a half must be entered as 1.5, not as 1:30 or 90 minutes, with no choices to confirm the form. (Enright, Morley & Sheehan, 2002)

Strategy: Take the one-time trip charge off the budget first, divide what is left by the hourly charge, and give the time in hours as a decimal rather than in hours and minutes.

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

The wrong answers, classified

2.25 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole $135 budget by $60 an hour, 135 ÷ 60 = 2.25, leaving out the one-time $45 trip charge that must come off first.
1.3 Units or scale (King et al., 2004)
Reveals a misconception Expected pull: moderate
Finds that the $90 left pays for 1 hour 30 minutes and writes that as 1.3, as if minutes were tenths of an hour rather than sixtieths.
90 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Subtracts the trip charge and enters the $90 left for labor, stopping before dividing by the $60 hourly charge to turn dollars into hours.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a one-time charge: $C = f + r \times h$; C is the total cost. f is the one-time charge, paid once no matter how long the job takes. r is the rate, the cost for each hour. h is the number of hours. The hourly part grows with time, but the one-time charge does not.
Keeping an inequality balanced: $\text{if } a \leq b \text{, then } a - c \leq b - c \text{ and } a \div c \leq b \div c$; The sign ≤ says the left side is less than or equal to the right side. a and b are the two sides. c is a positive number. Subtracting c from both sides, or dividing both sides by c, keeps the statement true.

Sort out the question
The homeowner can spend $135 at most. That is the budget. The electrician has a one-time trip charge of $45. One-time means it is paid once, not every hour. On top of that, the electrician charges $60 for each hour. You need the greatest number of hours that keeps the total at $135 or less. The question asks for a decimal.
Write an inequality
Let h be the number of hours. Use the total cost formula with f = 45 and r = 60. The total must be less than or equal to $135. An inequality is like an equation, but it uses a sign such as ≤ instead of =.
$C = f + r \times h$
$45 + 60 \times h \leq 135$
Take off the trip charge
Subtract 45 from both sides. This keeps the inequality balanced.
$45 + 60 \times h - 45 \leq 135 - 45$
$60 \times h \leq 90$
So $90 is left to pay for the hours. Watch out: don’t divide the whole $135 by $60. That leaves out the trip charge, and you would grid 2.25.
Divide by the hourly rate
60 × h means 60 times h. Divide both sides by 60 to get h by itself.
$h \leq \frac{90}{60} = \frac{3}{2} = 1.5$
The electrician can work at most 1.5 hours. Watch out: 1.5 hours is 1 hour 30 minutes, but don’t grid 1.3. Thirty minutes is half an hour, 0.5, not 0.3. Also don’t grid 90. That is dollars, not hours. So you would grid 1.5.

Check your answer

Put 1.5 hours back in: $45 + 60 \times 1.5 = 45 + 90 = 135$ . That is exactly the budget. Now try 2 hours: $45 + 60 \times 2 = 165$ . That is over $135. So 1.5 hours is the most.

More questions like this

The original DOE question

Question 59 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 59. Open page 79 of the handbook (PDF)

Question 3 of 57

2022 Test A Math, question 3

Question

The list shows the daily low temperatures in a city during six days in the winter.

7° F, 4° F, 1° F, −2° F, −5° F, 3° F

What is the difference between the lowest and highest temperatures, in degrees Fahrenheit, for the six days?

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

Show the answer

Correct answer: 12

Determine the greatest temperature listed for the six days:

Of 7, 4, 1, (−2), (−5), and 3, the greatest value is 7 because its location is farthest to the right on a number line.

Greatest temperature = 7° F

Determine the lowest temperature listed for the six days:

Of 7, 4, 1, (−2), (−5), and 3, the lowest value is (−5) because its location is farthest to the left on a number line.

Lowest temperature = (−5)° F

Determine the difference between the two temperatures:

$7^{\circ}\text{ F} - (-5)^{\circ}\text{ F} = 7^{\circ}\text{ F} + 5^{\circ}\text{ F} = 12^{\circ}\text{ F}$

Question analysis

A list gives a city's daily low temperatures over six winter days, some above zero and some below. This grid-in asks for the difference between the highest and the lowest of them, in degrees Fahrenheit. Students order positive and negative numbers and subtract integers; picking the right negative reading and handling the double minus sign are the tricky steps.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7b order statements in context
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Subtracting a negative number to find the distance between two temperatures is core grade 7 integer work. Picking the extremes from six one-digit readings is quick, so this is a typical, short grade 7 exercise.
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
An answer in an unexpected form minor
Solution: 2 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

Six readings, two of them below zero, must be sorted before any subtraction, and −5 has to be seen as colder than −2 even though 5 is the bigger number. The subtraction is where most points go: the gap runs 5 degrees up to zero and 7 more above it. A quick sketch of a thermometer from −5 to 7 shows the full span and catches a sign dropped along the way.

Error-prone arithmetic moderate Working it out
The span is 7 − (−5), and subtracting a negative means adding 5, so the five degrees below zero are easy to drop, leaving only 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
With no choices to compare, nothing flags an entry of 2 or −12; the student must see alone that the gap between the extremes is a positive count of degrees. (Enright, Morley & Sheehan, 2002)

Strategy: Picture the six readings on a number line: the lowest is the one farthest to the left. Subtract the lowest from the highest, remembering that subtracting a negative number means adding.

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.)
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

2 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 7 − 5 = 2, as if both readings were above zero, leaving out the 5 degrees between −5° F and zero.
9 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Takes −2° F as the lowest reading because 2 is less than 5, then subtracts 7 − (−2) = 9.
-12 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −5 − 7 = −12, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between two values: $D = \text{highest} - \text{lowest}$; D is the difference. Subtract the lowest value from the highest, in that order, so D is not negative. It tells how far apart the two values are on a number line.
Subtracting a negative number: $a - (-b) = a + b$; a and b are numbers, and −b is the negative of b. Taking away a negative is the same as adding its positive. On a number line, going from −b up to a passes b steps to zero, then a more.

Sort out the question
The list has six low temperatures, in degrees Fahrenheit (° F). Some are below zero. A minus sign means below zero. You need the highest and the lowest temperatures. Then find the difference between them. The difference tells how many degrees apart they are.
Find the highest and lowest
Picture a number line, or a thermometer. Numbers farther to the right are higher. The highest temperature is 7° F. The lowest is the one farthest to the left. That is −5° F. Watch out: −5 is colder than −2. The number 5 is bigger than 2, but 5 below zero is lower than 2 below. Using −2 as the lowest leads to 9.
Subtract the lowest from the highest
$D = \text{highest} - \text{lowest}$
$D = 7 - (-5) = 7 + 5 = 12$
Subtracting a negative is the same as adding. The temperatures are 12 degrees apart. Watch out: $7 - 5 = 2$ leaves out the 5 degrees below zero. Also, a difference here is not negative, so don’t grid −12. So you would grid 12.

Check your answer

Count on a thermometer. From −5° F up to 0° F is 5 degrees. From 0° F up to 7° F is 7 more degrees. $5 + 7 = 12$ degrees. That matches.

More questions like this

The original DOE question

Question 60 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 60. Open page 79 of the handbook (PDF)

Question 4 of 57

2022 Test A Math, question 4

Question

A grocery store sells 1-gallon containers of milk for $3.99. The store also sells orange juice in a 6-pack of 5.5-fluid-ounce bottles for $1.79. Suppose the store wants to sell its orange juice in gallon containers instead. To the nearest dollar, how much more would a gallon of orange juice cost than a gallon of milk?

Enter your answer in the space.

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

Show the answer

Correct answer: 3

To determine the cost of a gallon of orange juice, begin by finding the total number of ounces in the 6-pack of bottles:

$5.5\text{ oz} \times 6 = 33\text{ oz}$

So, 33 oz of orange juice costs $1.79.

There are 128 oz in 1 gallon. Set up a proportion using the $1.79 cost per 33 oz of orange juice and the unknown cost (x) per 1 gallon of orange juice:

$\frac{33\text{ oz}}{\$1.79} = \frac{128\text{ oz}}{x}$

Cross multiply to solve for x in the proportion:

$(33) \times (x) = (\$1.79) \times (128)$

$33x = \$229.12$

$x \approx \$6.94$

So, the orange juice costs $6.94 per gallon, and the milk costs $3.99 per gallon.

Subtract the cost per gallon of milk from the cost per gallon of orange juice.

$\$6.94 - \$3.99 = \$2.95$

Round the difference to the nearest dollar.

$2.95 rounds to $3.

Question analysis

A store sells milk by the gallon for $3.99 and orange juice in six-packs of 5.5-fluid-ounce bottles for $1.79. Students work out what a gallon of the juice would cost at the same price per ounce and how much more that is than the milk, to the nearest dollar. The ounces in a gallon must be known, and several steps involve decimals.

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: Above grade 7 Unit prices are grade 6 and 7 work, but this chains five steps with decimals and needs a fact the question does not give, the number of fluid ounces in a gallon. Typing a rounded difference, with no choices, adds risk.
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
Error-prone arithmetic minor
An answer in an unexpected form minor
Solution: 5 steps
Reading level: Question and choices: about grade 5, 60 words.
Corrected from the DOE handbook: A rounded value in the explanation is now marked as approximate (≈ instead of =).
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 calculation is long, and its key fact is missing from the page. Students who know the 128 fluid ounces still face a division by 33 that does not come out evenly, which tempts them to round the price per ounce to 5 cents and lose about 50 cents a gallon. Keeping the exact quotient to the end, or scaling 33 ounces up by about 3.9, keeps the difference near $2.95, which rounds to 3.

Something you must already know moderate Working it out
The question never says that a gallon holds 128 fluid ounces, yet the whole comparison depends on it; the 6-pack holds only 6 × 5.5 = 33 ounces. (Anderson & Krathwohl, 2001)
Many steps moderate Working it out
Five moves follow one another: the ounces in a pack, the scale-up to 128 ounces, the division by 33, the subtraction of $3.99 and the rounding to a whole dollar. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
1.79 × 128 = 229.12 and 229.12 ÷ 33 are long decimal calculations without a calculator, and a slip in either moves the price of a gallon. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The rounded difference is typed with no choices, so an unrounded $2.95, or a figure rounded too soon, looks just as finished as the whole-dollar answer. (Enright, Morley & Sheehan, 2002)

Strategy: Find the ounces in one six-pack, scale its price up to the 128 fluid ounces in a gallon, subtract the milk's price, and round only at the end.

Traps

Stops one step early: A choice is a number from partway through the solution.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

7 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the juice's cost per gallon, about $6.94, rounds it to $7, and enters that without subtracting the $3.99 price of the milk.
2 Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds the price per ounce, 1.79 ÷ 33 ≈ 0.054, to $0.05, so a gallon costs $6.40 and the difference, $2.41, rounds to 2.
2.95 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Works out the difference of $2.95 correctly but enters it as it stands, though the question asks for it to the nearest dollar.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation goes from 33x = $229.12 to “x = $6.94”. But 229.12 ÷ 33 = 6.9430…, so $6.94 is the cost per gallon rounded to the cent. The site writes x ≈ $6.94. Nothing else changes: $6.94 − $3.99 = $2.95, which rounds to $3, and the unrounded cost gives the same answer.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fluid ounces in a gallon: $\text{fluid ounces} = 128 \times \text{gallons}$; A gallon holds 128 fluid ounces. This is a fact you must know, because the question does not give it. A fluid ounce measures an amount of liquid. Multiply gallons by 128 to change them into fluid ounces.
Cross-multiplying a proportion: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; A proportion says two ratios are equal. Here each ratio is ounces over dollars. a and c are the tops, and b and d are the bottoms. Multiply each top by the other bottom, and the two products are equal. That turns the proportion into an equation.

Sort out the question
Milk costs $3.99 for 1 gallon. Orange juice comes in a 6-pack of 5.5-fluid-ounce bottles for $1.79. You need how much more a gallon of juice would cost than a gallon of milk. Round to the nearest dollar.
The question doesn’t say how big a gallon is. You must know this fact: 1 gallon is 128 fluid ounces.
Find the ounces in a 6-pack
Each of the 6 bottles holds 5.5 fluid ounces:
$6 \times 5.5 = 33$
So $1.79 buys 33 fluid ounces of juice.
Set up a proportion
Let x be the cost in dollars of a gallon, 128 fluid ounces, of juice. The price per ounce stays the same. So ounces over dollars is the same for the pack and the gallon:
$\frac{33}{1.79} = \frac{128}{x}$
Cross-multiply:
$33 \times x = 1.79 \times 128 = 229.12$
Find the cost of a gallon
Divide both sides by 33:
$x = 229.12 \div 33 \approx 6.94$
The sign ≈ means about equal. The exact value is 6.943…, so a gallon of juice costs about $6.94.
Watch out: don’t round too early. One ounce costs about $0.054. Rounding that to $0.05 makes a gallon $6.40, and your answer becomes 2.
Subtract the milk’s price
Find how much more the juice costs:
$6.94 - 3.99 = 2.95$
A gallon of juice costs $2.95 more than a gallon of milk.
Watch out: entering 7, the juice’s price, forgets to subtract the milk.
Round and enter it
The question asks for the nearest dollar. $2.95 is between $2 and $3. Its cents, 95, are 50 or more, so it rounds up to $3. Do not enter 2.95.
Type only the number in the box. So you would enter 3.

Check your answer

Estimate. 128 ounces is almost 4 packs, since $4 \times 33 = 132$ . Four packs cost $4 \times 1.79 = 7.16$ dollars, so a gallon costs a bit less than that. Milk costs about $4. About $7 minus $4 is $3, which matches.

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 C Math, page 88, question 15. This is the DOE’s version, before the corrections listed in the analysis.

Question 5 of 57

2022 Test A Math, question 5

Question

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 15

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

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

$84^{\circ}\text{ F} - 39^{\circ}\text{ F} = 45^{\circ}\text{ F}$

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

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

$78^{\circ}\text{ F} - 48^{\circ}\text{ F} = 30^{\circ}\text{ F}$

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

$45^{\circ}\text{ F} - 30^{\circ}\text{ F} = 15^{\circ}\text{ F}$

Question analysis

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

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Reading a box plot and finding its interquartile range are grade 7 skills. Here two of the five values fall between ticks spaced 4 apart, and the range, the IQR and their difference must all be found and typed with no choices.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Many steps minor
Something you must already know minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph major Understanding the question
Two of the five values fall between ticks spaced 4 degrees apart: the left whisker ends just short of the 40 tick, at 39, and the box’s right edge sits midway between 76 and 80, at 78. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Four values are read and combined in three subtractions, the range, the interquartile range and then their difference, so one misread value changes the final entry. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Neither measure is defined on the page: the range spans the whiskers, and the interquartile range is the width of the box, the upper quartile minus the lower. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The result is typed with no choices to compare, so an entry one or two degrees off, from rounding a value to the nearest tick, looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

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

Traps

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

14 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Reads the left whisker’s end as 40, the nearest labeled tick, instead of 39, so the range becomes 84 − 40 = 44 and 44 − 30 = 14.
45 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the range, 84 − 39 = 45 degrees, and enters it, stopping before the interquartile range of 30 is subtracted from it.
17 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads the box’s right edge as 76, the tick just left of it, instead of 78, so the IQR becomes 76 − 48 = 28 and 45 − 28 = 17.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
A box plot shows data with five numbers. The two whiskers are the lines on each side of the box. Their ends are the minimum, the least value, and the maximum, the greatest. The box runs from the lower quartile to the upper quartile. These are the middles of the lower half and the upper half of the data. The line inside the box is the median, the middle value.
You need the range minus the interquartile range (IQR), in degrees.
Read the values between the ticks
The ticks on the number line are 4 degrees apart. So each degree is one quarter of the space between two ticks. Three values sit right on ticks: the box’s left edge at 48, the median at 64, and the right whisker’s end at 84.
- The left whisker starts just left of the 40 tick. It sits about one quarter of the way back toward 36. So the minimum is $40 - 1 = 39$ degrees.
- The box’s right edge is halfway between 76 and 80. Halfway is 78 degrees.
Find the range
Use the range formula:
$\text{range} = 84 - 39 = 45$
The data spread over 45 degrees.
Watch out: rounding the minimum to the 40 tick gives a range of 44. Then the final answer comes out 14, which is wrong.
Find the IQR
Use the IQR formula with the two edges of the box:
$\text{IQR} = 78 - 48 = 30$
The middle half of the data spreads over 30 degrees.
Watch out: reading the box’s right edge as 76 gives an IQR of 28. Then the final answer comes out 17.
Subtract the IQR from the range
The question asks for the difference between the two:
$45 - 30 = 15$
Watch out: 45 is only the range. Do not stop there. Type only the number, with no degree sign.
So you would enter 15.

Check your answer

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

More questions like this

The original DOE question

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

Question 6 of 57

2022 Test A Math, question 6

Question

What is the prime factorization of 756?

A $2^{2} \cdot 3^{2} \cdot 21$
B $2^{2} \cdot 3 \cdot 7 \cdot 9$
C $2^{2} \cdot 7 \cdot 27$
D $2^{2} \cdot 3^{3} \cdot 7$

Show the answer

Correct answer: D

Determine the prime factors of 756 using a factor tree:

$A factor tree for 756: 756 splits into 2 and 378; 378 into 2 and 189; 189 into 3 and 63; 63 into 3 and 21; and 21 into 3 and 7.$

The prime factors are 2, 2, 3, 3, 3, and 7, or $2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7$ , which can be expressed using exponents as $2^{2} \cdot 3^{3} \cdot 7$ .

Question analysis

This number question asks for the prime factorization of 756, written with exponents. Students break the number into prime factors, with a factor tree or repeated division, and match the result to one of four products. Every choice multiplies back to 756, so the real test is whether each factor shown is actually prime.

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 Factor trees and exponent notation are grade 6 skills, and 756 splits quickly by small divisors. What adds a little is that every choice multiplies out to 756, so checking the product alone settles nothing.
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

The factoring itself is quick: halving twice and dividing by 3 three times leaves 7. The question is built so that multiplying back proves nothing, since every choice equals 756, and three of them simply stop the factor tree at a composite number. Students who check only the product, or who take an odd factor such as 21 or 27 for a prime, pick a factorization that is unfinished.

A very close wrong answer moderate Choosing the answer
2² · 3² · 21 shares the right answer's 2² and differs only in 3² · 21 against 3³ · 7, and both equal 756, so neither an estimate nor a product check 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 prime factorization may hold only primes, and must spot that 21, 9 and 27 are composite even though none of them is even. (Anderson & Krathwohl, 2001)

Strategy: Divide 756 by 2 while the result is even, then by 3 while it divides evenly, and keep going until every factor left is prime; then compare factor by factor, not by product.

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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Divides out 2 twice and 3 twice, then stops at 21, which is 3 × 7, so one composite factor remains in a product that still equals 756.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Splits off a 3 and a 7 but keeps 9 as if it were prime, perhaps because it is odd, though 9 = 3 × 3.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Splits 756 into 4 × 189 and 189 into 7 × 27, then stops, not seeing that 27 is 3 × 3 × 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Writing a repeated factor with an exponent: $a^{n} = a \times a \times \cdots \times a$; a is the number being multiplied, called the base. n is the exponent, the small raised number. It tells how many times a appears in the product. So 2² means 2 × 2, and 3³ means 3 × 3 × 3. Exponents are a short way to write a factor that repeats.

Sort out the question
You need the prime factorization of 756. That means writing 756 as a product of prime numbers only. A product is the answer to a multiplication. A prime number has exactly two factors, 1 and itself. Examples are 2, 3, 5 and 7.
All four choices multiply to 756. So multiplying a choice back does not pick the answer. You must check that every factor is prime.
Divide by 2 while you can
A factor tree splits a number into two factors, again and again. Start with the smallest prime, 2. 756 is even, so 2 goes into it:
$756 = 2 \times 378$
378 is even too:
$378 = 2 \times 189$
189 is odd, so 2 does not go into it. You have pulled out two 2s.
Divide by 3 while you can
Add the digits of 189: $1 + 8 + 9 = 18$ . A number can be divided by 3 when its digits add up to a number that 3 goes into. 3 goes into 18, so it goes into 189:
$189 = 3 \times 63$
$63 = 3 \times 21$
$21 = 3 \times 7$
7 is prime, so the tree stops. You have pulled out three 3s and one 7.
Write the factors with exponents
Collect every prime you pulled out:
$756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7$
Use the exponent rule. Two 2s make $2^{2}$ , and three 3s make $3^{3}$ . The choices use a raised dot, ·, for times. So 756 is $2^{2} \cdot 3^{3} \cdot 7$ .
Choose the answer
Watch out: each wrong choice is a factor tree that stopped too soon. A ends in 21, which is 3 × 7. B has 9, which is 3 × 3. C has 27, which is 3 × 3 × 3. Being odd does not make a number prime.
Only D uses primes alone. So the answer is D, $2^{2} \cdot 3^{3} \cdot 7$ .

Check your answer

Multiply your answer back: $2^{2} \times 3^{3} = 4 \times 27 = 108$ , and $108 \times 7 = 756$ . Then look at each factor. 2, 3 and 7 are all prime. D passes both tests.

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 C Math, page 74, question 1.

Question 7 of 57

2022 Test A Math, question 7

Question

$\frac{x^{2}y^{6}}{xy^{2}}$

What is the value of the expression above if $x = -2$ and $y = 2$ ?

A−64
B−32
C16
D32

Show the answer

Correct answer: B

First, substitute −2 for x and 2 for y.

$\frac{(-2)^{2}(2)^{6}}{(-2)(2)^{2}}$

Then, simplify.

$\frac{(-2)(-2)(2)^{2}(2)^{4}}{(-2)(2)^{2}}$

$\frac{(-2)(2)^{4}}{1}$

$\frac{-2 \times 16}{1}$

$-32$

Question analysis

This question asks for the value of a fraction built from powers of x and y when x is −2 and y is 2. Students evaluate an expression with exponents and a negative number, or simplify it first by cancelling common factors. Keeping track of the negative sign through the squaring and the division is the tricky part.

NYS standards: 6.EE.2c evaluating expressions and order of operations
7.NS.2a rational number multiplication
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students evaluate expressions with negative values and whole-number exponents. This one is a single fraction of powers, routine in method, but raising a negative base to powers and then dividing leaves several chances to lose the sign.
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
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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 routes work: substitute and compute 256 ÷ (−8), or simplify to xy⁴ first and evaluate (−2)(16). Simplifying first is safer, because it leaves a single negative factor to track. Substituting first brings in (−2)², where treating the minus sign as outside the parentheses flips the final sign, and each listed wrong value comes from an exponent or a sign handled wrongly.

Error-prone arithmetic moderate Working it out
Substituting first gives (−2)²(2)⁶ over (−2)(2)², so the work runs through a squared negative, 2⁶ = 64 and a negative divisor, and one factor of −2 decides the sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The quotient rule for powers is not stated; students who know that x²/x = x and y⁶/y² = y⁴ can reduce the expression to xy⁴ before any number goes in. (Anderson & Krathwohl, 2001)

Strategy: Cancel the factors of x and y that the top and bottom share before substituting, so there is less to compute, and track the sign of every factor of −2 through to the end.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
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 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a slip Expected pull: weak
Drops the square on y in the denominator, dividing 4 × 64 = 256 by (−2)(2) = −4 instead of by (−2)(4) = −8, giving −64.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Cancels x² against x as if x²/x were 1 rather than x, so only y⁶/y² = y⁴ = 16 is left, with no factor of −2.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−2)² as −4, so −4 × 64 = −256 and −256 ÷ (−8) = 32.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A power: $a^{4} = a \times a \times a \times a$; a is the base, the number being multiplied. The small number is the exponent. It says how many times the base is used as a factor, a number being multiplied. So y⁶ means six y’s multiplied together. It does not mean y times 6.
Dividing powers with the same base: $\frac{a^{m}}{a^{n}} = a^{m - n}$; a is the base. m and n are exponents. The top has m factors of a and the bottom has n. Each matching pair cancels, because a ÷ a = 1. So m − n factors are left. A letter with no exponent, like x, has exponent 1.

Sort out the question
You need the value of $\frac{x^{2}y^{6}}{xy^{2}}$ when x = −2 and y = 2. The top means $x^{2} \times y^{6}$ . The bottom means $x \times y^{2}$ . You could put the numbers in right away. But simplifying first is safer. It leaves only one negative number to track.
Simplify the x part
Use the rule for dividing powers. The x on the bottom has exponent 1.
$\frac{x^{2}}{x} = x^{2 - 1} = x^{1} = x$
You can see it by writing it out: $\frac{x \times x}{x}$ leaves one x. Watch out: x² ÷ x is x, not 1. If you cancel every x, you lose the −2 and get 16, choice C.
Simplify the y part
$\frac{y^{6}}{y^{2}} = y^{6 - 2} = y^{4}$
So the whole fraction becomes a simpler expression:
$\frac{x^{2}y^{6}}{xy^{2}} = x \times y^{4}$
Watch out: the bottom has y², not y. Using y leaves y⁵ and leads to −64, choice A.
Put in the numbers
Now substitute x = −2 and y = 2.
$x \times y^{4} = (-2) \times 2^{4}$
$2^{4} = 2 \times 2 \times 2 \times 2 = 16$
$(-2) \times 16 = -32$
A negative times a positive is negative. So the answer is B, −32.

Check your answer

Substitute first, the long way. Top: $(-2)^{2} \times 2^{6} = 4 \times 64 = 256$ . Bottom: $(-2) \times 2^{2} = (-2) \times 4 = -8$ . Then $256 \div (-8) = -32$ . Watch out: (−2)² is +4, since two negatives multiply to a positive. Writing −4 leads to 32, choice D.

More questions like this

The original DOE question

Question 64 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 81, question 64. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 8 of 57

2022 Test A Math, question 8

Question

If $(8 \div m) + 4 = 20$ , what is the value of m?

A $\frac{1}{3}$
B $\frac{1}{2}$
C2
D3

Show the answer

Correct answer: B

The quantity $(8 \div m)$ means the same as $\frac{8}{m}$ , so the equation can be rewritten:

$\left(\frac{8}{m}\right) + 4 = 20$

Subtract 4 from both sides:

$\left(\frac{8}{m}\right) = 16$

Multiply both sides of the equation by m:

$8 = 16m$

Solve for m:

$\left(\frac{8}{16}\right) = m$

Simplify:

$\left(\frac{8(\div 8)}{16(\div 8)}\right) = \left(\frac{1}{2}\right)$

Question analysis

This equation question asks for the value of m when 8 divided by m, plus 4, equals 20. Students undo the addition, then work out what m must be for 8 divided by it to give the value left. Because m is the divisor rather than the number being divided, the last step needs care.

NYS standards: 7.EE.4a two-step equations
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Undoing an addition and then a division is typical two-step grade 7 equation work. The unknown sitting in a denominator is less common than px + q = r, which makes the last step easier to get backward.
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
Something you must already know minor
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

Undoing the + 4 is routine, so nearly all the risk sits in the last step, where m divides 8 rather than being divided by it. Students who read 8 ÷ m = 16 as asking how many 8s make 16 answer 2, and substituting it back, 8 ÷ 2 + 4 = 8, shows at once that it fails. The right value makes 8 ÷ 1/2 = 16, a division that gives a larger number, which feels wrong but is correct.

Error-prone arithmetic moderate Working it out
The unknown is the divisor, so 8 ÷ m = 16 ends in m = 8 ÷ 16, a quotient less than 1; the familiar order, 16 ÷ 8, gives a whole number instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must see that (8 ÷ m) is the fraction 8/m and that freeing m from a denominator means multiplying both sides by it, a move grade 7 equations rarely need. (Anderson & Krathwohl, 2001)

Strategy: Subtract 4 from both sides, multiply both sides by m so the unknown is no longer in a denominator, then divide; check by substituting your value back into the equation.

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 slip Expected pull: weak
Adds 4 to 20 instead of subtracting it, so 8 ÷ m = 24 and m = 8/24, which simplifies to 1/3.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Gets 8 ÷ m = 16 correctly but then divides 16 by 8, putting the divisor and the dividend the wrong way round, so m comes out as 2.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Makes both mistakes, adding 4 to get 8 ÷ m = 24 and then dividing 24 by 8 instead of 8 by 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \times c = b \times c$; a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. Subtract the same number from both sides, or multiply both sides by the same number, and it stays balanced. That is how you get the unknown by itself.

Sort out the question
The equation is $(8 \div m) + 4 = 20$ . You need the value of m. Here m is the number you divide 8 by. In words: 8 divided by some number, plus 4, makes 20.
Undo the plus 4
A fraction bar means divide, so $8 \div m$ is the same as $\frac{8}{m}$ . Subtract 4 from both sides to undo the plus 4:
$\frac{8}{m} + 4 - 4 = 20 - 4$
$\frac{8}{m} = 16$
Now 8 divided by m equals 16.
Watch out: adding 4 instead gives 24 on the right. That leads to choice A, 1/3, or to D, 3.
Get m out of the bottom
m is on the bottom of the fraction. Multiply both sides by m to free it:
$8 = 16 \times m$
Now divide both sides by 16:
$m = \frac{8}{16} = \frac{8 \div 8}{16 \div 8} = \frac{1}{2}$
8 goes into both 8 and 16, so 8/16 simplifies to 1/2.
Choose the answer
Watch out: it is tempting to divide 16 by 8 and get 2, choice C. But ask yourself: 8 divided by what number makes 16? To get an answer bigger than 8, you must divide by a number less than 1.
So the answer is B, 1/2.

Check your answer

Put 1/2 back into the equation. How many halves are in 8? There are 16, so $8 \div \frac{1}{2} = 16$ . Then $16 + 4 = 20$ . That matches the equation, so m = 1/2 is right.

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 C Math, page 75, question 2.

Question 9 of 57

2022 Test A Math, question 9

Question

$6x = x - 1{,}680$

What is the value of x in the equation shown above?

A336
B240
C−240
D−336

Show the answer

Correct answer: D

Simplify $6x = x - 1{,}680$ . First, subtract x from both sides of the equation. Then, divide both sides by 5.

$6x - x = x - x - 1{,}680$

$5x = -1{,}680$

$\frac{5x}{5} = \frac{-1{,}680}{5}$

$x = -336$

Question analysis

This algebra question gives the equation 6x = x − 1,680, with x on both sides, and asks for the value of x. Students collect the x terms on one side and divide by the coefficient that remains. The choices pair two sizes with both signs, so every move across the equals sign has to keep its sign right.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.NS.2b rational number division
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 An unknown on both sides of the equals sign is a grade 8 topic, yet gathering the x terms here takes one subtraction and one division, the same work as a two-step grade 7 equation.
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
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 5, 12 words.
Corrected from the DOE handbook: A line of the explanation’s working, printed without its fraction bars and equals sign, was restored.
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

Collecting the x terms takes one subtraction, and the only way the method goes wrong is moving x across the equals sign, which turns the 5 into a 7. The sign is the bigger hazard: 6x equals x minus a large number, so x must be negative, yet the division 1,680 ÷ 5 is done on positive numbers. Substituting −336 back gives −2,016 on each side, which confirms both the size and the sign.

Error-prone arithmetic moderate Working it out
Subtracting x from both sides leaves 5x = −1,680, and 1,680 ÷ 5 = 336 has to be worked out by hand while the negative sign is carried through to the result. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
The choices are two sizes, 336 and 240, each offered with both signs, so 336 and −336 are written the same except for the minus; only tracking the sign of 1,680 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract x from both sides so the unknown appears once, divide by the coefficient that is left, and substitute the result into both sides to check that they match.

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 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches 5x = −1,680 correctly but drops the negative sign when dividing by 5, giving a positive 336 of the right size.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves x to the left without changing its sign, getting 7x, and also loses the minus on 1,680, so 1,680 ÷ 7 = 240.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Moves x to the left side without changing its sign, writing 6x + x = −1,680, so divides by 7 and gets −240.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation page for this question is one of the pages replaced in the appendix, and on it the step between 5x = −1,680 and x = −336 shows only 5x over 5 and −1,680 over 5: the two fraction bars and the equals sign between them are missing, so the line is not an equation. The sentence above it says to divide both sides by 5, and the 2022–23 guide, which first printed this explanation, prints the step as 5x/5 = −1,680/5. The site shows 5x/5 = −1,680/5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. Subtract the same thing from both sides, or divide both sides by the same number, and it stays balanced.
Dividing a negative by a positive: $(-a) \div b = -(a \div b)$; a and b are positive numbers, so −a is negative. Divide the sizes as usual. Then give the answer a minus sign, because a negative divided by a positive is negative.

Sort out the question
The equation is $6x = x - 1{,}680$ . You need the value of x. Here 6x means 6 times x. The unknown x is on both sides. So first gather it on one side.
Subtract x from both sides
Take x away from both sides. On the right, x − x is 0, so only −1,680 is left:
$6x - x = x - x - 1{,}680$
$5x = -1{,}680$
Six x’s take away one x leaves five x’s.
Watch out: moving x to the left without changing its sign gives 7x. Dividing by 7 then leads to choice C, −240, or B, 240.
Divide both sides by 5
5x means 5 times x. Divide both sides by 5 to undo it:
$\frac{5x}{5} = \frac{-1{,}680}{5}$
Find the size first. Split 1,680 into 1,500 + 180:
$1{,}500 \div 5 + 180 \div 5 = 300 + 36 = 336$
A negative divided by a positive is negative:
$x = -336$
Choose the answer
Watch out: the minus sign is easy to drop. Positive 336 is choice A. Carry the sign of −1,680 all the way to the end.
So the answer is D, −336.

Check your answer

Put −336 into both sides. The left side is $6 \times (-336) = -2{,}016$ . The right side is $-336 - 1{,}680 = -2{,}016$ . The two sides match, so x = −336 is right.

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 C Math, page 76, question 3. This is the DOE’s version, before the corrections listed in the analysis.

Question 10 of 57

2022 Test A Math, question 10

Question

A person buys a used car with a down payment of $1,200 and makes monthly payments of $275 for 3 years. What is the total amount the person pays for the car?

A$2,025
B$9,900
C$11,100
D$13,500

Show the answer

Correct answer: C

Determine the total cost of the car by adding the down payment to the total monthly payments over 3 years.

There are 12 months in a year, so multiply (12 months/year × 3 years) to get a total of 36 months. Determine the total monthly payments by multiplying the monthly payment ($275) × the total number of months (36):

$275 \times 36 = 9{,}900$

Determine the total cost of the car by adding the down payment ($1,200) to the total monthly payments ($9,900):

$9{,}900 + 1{,}200 = 11{,}100$

The total cost of the car is $11,100.

Question analysis

A used car is bought with a down payment of $1,200 followed by monthly payments of $275 for 3 years, and students find the total paid. The work is to count the payments, multiply, and add the down payment. Mixing up years and months, or forgetting the money paid up front, leads to the wrong totals among the choices.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 A total made of one payment up front plus equal monthly payments is a short whole-number calculation, simpler than grade 7 word problems with percents or negatives. The only conversion is from years to months.
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: Much to keep track of 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 situation is familiar and the method direct, so this is one of the easier questions on the test. Points are lost in two places: turning 3 years into 36 payments, and remembering the money paid up front after the long multiplication. Each slip, like paying the down payment once a year, leads to a listed total, so checking that the answer holds exactly one $1,200 and 36 payments of $275 is worth the seconds.

Much to keep track of minor Working it out
The $1,200 down payment is paid once while the $275 repeats every month, so it has to be set aside while 36 payments are totaled and then added at the end. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Error-prone arithmetic minor Working it out
The product 275 × 36 = 9,900 is a two-digit multiplication done by hand; splitting it as 275 × 4 = 1,100 and then 1,100 × 9 makes it quick and safe. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn 3 years into a number of months, multiply the monthly payment by it, and add the down payment once at the end.

Traps

Stops one step early: A choice is a number from partway through the solution.
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 Units or scale (King et al., 2004)
Reveals a slip Expected pull: moderate
Multiplies the $275 by 3, the number of years, instead of by 36 months, then adds the $1,200 down payment: 825 + 1,200 = 2,025.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 36 monthly payments, 36 × $275 = $9,900, and stops without adding the $1,200 down payment made at the start.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Works out one year as $1,200 + 12 × $275 = $4,500 and multiplies by 3, so the down payment is paid three times instead of once.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing years into months: $\text{months} = 12 \times \text{years}$; Each year has 12 months. Multiply the number of years by 12 to get the number of months. You need months here because a payment is made every month.
Total cost with a down payment: $\text{total} = \text{down payment} + \text{monthly payment} \times \text{months}$; The down payment is money paid once, at the start. The monthly payment is paid every month. Multiply it by the number of months to get all the monthly payments. Then add the down payment only once, because it is paid only once.

Sort out the question
The person pays $1,200 once, at the start. This is the down payment. Then the person pays $275 every month for 3 years. You need the total amount paid for the car.
The payments come once a month, but the time is given in years. So change the years into months first.
Change 3 years into months
Use the months formula:
$\text{months} = 12 \times 3 = 36$
So the person makes 36 monthly payments.
Watch out: multiplying $275 by 3 counts only 3 payments. That leads to choice A, $2,025.
Add up the monthly payments
Multiply $275 by 36. A quick way is to split 36 into 4 × 9:
$275 \times 4 = 1{,}100$
$1{,}100 \times 9 = 9{,}900$
So the 36 payments add up to $9,900.
Watch out: stopping here gives choice B, $9,900. The down payment is still missing.
Add the down payment once
Use the total-cost formula:
$\text{total} = 1{,}200 + 9{,}900 = 11{,}100$
The person pays $11,100 in all. Adding $1,200 for every year counts it three times. That gives choice D, $13,500.
So the answer is C, $11,100.

Check your answer

Estimate. $275 is a little less than $300, and $36 \times 300 = 10{,}800$ . Add the $1,200: $10{,}800 + 1{,}200 = 12{,}000$ . The real total should be a little less than $12,000. $11,100 fits.

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 C Math, page 77, question 4.

Question 11 of 57

2022 Test A Math, question 11

Question

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

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

Show the answer

Correct answer: A

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

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

$3x + 2x = 180$

Solve for x:

$5x = 180$

$\frac{5x}{5} = \frac{180}{5}$

$x = 36$

Question analysis

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

NYS standards: 7.G.5 angle relationships and equations
8.G.5 angle sums, transversals, angle-angle similarity
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 The angle fact needed, about consecutive angles of a parallelogram, rests on grade 8 parallel-line reasoning. Once it is known, the work is one short equation, like a typical grade 7 problem about angle sums.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Something you must already know moderate Working it out
The figure gives no number of degrees, only 3x° and 2x°, so everything rests on recalling that the two angles at the ends of one side of a parallelogram add to 180°. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The marked angles sit at the upper and lower ends of the left side, consecutive rather than opposite, which is what makes them supplementary instead of equal. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

Traps

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

The wrong answers, classified

B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses 90° as the total of the two angles, the complementary total, so 5x = 90 gives 18, though consecutive angles of a parallelogram add to 180°.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the two coefficients from the labels, 3 × 2 = 6, without using any angle fact or writing an equation.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Combines 3x + 2x into 5x and gives the coefficient 5, stopping before dividing the 180° total by it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The figure is a parallelogram. That is a four-sided shape whose opposite sides are parallel, so they never meet. Two angles are marked: 3x° at the upper left and 2x° at the lower left. You need the value of x.
Write an equation for the angles
The two marked angles are at the two ends of the left side. So they are neighbors, not opposite angles. Use the rule that neighbors add to 180°:
$3x + 2x = 180$
Watch out: the total is 180°, not 90°. Using 90 gives 5x = 90 and x = 18, choice B.
Solve for x
Three x’s plus two x’s make five x’s:
$5x = 180$
Divide both sides by 5 to get x alone:
$x = \frac{180}{5} = 36$
Watch out: stopping at 5, the number in front of x, gives choice D. Multiplying 3 × 2 = 6 gives choice C, but that uses no angle fact.
Choose the answer
The value of x is 36. It is a number, not the size of an angle. The angles themselves are 3 × 36 and 2 × 36 degrees.
So the answer is A, 36.

Check your answer

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

More questions like this

The original DOE question

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

Question 12 of 57

2022 Test A Math, question 12

Question

A bus begins its trip with R occupants (including the driver). At each of the 20 stops, 3 people get on and no one leaves. If there are 3 times as many bus occupants at the end of the 20th stop as there are at the end of the 4th stop, how many occupants did the bus have immediately after the 10th stop?

Show the answer

Correct answer: D

Write an expression to represent the number of occupants after 20 stops. Then, simplify the expression.

$R + (20 \times 3) = R + 60$

Write an expression to represent the number of occupants after 4 stops. Then, simplify the expression.

$R + (4 \times 3) = R + 12$

Write an equation that shows that the number of occupants after 20 stops is 3 times the number of occupants after 4 stops. Then, solve for R.

$R + 60 = 3(R + 12)$

$R + 60 = 3R + 36$

$60 = 2R + 36$

$24 = 2R$

$12 = R$

Write an expression to represent the number of occupants after 10 stops. Then, substitute 12 for R and evaluate.

$R + (10 \times 3)$

$12 + (10 \times 3) = 12 + 30 = 42$

Question analysis

A bus starts with an unknown number of riders, and 3 people board at each of 20 stops while no one gets off. After the 20th stop it holds three times as many as after the 4th. Students write and solve an equation for the starting number, then find how many were aboard after the 10th stop, a stop the condition never mentions.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.4 equations and inequalities from problems
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 equations usually have the unknown on one side, as in px + q = r. Here students build two expressions from a story, set one equal to three times the other, solve with the unknown on both sides, then answer a third question.
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 major
Many steps moderate
Much to keep track of minor
Solution: 5 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

The question hides a linear equation in a story, and the comparison sentence is the hard part: the count that gets tripled is the one at the 4th stop, parentheses and all. Solving gives the starting count, but the question asks about a different moment, the end of the 10th stop, so one more step remains. Labeling each number with its stop, start, 4th, 10th or 20th, keeps the right count from being given for the wrong time.

Turning words into math major Understanding the question
“3 times as many bus occupants at the end of the 20th stop as there are at the end of the 4th stop” must become R + 60 = 3(R + 12), the 3 multiplying the whole smaller count. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Two expressions, an equation solved for R = 12, and then a third count at the 10th stop: four stages, and the number asked for appears only at the last one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Three different stops matter, the 4th and the 20th for the equation and the 10th for the question, and each count has to keep the starting R. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let R be the starting count, write the counts after the 4th and the 20th stops in terms of R, make the larger equal to 3 times the smaller and solve, then add the riders from 10 stops.

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: strong
Solves R + 60 = 3(R + 12) correctly and stops at R = 12, the occupants at the start, before adding the 30 riders from 10 stops.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds R = 12 but gives the count at the end of the 4th stop, 12 + 4 × 3 = 24, instead of the count after the 10th.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Counts only the riders who board at 10 stops, 10 × 3 = 30, leaving out the R occupants aboard from the start.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

People on the bus after some stops: $N = R + 3 \times s$; N is the number of people on the bus. R is the number at the start, driver included. s is the number of stops so far. 3 people get on at each stop and no one leaves, so each stop adds 3.
Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. Multiply each term inside by a. Use it when a number multiplies a whole sum.

Sort out the question
Occupants are the people on the bus. R is how many there are at the start, counting the driver. At each stop, 3 people get on and nobody gets off. After the 20th stop, there are 3 times as many people as after the 4th stop. You need the number after the 10th stop. That is a third moment, so one more step waits at the end.
Write the counts for two stops
$N = R + 3 \times s$
$\text{after stop 4: } R + 3 \times 4 = R + 12$
$\text{after stop 20: } R + 3 \times 20 = R + 60$
Keep R in each count. The people who started are still on the bus.
Write the equation
The count after stop 20 is 3 times the count after stop 4. The 3 multiplies the whole count after stop 4, so use parentheses.
$R + 60 = 3(R + 12)$
Use the distributive property: 3 times R, plus 3 times 12.
$R + 60 = 3R + 36$
Solve for R
Subtract R from both sides. Then subtract 36 from both sides. Then divide by 2.
$60 = 2R + 36$
$24 = 2R$
$R = 24 \div 2 = 12$
So 12 people were on the bus at the start. Watch out: 12 is choice A, but it is the count at the start. The question asks about the 10th stop.
Count after the 10th stop
$N = 12 + 3 \times 10 = 12 + 30 = 42$
Watch out: 24 is the count after the 4th stop, choice B. 30 counts only the people who got on, choice C. It leaves out the 12 who started. So the answer is D, 42.

Check your answer

Test R = 12. After the 4th stop: $12 + 12 = 24$ people. After the 20th stop: $12 + 60 = 72$ people. And $3 \times 24 = 72$ . That matches the question, so R = 12 is right, and so is 42.

More questions like this

The original DOE question

Question 69 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 82, question 69. Open page 82 of the handbook (PDF)

Question 13 of 57

2022 Test A Math, question 13

Question

Jay sold n tickets. Pilar sold three times as many tickets as Jay did. Together, Jay and Pilar sold ten more tickets than Amie sold. Which expression shows how many tickets Amie sold?

A $10 - 4n$
B $3n^{2} + 10$
C $4n + 10$
D $4n - 10$

Show the answer

Correct answer: D

Since Jay sold n tickets and Pilar sold 3 times the number of tickets Jay sold, Pilar sold $3n$ tickets.

Then, since together Jay and Pilar sold 10 more tickets than Amie, subtract 10 from $(n + 3n)$ to get the expression that shows how many tickets Amie sold.

$(n + 3n) - 10$

Then combine like terms.

$4n - 10$

Question analysis

Jay sells n tickets, Pilar sells three times as many, and together they sell ten more tickets than Amie. Students choose the expression in n for the number of tickets Amie sold. It tests writing and combining algebraic expressions from words, and the last sentence needs a careful reading, because it describes Amie only through the other two.

NYS standards: 6.EE.6 variables representing unknown numbers
6.EE.2a writing algebraic expressions
6.EE.3 generating equivalent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Writing an expression from words is grade 6 content, but this one links three people in a chain, and the last sentence compares the pair's total with Amie instead of describing Amie directly, which grade 7 students often misread.
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
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 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

Building the pair's total, n + 3n = 4n, is routine; the question is decided by which way the last comparison runs. Because the pair sold more than Amie, her expression must come out smaller than 4n, yet the most tempting choice adds the 10. Trying n = 5 settles it: the pair sold 20 tickets, so Amie sold 10, and only one expression gives 10.

Turning words into math moderate Understanding the question
“Together, Jay and Pilar sold ten more tickets than Amie sold” compares the pair's 4n with Amie, so her count is 10 less, though the word more suggests adding. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
Every count is written in terms of n, so no ticket total can be checked unless the student substitutes a value such as n = 5 into each expression. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Write Pilar's tickets and the pair's total in terms of n, then test each choice with a small value such as n = 5 against the sentence comparing the pair with Amie.

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.
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.)
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 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Sees that 10 and the pair's 4n must be subtracted but writes them in the wrong order, 10 − 4n, which is negative once n is 3 or more.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Multiplies Jay's n by Pilar's 3n instead of adding them, getting 3n², and then adds the 10 as the word more suggests.
C Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Follows the word more and adds 10 to the pair's 4n, though the pair sold 10 more than Amie, so Amie sold fewer.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms: $a \times n + b \times n = (a + b) \times n$; n is the unknown number. a and b are the numbers that multiply it. Like terms are terms with the same letter. Add the numbers in front: one n plus three n’s is four n’s, just as 1 apple plus 3 apples is 4 apples.
Ten more than: $\text{if } A = B + 10 \text{, then } B = A - 10$; A and B are two amounts. If A is 10 more than B, A is the bigger one. To get the smaller amount, B, take 10 away from A. The words more than tell you which amount is bigger. They don’t always mean add.

Sort out the question
Jay sold n tickets. The letter n stands for a number you don’t know. Pilar sold three times as many as Jay. Jay and Pilar together sold ten more than Amie. You need an expression for Amie’s tickets. An expression is a math phrase with numbers and letters, like 2n + 1.
Write Pilar’s and the pair’s tickets
Three times as many as n is 3n. So Pilar sold 3n tickets. Together, Jay and Pilar sold n + 3n. Combine like terms:
$n + 3n = 1 \times n + 3 \times n = 4n$
So the pair sold 4n tickets.
Turn the last sentence into math
The pair sold ten more than Amie. So the pair has the bigger number. Amie’s tickets plus 10 make the pair’s 4n.
$\text{Amie} + 10 = 4n$
To find Amie’s tickets, take 10 away from 4n.
$\text{Amie} = 4n - 10$
Watch out: the word more tempts you to add 10. That gives $4n + 10$ , choice C. But Amie sold fewer tickets than the pair, not more.
Rule out the other choices
Choice A, $10 - 4n$ , subtracts in the wrong order. It is negative once n is 3 or more. Choice B, $3n^{2} + 10$ , multiplies n by 3n instead of adding them. So the answer is D, $4n - 10$ .

Check your answer

Try n = 5. Jay sold 5 and Pilar sold 15, so together they sold 20. Amie sold 10 fewer, which is 10. Choice D gives $4 \times 5 - 10 = 10$ . That matches. Choice C gives 30, more than the pair sold.

More questions like this

The original DOE question

Question 70 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 82, question 70. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 14 of 57

2022 Test A Math, question 14

Question

$-12 > 6x$

For what values of x is the above inequality true?

A $x > -6$
B $x < -6$
C $x < -2$
D $x > -2$

Show the answer

Correct answer: C

Divide to solve for x:

$\frac{(-12)}{6} > \frac{6x}{6}$

$-2 > x$

Rewriting $-2 > x$ with the x term first requires reversing the inequality symbol in order to keep the meaning “x is less than $(-2)$ ”.

$x < -2$

Question analysis

This inequality question asks which values of x make −12 > 6x true. Students solve a one-step inequality by dividing and then state the solution with x written first. A negative number on one side and the variable on the other make the direction of the symbol the main thing to get right.

NYS standards: 7.EE.4b two-step inequalities
6.NS.7a inequality as relative position
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 inequalities usually take two steps, such as 3x + 4 > 19. This one needs a single division, and the care goes into reading the result correctly when the variable ends up on the right of the symbol.
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: 2 steps
Reading level: Question and choices: about grade 10, 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

One division finds the boundary, so the question is really about direction. Two things pull students the wrong way: the −12 on the left makes the inequality look like one that must be flipped, and the result −2 > x has x on the right, where it is easy to read backward. Testing x = −3, which gives −12 > −18, shows that values below −2 work.

A very close wrong answer moderate Choosing the answer
x < −2 and x > −2 share the boundary −2 and differ only in direction, so finding the right number settles nothing until the direction is read correctly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that the symbol flips only when both sides are multiplied or divided by a negative number, and that −2 > x says the same as x < −2. (Anderson & Krathwohl, 2001)

Strategy: Divide both sides by 6, asking whether you divided by a negative number, then rewrite the result with x on the left so that the symbol still opens toward the larger side.

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.)
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
Adds 6 to −12 instead of dividing by 6, getting −6, and also reverses the direction, so both the boundary and the symbol are wrong.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Undoes the 6 by adding it to −12 rather than dividing, as if 6x meant x − 6, getting −6 > x and so x < −6.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Finds −2 > x but flips the symbol because −12 is negative, or reads −2 > x as x > −2, though dividing by a positive 6 keeps the direction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing both sides of an inequality: $\text{if } a > b \text{ and } c > 0 \text{, then } \frac{a}{c} > \frac{b}{c}$; a and b are the two sides of an inequality, and c is the number you divide by. If c is positive, the symbol stays the same. It flips only when you multiply or divide both sides by a negative number. Here c is 6, which is positive.
Reading an inequality the other way: $a > b \text{ means the same as } b < a$; a and b are any two numbers. Saying a is greater than b is the same as saying b is less than a. When you swap the two sides, turn the symbol around. Its point still aims at the smaller number.

Sort out the question
The inequality is $-12 > 6x$ . The symbol > means is greater than. So it says −12 is greater than 6 times x. You need every value of x that makes this true.
Divide both sides by 6
To get x alone, divide both sides by 6:
$\frac{-12}{6} > \frac{6x}{6}$
$-2 > x$
You divided by 6, which is positive. So the symbol does not flip.
Watch out: −12 is negative, but that does not matter. Only dividing by a negative number flips the symbol.
Put x first
$-2 > x$ says −2 is greater than x. That means x is less than −2. Swap the sides and turn the symbol around:
$x < -2$
Watch out: keeping > when you swap the sides gives choice D, $x > -2$ . That is the opposite meaning.
Choose the answer
Choices A and B use −6. That comes from adding 6 to −12 instead of dividing by 6. The right boundary is −2, and x is below it.
So the answer is C, $x < -2$ .

Check your answer

Test a number below −2, such as −3. Then $6 \times (-3) = -18$ , and $-12 > -18$ is true. Now test 0, which is above −2. Then $6 \times 0 = 0$ , and $-12 > 0$ is false. So only numbers below −2 work.

More questions like this

The original DOE question

Question 7 as shown on the DOE’s digital test, page 80 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 80, question 7.

Question 15 of 57

2022 Test A Math, question 15

Question

If $5(x - 6y) = 50y$ , what is the value of x in terms of y?

A $16y$
B $15y$
C $11\frac{2}{5}y$
D $11y$

Show the answer

Correct answer: A

To find the value of x in terms of y, solve $5(x - 6y) = 50y$ for x. Begin by distributing 5 on the left side of the equation:

$(5 \times x) - (5 \times 6y) = 50y$

$5x - 30y = 50y$

Solve for the value of x:

$5x = 80y$

$\frac{5x}{5} = \frac{80y}{5}$

$x = 16y$

Question analysis

This algebra question gives the equation 5(x − 6y) = 50y and asks for x in terms of y. Students use the distributive property, or divide both sides by 5, and then solve a literal equation with inverse operations. Forgetting that the 5 multiplies both terms inside the parentheses is the classic slip here.

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 equations such as 5(x − 6) = 50 with numbers. Solving for x in terms of another letter, through the distributive property, is grade 8 algebra, though the choices can be checked by substitution.
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: 3 steps
Reading level: Question and choices: about grade 6, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The equation needs only three moves, distribute or divide by 5, add the y term, divide, so the risk lies in the first move. Multiplying only the x by 5 is the classic slip, and it gives 56/5 y, a value that is not listed but sits between two of the choices. Substituting y = 1 and the chosen x back into 5(x − 6) = 50 checks any choice in seconds.

Variables instead of numbers moderate Understanding the question
Both sides hold y and the result must be a multiple of y, so there is no number to check against unless the student substitutes, for instance y = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
The 5 has to multiply both terms in the parentheses, giving 5x − 30y, and then 80y ÷ 5 must come out exactly; dividing both sides by 5 first keeps the numbers small. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Distribute the 5 to both terms in the parentheses, or divide both sides by 5 first, then add the y term to both sides; substitute your choice back for x to check.

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.

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 15y; it sits one y below the correct value and would need a slip such as 80 ÷ 5 = 15, so it mostly draws guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 57/5 y; multiplying only the x by 5 leads to 56/5 y, or 11 1/5 y, so this mostly draws students who land near it.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Multiplies only the x by 5, so 5x − 6y = 50y and 5x = 56y, then drops the remainder of 56 ÷ 5 = 11 R1, keeping 11y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. Multiply each term inside by a. Use it when a number multiplies a whole sum.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
The equation is $5(x - 6y) = 50y$ . It has two letters, x and y. In terms of y means get x alone on one side. The other side may still have y in it. So the answer will look like a number times y.
Multiply out the parentheses
Use the distributive property. The 5 multiplies both terms inside.
$5(x - 6y) = 5 \times x - 5 \times 6y = 5x - 30y$
$5x - 30y = 50y$
Watch out: the 5 must multiply the 6y too. Multiplying only the x gives $5x = 56y$ . Then x is $11\frac{1}{5}y$ . That is not a choice, but it sits close to C and D. Don’t pick one of them just because it is close.
Get the x term alone
Add 30y to both sides. This keeps the equation balanced.
$5x - 30y + 30y = 50y + 30y$
$5x = 80y$
Like terms add: 50 y’s plus 30 y’s is 80 y’s.
Divide by 5
5x means 5 times x. Divide both sides by 5 to get x by itself.
$x = \frac{80y}{5} = 16y$
So the answer is A, $16y$ .

Check your answer

Pick y = 1. Then x = 16. Put both into the first equation: $5(16 - 6 \times 1) = 5 \times 10 = 50$ . The right side is $50 \times 1 = 50$ . Both sides match. A faster route: divide both sides by 5 first. Then $x - 6y = 10y$ , so x = 16y.

More questions like this

The original DOE question

Question 72 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 83, question 72. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 16 of 57

2022 Test A Math, question 16

Question

$An L-shaped outline with all right angles. The top side is labeled 6 in. and the right side, the full height of the figure, is labeled 5 in. The left side, running down from the top-left corner, is labeled 2 in.; from its lower end a horizontal side labeled 3 in. runs to the right, and from there an unlabeled side runs down to the bottom, which is also unlabeled and meets the right side.$

Scale: $\frac{1}{2}\text{ in.} = 25\text{ ft}$

A scale drawing of a new pool is shown above. All the corners are right angles. What is the perimeter of the actual pool?

A550 ft
B800 ft
C1,100 ft
D1,500 ft

Show the answer

Correct answer: C

Determine the two missing side lengths on the scale drawing:

All corners are right angles, so the figure can be divided into rectangles with opposite sides that are congruent in length. This means the sum of the shorter vertical sides on the left and in the center of the figure ( $2\text{ in.} + x\text{ in.}$ ) must equal the length of the longer vertical side on the right side of the figure (5 in.):

$2 + x = 5$

$x = 3$

So, the missing vertical side length of the figure is 3 in.

Again, because opposite sides of a rectangular figure are congruent in length, the sum of the shorter horizontal sides in the middle and bottom of the figure ( $3\text{ in.} + x\text{ in.}$ ) must equal the length of the longer horizontal side on the top of the figure (6 in.):

$3 + x = 6$

$x = 3$

So, the missing horizontal side length of the figure is 3 in.

Determine the perimeter by adding all the side lengths of the figure:

$6 + 2 + 3 + 3 + 3 + 5 = 22$

This is a scale drawing in which $\frac{1}{2}\text{ in.} = 25\text{ ft}$ . Write a proportion to determine the value of x, the unknown number of feet that is represented by 22 in.:

$\frac{\left(\frac{1}{2}\right)\text{ in.}}{25\text{ ft}} = \frac{22\text{ in.}}{x}$

Cross multiply and solve the equation for x:

$\left(\frac{1}{2}\right)(x) = (22)(25)$

$\frac{1}{2}x = 550$

$x = 1{,}100$

Question analysis

A scale drawing shows an L-shaped pool with all right angles, four labeled sides and two unlabeled ones, and a scale of 1/2 inch to 25 feet. Students find the perimeter of the actual pool. It combines the perimeter of a composite figure with a scale given for half an inch, so the missing sides and the fractional scale are both easy to mishandle.

NYS standards: 7.G.1 scale drawings
7.RP.1 unit rates with fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are grade 7 content, and a typical task converts one given length. Here two sides are unlabeled and must be worked out, all six added, and a scale given for half an inch turned into feet per inch.
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: Reading a figure, table or graph moderate
Combines ideas moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 2, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The figure gives four of the six sides and the scale gives feet per half inch, so two conversions hide behind a simple-looking perimeter question. Students who add only the labeled sides, or who read the scale as 25 feet per inch, reach totals that look reasonable for a pool and are listed. Tracing the outline to count six sides, and doubling 25 before multiplying, avoids both.

Reading a figure, table or graph moderate Understanding the question
Two sides of the notch carry no labels, and the 5 in. arrow spans the full right side, so the missing lengths come from 5 − 2 = 3 and 6 − 3 = 3. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
The perimeter of an L-shaped drawing has to be joined to a scale in which half an inch, not a whole inch, stands for 25 feet. (Embretson, 1998; Daniel & Embretson, 2010)
Many steps minor Working it out
Four stages follow in order, the two missing sides, the 22-inch perimeter, the feet per inch and the product, and a slip at most of them reaches a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use opposite sides to fill in the two missing lengths, add all six sides of the drawing, then double the 25 feet to get the feet one inch stands for, and multiply.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

A Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds the drawing's 22-inch perimeter but multiplies by 25, using the scale as if 1 inch, not 1/2 inch, stood for 25 feet.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: moderate
Adds only the four labeled sides, 6 + 2 + 3 + 5 = 16 inches, and scales that to 800 feet, missing the two unlabeled sides of the notch.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Multiplies the 6-inch and 5-inch sides to get 30, the outer rectangle's area, and scales that by 50 feet per inch as if it were a length.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter: $P = \text{the sum of all the side lengths}$; P is the perimeter, the distance all the way around a shape. Add the length of every side, even sides with no label. It is like walking around the edge of the pool and adding up each part of the walk.
Using a scale: $\text{real length} = \text{drawing length} \times \text{feet per inch}$; The drawing length is measured on the drawing, in inches. Feet per inch is how many real feet one inch of the drawing stands for. Multiplying turns a length on the drawing into a real length.

Sort out the question
The drawing is an L shape with six sides. Four sides have labels. The top is 6 in. long, and the right side is 5 in. long. The left side is 2 in. long, and a middle flat side is 3 in. long. Two sides have no label. The scale says 1/2 inch on the drawing stands for 25 feet. You need the perimeter of the real pool, in feet.
Find the missing up-and-down side
Call it s. It runs down from the right end of the 3 in. side to the bottom. The side labeled 2 in. and side s together go from the top to the bottom. All corners are right angles, so they match the full height on the right, 5 in.
$2 + s = 5$
$s = 5 - 2 = 3$
Side s is 3 inches.
Find the bottom side
Call the bottom side b. The middle flat side, 3 in., and side b together are as wide as the top, labeled 6 in.
$3 + b = 6$
$b = 6 - 3 = 3$
The bottom side is 3 inches.
Add all six sides
Go around the shape: top, right side, bottom, side s, the 3 in. side, then the 2 in. side.
$P = 6 + 5 + 3 + 3 + 3 + 2 = 22$
The perimeter on the drawing is 22 inches. Watch out: count the two unlabeled sides. Adding only the four labeled sides gives 16 inches. That leads to 800 feet, choice B.
Find the feet per inch
The scale says 1/2 inch stands for 25 feet. A whole inch is two halves, so 1 inch stands for twice as much.
$25 \times 2 = 50$
Each inch of the drawing is 50 real feet. Watch out: don’t use 25 feet per inch. That gives $22 \times 25 = 550$ feet, choice A.
Change the perimeter to feet
Call the real perimeter F, for feet. Use the scale formula with 22 inches and 50 feet per inch.
$\text{real length} = \text{drawing length} \times \text{feet per inch}$
$F = 22 \times 50 = 1{,}100$
The real pool’s perimeter is 1,100 feet. So the answer is C, 1,100 ft.

Check your answer

Change each side to feet first. At 50 feet per inch, the sides are 300, 250, 150, 150, 150 and 100 feet. $300 + 250 + 150 + 150 + 150 + 100 = 1{,}100$ . That matches. Choice D, 1,500, comes from $6 \times 5 \times 50$ . But 6 × 5 is an area, not a distance around.

More questions like this

The original DOE question

Question 73 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 83, question 73. Open page 83 of the handbook (PDF)

Question 17 of 57

2022 Test A Math, question 17

Question

Maria rides her bicycle to school at a constant speed of 15 miles per hour. If the distance to school is 6 miles, how many minutes will it take Maria to get to school?

Show the answer

Correct answer: C

Maria’s rate of speed is given as 15 miles per hour. Since 1 hour is equivalent to 60 minutes, her rate of speed, in miles per minute, can be determined.

$\frac{15\text{ miles}}{1\text{ hour}} = \frac{15\text{ miles}}{60\text{ minutes}} = \frac{1\text{ mile}}{4\text{ minutes}}$ , or $\frac{1}{4}$ mile per minute.

Therefore, it takes Maria 4 minutes to ride 1 mile.

The distance to school is 6 miles, so it will take $6 \times 4 = 24$ minutes to ride to school.

Question analysis

Maria bikes 6 miles to school at a constant 15 miles per hour, and students find how many minutes the ride takes. It tests using a rate to find a time and converting hours to minutes. With the speed given per hour and the question asking, in bold, for minutes, keeping track of units is the main hurdle.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Distance, speed and time with whole numbers is grade 6 rate work. The only extra demand is an answer in minutes when the speed is per hour, a conversion grade 7 students make routinely.
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: Something you must already know minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The numbers are small, but the speed is per hour and the answer is wanted in minutes, so students have to choose a route through the units. Finding the minutes per mile first, 4 at 15 miles an hour, turns the rest into a single multiplication. The wrong choices come from using a number in the stem as the answer, from mixing the 60 minutes with the 6 miles, or from rounding 0.4 hour to a half hour.

Something you must already know minor Working it out
Nothing on the page says that time is distance divided by speed or that an hour has 60 minutes, and the bold word minutes requires both. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
6 ÷ 15 is less than 1, 0.4 hour, and turning it into minutes means 0.4 × 60; the cleaner route, 60 ÷ 15 = 4 minutes a mile, avoids decimals. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find how many minutes Maria needs to ride one mile at 15 miles per hour, then multiply by the number of miles to school.

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.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the 60 minutes in an hour by the 6 miles, 60 ÷ 6 = 10, as if the ride took a full hour, and ignores the 15 mph speed.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 15 in 15 miles per hour as the number of minutes, a number copied from the question without any calculation.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Sees that 6 miles is a little under half of 15, rounds the time of 0.4 hour up to half an hour, and gives 30 minutes.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Minutes for each mile: $\text{minutes per mile} = 60 \div \text{miles per hour}$; Miles per hour tells how many miles you ride in one hour, which is 60 minutes. Sharing the 60 minutes evenly among those miles gives the minutes for each mile. This works when the speed stays the same the whole time.
Distance, speed and time: $\text{time} = \text{distance} \div \text{speed}$; Distance is how far you go, here in miles. Speed is how far you go in one hour. Dividing the distance by the speed gives the time in hours.

Sort out the question
Maria rides at a constant speed of 15 miles per hour. Constant means it never changes. School is 6 miles away. You need the time in minutes, not hours. You must know that one hour has 60 minutes.
Find the minutes for one mile
15 miles per hour means Maria rides 15 miles in 60 minutes. Use the minutes-per-mile formula:
$60 \div 15 = 4$
So each mile takes Maria 4 minutes.
Find the minutes for 6 miles
Each of the 6 miles takes 4 minutes:
$6 \times 4 = 24$
So the ride takes 24 minutes.
Watch out: 15 is the speed, not a time, so choice B is a trap. And $60 \div 6 = 10$ , choice A, ignores the speed.
Choose the answer
Another way is the time formula. $6 \div 15 = 0.4$ , so the ride takes 0.4 of an hour. In minutes, that is $0.4 \times 60 = 24$ .
Watch out: rounding 0.4 hour up to half an hour gives choice D, 30.
So the answer is C, 24.

Check your answer

Check with the speed. 24 minutes is $\frac{24}{60} = \frac{2}{5}$ of an hour. In a full hour Maria rides 15 miles. So in 2/5 of an hour she rides $\frac{2}{5} \times 15 = 6$ miles. That is exactly the distance to school.

More questions like this

The original DOE question

Question 9 as shown on the DOE’s digital test, page 82 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 82, question 9.

Question 18 of 57

2022 Test A Math, question 18

Question

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

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

A17
B117
C163
D173

Show the answer

Correct answer: C

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

$180^{\circ} = 73^{\circ} + 90^{\circ} + x$

$180^{\circ} = 163^{\circ} + x$

$17^{\circ} = x$

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

$s = 180^{\circ} - 17^{\circ}$

$s = 163^{\circ}$

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

Question analysis

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

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle problems use supplementary, vertical and adjacent angles. This one also needs the angle sum of a triangle and the angles formed by parallel lines and a transversal, grade 8 facts, chained across a busy figure.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Combines ideas moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph moderate Understanding the question
The 73° sits between line p and the vertical segment, not between p and m, and r is the obtuse angle above m on the left of line n, far from every marked angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
Three facts chain together: the triangle's angles sum to 180°, the angles along straight line m sum to 180°, and corresponding angles at the parallel lines n and p are equal. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
The angle pairs formed by a line crossing two parallel lines are a grade 8 topic, and the question uses them without naming them. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the triangle's third angle, 180 − 90 − 73 = 17, and stops, though r is the obtuse angle next to the 17° angle, not that angle itself.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No angle fact in the figure gives 117; even 180 − 73 is 107, so this choice mostly catches guesses at an obtuse angle.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds the 17° angle and subtracts it from 180 but borrows without reducing the 8, so 180 − 17 comes out as 173.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Lines n and p are parallel, and line m crosses both. Near line p, a straight-up segment meets line m. The small square shows a right angle, 90°. Line p, line m and that segment make a small triangle. Its top angle is 73°.
You need r, the wide angle where line n crosses line m.
Find the triangle’s third angle
The third angle is where line p crosses line m. Call it x. Use the triangle rule:
$73 + 90 + x = 180$
$x = 180 - 163 = 17$
So that angle is 17°.
Watch out: this is not r. Stopping here gives choice A, 17.
Use the straight line m
The 17° angle sits above line m, on the right of line p. Beside it, on the left of line p, is a wide angle. Call it s. The two angles fill the straight line m, so they add to 180°:
$s = 180 - 17 = 163$
So s is 163°.
Watch out: borrow carefully. 180 − 17 is 163, not 173, choice D.
Carry the angle to line n
Angle s is above line m and to the left of line p. Angle r is above line m and to the left of line n. They are in the same position, so they are corresponding angles. Since n and p are parallel:
$r = s = 163$
So the answer is C, 163.

Check your answer

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

More questions like this

The original DOE question

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

Question 19 of 57

2022 Test A Math, question 19

Question

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

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

A112 sq cm
B396 sq cm
C612 sq cm
D748 sq cm

Show the answer

Correct answer: C

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

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

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

Question analysis

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

NYS standards: 6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The area of a parallelogram is grade 6 content, used here once with the height already drawn and labeled. Grade 7 area tasks usually involve circles, composite shapes or scale drawings, so this is easier.
Estimated difficulty: Routine: one familiar step, little to misread.
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 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One multiplication answers the question once the right pair of lengths is chosen. Each wrong choice pairs or combines the labeled lengths in another way: the two sides as for a rectangle, the height with the slanted side, or all four sides as a perimeter. Following the right-angle mark at T from QT down to PS identifies the base and height that belong together.

Reading a figure, table or graph minor Understanding the question
Three lengths are labeled, 34 cm, 18 cm and 22 cm, but only the base PS and the height QT meet at the right-angle mark; the slanted side RS is not needed. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The formula, base times height, is not given, and the height must be the perpendicular distance between two sides, not the length of a slanted side. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds all four sides, 2 × (34 + 22) = 112, finding the perimeter of the parallelogram instead of its area.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Multiplies the 18 cm height by the 22 cm slanted side, a pair that does not meet at a right angle, getting 396.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies the two side lengths, 34 × 22 = 748, as if a parallelogram's area were side times side like a rectangle's.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
PQRS is a parallelogram, a four-sided shape whose opposite sides are parallel. The figure shows three lengths. Side PS, the bottom, is 34 cm. Segment QT goes straight down from Q to PS and is 18 cm. Side RS is slanted and is 22 cm. You need the area, in square centimeters.
Pick the base and the height
The base is PS, 34 cm. The height must meet the base at a right angle. The small square at T shows that QT does. So the height is QT, 18 cm.
Watch out: the 22 cm side is slanted, so it is not the height. Multiplying 22 × 18 gives 396, choice B.
Multiply the base by the height
Use the area formula:
$A = 34 \times 18$
Break it up to multiply:
$34 \times 18 = 34 \times 20 - 34 \times 2 = 680 - 68 = 612$
The area is 612 square centimeters.
Choose the answer
Watch out: 34 × 22 = 748, choice D, treats the shape like a rectangle. Adding all four sides gives 112, choice A. That is the perimeter, the distance around.
So the answer is C, 612 sq cm.

Check your answer

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

More questions like this

The original DOE question

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

Question 20 of 57

2022 Test A Math, question 20

Question

Allison has 5 stamp albums with 576 stamps in each album. If she transfers her stamp collection to 6 albums, each holding 378 stamps, how many stamps will be left over?

A198
B612
C620
D632

Show the answer

Correct answer: B

Determine the total number of stamps in 5 albums holding 576 stamps:

$5 \times 576 = 2{,}880$

Determine the total number of stamps in 6 albums holding 378 stamps:

$6 \times 378 = 2{,}268$

Subtract 2,268 from 2,880 to find the number of stamps that will be left over:

$2{,}880 - 2{,}268 = 612$

Question analysis

Allison moves a stamp collection from 5 albums of 576 stamps into 6 albums that hold 378 each, and students find how many stamps are left over. It tests multiplying and subtracting large whole numbers in a word problem. The question gives no total, so the collection has to be counted before the new albums' capacity is taken away.

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: Below grade 7 Whole-number multiplication and subtraction in a three-step word problem is easier than grade 7 work with fractions, decimals or negatives. Because the products are large, a slip or a rounded estimate easily gives another choice.
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 moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The plan is clear from the stem: count the stamps, then take away what the six new albums hold, so the question tests careful arithmetic. Its choices punish the estimating that saves time elsewhere, since rounding either factor shifts the result by only a few tens, onto another listed number. Students who skip the album counts altogether and compare one album with one album find 198 waiting.

A very close wrong answer moderate Choosing the answer
612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply to find the total number of stamps, multiply again to find how many the six new albums hold, subtract, and check the last digit of each product.

Traps

Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
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: moderate
Subtracts 378 from 576, comparing one old album with one new album, and ignores that there are 5 old albums and 6 new ones.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Rounds 576 to 580 and 378 to 380 before multiplying, so 2,900 − 2,280 = 620, an estimate rather than the exact number left over.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: weak
Rounds only 576 up to 580, getting 2,900 stamps, then subtracts the exact 2,268, so the count comes out 20 too high.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total in equal groups: $\text{total} = \text{number of groups} \times \text{amount in each group}$; Here the groups are albums. The amount in each group is how many stamps one album holds. When every group is the same size, multiplying is a fast way to add them all up.

Sort out the question
Allison has 5 albums with 576 stamps in each. She moves all her stamps into 6 new albums. Each new album holds 378 stamps. Some stamps will not fit. You need the number left over.
Count all her stamps
Use the equal-groups formula. Split 576 into 500 + 76 to make it easy:
$5 \times 576 = 5 \times 500 + 5 \times 76$
$2{,}500 + 380 = 2{,}880$
Allison has 2,880 stamps.
Count what the new albums hold
Do the same for the 6 new albums. Split 378 into 300 + 78:
$6 \times 378 = 6 \times 300 + 6 \times 78$
$1{,}800 + 468 = 2{,}268$
The new albums hold 2,268 stamps in all.
Subtract to find the leftovers
The stamps left over are the ones she has minus the ones that fit:
$2{,}880 - 2{,}268 = 612$
Watch out: $576 - 378 = 198$ , choice A, compares just one old album with one new one. Rounding to 580 and 380 first gives 620, choice C. Rounding only 576 gives 632, choice D. Use the exact numbers.
So the answer is B, 612.

Check your answer

Add the leftovers to the stamps that fit: $2{,}268 + 612 = 2{,}880$ . That is exactly the number Allison started with. So no stamps were lost or counted twice.

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 C Math, page 84, question 11.

Question 21 of 57

2022 Test A Math, question 21

Question

Start with the number 135.28 and perform the following steps in order:

Step 1: Multiply by 10.
Step 2: Add 0.5 to the result of Step 1.
Step 3: Drop the digits after the decimal point.
Step 4: Divide by 10.

What is the result?

A13.5
B135
C135.2
D135.3

Show the answer

Correct answer: D

Begin by multiplying 135.28 by 10:

$135.28 \times 10 = 1{,}352.8$

Add 0.5:

$1{,}352.8 + 0.5 = 1{,}353.3$

Drop the digits after the decimal point, so $1{,}353.3$ becomes $1{,}353$ .

Divide by 10:

$1{,}353 \div 10 = 135.3$

Question analysis

Students start with 135.28 and carry out four listed steps: multiply by 10, add 0.5, drop the digits after the decimal point, and divide by 10. It tests place value and operations with decimals. Skipping a step, or doing two of them in the wrong order, changes the result, and the choices are built from those slips.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Multiplying and dividing a decimal by 10 and adding 0.5 are grade 5 and 6 skills. The four steps are spelled out, so the demand is following them exactly and in order, not choosing a method.
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
Many steps minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 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 steps spell out how to round 135.28 to the nearest tenth, but nothing says so, and they have to be followed exactly. The decisive moment is step 2, where adding 0.5 pushes 1,352.8 past 1,353; skip it and the digits are simply cut, not rounded. Students who recognize the routine can check the result directly, since 135.28 lies closer to 135.3 than to 135.2.

A very close wrong answer moderate Choosing the answer
135.3 and 135.2 differ only in the tenths digit, and 135 shares their whole-number part, so a rough sense of the result's size rules out nothing. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
Four listed steps move the decimal point twice, and doing them in another order, or skipping one, still produces a believable number near 135. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Adding 0.5 to 1,352.8 carries into the ones place, since 8 tenths and 5 tenths make 1.3, which turns 1,352 into 1,353 before the decimals are dropped. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write down the number after each step instead of doing them in your head, and carry out the steps exactly in the order listed.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Skips step 1, so adds 0.5 to 135.28, drops the decimals to get 135, and then divides by 10 to reach 13.5.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Divides by 10 before dropping the digits after the decimal point, so 1,353.3 becomes 135.33 and then 135, doing steps 3 and 4 in reverse order.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Leaves out step 2, or adds 0.5 without carrying, so 1,352 is kept and 135.2 results: 135.28 cut down instead of rounded up.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying and dividing by 10: $a \times 10 \quad \text{and} \quad a \div 10$; a is any decimal number. Multiplying by 10 moves its decimal point one place to the right, so each digit is worth ten times as much. Dividing by 10 moves the point one place to the left. The two steps undo each other.

Sort out the question
You start with 135.28 and do four steps, in order. Write down the number after each step. Doing them in your head, or in another order, leads to a wrong choice. You need the number at the end.
Multiply by 10
Step 1. Multiplying by 10 moves the decimal point one place to the right:
$135.28 \times 10 = 1{,}352.8$
The digits stay the same. Only the decimal point moves.
Add 0.5
Step 2. Line up the decimal points. 8 tenths plus 5 tenths is 13 tenths. That is 1 whole and 3 tenths, so carry the 1 into the ones place:
$1{,}352.8 + 0.5 = 1{,}353.3$
Watch out: skipping this step, or forgetting to carry, keeps 1,352. That leads to choice C, 135.2.
Drop the digits after the point
Step 3. Cut off the digits after the decimal point. Do not round. So 1,353.3 becomes 1,353.
Watch out: do step 3 before step 4. Dividing first gives 135.33. Dropping its decimals then gives 135, choice B.
Divide by 10
Step 4. Dividing by 10 moves the decimal point one place to the left:
$1{,}353 \div 10 = 135.3$
Skipping step 1 gives 13.5, choice A. So the answer is D, 135.3.

Check your answer

These four steps are a recipe for rounding to the nearest tenth. 135.28 is between 135.2 and 135.3. Its hundredths digit is 8, which is 5 or more, so it rounds up to 135.3. The recipe and the rounding agree.

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 C Math, page 86, question 13.

Question 22 of 57

2022 Test A Math, question 22

Question

If $x = -2$ , what is the value of $|x + x^{2} + x^{3}| - |x + 2x + 3x|$ ?

A−18
B−6
C−2
D2

Show the answer

Correct answer: B

Substitute $-2$ for each value of x in the expression:

$\left|(-2) + (-2)^{2} + (-2)^{3}\right|$
minus
$|(-2) + 2(-2) + 3(-2)|$

Evaluate the terms involving exponents:

$|(-2) + 4 + (-8)|$
minus
$|(-2) + 2(-2) + 3(-2)|$

Evaluate the terms involving multiplication:

$|(-2) + 4 + (-8)|$
minus
$|(-2) + (-4) + (-6)|$

Simplify the expression:

$|2 + (-8)| - |(-2) + (-10)|$

$\lvert -6 \rvert - \lvert -12 \rvert$

The absolute value of $(-6)$ is 6, and the absolute value of $(-12)$ is 12, so $\lvert -6 \rvert - \lvert -12 \rvert$ is the equivalent of subtracting 12 from 6:

$6 - 12 = (-6)$

Question analysis

This substitution question sets x = −2 in an expression that subtracts one absolute value from another: the first holds x, x squared and x cubed, the second x, 2x and 3x. Students evaluate the expression. It tests powers of negative numbers, absolute value and integer subtraction, and the signs of the square and the cube are where answers go wrong.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Substituting a negative number and taking absolute values is typical grade 7 sign practice. This expression puts a square and a cube of a negative inside one pair of bars and then subtracts two absolute values, so there are several places to slip.
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
Solution: 4 steps
Reading level: Question and choices: about grade 7, 8 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every step is routine, but the expression stacks sign decisions, and each wrong choice records a different one going wrong. The powers of −2 alternate in sign, which students who treat every power as positive, or the square as negative, miss. The last step catches students who know the rules: both absolute values are positive, yet 6 − 12 is still negative, so the answer keeps its minus sign.

Error-prone arithmetic moderate Working it out
With x = −2 the square is 4 but the cube is −8, so the first bars hold −2 + 4 − 8 = −6, and the sign changes from each term to the next. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Six terms are substituted, two pairs of bars are simplified and made positive, and only then is 12 subtracted from 6, four stages that must come in a fixed order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out each power of −2 separately, keeping its sign, add the terms inside each pair of bars, take both absolute values, and subtract only at the end.

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 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Simplifies both insides correctly but treats the first bars like parentheses, keeping −6, so −6 − 12 = −18.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes (−2)³ as positive 8, as if every power of a negative were positive, so the first bars hold −2 + 4 + 8 = 10, and 10 − 12 = −2.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Takes (−2)² as −4, so the first bars hold −2 − 4 − 8 = −14, and 14 − 12 = 2; dropping the minus from x throughout also gives 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; a is any number. The bars mean absolute value: how far a is from 0 on the number line. A distance is never negative, so |−6| = 6 and |6| = 6. Work out what is inside the bars first, then take the absolute value.
Powers of a negative number: $(-2)^{2} = (-2) \times (-2) \quad (-2)^{3} = (-2) \times (-2) \times (-2)$; The small raised number is the exponent. It tells how many times −2 is multiplied by itself. Two negatives multiply to a positive. So an even number of negative factors gives a positive answer, and an odd number gives a negative one.

Sort out the question
You need the value of $|x + x^{2} + x^{3}| - |x + 2x + 3x|$ when x = −2. The bars | | mean absolute value. Work out each pair of bars on its own. Subtract only at the end.
Find the powers of −2
Put −2 in for x. Keep each negative sign:
$(-2)^{2} = (-2) \times (-2) = 4$
$(-2)^{3} = 4 \times (-2) = -8$
The square is positive, but the cube is negative.
Watch out: taking $(-2)^{3}$ as +8 makes the first bars hold 10. That leads to choice C, −2.
Work out the first bars
Add the three terms inside the first pair of bars:
$|(-2) + 4 + (-8)| = |-6| = 6$
−2 + 4 is 2, and 2 + (−8) is −6. The distance from −6 to 0 is 6.
Work out the second bars
Here 2x means 2 times x, so it is −4. And 3x is −6:
$|(-2) + (-4) + (-6)| = |-12| = 12$
All three terms are negative, so they add to −12. Its absolute value is 12.
Subtract the second value
Now subtract the second value from the first:
$6 - 12 = -6$
Watch out: the bars make each part positive, but the difference can still be negative. Keeping −6 inside the first bars gives −6 − 12 = −18, choice A.
So the answer is B, −6.

Check your answer

Check the second bars a faster way. $x + 2x + 3x = 6x$ , and $6 \times (-2) = -12$ . That matches. Also, the second value, 12, is bigger than the first, 6. So the difference must be negative, as −6 is.

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 C Math, page 87, question 14.

Question 23 of 57

2022 Test A Math, question 23

Question

$\frac{3+(-3)}{3-(-3)} =$

A−1
B0
C $\frac{1}{6}$
D1

Show the answer

Correct answer: B

Simplify the expression.

$3 + (-3)$ is the equivalent of subtracting 3 from 3, and $3 - (-3)$ is the equivalent of adding 3 to 3:

$\frac{3+(-3)}{3-(-3)} = \frac{3-3}{3+3} = \frac{0}{6}$

The zero property of division states that zero divided by any nonzero number equals zero, so $\frac{0}{6} = 0$ .

Question analysis

This fraction has 3 plus negative 3 on top and 3 minus negative 3 on the bottom, and students find its value. It tests adding and subtracting negative numbers and dividing integers. The top and the bottom look almost alike, so each sign has to be read carefully before anything is divided.

NYS standards: 7.NS.1c rational number subtraction
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding and subtracting a negative integer and dividing are grade 7 basics, and here the numbers are small. It is a short routine calculation, easier than typical grade 7 work, though the two signs invite a quick misreading.
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 16+, 2 words.
Corrected from the DOE handbook: A grammatical error in the explanation was corrected.
A rule stated too broadly in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is two one-step sign calculations and a division by 6. The danger is reading the top and the bottom as alike: they differ in a single sign, and treating them the same way gives 1 or −1. Students who rewrite 3 − (−3) as 3 + 3 and 3 + (−3) as 3 − 3 see 0 over 6 at once, and zero divided by a nonzero number is zero.

Error-prone arithmetic minor Working it out
The top adds −3 and the bottom subtracts −3, signs that look nearly the same but give 0 and 6, so each has to be rewritten before anything is divided. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Simplify the top and the bottom separately, rewriting the subtraction of a negative as an addition, and only then divide the top by the bottom.

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 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Adds the sizes on top and keeps the minus, taking 3 + (−3) as −6, so the fraction becomes −6/6 = −1.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No sign rule gives a top of 1, so 1/6 mostly draws students who find the bottom, 6, and are unsure what a top of 0 means.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Treats the top like the bottom, adding 3 and 3 in both places, so the fraction becomes 6/6 = 1.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “3 − (−3) is equivalent of adding 3 to 3”, without the article that the same sentence uses in “3 + (−3) is the equivalent of subtracting 3 from 3”. The site now says “is the equivalent of adding 3 to 3”.
Explanation: The DOE’s explanation says “The zero property of division states that zero divided by any number equals zero.” Zero divided by zero is undefined, not zero, so the rule holds only when the divisor is not zero. The site now says “zero divided by any nonzero number equals zero”. Here the divisor is 6, so 0/6 = 0, as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding a negative: $a + (-b) = a - b$; a and b are numbers. Adding a negative is the same as subtracting. On a number line, adding −3 moves you 3 steps left, just like subtracting 3.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding. Taking away a debt of 3 dollars leaves you 3 dollars better off. On a number line, you move right.
Zero divided by a number: $\frac{0}{b} = 0 \text{ when } b \neq 0$; b is any number except 0. Zero split into b equal parts leaves 0 in each part. The rule needs b not to be 0, because dividing by 0 has no answer.

Sort out the question
The top of the fraction is $3 + (-3)$ , and the bottom is $3 - (-3)$ . A fraction bar means divide. Work out the top and the bottom separately first. Then divide the top by the bottom.
Work out the top
Adding −3 is the same as subtracting 3:
$3 + (-3) = 3 - 3 = 0$
A number plus its opposite is always 0.
Watch out: do not add the 3s and keep the minus sign. That makes the top −6 and leads to choice A, −1.
Work out the bottom
Subtracting −3 is the same as adding 3:
$3 - (-3) = 3 + 3 = 6$
Watch out: the top and bottom look alike, but they are not. Making both of them 6 gives 6/6 = 1, choice D.
Divide the top by the bottom
Now the fraction is 0 over 6. Zero divided by any number other than zero is zero:
$\frac{3+(-3)}{3-(-3)} = \frac{0}{6} = 0$
A top of 0 does not make the fraction 1/6, choice C. So the answer is B, 0.

Check your answer

Check by multiplying. A division is right when the answer times the bottom gives the top. Here $0 \times 6 = 0$ , which is the top. So 0/6 really is 0.

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 C Math, page 89, question 16. This is the DOE’s version, before the corrections listed in the analysis.

Question 24 of 57

2022 Test A Math, question 24

Question

$12.96 \div 0.08 =$

A1.62
B16.2
C162
D1,620

Show the answer

Correct answer: C

Since the greatest number of decimal places is 2 (hundredths), multiply both terms by 100, and then divide:

$12.96(100) \div 0.08(100) =$

$1{,}296 \div 8 =$

$162$

Question analysis

This question asks students to divide 12.96 by 0.08, a decimal division with no story around it. It tests dividing multi-digit decimals with the standard algorithm. The choices differ only in where the decimal point falls, so moving the point the same number of places in both numbers is the whole challenge.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing multi-digit decimals is grade 6 fluency that grade 7 work builds on rather than teaches. The division comes out exactly, so the real demand is placing the decimal point correctly, a single routine skill.
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
Solution: 2 steps
Reading level: Question and choices: about grade 16+, 1 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 choice uses the digits 1, 6 and 2, so the question tests only where the decimal point goes. Dividing by a number less than one gives a result larger than 12.96, which feels wrong to students who expect division to shrink a number. Estimating first, 12.96 ÷ 0.08 is about 13 × 12.5, puts the result in the low hundreds and rules out the other three at once.

Error-prone arithmetic moderate Working it out
The divisor 0.08 has two decimal places, so both numbers must be multiplied by 100, turning the problem into 1,296 ÷ 8, and the quotient comes out larger than 12.96. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply the dividend and the divisor by the same power of 10 so that the divisor becomes a whole number, then carry out the long division.

Traps

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 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Divides 12.96 by 8, leaving out the decimal point in the divisor 0.08, so the quotient, 1.62, is 100 times too small.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Moves the decimal point one place instead of two, in effect dividing 12.96 by 0.8, so the quotient, 16.2, is ten times too small.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Shifts the decimal point three places in 12.96 but only two in 0.08, dividing 12,960 by 8, so the quotient is ten times too big.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a decimal: $a \div b = (a \times 100) \div (b \times 100)$; a is the number being divided. b is the divisor, the number you divide by. Multiplying both by the same number, like 100, does not change the answer. Use 100 when the divisor has two decimal places, so it becomes a whole number.

Sort out the question
You need $12.96 \div 0.08$ . The divisor, the number you divide by, is 0.08. It has two decimal places. Dividing by a decimal is hard, so make the divisor a whole number first. All four choices use the digits 1, 6 and 2. So the real question is where the decimal point goes.
Make the divisor a whole number
Multiply both numbers by 100. This moves each decimal point two places to the right.
$12.96 \div 0.08 = (12.96 \times 100) \div (0.08 \times 100)$
$= 1{,}296 \div 8$
The new problem has the same answer. Watch out: move both points the same number of places. Moving the point three places in 12.96 but two in 0.08 gives 1,620, choice D.
Divide by 8
Split 1,296 into parts that 8 divides easily.
$1{,}200 \div 8 = 150 \qquad 96 \div 8 = 12$
$1{,}296 \div 8 = 150 + 12 = 162$
Watch out: ignoring the point in 0.08 means dividing 12.96 by 8. That gives 1.62, choice A. Moving the point only one place gives 16.2, choice B. So the answer is C, 162.

Check your answer

Multiply back: $162 \times 0.08 = 12.96$ . That matches. Also, 0.08 is less than 1. Dividing by a number less than 1 makes the answer bigger. And $1{,}620 \times 0.08 = 129.6$ , which is far too big.

More questions like this

The original DOE question

Question 81 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 85, question 81. Open page 85 of the handbook (PDF)

Question 25 of 57

2022 Test A Math, question 25

Question

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

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

A $\frac{1}{2}$ unit
B1 unit
C $1\frac{1}{2}$ units
D2 units

Show the answer

Correct answer: A

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

$7 - 1 = 6$

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

$6 \div 2 = 3$

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

$1 + 3 = 4$

$7 - 3 = 4$

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

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

$6 - 3 = 3$

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

$3 \div 2 = 1.5$

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

$3 + 1.5 = 4.5$

$6 - 1.5 = 4.5$

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

Determine the distance between the two midpoints by subtracting:

$4.5 - 4 = 0.5$

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

Question analysis

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

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding distances between points on a number line is grade 7 work, here with whole-number positions. Two midpoints must be located before their distance can be found, so the task takes three short steps rather than one.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph minor Understanding the question
Segment RS lies inside MN, from 3 to 6, and its midpoint, 4 1/2, falls between two ticks of the number line rather than on a labeled point. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Two midpoints have to be found, at 4 and 4 1/2, before their distance is taken, and the half-length of RS met on the way, 1 1/2, is among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Sees that RS starts 2 units in from M and ends 1 unit in from N, and takes the difference of those gaps, 1, without halving it.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Halves each segment, 3 units for MN and 1 1/2 for RS, and subtracts the half-lengths instead of comparing the midpoints' positions.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the distance from M to R, 2 units, the gap between the left ends of the two segments, without locating any midpoint.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The number line shows M at 1, R at 3, S at 6 and N at 7. Segment MN goes from 1 to 7. Segment RS goes from 3 to 6. A midpoint is the point exactly halfway along a segment. You need the distance between the two midpoints.
Find the midpoint of MN
Use the midpoint formula with the ends 1 and 7:
$\frac{1 + 7}{2} = \frac{8}{2} = 4$
The midpoint of MN is at 4, which is 3 units from each end.
Find the midpoint of RS
Use the same formula with the ends 3 and 6:
$\frac{3 + 6}{2} = \frac{9}{2} = 4.5$
The midpoint of RS is at 4.5. That is halfway between the marks for 4 and 5, where no point is labeled.
Subtract the two midpoints
Use the distance formula:
$4.5 - 4 = 0.5$
0.5 is one half, so the midpoints are 1/2 unit apart.
Watch out: half of MN is 3, and half of RS is 1 1/2. Subtracting those gives 1 1/2, choice C. But they are half-lengths, not positions. And 2, choice D, is only the distance from M to R.
Choose the answer
The midpoints, 4 and 4.5, are close together because RS sits near the middle of MN. So the answer is A, 1/2 unit.

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 57

2022 Test A Math, question 26

Question

I. $s - t$
II. $st^{2}$
III. $s^{t}$

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

AII only
BI and II only
CII and III only
DI, II, and III

Show the answer

Correct answer: D

Each of the three expressions must be considered:

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

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

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

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

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

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

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

Question analysis

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

NYS standards: 8.EE.1 integer exponent properties
7.NS.1c rational number subtraction
7.NS.2a rational number multiplication
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students learn sign rules for subtracting and multiplying signed numbers. Here they reason about letters rather than numbers, decide what must always hold, and handle a negative exponent, a grade 8 idea.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Something you must already know moderate
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 42 words.
Corrected from the DOE handbook: Two steps of the explanation that treated t as a positive number were corrected.
A punctuation error in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know moderate Working it out
Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)
Every choice must be checked minor Choosing the answer
Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Trusts only the squared expression, believing that subtracting t must make s smaller and that a negative exponent makes a negative number.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Handles the subtraction and the square correctly but believes a negative exponent makes the result negative, though s to the power t is a positive fraction such as 1/8.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads s − t as taking a number away from s, so thinks it could be negative, missing that subtracting the negative t adds to s.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.
Try one pair of numbers
Try s = 2 and t = −3 in each expression:
- I: $2 - (-3) = 2 + 3 = 5$
- II: $2 \times (-3)^{2} = 2 \times 9 = 18$
- III: $2^{-3} = \frac{1}{2^{3}} = \frac{1}{8}$
All three are positive this time. Now see why that always happens.
Explain expression I
Use the rule for subtracting:
$s - t = s + (-t)$
t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.
Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.
Explain expression II
$t^{2}$ means t × t. A negative times a negative is positive, so $t^{2}$ is always positive. Then $s \times t^{2}$ is a positive times a positive. That is always positive too.
Explain expression III
Use the rule for a negative exponent:
$s^{t} = \frac{1}{s^{-t}}$
−t is a positive integer, so $s^{-t}$ is a positive integer. So $s^{t}$ is 1 over a positive number. That is a positive fraction.
Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.
Choose the answer
Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.
So the answer is D, I, II, and III.

Check your answer

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

More questions like this

The original DOE question

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

Question 27 of 57

2022 Test A Math, question 27

Question

What is the value of $\left(-\frac{3}{4}\right)\left(-\frac{2}{5}\right)$ ?

A $-\frac{7}{20}$
B $-\frac{3}{10}$
C $\frac{3}{10}$
D $\frac{7}{20}$

Show the answer

Correct answer: C

Evaluate the expression by multiplying the terms in the numerator and the terms in the denominator. The product of two negative values is positive:

$\frac{(-3) \times (-2)}{4 \times 5} = \frac{6}{20}$

Simplify the result:

$\frac{6 \div 2}{20 \div 2} = \frac{3}{10}$

Question analysis

This question asks for the product of negative three fourths and negative two fifths. Students multiply the fractions, settle the sign of the result, and simplify. The choices mix positive and negative answers with results from other operations, so both the sign and the operation have to be right.

NYS standards: 7.NS.2a rational number multiplication
7.NS.2c properties for multiplying and dividing
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multiplying two negative fractions is a standard grade 7 exercise, done here in one product and one simplification. It matches typical classroom work, with the sign rule and lowest terms as the only points of care.
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: A very close wrong answer minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a single product of two fractions, and most students finish it quickly. The choices are built around two decisions, the sign of the product and the operation itself: two of them come from subtracting the fractions, and one keeps a minus sign that two negative factors cancel. Settling the sign first, positive, rules out half of the choices before any multiplying.

A very close wrong answer minor Choosing the answer
3/10 and −3/10 are identical except for the sign, so the size alone does not separate them; only the rule for multiplying two negatives does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Two negative fractions are multiplied, 3 × 2 over 4 × 5, and the product 6/20 has to be reduced to 3/10 before it matches a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Decide the sign of the product first, then multiply the numerators and the denominators and reduce the fraction to lowest terms.

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.)
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
Subtracts the fractions instead of multiplying them, −3/4 − (−2/5) = −15/20 + 8/20, which gives −7/20.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Multiplies the sizes correctly, 3/4 × 2/5 = 6/20 = 3/10, but keeps a minus sign, though two negative factors give a positive product.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Drops both signs and subtracts the sizes, 3/4 − 2/5 = 15/20 − 8/20 = 7/20, instead of multiplying the fractions.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops (numerators). b and d are the bottoms (denominators). Multiply the tops together and the bottoms together. You do not need a common denominator to multiply. That is only for adding and subtracting.
Multiplying two negatives: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. So when both factors are negative, the minus signs cancel.

Sort out the question
The two fractions sit side by side in parentheses, with nothing between them. That means multiply. Both fractions are negative. You need their product, the answer to the multiplication.
Decide the sign first
A negative times a negative is positive. So the answer will be positive. That rules out choices A and B.
Watch out: choice B, −3/10, has the right size but the wrong sign. The two minus signs cancel.
Multiply tops and bottoms
Now multiply the sizes. Multiply the tops, then the bottoms:
$\frac{3}{4} \times \frac{2}{5} = \frac{3 \times 2}{4 \times 5} = \frac{6}{20}$
Watch out: do not find a common denominator. That is for adding and subtracting. Subtracting the fractions gives 7/20 or −7/20, choices D and A.
Simplify the fraction
2 goes into both 6 and 20. Divide the top and the bottom by 2:
$\frac{6 \div 2}{20 \div 2} = \frac{3}{10}$
The product is positive 3/10. So the answer is C, 3/10.

Check your answer

Use decimals. 3/4 is 0.75 and 2/5 is 0.4. Then $0.75 \times 0.4 = 0.3$ , which is 3/10. Two negatives give a positive, so the product is positive 3/10.

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 C Math, page 92, question 19.

Question 28 of 57

2022 Test A Math, question 28

Question

The table shows the income and expenses in a bicycle shop on one day.

BICYCLE SHOP TRANSACTIONS

Description	Income	Expense
Paid employees		$\$120 \times 4$
Sold 2 bicycles	$\$400 \times 2$
Sold clothing	$95
Bought new door		$500
Sold bike parts	p

The sum of all transactions was $70. What is the value of p?

A−$275
B−$85
C$70
D$155

Show the answer

Correct answer: D

The sum of all transactions ($70) includes the income data (positive values) and the expense data (negative values). Write an equation to determine the value of p by adding all the income data and subtracting all the expense data:

$400 \bullet 2 + 95 + p - 120 \bullet 4 - 500 = 70$

Simplify the expression to determine the value of p:

$800 + 95 + p - 480 - 500 = 70$

$895 + p - 980 = 70$

$p - 85 = 70$

$p = 155$

Question analysis

A table lists one day of income and expenses at a bicycle shop, including two products to work out and an unknown amount, p, from selling bike parts. Given that all the transactions add up to $70, students find p. They must treat income as positive and expenses as negative, then solve an equation.

NYS standards: 7.EE.3 multi-step rational number problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve multi-step money problems with positive and negative amounts. This table adds an unknown income entry, so students treat income as positive and expenses as negative before solving a one-step equation.
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
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The table turns a one-variable equation into bookkeeping: each entry must be multiplied out and given a sign by its column before anything is added. The known entries come to a loss of $85, so the bike parts must cover that loss and still leave $70, which is why p is larger than the day's total. Students who stop at the running total, or who copy the $70, never use the equation at all.

Turning words into math moderate Understanding the question
“The sum of all transactions was $70” has to be read with income positive and expenses negative, giving 800 + 95 + p − 480 − 500 = 70. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
Two cells hold products, $120 × 4 and $400 × 2, that must be worked out, and p sits in the Income column beside the sales, so it is added. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
The known entries total 895 − 980 = −85, a negative running sum, so p must exceed 70 by 85 rather than fall short of it. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the two products in the table, add the income entries and subtract the expenses, set that total plus p equal to 70, and solve for p.

Traps

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

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single mistake gives −275; it takes counting the $95 clothing sale as an expense and then stopping at the known total, 800 − 95 − 480 − 500.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Totals the known entries correctly, 895 − 980 = −85, and gives that, stopping before solving for the p that brings the sum to $70.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Picks $70, the stated sum of all the transactions, as the value of the bike parts, without setting up an equation.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of all transactions: $\text{sum} = \text{income} - \text{expenses}$; A transaction is money coming in or going out. Income is money that comes in, so it counts as plus. Expenses are money that goes out, so they count as minus. The sum is what is left, like money added to and taken from a wallet.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
The table lists money that came into the shop, called income. It also lists money that went out, called expenses. The bike parts brought in p dollars, an amount you don’t know. All the income minus all the expenses is $70. You need p.
Work out the products in the table
Two cells in the table show multiplication.
$120 \times 4 = 480$
$400 \times 2 = 800$
So paying the employees cost $480. The 2 bicycles brought in $800.
Write the equation
The income is $800, $95 and p. The expenses are $480 and $500. p sits in the Income column, so it is added.
$\text{income} - \text{expenses} = 70$
$800 + 95 + p - 480 - 500 = 70$
Combine the known numbers
$800 + 95 = 895 \qquad 480 + 500 = 980$
$895 - 980 = -85$
The known entries come to −85 dollars. That is a loss of $85. Now the equation is shorter:
$p - 85 = 70$
Watch out: −85 is choice B, but it is not p. You still have to solve for p.
Solve for p
Add 85 to both sides. This keeps the equation balanced.
$p - 85 + 85 = 70 + 85$
$p = 155$
The bike parts brought in $155. Watch out: $70 is the total for the day, not the bike parts. Don’t just copy it, choice C. So the answer is D, $155.

Check your answer

Put 155 back in. Income: $800 + 95 + 155 = 1{,}050$ dollars. Expenses: $480 + 500 = 980$ dollars. $1{,}050 - 980 = 70$ . That matches. And p should be more than 70, because the bike parts had to cover the $85 loss first.

More questions like this

The original DOE question

Question 85 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 86, question 85. Open page 86 of the handbook (PDF)

Question 29 of 57

2022 Test A Math, question 29

Question

Dakota has four yardsticks and two 12-inch rulers. If she lays all four of the yardsticks and both rulers end-to-end, what is the maximum number of feet she can measure accurately at once?

A12 ft
B14 ft
C16 ft
D18 ft

Show the answer

Correct answer: B

A yardstick is 1 yard in length.

$1\text{ yd} = 3\text{ ft}$

$4\text{ yd} = 12\text{ ft}$

A 12-inch ruler is 1 foot in length.

Two 12-inch rulers are 2 feet in length.

$12\text{ ft} + 2\text{ ft} = 14\text{ ft}$

Question analysis

Dakota lays four yardsticks and two 12-inch rulers end to end and wants to know how many feet she can measure at once. Students convert yards and inches to feet and add the lengths. Mixing up the units, or leaving out one kind of measuring stick, is the main risk in this unit conversion question.

NYS standards: 6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting yards and inches to feet is grade 6 measurement work. With friendly numbers and only two conversions to add, this is easier than typical grade 7 tasks, though the wording about measuring at once may slow some readers.
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: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 37 words.
Corrected from the DOE handbook: A grammatical slip in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a unit conversion dressed as a measuring story: four yardsticks and two rulers are laid end to end, so the total is simply a sum of lengths. Points are lost by forgetting that the rulers count too, or by giving the rulers the yardsticks' length. Writing every piece in feet before adding, 3 + 3 + 3 + 3 + 1 + 1, shows the total at a glance.

Something you must already know minor Working it out
Neither conversion is given: a yardstick is 1 yard, or 3 feet, and each 12-inch ruler is 1 foot, so 4 yards and 24 inches become 12 + 2 feet. (Anderson & Krathwohl, 2001)

Strategy: Change every length to feet before adding: a yard is 3 feet and 12 inches is 1 foot. Then total the six pieces laid end to end.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Converts the four yardsticks to 12 feet and stops, forgetting to add the 2 feet of the two 12-inch rulers laid at the end.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 16 feet; it would need each 12-inch ruler to be 2 feet long, so it mostly draws guesses between nearby totals.
D Units or scale (King et al., 2004)
Reveals a slip Expected pull: weak
Converts the rulers with the yard factor too, counting all six sticks as 3 feet long, 6 × 3 = 18, though a 12-inch ruler is 1 foot.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “Two 12-inch rulers is 2 feet in length.” (p. 142, question 86), repeating the verb of the line before it, “A 12-inch ruler is 1 foot in length.” The site now reads “Two 12-inch rulers are 2 feet in length.” The working and the answer, 14 ft, are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Yards to feet: $f = 3 \times y$; y is a length in yards. f is the same length in feet. One yard is 3 feet, so each yard adds 3 feet. A yardstick is exactly 1 yard long.
Inches to feet: $f = \frac{i}{12}$; i is a length in inches. f is the same length in feet. One foot is 12 inches, so divide the inches by 12. A 12-inch ruler is 1 foot long.

Sort out the question
Dakota lays 4 yardsticks and 2 rulers end to end, in one long line. End to end means each piece starts where the last one stops. So the total length is the sum of all six pieces. You need it in feet. A yardstick is 1 yard long. Each ruler is 12 inches long.
Change the yardsticks to feet
The four yardsticks make 4 yards.
$f = 3 \times y$
$f = 3 \times 4 = 12$
The yardsticks make 12 feet. Watch out: don’t stop here. 12 feet is choice A, but the rulers count too.
Change the rulers to feet
Two rulers of 12 inches make 24 inches.
$f = \frac{i}{12} = \frac{24}{12} = 2$
The rulers make 2 feet. Watch out: a ruler is not as long as a yardstick. Counting all six pieces as 3 feet gives $6 \times 3 = 18$ feet, choice D.
Add the lengths
$12 + 2 = 14$
Dakota can measure 14 feet at once. So the answer is B, 14 ft.

Check your answer

List each piece in feet: $3 + 3 + 3 + 3 + 1 + 1 = 14$ . That matches. Or use inches. A yard is 36 inches, so 4 yards is $4 \times 36 = 144$ inches. Add the rulers: $144 + 24 = 168$ inches. Then $168 \div 12 = 14$ feet.

More questions like this

The original DOE question

Question 86 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 86, question 86. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 30 of 57

2022 Test A Math, question 30

Question

In a fruit basket, the ratio of apples to oranges is the same as the ratio of cherries to walnuts. If there are 6 oranges, 16 cherries, and 48 walnuts, how many apples are there?

Show the answer

Correct answer: A

The ratios of the items are equal, so they can be expressed using a proportion.

Let a represent the number of apples, o the number of oranges, c the number of cherries, and w the number of walnuts. Write a proportion showing apples:oranges is equal to cherries:walnuts:

$\frac{a}{o} = \frac{c}{w}$

Substitute the given values for oranges (6), cherries (16), and walnuts (48):

$\frac{a}{6} = \frac{16}{48}$

Cross multiply to form an equation to solve for the number of apples (a):

$(a \times 48) = (16 \times 6)$

Solve for the value of a:

$48a = 96$

$a = 2$

Question analysis

In a fruit basket, the ratio of apples to oranges equals the ratio of cherries to walnuts, and students find the number of apples from the counts of oranges, cherries and walnuts. It tests setting up and solving a proportion. Matching each quantity to its place in the ratio matters, since the wrong order gives a very different count.

NYS standards: 6.RP.3 ratio and rate reasoning
6.RP.1 ratio concepts and language
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Equal ratios with whole numbers are grade 6 work, and the numbers divide evenly. Grade 7 proportion problems are usually longer; here the only care is keeping apples with oranges in the same order as cherries with walnuts.
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
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The numbers divide evenly, since 16/48 is 1/3, so the work is in the setup. Writing the two ratios in the order the sentence gives them makes the apples a third of the oranges; writing either one upside down makes them three times the oranges. A quick sense check helps: cherries are fewer than walnuts, so apples must be fewer than oranges.

Turning words into math minor Understanding the question
The sentence pairs apples with oranges and cherries with walnuts in the same order, so a/6 = 16/48; reading either pair the other way turns one ratio upside down. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Four kinds of fruit are named and three counts given, and the unknown apples must be matched with the smaller count, the cherries, not the larger, the walnuts. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write apples over oranges equal to cherries over walnuts, simplify the known ratio, and scale it to match the 6 oranges.

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).
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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Divides the 48 walnuts by the 16 cherries to get the factor 3 and stops before using it to divide the 6 oranges.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Turns one ratio upside down, setting apples over oranges equal to 48 over 16, so the apples come out as 3 × 6 = 18.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reasons with differences instead of ratios, taking the gap of 32 between walnuts and cherries and adding it to the 6 oranges.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios: $\frac{\text{apples}}{\text{oranges}} = \frac{\text{cherries}}{\text{walnuts}}$; A ratio compares two amounts. Writing it as a fraction puts the first thing named on top. The question says the two ratios are the same, so the two fractions are equal. Keep the same order in both.

Sort out the question
A ratio compares two amounts. The ratio of apples to oranges equals the ratio of cherries to walnuts. You know there are 6 oranges, 16 cherries and 48 walnuts. You need the number of apples.
Write the proportion
Put the first thing of each pair on top. Apples go with oranges, and cherries go with walnuts. Let a be the number of apples:
$\frac{a}{6} = \frac{16}{48}$
Watch out: keep the order the same. Writing 48 over 16 flips the second ratio. That gives 18, choice C.
Simplify the known ratio
16 goes into both 16 and 48. Divide the top and the bottom by 16:
$\frac{16 \div 16}{48 \div 16} = \frac{1}{3}$
So there is 1 cherry for every 3 walnuts. There must also be 1 apple for every 3 oranges.
Find the apples
Now the proportion is:
$\frac{a}{6} = \frac{1}{3}$
The bottom 6 is 3 × 2, so the top is 1 × 2. That makes $a = 2$ .
Watch out: $48 \div 16 = 3$ is choice B, but that is not the number of apples. And 38 comes from using differences instead of ratios.
So the answer is A, 2.

Check your answer

Compare the ratios. Apples to oranges is 2 to 6. Cherries to walnuts is 16 to 48. In both, the second number is 3 times the first: $2 \times 3 = 6$ and $16 \times 3 = 48$ . So the ratios are equal.

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 C Math, page 94, question 21.

Question 31 of 57

2022 Test A Math, question 31

Question

In a showing of an artist’s works, the ratio of the number of paintings, drawings, and photographs shown is 3:5:4, respectively. If the number of drawings shown was 45, what is the total number of photographs and paintings that were shown?

A36
B63
C84
D108

Show the answer

Correct answer: B

The ratio of paintings to drawings to photographs is $3 : 5 : 4$ .

First, determine the number of actual paintings and photographs.

Since the number of actual drawings is 45, each ratio is multiplied by 9, since $5 \times 9 = 45$ .

Type	Ratio	Actual Number
Paintings	3	27
Drawings	5	45
Photographs	4	36

Then, add the actual number of paintings and photographs.

$27 + 36 = 63$

Question analysis

An art showing has paintings, drawings and photographs in the ratio 3:5:4, and 45 drawings were shown. Students find the total number of photographs and paintings. It tests using a three-part ratio to find actual amounts, and the question asks for two of the three groups together, not any single group or the whole show.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scaling a ratio from one known part is grade 6 and 7 work. A three-part ratio, and a question about two of the three groups combined, make it a three-step task typical of grade 7 ratio problems.
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
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The scale factor comes from one division, 45 ÷ 5 = 9, so the risk lies in what to do with it. Two tempting results are listed: the photographs alone, and the whole show with the drawings added in. Multiplying the 3 + 4 = 7 parts for paintings and photographs by 9 gives the total in one step and avoids both.

Much to keep track of moderate Working it out
The ratio has three parts, the known count belongs to the middle part, drawings, and the question wants the first and third parts combined, so three roles must be tracked. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
The word respectively assigns 3, 5 and 4 to paintings, drawings and photographs in that order, and the question then names two of them in the other order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many works each ratio part stands for from the drawings, multiply it by the parts for paintings and for photographs, and add only those two.

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: strong
Finds the photographs, 4 × 9 = 36, and stops before adding the 27 paintings that the question also asks about.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 7 parts for paintings and photographs by 12, the total of the ratio's parts, instead of by 9.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds every group correctly but adds the 45 drawings too, giving the whole show, 108, rather than the paintings and photographs.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Using the parts of a ratio: $\text{actual number} = \text{parts} \times \text{value of one part}$; The ratio 3:5:4 splits the show into equal parts. Parts is how many parts a group has. Each part stands for the same number of works. Find that number from the group you know, then multiply each group’s parts by it.

Sort out the question
The ratio of paintings to drawings to photographs is 3:5:4. Respectively means in that order: 3 for paintings, 5 for drawings and 4 for photographs. There were 45 drawings. You need the number of photographs and paintings together.
Find the value of one part
Drawings are 5 parts, and there were 45 drawings:
$45 \div 5 = 9$
So each part of the ratio stands for 9 works.
Find each group
Multiply each group’s parts by 9:
Type Parts Actual number
Paintings 3 $3 \times 9 = 27$
Drawings 5 $5 \times 9 = 45$
Photographs 4 $4 \times 9 = 36$
The drawings row gives 45, which matches the question.
Add only the two groups asked for
Add the paintings and the photographs:
$27 + 36 = 63$
Watch out: 36 alone, choice A, leaves out the paintings. Adding the 45 drawings too gives 108, choice D, the whole show.
So the answer is B, 63.

Type	Parts	Actual number
Paintings	3	$3 \times 9 = 27$
Drawings	5	$5 \times 9 = 45$
Photographs	4	$4 \times 9 = 36$

Check your answer

Paintings and photographs are $3 + 4 = 7$ parts. Each part is 9 works, so $7 \times 9 = 63$ . That matches. Also, 63 to 45 is the same as 7 to 5, just as the ratio says.

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 C Math, page 97, question 24.

Question 32 of 57

2022 Test A Math, question 32

Question

For each $\frac{3}{8}$ foot of curtain width, $\frac{5}{6}$ yard of fabric is needed to make ruffles. How many yards of fabric are needed per foot of curtain width?

A $\frac{5}{16}$ yd
B $\frac{29}{24}$ yd
C $\frac{20}{9}$ yd
D $\frac{7}{2}$ yd

Show the answer

Correct answer: C

It takes $\frac{5}{6}$ yard of fabric to make $\frac{3}{8}$ foot of curtain width.

To find the ratio of 1 foot of curtain width to $\frac{3}{8}$ foot of curtain width $\left(1:\frac{3}{8}\right)$ , divide.

$1 \div \frac{3}{8} = 1 \times \frac{8}{3} = \frac{8}{3}$

So the amount of fabric needed for 1 foot of curtain width is $\frac{8}{3}$ times the amount needed for $\frac{3}{8}$ foot of curtain width.

$\frac{8}{3} \times \frac{5}{6} = \frac{40}{18}$

$\frac{40 \div 2}{18 \div 2} = \frac{20}{9}$

The amount of fabric needed for 1 foot of curtain width is $\frac{20}{9}$ yards.

Question analysis

Making ruffles takes 5/6 yard of fabric for every 3/8 foot of curtain width, and students find how many yards of fabric one foot of curtain width needs. It tests finding a unit rate when both quantities are fractions. The two units differ, and choosing whether to multiply, divide or add the fractions is the crucial decision.

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 two fractions is the grade 7 rate standard itself. Two units, feet and yards, and an unfamiliar sewing setting make it harder to see which fraction divides which.
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
Error-prone arithmetic 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

Two fractions and two units, feet and yards, leave students to decide whether to multiply, divide or add, and each wrong choice records one of those decisions. A sense check rules out the most tempting one before any arithmetic: a whole foot is more than 3/8 foot, so it needs more than 5/6 yard, yet multiplying gives only 5/16. Dividing the yards by the feet then picks out the answer among the rest.

Turning words into math moderate Understanding the question
“For each 3/8 foot of curtain width, 5/6 yard of fabric” gives the rate as a pair of fractions, and “per foot” means dividing the yards by the feet. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing by 3/8 means multiplying by 8/3, and the product 40/18 has to be reduced to 20/9, an improper fraction rather than a whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide the yards of fabric by the feet of curtain width, since a rate per foot puts feet in the denominator, then simplify the fraction.

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: strong
Multiplies 5/6 by 3/8 instead of dividing, getting 15/48 = 5/16, so a whole foot seems to need less fabric than 3/8 foot.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Adds 5/6 and 3/8 over a common denominator, 20/24 + 9/24 = 29/24, combining yards with feet instead of forming a rate.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: weak
Finds that a foot is 8/3 times 3/8 foot, then adds 8/3 and 5/6 instead of multiplying them: 16/6 + 5/6 = 21/6 = 7/2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{yards per foot} = \text{yards} \div \text{feet}$; A unit rate tells how much for exactly 1 of something. Here it is yards of fabric for 1 foot of curtain. Per foot means for each foot, so feet go on the bottom. Divide the yards by the feet.
Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b/c is a fraction. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down. It works because b/c fits into 1 exactly c/b times.

Sort out the question
For every 3/8 foot of curtain width, you need 5/6 yard of fabric. You need the yards of fabric for 1 whole foot. That is a unit rate, an amount for exactly 1 foot.
3/8 foot is less than a whole foot. So a whole foot needs more than 5/6 yard.
Set up the division
Per foot means divide the yards by the feet:
$\frac{5}{6} \div \frac{3}{8}$
Watch out: multiplying 5/6 by 3/8 gives 5/16, choice A. That is less fabric for more curtain, which makes no sense. Adding them gives 29/24, choice B, but you can’t add yards and feet.
Multiply by the reciprocal
Flip 3/8 to get 8/3, and multiply:
$\frac{5}{6} \times \frac{8}{3} = \frac{5 \times 8}{6 \times 3} = \frac{40}{18}$
The 8/3 says a whole foot is 8/3 times as long as 3/8 foot. So it needs 8/3 times as much fabric.
Watch out: adding 8/3 and 5/6 gives 7/2, choice D. Multiply them instead.
Simplify the fraction
2 goes into both 40 and 18:
$\frac{40 \div 2}{18 \div 2} = \frac{20}{9}$
So 1 foot of curtain width needs 20/9 yards of fabric, a little more than 2 yards. So the answer is C, 20/9 yd.

Check your answer

Multiply back. 3/8 foot should need 3/8 of the fabric for a whole foot: $\frac{20}{9} \times \frac{3}{8} = \frac{60}{72} = \frac{5}{6}$ yard. That is the amount the question gives. Also, 20/9 is more than 5/6, as it should be.

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 C Math, page 98, question 25.

Question 33 of 57

2022 Test A Math, question 33

Question

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

Which straight line is the graph of a proportional relationship?

A $\overleftrightarrow{\mathrm{MS}}$
B $\overleftrightarrow{\mathrm{VP}}$
C $\overleftrightarrow{\mathrm{VQ}}$
D $\overleftrightarrow{\mathrm{TR}}$

Show the answer

Correct answer: D

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

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

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

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

Question analysis

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

NYS standards: 7.RP.2a testing for proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Deciding from a graph whether a relationship is proportional is a named grade 7 skill, and this is a typical case: four lines on a clean grid, each checked against the same rule, with no calculation needed.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Something you must already know minor
Reading a figure, table or graph minor
Solution: 1 step
Reading level: Question and choices: about grade 9, 14 words.
Corrected from the DOE handbook: Coordinates and an equation in the explanation were corrected to match the graph.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Something you must already know minor Working it out
The rule that a proportional graph is a straight line through the origin is not printed, and all four lines are straight, so straightness alone rules none of them out. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
Line VP runs off the grid just right of O, crossing the x-axis at (1, 0), so whether it passes through the origin has to be read from the gridlines, not from its general direction. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes MS as proportional because it is straight and meets both axes, though it starts at (0, 8) and its y-values fall as x rises.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Sees the steep line’s lower end vanish near O and assumes it passes through the origin, though VP crosses the x-axis at (1, 0).
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Mistakes the steady y-value of 3 along VQ for a constant of proportionality, though y ÷ x falls from 3/2 at V to 1/2 at Q.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The grid shows four straight lines. Each one is named by two labeled points on it: line MS, line VP, line VQ and line TR.
You need the line that shows a proportional relationship. In that kind of relationship, y is always the same number times x. For example, y might always be 3 times x.
Recall what a proportional graph looks like
Use the rule $y = kx$ . Put in 0 for x:
$y = k \times 0 = 0$
So every proportional graph goes through the origin. That is the corner point (0, 0), labeled O, where the two axes meet. The graph must also be a straight line.
All four lines here are straight. So being straight does not settle it. Look for the line through O.
Check where each line starts
- Line MS meets the y-axis at M (0, 8). When x is 0, y is 8, not 0.
- Line VP is steep. Its lower end crosses the x-axis at (1, 0), one unit to the right of O.
- Line VQ is flat at a height of 3. It never comes down to 0.
- Line TR starts at O and runs up through T and R.
Watch out: the lower end of line VP looks close to O. Follow it along the grid lines. It misses O by one unit, so choice B is wrong.
Test the ratio on line TR
In a proportional relationship, the ratio y ÷ x is the same at every point. T is at (4, 1) and R is at (8, 2):
$\text{T: } \frac{1}{4} \qquad \text{R: } \frac{2}{8} = \frac{1}{4}$
Both give $\frac{1}{4}$ . So each y-value is $\frac{1}{4}$ of its x-value, and the line is $y = \frac{1}{4}x$ .
Watch out: line VQ keeps the same y-value, 3, but not the same ratio. At V it is $\frac{3}{2}$ , and at Q it is $\frac{3}{6} = \frac{1}{2}$ . So choice C is wrong.
Give the answer
Line MS starts at 8 on the y-axis, so choice A is out too. Only line TR goes through the origin and keeps y ÷ x the same.
So the answer is D, line TR.

Check your answer

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

More questions like this

The original DOE question

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

Question 34 of 57

2022 Test A Math, question 34

Question

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

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

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

Show the answer

Correct answer: C

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

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

Question analysis

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

NYS standards: 7.RP.2d points on proportional graphs
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on a proportional graph means in its situation is a named grade 7 skill. The graph is clear and starts at the origin, but each statement has to be read closely against the axis labels.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Reading a figure, table or graph moderate
Every choice must be checked minor
Solution: 2 steps
Reading level: Question and choices: about grade 2, 77 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph moderate Understanding the question
Each statement must be checked against the axis labels: x is the weight of ice in pounds and y the cost in dollars, on a grid ruled every 2 pounds and every $0.30. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Every choice must be checked minor Choosing the answer
All four statements name a point that really lies on the line, (1, 0.15) included, so being on the graph rules nothing out; only the meaning given to each coordinate does. (Embretson & Wetzel, 1987)

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

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads the point (20, 3) backward, taking 20 as the price and 3 as the weight, though x counts pounds of ice and y counts dollars.
B Units or scale (King et al., 2004)
Reveals a slip Expected pull: strong
Confirms that (1, 0.15) is on the line, the cost of 1 pound, but accepts “1 bag” because the friends buy bags, though the x-axis measures pounds.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Knows (0, 0) is on every proportional graph and accepts a true-sounding story about people, though the graph never counts people; the point means 0 pounds cost $0.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The graph is called ICE COSTS. Read both axes first:
- The x-axis, across the bottom, is the weight of the ice in pounds (lb).
- The y-axis, up the side, is the cost in dollars.
A point is written (x, y), weight first and cost second. So a point means: x pounds of ice cost y dollars. You need the statement that says this correctly.
Find the cost of one pound
The line reaches $3.00 at 20 pounds. Use the unit rate formula:
$\text{cost per pound} = \frac{3}{20} = 0.15$
So 1 pound of ice costs $0.15. This helps you test each point.
Test the point in choice C
Find 10 on the weight axis and go straight up to the line. The line crosses there at 1.5 on the cost axis. The unit rate agrees:
$10 \times 0.15 = 1.5$
So the point (10, 1.5) means 10 pounds of ice cost $1.50. That is exactly what choice C says.
Check the other statements
Each one names a real point on the line. The problem is the meaning it gives the numbers.
Watch out: choice A reads (20, 3) backward. It means 20 pounds cost $3, not 3 pounds cost $20.
Watch out: choice B is tricky. The point (1, 0.15) is on the line, but the x-axis counts pounds, not bags. So it means 1 pound costs $0.15.
Choice D talks about people, but the graph never counts people. The point (0, 0) means 0 pounds cost $0.
Give the answer
Only choice C gives the right meaning to both numbers.
So the answer is C, 10 lbs of ice costs $1.50.

Check your answer

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

More questions like this

The original DOE question

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

Question 35 of 57

2022 Test A Math, question 35

Question

$A graph titled INFORMATION ENTERED, with Time (hr) on the x-axis from 0 to 2, labeled at 0.5, 1, 1.5 and 2, with a grid line every 0.25 hour, and Keystrokes on the y-axis from 0 to 12,000, labeled and with a grid line every 1,000. The origin is labeled O. A straight line starts at the origin and rises to the right, passing through (0.5, 3,500), (1, 7,000) and (1.5, 10,500), and ends in an arrow just past 12,000.$

A clerk enters information on a computer. The graph shows the relationship between time, in hours, and the number of keystrokes the clerk makes. What is the meaning of the point (1.5, 10500) on the graph?

AThe clerk makes 10,500 keystrokes in each 1 hr of entering information.
BThe clerk makes 10,500 keystrokes in each 1.5 hr of entering information.
CThe clerk makes 15,750 keystrokes in each 1 hr of entering information.
DThe clerk makes 15,750 keystrokes in each 1.5 hr of entering information.

Show the answer

Correct answer: B

The x-axis represents the time, in hours, the clerk enters information into the computer, and the y-axis represents the number of keystrokes the clerk makes.

The x-value in (1.5, 10500) gives 1.5 hours as the amount of time the clerk enters information.

The y-value gives 10,500 as the number of keystrokes made by the clerk.

Since the graph is a proportional line, the clerk’s rate is 10,500 keystrokes in each 1.5 hours of entering information.

Question analysis

A straight-line graph through the origin shows how many keystrokes a clerk makes over time, in hours. Students choose the statement that explains what the point (1.5, 10500) means in this situation. Each choice pairs a number of keystrokes with a span of time, so reading both coordinates and their units correctly decides it.

NYS standards: 7.RP.2d points on proportional graphs
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on the graph of a proportional relationship means in its situation is a named grade 7 skill. This is a short, direct example with a clear graph, though the four statements look alike and must be read closely.
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 moderate
A very close wrong answer minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 74 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Nothing needs computing; the question checks that a point on a proportional graph is read as a pair, a time and a count together. Students trained to look for a unit rate expect every statement about a graph to be per hour, and the two statements with 15,750 come from forcing that idea with a product. The graph itself shows 7,000 keystrokes at 1 hour, a figure no statement offers, so every per-hour statement fails.

Reading a figure, table or graph moderate Understanding the question
The stem gives the point (1.5, 10500), and the axis labels must say what each number is: 1.5 hours of time along x and 10,500 keystrokes up y. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer minor Choosing the answer
The two statements with 10,500 keystrokes differ only in the time, each 1 hr or each 1.5 hr, so the x-coordinate alone separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Use the axis labels to say in words what each coordinate of the point stands for, the x-value in hours and the y-value in keystrokes, then find the statement that says the same.

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).

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
Reads 10,500 from the point but assumes any point on a rate graph gives the amount for 1 hour, ignoring the x-coordinate, 1.5.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies 10,500 by 1.5 to get 15,750 and calls it an hourly rate, though the hourly rate is 10,500 ÷ 1.5 = 7,000.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the two coordinates, 10,500 × 1.5 = 15,750, and pairs that product with the 1.5 hours from the point, combining numbers without a model.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A point on a graph: $(x, y)$; A point is written as a pair in parentheses. x, the first number, tells how far across, along the bottom axis. y, the second number, tells how far up, along the side axis. The axis labels tell what each number measures.
Unit rate from a point: $\text{unit rate} = \frac{y}{x}$; Here y is the number of keystrokes and x is the time in hours. Dividing gives the keystrokes for 1 hour. On a straight line that starts at (0, 0), every point gives the same unit rate.

Sort out the question
The graph shows how many keystrokes a clerk makes over time. A keystroke is one press of a key. You need to say what the point (1.5, 10500) means in words. Each choice pairs a number of keystrokes with an amount of time. The axis labels tell you what each number in the point stands for.
Read the x-value
The first number in a point is the x-value. The bottom axis is labeled Time (hr). So x = 1.5 means 1.5 hours of work. On the graph, 1.5 is halfway between 1 and 2.
Read the y-value
The second number is the y-value. The side axis is labeled Keystrokes. So y = 10,500 means 10,500 keystrokes. Put the two together. In 1.5 hours, the clerk makes 10,500 keystrokes.
Use the straight line
The line is straight and starts at O, the point (0, 0). That means the clerk types at a steady speed. So every 1.5 hours of work brings 10,500 keystrokes. Watch out: the point does not give a count for 1 hour. Its x-value is 1.5, not 1. Choice A treats 10,500 as the count for 1 hour.
Rule out the 15,750 choices
15,750 is $10{,}500 \times 1.5$ . Multiplying the two numbers in a point means nothing here. The count for 1 hour comes from dividing:
$\text{unit rate} = \frac{10{,}500}{1.5} = 7{,}000$
That is not 15,750, so C is wrong. D is wrong too, since the graph shows 10,500 at 1.5 hours. So the answer is B: 10,500 keystrokes in each 1.5 hr.

Check your answer

Look at the graph at 1 hour. The line crosses 7,000 keystrokes there. At 1.5 hours it is at 10,500. Since $7{,}000 \times 1.5 = 10{,}500$ , the line fits choice B. No choice gives 7,000 for 1 hour, so every per-hour choice is wrong.

More questions like this

The original DOE question

Question 92 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 89, question 92. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 57

2022 Test A Math, question 36

Question

When the three-digit whole number 27 ? is divided by 3 or by 4, the remainder is zero. What is the missing digit?

Show the answer

Correct answer: D

Each of the options 0, 2, 3, and 6 must be considered:

Option A: In the case of 0, the three-digit whole number is 270. Divide 270 by both 3 and 4:

$270 \div 3 = 90 \text{ (or } 90R0\text{)}$

$270 \div 4 = 67\frac{1}{2} \text{ (or } 67R2\text{)}$

Dividing 270 by 3 results in a remainder of 0, but dividing 270 by 4 results in a remainder of 2, so 0 cannot be the missing digit.

Option B: In the case of 2, the three-digit whole number is 272. Divide 272 by both 3 and 4:

$272 \div 3 = 90\frac{2}{3} \text{ (or } 90R2\text{)}$

$272 \div 4 = 68 \text{ (or } 68R0\text{)}$

Dividing 272 by 4 results in a remainder of 0, but dividing 272 by 3 results in a remainder of 2, so 2 cannot be the missing digit.

Option C: In the case of 3, the three-digit whole number is 273. Divide 273 by both 3 and 4:

$273 \div 3 = 91 \text{ (or } 91R0\text{)}$

$273 \div 4 = 68\frac{1}{4} \text{ (or } 68R1\text{)}$

Dividing 273 by 3 results in a remainder of 0, but dividing 273 by 4 results in a remainder of 1, so 3 cannot be the missing digit.

Option D: In the case of 6, the three-digit whole number is 276. Divide 276 by both 3 and 4:

$276 \div 3 = 92 \text{ (or } 92R0\text{)}$

$276 \div 4 = 69 \text{ (or } 69R0\text{)}$

Dividing 276 by 3 results in a remainder of 0, and dividing 276 by 4 results in a remainder of 0, so 6 is the missing digit.

Question analysis

A three-digit number begins with 27 and is missing its last digit, and it divides evenly by both 3 and 4. Students find the missing digit. It tests factors, multiples and divisibility rules; a digit that works for one divisor may fail the other, so both tests must be applied.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Divisibility by 3 and by 4 is grade 6 number sense. Testing four possible digits against two conditions is quick, much like a grade 6 exercise, as long as both divisors are actually checked.
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
Every choice must be checked minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every candidate passes at least one of the two tests, which makes the question a matter of finishing the checks rather than finding a method. Students who test only by 3, or only by 4, find a fitting number and stop. Seeing that a number divisible by both 3 and 4 is divisible by 12, and that 276 = 12 × 23, settles it in one step.

Turning words into math minor Understanding the question
The words “divided by 3 or by 4, the remainder is zero” mean that both divisions come out even, though or can be read as either one being enough. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
Each candidate, 270, 272, 273 and 276, has to be tested against two divisors, and three of them pass one test, so stopping after one check is costly. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
The divisibility tests are not given: a number is divisible by 3 when its digit sum is, and by 4 when the number formed by its last two digits is. (Anderson & Krathwohl, 2001)

Strategy: Use the divisibility tests: a number is divisible by 4 when its last two digits are, and by 3 when its digit sum is. Keep only a digit that passes both.

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 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Checks only division by 3, which 270 passes with a digit sum of 9, and never tests 4, though 270 ÷ 4 leaves a remainder of 2.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Checks only division by 4, which 272 passes because 72 is a multiple of 4, but its digit sum is 11, so 272 ÷ 3 leaves 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the missing digit to the divisor 3 in the question; 273 is a multiple of 3 but leaves a remainder of 1 when divided by 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Test for dividing by 3: $\text{digit sum divides by 3} \rightarrow \text{number divides by 3}$; The digit sum is what you get when you add up a number’s digits. Divides by 3 means it divides evenly, with a remainder of 0. If the digit sum divides evenly by 3, so does the whole number. Use it to test a big number quickly.
Test for dividing by 4: $\text{last two digits divide by 4} \rightarrow \text{number divides by 4}$; Look only at the number made by the last two digits. If it divides evenly by 4, so does the whole number. This works because every 100 divides evenly by 4, so only the last two digits matter.

Sort out the question
The number is 27 followed by a missing digit. Remainder zero means the division comes out even, with nothing left over. The remainder is zero when the number is “divided by 3 or by 4.” That means both divisions come out even. So the number must pass both tests. The choices give four digits to try: 0, 2, 3 and 6.
Test each number for 3
Add the digits of each number.
Number Digit sum Divides by 3?
270 $2 + 7 + 0 = 9$ yes
272 $2 + 7 + 2 = 11$ no
273 $2 + 7 + 3 = 12$ yes
276 $2 + 7 + 6 = 15$ yes
So 270, 273 and 276 pass. 272 fails.
Test each number for 4
Look at the last two digits: 70, 72, 73 and 76.
$70 \div 4 = 17 \text{ remainder } 2 \qquad 72 \div 4 = 18$
$73 \div 4 = 18 \text{ remainder } 1 \qquad 76 \div 4 = 19$
So 272 and 276 pass. 270 and 273 fail.
Keep the digit that passes both
Only 276 passes both tests. Check by dividing:
$276 \div 3 = 92 \qquad 276 \div 4 = 69$
Watch out: 0 and 3 pass only the test for 3, so A and C fail. 2 passes only the test for 4, so B fails. And 3 is not right just because the question says 3. So the answer is D, 6.

Number	Digit sum	Divides by 3?
270	$2 + 7 + 0 = 9$	yes
272	$2 + 7 + 2 = 11$	no
273	$2 + 7 + 3 = 12$	yes
276	$2 + 7 + 6 = 15$	yes

Check your answer

A number that divides evenly by both 3 and 4 also divides evenly by 12, since $3 \times 4 = 12$ . And $12 \times 23 = 276$ . So 276 is a multiple of 12. None of 270, 272 or 273 is.

More questions like this

The original DOE question

Question 93 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 89, question 93. Open page 89 of the handbook (PDF)

Question 37 of 57

2022 Test A Math, question 37

Question

$A circle graph titled FOOD CONSUMPTION OF THE AVERAGE AMERICAN PER YEAR, with six sections: Fruits and vegetables 38%, Cereals 10%, Meat and fish 10%, Sugar 8%, Fats and oils 4%, and Milk and eggs x%. Below the graph: Total consumption = 1,840 pounds per person.$

Based on the circle graph, how many pounds of milk and eggs does the average American consume each year?

A300
B368
C460
D552

Show the answer

Correct answer: D

First, determine the missing percentage for “milk and eggs” consumed, based on the circle graph.

$100 - (38 + 10 + 10 + 8 + 4) =$

$100 - 70 =$

$30$

Then, determine 30% of 1,840 pounds. Multiply:

$1{,}840 \times \frac{30}{100} = 552$

Question analysis

A circle graph shows how the 1,840 pounds of food the average American eats in a year divide among six groups, with the percent for milk and eggs left as x. Students find how many pounds of milk and eggs that is. They first work out the missing percent, then take that percent of the total.

NYS standards: 6.RP.3c percent as a rate per 100
6.SP.5 summarizing data sets
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading a circle graph and finding a percent of a total are grade 6 skills. Finding the missing sector from 100% adds one easy step, so this is shorter and simpler than typical grade 7 multistep percent 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: Reading a figure, table or graph moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two short steps decide the question: find the missing percent from the other five, then take that percent of 1,840 pounds. The drawing does not help, since the milk-and-eggs sector is drawn larger than the 38% sector beside it, so the share must come from the labels. Students who find the 30 and then pick the choice whose digits match it skip the second step.

Reading a figure, table or graph moderate Understanding the question
Milk and eggs is labeled only x%, so its share comes from the five labeled sectors, 100 − 70 = 30, and the 1,840-pound total sits in a caption below the circle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Taking 30% of 1,840 by hand, 0.3 × 1,840 = 552, means a decimal times a four-digit number, best done as 184 × 3. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the five labeled percents from 100% to find x, then multiply the 1,840-pound total by that percent written as a decimal.

Traps

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 the missing share, 30%, and stops; 300 is the choice whose digits match it, though 30% of 1,840 pounds is still to be taken.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 368, which is 20% of 1,840; it would need a slip such as 80 for the labeled percents' sum, so it mostly draws guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 460, a quarter of 1,840; the sector is drawn larger than a quarter, so even judging by eye does not lead here.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a circle graph: $\text{sum of all the percents} = 100$; A circle graph shows one whole split into parts. Each part, or section, shows its percent. The whole is 100%, so all the percents add up to 100. Use it to find a missing section.
Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the full amount. The part is the piece that matches the percent. Percent means out of 100, so p% is p hundredths of the whole.

Sort out the question
The circle graph splits all the food an average American eats in a year into six sections. The total is 1,840 pounds per person. Five sections show a percent. Milk and eggs shows only x%. You need the pounds of milk and eggs. First find x. Then find x% of 1,840 pounds.
Trust the labels, not the drawing
This graph is not drawn to scale, though it does not say so. The milk and eggs section looks bigger than the fruits and vegetables section, which is labeled 38%. But you will see that milk and eggs is smaller. So don’t judge the size of any section by eye here. Work only from the printed numbers.
Find the missing percent
All six sections add to 100%. Put in the five labeled percents and x.
$\text{sum of all the percents} = 100$
$38 + 10 + 10 + 8 + 4 + x = 100$
$70 + x = 100 \qquad x = 100 - 70 = 30$
Milk and eggs is 30% of the total. That is less than 38%, though it is drawn bigger.
Find 30% of 1,840 pounds
$\text{part} = \frac{p}{100} \times \text{whole}$
$\text{part} = \frac{30}{100} \times 1{,}840 = 0.3 \times 1{,}840 = 552$
Another way: 10% of 1,840 is 184. 30% is 3 times that, $3 \times 184 = 552$ . Watch out: don’t stop at 30. Choice A, 300, only looks like 30. So the answer is D, 552 pounds.

Check your answer

Estimate. 30% is a bit less than one third. One third of 1,840 is about 613. So the answer should be a bit under 613, and 552 is. Also, $552 \div 1{,}840 = 0.3$ , which is 30%. Choice C, 460, is only one quarter, 25%.

More questions like this

The original DOE question

Question 94 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 89, question 94. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 38 of 57

2022 Test A Math, question 38

Question

Out of 2,000 high school seniors, 191 were enrolled in the Honor Society. What percentage of the entire group of 2,000 seniors were in the Honor Society?

A0.955%
B1.91%
C9.55%
D19.1%

Show the answer

Correct answer: C

The fraction $\frac{191}{2000}$ represents the 191 high school seniors out of 2,000 enrolled in the Honor Society.

To determine the percentage, convert the fraction to a decimal. Then, rewrite the decimal as a percentage.

$\frac{191}{2000} = 0.0955 = 9.55\%$

Question analysis

Out of 2,000 high school seniors, 191 belong to the Honor Society, and students find what percent of the whole group that is. It tests writing a part of a whole as a percent. Several choices share the same digits, so the decimal point has to be placed with care when the fraction becomes a percent.

NYS standards: 6.RP.3c percent as a rate per 100
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Writing a part of a whole as a percent is grade 6 work, done here with one division and one conversion. The awkward numbers, 191 out of 2,000, make placing the decimal point the only real demand.
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
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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 method is a single fraction, 191 over 2,000, so the question tests decimal places. The four choices share the digits 955 or 191 and differ only in where the point falls, which means a slip of one place still finds a choice waiting. A rough check settles it: 191 is a little under 200, and 200 is 10% of 2,000, so the share must be a little under 10%.

Error-prone arithmetic moderate Working it out
Dividing 191 by 2,000 gives 0.0955, with a zero right after the decimal point, and writing it as a percent moves the point two more places; each move is a chance to land one place off. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Honor Society members as a fraction of all 2,000 seniors, divide, and move the decimal point two places to the right to turn the decimal into a percent.

Traps

An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Finds 0.0955 but moves the decimal point only one place when writing the percent, giving 0.955%, ten times too small.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Divides 191 by 100, as if a percent were always out of 100, and never uses the 2,000 seniors, so 191 ÷ 100 = 1.91 becomes 1.91%.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Divides 191 by 1,000 instead of 2,000, dropping the 2 while moving the decimal point, so 0.191 becomes 19.1%, twice the true share.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a whole: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the smaller group you care about. The whole is the entire group. The fraction tells what share of the whole the part is. Multiplying by 100 turns that share into a percent, which means out of 100.

Sort out the question
There are 2,000 seniors in all. Of them, 191 are in the Honor Society. You need the percent of all 2,000 seniors who are in it.
So 191 is the part, and 2,000 is the whole.
Write the share as a fraction
Put the part over the whole:
$\frac{\text{part}}{\text{whole}} = \frac{191}{2{,}000}$
This fraction says 191 out of every 2,000 seniors. Next, turn it into a decimal by dividing.
Divide to get a decimal
Dividing by 2,000 is the same as dividing by 2 and then by 1,000:
$191 \div 2 = 95.5$
$95.5 \div 1{,}000 = 0.0955$
Dividing by 1,000 moves the decimal point 3 places to the left. So the share is 0.0955.
Change the decimal to a percent
Multiply by 100. That moves the decimal point 2 places to the right:
$0.0955 \times 100 = 9.55$
So the share is 9.55%.
Watch out: moving the point only 1 place gives 0.955%, choice A. Dividing by 1,000 instead of 2,000 gives 19.1%, choice D. Dividing 191 by 100 gives 1.91%, choice B.
Give the answer
The 191 Honor Society members are 9.55% of all 2,000 seniors. That is a little less than one tenth of them.
So the answer is C, 9.55%.

Check your answer

Estimate. 191 is a little less than 200. And 200 out of 2,000 is one tenth, which is 10%. So the answer should be a little less than 10%. Only 9.55% fits.

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 C Math, page 103, question 30.

Question 39 of 57

2022 Test A Math, question 39

Question

NUMBER OF VISITORS
TO BROOKSIDE ZOO

	Male	Female
Children	360	340
Adults	210	290

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

A30%
B28%
C24%
D17%

Show the answer

Correct answer: A

First, determine the total number of visitors.

$360 + 340 + 210 + 290 = 1200$

The largest group is 360 (Male Children).

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

$\frac{360}{1200} \times \frac{100}{1}\% = \frac{360}{12} \times \frac{1}{1}\% = 30\%$

Question analysis

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

NYS standards: 6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding what percent one count is of a total is grade 6 work. A two-way table adds two quick steps, totaling the four groups and spotting the largest, but the numbers divide evenly, so it stays below grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph minor Understanding the question
The two-way table gives four separate counts and no totals, so the 1,200 visitors must be added up from the cells and the largest group picked out of a grid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three short steps run in order: total the four cells, pick the largest, and divide; using the wrong cell in the last step still gives a believable percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Takes 340, the female children, as the largest cell, misjudging it against the 360 male children beside it, so 340 ÷ 1,200 ≈ 28%.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Answers for the larger adult group, the 290 female adults, as if the question were about adults, so 290 ÷ 1,200 ≈ 24%.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Reads largest as smallest and uses the 210 male adults, so 210 ÷ 1,200 = 17.5%, and takes the nearest choice.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The table splits one day’s visitors into four groups:
Male Female
Children 360 340
Adults 210 290
You need the largest group as a percent of all the visitors. The table gives no total, so you must find it.
Add up all the visitors
Add all four groups to get the whole:
$360 + 340 + 210 + 290 = 1{,}200$
So 1,200 people visited the zoo that day.
Find the largest group
Compare the four numbers. The biggest is 360, the male children.
Watch out: 340 is close to 360, but it is smaller. Using 340 gives about 28%, choice B. Using the 290 female adults gives about 24%, choice C. And 210 is the smallest group, which gives 17.5%, near choice D.
Find the percent
Put the part over the whole. Then simplify by dividing the top and bottom by 120:
$\frac{360}{1{,}200} = \frac{3}{10}$
Now multiply by 100 to get a percent:
$\frac{3}{10} \times 100 = 30$
So the male children made up 30% of the visitors.
Give the answer
The largest group, 360 male children, is 30% of all 1,200 visitors.
So the answer is A, 30%.

Check your answer

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

More questions like this

The original DOE question

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

Question 40 of 57

2022 Test A Math, question 40

Question

A bag contains 40 tiles that are either red, green, or blue. A tile is selected at random, the color is recorded, and the tile is put back in the bag. This process is carried out 120 times, and 18 of those times a red tile is selected. Based on the information, what is the most likely number of red tiles in the bag?

Show the answer

Correct answer: A

First, determine what fraction of the sample were red tiles, given 18 of the 120 tiles drawn were red.

$\frac{18 \div 6}{120 \div 6} = \frac{3}{20}$

To predict how many of the actual 40 tiles in the bag are red, multiply 40 by the fraction of red tiles in the sample.

$\frac{3}{20}(40) = \frac{120}{20} = 6$

Question analysis

A bag holds 40 red, green and blue tiles, and a tile is drawn and put back 120 times, with red coming up 18 times. Students predict the most likely number of red tiles in the bag. It tests using experimental results to estimate what is in the bag, and the draws outnumber the tiles, which can confuse the scaling.

NYS standards: 6.SP.7 long-run relative frequency
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Predicting from repeated trials is grade 6 probability, and grade 7 students make similar estimates from samples. The twist is scaling down, from 120 draws to a bag of 40 tiles, rather than up to a larger group.
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 6, 61 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

There are more draws than tiles, which is what makes this unusual: the 120 draws must be scaled down to a bag of 40, not up to a bigger group. Red came up 18 times because each tile went back in the bag, so 18 is not a count of red tiles. Since there are three draws for every tile, dividing the 18 red draws by 3 gives the same estimate as taking 3/20 of 40.

Turning words into math minor Understanding the question
The 18 red draws come from 120 draws with the tile put back each time, so they give a share, 3/20, to take of the 40 tiles, not a count of red tiles. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Students must know that the share of red in repeated draws estimates the share of red tiles in the bag; the stem never says to multiply 40 by that share. (Anderson & Krathwohl, 2001)

Strategy: Write the red draws as a fraction of all 120 draws, simplify it, and take that same fraction of the 40 tiles in the bag.

Traps

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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that red came up in 18 of 120 draws, or 15%, and gives 15 as the count, stopping before taking that share of the 40 tiles.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives 18, the number of times red was drawn, as the number of red tiles, though each tile went back and could be drawn many times.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 18 red draws in 120 gives 30 red tiles, three quarters of the bag, when red came up in fewer than one draw in six.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from repeated draws: $\text{red tiles} \approx \frac{\text{red draws}}{\text{all draws}} \times \text{tiles in the bag}$; The fraction is the share of draws that came up red. That share is your best guess for the share of tiles that are red. Multiply it by the number of tiles in the bag to predict the red tiles. The sign ≈ means about equal.

Sort out the question
The bag has 40 tiles, each red, green or blue. A tile is drawn, its color is written down, and it goes back in the bag. This happens 120 times, and red comes up 18 times.
You need the most likely number of red tiles in the bag.
See why 18 is not the answer
Each tile goes back in the bag after it is drawn. So the same red tile can be drawn again and again. That means 18 counts draws, not tiles.
Watch out: choice C, 18, mixes up the number of red draws with the number of red tiles.
Find the share of red draws
Red came up 18 times out of 120. Simplify by dividing the top and bottom by 6:
$\frac{18}{120} = \frac{3}{20}$
As a decimal, that is 0.15, or 15%.
Watch out: 15 is a percent here, not a number of tiles. Stopping now gives choice B.
Take that share of the bag
Use the same share of the 40 tiles in the bag:
$\frac{3}{20} \times 40 = \frac{120}{20} = 6$
So about 6 of the 40 tiles are red.
Give the answer
The best prediction is 6 red tiles. The other 34 tiles are probably green or blue.
So the answer is A, 6.

Check your answer

There were 120 draws for 40 tiles. That is 3 draws for every tile. So divide the red draws by 3: $18 \div 3 = 6$ . Same answer. Choice D, 30 red tiles, would be three quarters of the bag. Then red would come up far more than 18 times.

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 C Math, page 106, question 33.

Question 41 of 57

2022 Test A Math, question 41

Question

LAST YEAR’S TOTAL ATTENDANCE

Attendees	Number
Children	1,750
Adults	2,250
Seniors	1,000
Total	5,000

If 100 individuals will attend the museum next week, what is the best estimate (based on last year’s attendance) of the number of children who will be in that group?

Show the answer

Correct answer: C

First, determine what percentage of the total attendees last year were children.

$\frac{1{,}750}{5{,}000} = \frac{35}{100} = 0.35,\ \text{or } 35\%$

The best estimate based on last year’s attendance is 35% of 100.

$0.35 \times 100 = 35$

Question analysis

A table gives last year's attendance by children, adults and seniors, 5,000 people in all. If 100 people attend the museum next week, students estimate how many of them will be children. It tests using a ratio or percent from data to make a prediction; only the children's row and the total matter.

NYS standards: 6.RP.3c percent as a rate per 100
6.SP.7 long-run relative frequency
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Using last year's share of visitors to predict a smaller group is grade 6 ratio and probability work. A group of exactly 100 makes the scaling easy, so the arithmetic is lighter than in typical grade 7 proportion 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: Turning words into math minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 38 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 share 1,750 out of 5,000 converts neatly to 35 out of 100, and next week's group happens to be exactly 100, so the arithmetic is nearly done once the proportion is written. Points slip away when a student works out the scale factor between the two totals and gives it, or estimates the share instead of computing it. Dividing 1,750 by 50 checks the result.

Turning words into math minor Understanding the question
The phrase “best estimate (based on last year’s attendance)” has to become a proportion: children keep the share of next week's 100 that they had of last year's 5,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
Only the children row, 1,750, and the 5,000 total are needed; the adults and seniors rows add numbers that play no part in the estimate. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find what fraction of last year's 5,000 visitors were children, rewrite it as an equivalent fraction out of 100, and apply it to next week's group.

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.

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 30, which would need 1,500 children last year; it mostly draws students who round the share down or guess a round number.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 34; splitting 100 evenly among three groups gives 33 1/3, the closest route, so it mostly draws guesses.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Divides 5,000 by 100 to get the scale factor, 50, and gives it, stopping before dividing the 1,750 children by that factor.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the number in one group, here children. The whole is the total of all groups. Dividing gives the fraction of the whole that the group makes up. Times 100 turns it into a percent, a number out of 100.
Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the full amount. The part is the piece that matches the percent. Percent means out of 100, so p% is p hundredths of the whole.

Sort out the question
Last year 5,000 people came in all. 1,750 of them were children. Next week 100 people will come. You need the best estimate of how many will be children. Best estimate means children make up the same share as last year. You don’t need the adults and seniors rows.
Find last year’s percent of children
$\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$
$\text{percent} = \frac{1{,}750}{5{,}000} \times 100 = 0.35 \times 100 = 35$
So 35% of last year’s visitors were children. That means 35 out of every 100.
Use that percent for 100 people
$\text{part} = \frac{35}{100} \times 100 = 35$
About 35 of the 100 people should be children. Watch out: $5{,}000 \div 100 = 50$ tells how many times bigger last year’s crowd was. It is not a number of children. Don’t pick 50, choice D. So the answer is C, 35.

Check your answer

Shrink both numbers the same way. Since $5{,}000 \div 50 = 100$ , divide the children by 50 too: $1{,}750 \div 50 = 35$ . That matches. Also, 1,750 is a bit more than one third of 5,000, and 35 is a bit more than one third of 100.

More questions like this

The original DOE question

Question 98 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 90, question 98. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 42 of 57

2022 Test A Math, question 42

Question

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

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

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

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

Show the answer

Correct answer: B

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

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

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

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

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

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

Question analysis

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

NYS standards: 7.SP.8b sample spaces for compound events
7.SP.8a compound event as fraction of outcomes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a compound probability from a tree diagram is a grade 7 standard, and here the tree is already drawn. Listing its paths and counting the ones that fit the event, in any order, is typical grade 7 classwork.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 65 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph moderate Understanding the question
The tree prints only A or B in each box, so the eight three-pick outcomes, such as A, B, A, have to be read by tracing each path across all three columns. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Much to keep track of minor Working it out
All eight paths must be tracked while counting those with exactly two A’s and one B, and “in any order” means three of them count, not just A, A, B. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Counts only the path A, A, B, missing the stem’s “in any order”, so it finds 1 of the 8 outcomes instead of 3.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Counts every outcome with at least two picks from Classroom A, adding A, A, A to the three right paths, so 4 of 8, or 1/2.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Counts every outcome that mixes the two classrooms, including those with two from B, so 6 of the 8 outcomes, or 3/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The teacher picks 3 students, one at a time. Each pick is from Classroom A or Classroom B.
You need the chance of getting 2 students from A and 1 from B. The words “in any order” matter. They mean the B can be the first, second or third pick.
List every path in the tree
Each path from left to right is one outcome, or possible result. The top path is A, then A, then A. Read all of them:
AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB
Each pick has 2 choices, so there are $2 \times 2 \times 2 = 8$ outcomes in all.
Mark the ones that fit
Look for exactly two A’s and one B. Three outcomes fit: AAB, ABA and BAA.
Watch out: counting only AAB misses “in any order” and gives $\frac{1}{8}$ , choice A. AAA has three A’s, not two. Counting it too gives $\frac{4}{8} = \frac{1}{2}$ , choice C. Counting every mixed outcome gives $\frac{6}{8} = \frac{3}{4}$ , choice D, but three of those have two B’s.
Find the probability
Use the probability formula. 3 outcomes fit, out of 8:
$P = \frac{3}{8}$
So the answer is B, $\frac{3}{8}$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 43 of 57

2022 Test A Math, question 43

Question

A bag contains 15 marbles: 11 blue marbles and 4 red marbles. Sarah picks one marble randomly, returns it to the bag, and then randomly picks one marble again. What is the probability that she picks a red marble both times?

A $\frac{16}{225}$
B $\frac{16}{121}$
C $\frac{8}{15}$
D $\frac{121}{225}$

Show the answer

Correct answer: A

The probability of drawing a red marble at random from a bag of 15 marbles (11 blue and 4 red) is shown:

$P(\text{Red}) = \frac{4\text{ red marbles}}{15\text{ total marbles}} = \frac{4}{15}$

When the red marble is replaced in the bag, the probability of picking a red marble is unchanged, $P(\text{Red}) = \frac{4}{15}$ .

The two events are independent, so the multiplication rule of probability applies.

$P(\text{Red}) \bullet P(\text{Red}) = \frac{4}{15} \bullet \frac{4}{15} = \frac{16}{225}$

Question analysis

A bag holds 11 blue and 4 red marbles. Sarah picks one at random, puts it back, and picks again, and students find the probability that she picks red both times. It tests the probability of a compound event made of two independent picks; the bold word both and the replacement shape the calculation.

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 Finding the probability of two independent picks is grade 7 compound-event work. Because the marble is put back, the two picks are alike, and the numbers are small, which makes this a typical grade 7 probability task.
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: Something you must already know moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 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

The arithmetic is two small fractions multiplied, so the question rests on knowing which rule applies. Adding the two chances gives more than either chance alone, which cannot be right for two reds in a row, and comparing red with blue instead of with all 15 marbles changes every denominator. A sense check helps: two reds in a row must be less likely than a single red, 4/15.

Something you must already know moderate Working it out
The multiplication rule for independent events is not given: because the marble goes back, each pick is 4 out of 15, and red both times means 4/15 × 4/15. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
Two phrases carry the method: “returns it to the bag” keeps the second pick at 4 out of 15, and “red marble both times” asks for both events, not either. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the chance of red on a single pick out of all the marbles. Since the marble goes back, the second pick has the same chance; then use the rule for two independent events both happening.

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).
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Uses red to blue, 4 to 11, as the chance of red instead of 4 out of all 15 marbles, then squares it, a part-to-part ratio.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Adds the two chances, 4/15 + 4/15 = 8/15, instead of multiplying, as if a red marble on either pick were enough.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the chance of blue both times, (11/15)² = 121/225, answering for the wrong color with the right method.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of ways it can happen}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens, from 0 (never) to 1 (always). An outcome is one possible result, here one marble. Divide the number of results you want by the number of all results.
Two independent events both happening: $P(A \text{ and } B) = P(A) \times P(B)$; A and B are two events. Independent means one does not change the chance of the other. P(A) is the chance of A, and P(B) is the chance of B. Multiply them to get the chance that both happen.

Sort out the question
The bag has 15 marbles: 11 blue and 4 red. Sarah picks one, puts it back, then picks again. You need the probability, or chance, that both picks are red. Putting the marble back matters. The bag is the same for the second pick.
Find the chance of red on one pick
$P = \frac{\text{number of ways it can happen}}{\text{total number of outcomes}}$
$P(\text{red}) = \frac{4}{15}$
There are 4 red marbles out of 15 in all. Watch out: compare red with all 15 marbles, not with the 11 blue ones. Using 4 out of 11 leads to $\frac{16}{121}$ , choice B.
See that the picks are independent
Sarah puts the marble back. So the bag again has 4 red marbles out of 15. The second pick has the same chance, 4/15. The first pick does not change the second one. That makes the two picks independent.
Multiply the two chances
$P(\text{red and red}) = \frac{4}{15} \times \frac{4}{15} = \frac{4 \times 4}{15 \times 15} = \frac{16}{225}$
To multiply fractions, multiply the tops and multiply the bottoms. Watch out: don’t add. $\frac{4}{15} + \frac{4}{15} = \frac{8}{15}$ , choice C, is a bigger chance than one red. And $\frac{121}{225}$ , choice D, is the chance of blue both times. So the answer is A, $\frac{16}{225}$ .

Check your answer

Two reds in a row should be less likely than one red. One red is $\frac{4}{15}$ , about 0.27. $\frac{16}{225}$ is about 0.07, much smaller. That makes sense. Adding gave $\frac{8}{15}$ , about 0.53, which is too big.

More questions like this

The original DOE question

Question 100 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 91, question 100. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 57

2022 Test A Math, question 44

Question

Report	Subject	Topic Choices
#1	Geography	Kenya, Thailand, Chile, Greece, New Zealand
#2	English	R. Frost, L. Hughes, N. Giovanni
#3	Science	1997 Mars mission
#4	Math	π, regular polyhedrons

Anita is assigned to report on 1 topic in each of the 4 subjects shown in the table above. How many different sets of 4 reports are possible, according to the choices available to her?

Show the answer

Correct answer: C

One report is assigned for each of 4 subjects, and each subject has a specified number of topic choices:

5 choices for Geography
3 choices for English
1 choice for Science
2 choices for Math

The basic counting principle (product rule) applies since the number of choices for each subject is independent of the choices for the other subjects. The number of different sets of 4 reports possible is the product.

$5 \times 3 \times 1 \times 2 = 30$

Question analysis

A table lists the topics Anita may choose from for reports in geography, English, science and math. She reports on one topic in each subject, and students count how many different sets of four reports are possible. It tests the counting principle; reading how many topics each row offers, including a row with only one, comes first.

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 Counting the outcomes of several independent choices is grade 7 work, usually done with a tree diagram or a list. Four subjects make a full tree impractical, so a counting shortcut is needed, but each count is small and read from the table.
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: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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

Once the four counts are read from the table, the question is a single product, so the real test is recognizing that independent choices multiply. Students who add the counts get 11, which is not offered, a useful signal that the method is wrong. Science's single topic multiplies by 1 and changes nothing, but it still has to be read as one choice, not as none.

Something you must already know moderate Working it out
The counting principle is not stated: choosing one topic from each subject means multiplying the numbers of choices, 5 × 3 × 1 × 2, not adding them. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The topic lists must be counted by their commas, and two rows mislead: Science offers a single topic, and Math's “π, regular polyhedrons” is two topics, not one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Count the topic choices in each row of the table, reading the commas carefully, then apply the counting principle, since Anita picks one topic from every subject.

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).
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 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Adds the topic counts instead of multiplying and leaves out Science's single topic as no real choice, 5 + 3 + 2 = 10.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Counts the orders in which four reports could be written, 4 × 3 × 2 × 1 = 24, instead of the ways to choose one topic per subject.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 4 subjects by the 11 topics listed in all, 4 × 11 = 44, combining numbers without a model of the choices.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{number of ways} = a \times b \times c \times d$; a, b, c and d are the numbers of choices for each separate pick. When you make one pick from each group, and the groups don’t affect each other, multiply the numbers of choices. Each choice in one group can go with every choice in the others.

Sort out the question
Anita writes 4 reports, one for each subject. For each subject she picks 1 topic from that row of the table. A set of 4 reports is one full pick: one topic for each subject. You need how many different sets are possible.
Count the topics in each row
Use the commas to count the topics.
- Geography: Kenya, Thailand, Chile, Greece, New Zealand. That is 5.
- English: Frost, Hughes, Giovanni. That is 3.
- Science: only the 1997 Mars mission. That is 1.
- Math: π and regular polyhedrons. That is 2.
New Zealand is two words but one country, so it is one topic.
Multiply the choices
Use the counting principle. Each of the 5 geography topics can go with each of the 3 English topics, and so on.
$5 \times 3 \times 1 \times 2 = 30$
Science has just one topic, so every set has it. Multiplying by 1 changes nothing. Watch out: don’t add the counts. Adding them without Science, $5 + 3 + 2 = 10$ , gives choice A.
Rule out the other choices
$4 \times 3 \times 2 \times 1 = 24$ counts the orders for writing 4 reports. The question does not ask about order, so 24, choice B, is wrong. $4 \times 11 = 44$ multiplies the subjects by all the topics. That is choice D, and it has no meaning here. So the answer is C, 30.

Check your answer

Build it up. Geography and English alone give $5 \times 3 = 15$ pairs, such as Kenya with Frost. Science adds only one topic, so there are still 15. Math doubles it, π or polyhedrons, so $15 \times 2 = 30$ sets.

More questions like this

The original DOE question

Question 101 as printed on page 92 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 92, question 101. Open page 92 of the handbook (PDF)

Question 45 of 57

2022 Test A Math, question 45

Question

A deck of 52 playing cards contains 13 hearts. Sara has 12 cards from this deck in her hand, 3 of which are hearts. The other cards remain in the deck. What is the probability that a card drawn at random from the remainder of the deck will be a heart?

A $\frac{1}{10}$
B $\frac{1}{4}$
C $\frac{3}{13}$
D $\frac{1}{3}$

Show the answer

Correct answer: B

First, determine the number of cards in the remaining deck:

$52 - 12 = 40$

Next, determine the number of hearts left in the remaining deck:

$13 - 3 = 10$

Last, determine the probability of drawing a heart from the remaining deck:

$\frac{10\text{ hearts remaining}}{40\text{ total cards remaining}} = \frac{10}{40} = \frac{1}{4}$

Question analysis

A standard deck of 52 playing cards has 13 hearts, and Sara holds 12 cards, 3 of them hearts. Students find the probability that a card drawn at random from the rest of the deck is a heart. Her hand changes both the number of hearts and the number of cards left, so both counts must be updated.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a simple probability is grade 6 content. This one adds a step many grade 7 problems skip: the cards in Sara's hand must be taken out of both the hearts and the whole deck first.
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: Much to keep track of minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 51 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 careful route takes two subtractions and a fraction: 10 hearts among 40 cards. The numbers are kind, because Sara’s hand is a quarter hearts, just like the full deck, so a student who ignores her hand and divides 13 by 52 also reaches 1/4. The wrong choices come from describing her hand instead of the deck, or from comparing hearts with the cards that are not hearts.

Much to keep track of minor Working it out
Four counts are in play, 52 cards, 13 hearts, Sara’s 12 cards and her 3 hearts, and both the hearts and the total must be reduced before the fraction is formed. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
“The other cards remain in the deck” and “the remainder of the deck” point to the 40 cards Sara is not holding, not to the full 52 or to her hand. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work out how many cards are left in the deck and how many of them are hearts, then write the hearts left over the cards left 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.
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 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the 10 hearts left in the deck to the 10 in 1/10, never dividing them by the 40 cards that remain.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Puts Sara’s 3 hearts over the deck’s 13, or her 12 cards over 52, both 3/13, which describes her hand rather than the cards left.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Compares the 10 hearts left with the 30 other cards left, a part-to-part ratio of 1 to 3, instead of with all 40 cards.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance something happens. The top counts the cards that are hearts. The bottom counts every card you could draw. Each card is equally likely to be drawn, so the fraction gives the chance.

Sort out the question
A full deck has 52 cards, and 13 of them are hearts. Sara holds 12 cards, and 3 of those are hearts. The rest of the cards stay in the deck.
You need the chance that a card drawn from the rest of the deck is a heart. Her hand changes both counts.
Count the cards left
Take Sara’s 12 cards out of the deck:
$52 - 12 = 40$
So 40 cards are left in the deck.
Count the hearts left
Sara holds 3 of the 13 hearts. Take those out:
$13 - 3 = 10$
So 10 hearts are left in the deck.
Find the probability
Use the probability formula. Put the hearts left over all the cards left, then simplify:
$P = \frac{10}{40} = \frac{1}{4}$
Watch out: comparing the 10 hearts with the 30 cards that are not hearts gives $\frac{1}{3}$ , choice D. The bottom must be all 40 cards. Choice C, $\frac{3}{13}$ , describes Sara’s hand, not the deck. Choice A, $\frac{1}{10}$ , uses the 10 hearts but forgets the 40 cards.
Give the answer
The chance of drawing a heart from the rest of the deck is 1 in 4.
So the answer is B, $\frac{1}{4}$ .

Check your answer

10 is one quarter of 40, because $4 \times 10 = 40$ . Here is another check. Sara’s hand is 3 hearts out of 12, which is also one quarter. Taking a hand that is one quarter hearts out of a deck that is one quarter hearts leaves one quarter hearts.

More questions like this

The original DOE question

Question 36 as shown on the DOE’s digital test, page 109 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 109, question 36.

Question 46 of 57

2022 Test A Math, question 46

Question

The table shows the probability of randomly picking each flavor of candy from a bowl.

PICKING CANDY FROM
A BOWL

Flavor	Probability
Butterscotch	0.57
Lemon	0.095
Maple	0.125
Strawberry	0.21

Which flavor is most likely to be randomly picked?

Abutterscotch
Blemon
Cmaple
Dstrawberry

Show the answer

Correct answer: A

The flavor most likely to be picked at random is butterscotch, because its probability is the greatest of the probabilities in the list. More than half the candies in the bowl are butterscotch.

Question analysis

A table lists the probability of picking butterscotch, lemon, maple or strawberry candy at random from a bowl, each as a decimal. Students decide which flavor is most likely to be picked. It tests the meaning of probability and comparing decimals, which have different numbers of digits here and must be compared by place value.

NYS standards: 6.SP.6 probability from 0 to 1
6.NS.7 ordering and absolute value
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
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: Error-prone arithmetic minor
Solution: 1 step
Reading level: Question and choices: about grade 9, 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

One idea answers the question: the most likely flavor has the greatest probability. The only obstacle is comparing decimals of different lengths, where 0.125 and 0.095 look bigger than 0.57 to students who count digits. Writing each as thousandths, 570, 95, 125 and 210, or noticing that 0.57 alone is more than one half, makes the order plain.

Error-prone arithmetic minor Working it out
The probabilities have two and three decimal places, 0.57 beside 0.095 and 0.125, so comparing them by length, or by their digits as whole numbers, ranks them wrongly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Line the four decimals up by place value, adding zeros so each has three decimal places, and compare them digit by digit starting from the tenths place.

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

B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads lemon’s 0.095 as 0.95, overlooking the 0 in the tenths place, so it looks almost certain and larger than 0.57.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the decimal with more digits as the larger, or compares 125 with 57 as whole numbers, so 0.125 seems bigger than 0.57.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common misreading makes 0.21 the largest, since it is neither the longest decimal nor the one with the biggest digits, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{candies of that flavor}}{\text{all candies in the bowl}}$; P is the probability of picking a flavor. It is a number from 0 to 1. The more candies of a flavor there are, the bigger its probability. So the flavor with the greatest probability is the most likely to be picked.

Sort out the question
The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.
The greater the probability, the more likely the pick. So you need the greatest of the four decimals.
Line up the decimals
The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:
Flavor Probability
Butterscotch 0.570
Lemon 0.095
Maple 0.125
Strawberry 0.210
Compare the tenths digits
Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.
Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.
Give the answer
Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.
So the answer is A, butterscotch.

Flavor	Probability
Butterscotch	0.570
Lemon	0.095
Maple	0.125
Strawberry	0.210

Check your answer

0.57 is more than 0.5, which is one half. So more than half the candies are butterscotch. The other three flavors add up to $0.095 + 0.125 + 0.21 = 0.43$ , less than half. So butterscotch must be the most likely.

More questions like this

The original DOE question

Question 37 as shown on the DOE’s digital test, page 110 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 110, question 37.

Question 47 of 57

2022 Test A Math, question 47

Question

This formula can be used to determine p, the number of pages that will be in a book that contains w words, with an average of r words per page, and f extra pages in the front for the title and publishing information.

$p = \frac{1}{r}w + f$

If the formula is solved for w, it can be used to find the number of words in a book with p pages. What will the formula be if it is solved for w?

A $w = r(p - f)$
B $w = \frac{1}{r}p + f$
C $w = f + rp$
D $w = r + p + f$

Show the answer

Correct answer: A

Apply the properties of equality to solve for w.

$p = \frac{1}{r}w + f$

Subtract f from both sides of the equation.

$p - f = \frac{1}{r}w$

Multiply both sides by r.

$r(p - f) = r\left(\frac{1}{r}w\right)$

$r(p - f) = w$

$w = r(p - f)$

Question analysis

A formula gives the number of pages p in a book from its number of words w, the average words per page r, and the extra front pages f. Students choose the formula solved for w. It tests rearranging a formula with inverse operations, and with only letters there are no numbers to check, so the order of the steps matters.

NYS standards: 8.EE.7 linear equations in one variable
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 such as 5 = x/4 + 2 with numbers. Rearranging a formula with four letters to solve for a different one is grade 8 work, with no numbers to check against.
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
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 82 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 formula is short, but solving it for w means undoing two operations in reverse order, with letters only. The tempting shortcut multiplies p by r and leaves the f untouched, which yields a formula that looks as finished as the right one. Swapping p and w without undoing anything is the other quick trap. Substituting a small book, 5,000 words at 250 a page with 4 front pages, shows which formula gives back 5,000.

Variables instead of numbers moderate Understanding the question
Four letters and no numbers mean each choice has to be judged by its structure, unless students pick values such as r = 250, w = 5,000 and f = 4 to test it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Two inverse operations must come in order, subtracting f before multiplying by r; multiplying first means the f would have to be multiplied by r as well. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Treat it like a two-step equation with numbers: undo what was done to w last, then the step before, applying each operation to the whole of the other side.

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.
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

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats solving for w as trading the places of p and w, keeping the 1/r and the + f exactly where they were.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies p by r to undo the 1/r but leaves the + f alone, neither subtracting it first nor multiplying it by r too.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the other three letters, r + p + f, keeping the formula’s letters but none of its multiplication or division.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying by the reciprocal: $r \times \frac{1}{r} = 1$; r is any number except 0. The reciprocal of r is 1/r, the number flipped upside down. A number times its reciprocal is always 1. So multiplying by r undoes the 1/r in front of w and leaves w by itself.

Sort out the question
The formula is $p = \frac{1}{r}w + f$ . Here p is the number of pages, w the number of words, r the words on each page, and f the extra front pages.
You need the formula solved for w. That means w alone on one side of the equal sign.
See what happens to w
Read the formula like a recipe. Start with w. First it is multiplied by $\frac{1}{r}$ , which is the same as dividing by r. Then f is added.
To get w back, undo the steps in reverse order. Take away f first. Then undo the $\frac{1}{r}$ .
Subtract f from both sides
Taking f from both sides keeps them equal:
$p - f = \frac{1}{r}w$
Now the only thing left to undo is the $\frac{1}{r}$ .
Multiply both sides by r
Use the reciprocal rule. Multiply both whole sides by r:
$r(p - f) = r \times \frac{1}{r}w = w$
So $w = r(p - f)$ .
Watch out: $w = f + rp$ , choice C, multiplies only p by r. The f must be taken away before you multiply. Or, if you multiply first, every term gets multiplied by r.
Give the answer
Choice B only trades the places of p and w, and it undoes nothing. Choice D just adds the letters.
So the answer is A, $w = r(p - f)$ .

Check your answer

Try a book with 5,000 words, 250 words a page and 4 front pages. The first formula gives $p = 5{,}000 \div 250 + 4 = 24$ pages. Choice A gives back the words: $250 \times (24 - 4) = 5{,}000$ . Choice C gives $4 + 250 \times 24 = 6{,}004$ , which is wrong.

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 C Math, page 111, question 38.

Question 48 of 57

2022 Test A Math, question 48

Question

A shipment contains 170 small boxes of medical supplies. In a random sample of 20 of these boxes, 8 have damage and the others are undamaged. Based on this sample, what is the best prediction of the number of undamaged boxes in the shipment, not including the boxes in the random sample?

A60
B68
C90
D102

Show the answer

Correct answer: C

Determine the fraction of undamaged boxes in the sample, given 8 of the 20 boxes are damaged.

$\frac{20-8}{20} = \frac{12}{20} = \frac{3}{5}\text{ or }0.6$

Subtract the number of boxes in the sample from 170.

$170 - 20 = 150$

Basing the prediction of undamaged boxes on the sample, multiply the number of boxes not in the sample by the fraction of undamaged boxes in the sample.

$150 \times 0.6 = 90$

Question analysis

A shipment holds 170 boxes of medical supplies, and in a random sample of 20, 8 are damaged. Students predict how many undamaged boxes are in the shipment, leaving out the boxes in the sample. It tests using a sample to make a prediction, and the question's two conditions each lead to a different tempting wrong choice.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.SP.1b samples and populations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Predicting from a random sample is grade 7 proportional reasoning. This version adds two conditions, undamaged rather than damaged and the shipment without the sample, so it takes three steps and a close reading.
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 8, 49 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The proportion itself is routine; the question is built around two conditions that each change the numbers. The stem reports the damaged boxes, but the question asks about undamaged ones, and its bold not removes the 20 sampled boxes from the shipment. Each wrong choice matches one way of mishandling the two conditions, so checking the result against both words, undamaged and not, is what the question rewards.

Much to keep track of moderate Working it out
Two conditions point away from the obvious numbers: the question wants undamaged boxes while the stem counts the 8 damaged, and it leaves out the 20 sampled, so 150 boxes, not 170. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Many steps minor Working it out
Three steps follow in order, the undamaged share 12/20, the 150 boxes outside the sample, and their product, and each wrong turn lands on a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the fraction of the sample that is undamaged, count the boxes that were not sampled, and multiply; then check that the answer describes the group of boxes asked about.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Uses the damaged share, 8 of 20 or 0.4, instead of the undamaged share, taking it of the 150 unsampled boxes: 0.4 × 150 = 60.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Takes the damaged share of the whole shipment, 0.4 × 170 = 68, answering for damaged boxes and keeping the 20 sampled ones as well.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the undamaged share, 0.6, but takes it of all 170 boxes, missing the bold not that leaves the 20 sampled boxes out: 0.6 × 170 = 102.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from a sample: $\text{prediction} = \frac{\text{part of the sample}}{\text{sample size}} \times \text{group size}$; A sample is a small group chosen at random from a bigger group. The fraction is the share of the sample with some feature. A random sample is likely to look like the bigger group. So use the same share of the group you are predicting for.

Sort out the question
The shipment has 170 boxes. A random sample of 20 boxes was checked, and 8 of them were damaged.
You need to predict the undamaged boxes in the shipment, not counting the 20 in the sample. Two words matter: undamaged and not.
Find the undamaged share
The sample has $20 - 8 = 12$ undamaged boxes. Write that as a share of the sample, and simplify:
$\frac{12}{20} = \frac{3}{5} = 0.6$
So 0.6, or 60%, of the sample is undamaged.
Watch out: $\frac{8}{20} = 0.4$ is the damaged share. Using it leads to choice A or B.
Count the boxes outside the sample
The question leaves out the 20 boxes in the sample. Take them away from the shipment:
$170 - 20 = 150$
So the prediction is about 150 boxes.
Make the prediction
Use the same undamaged share for the 150 boxes:
$0.6 \times 150 = 90$
About 90 of those boxes should be undamaged.
Watch out: $0.6 \times 170 = 102$ , choice D, forgets the bold not. It counts the sample boxes again.
Give the answer
The best prediction is 90 undamaged boxes outside the sample.
So the answer is C, 90.

Check your answer

Predict the damaged boxes outside the sample too: $0.4 \times 150 = 60$ . Then $90 + 60 = 150$ , every box outside the sample. Also, 12 out of 20 is a bit more than half. And 90 is a bit more than half of 150.

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 C Math, page 112, question 39.

Question 49 of 57

2022 Test A Math, question 49

Question

In basketball, a player can earn 1 point for making a free throw and 2 points for making another basket. Marcus makes a mean of 8 free throws per game, and he scores a total of at least 25 points per game. The inequality below shows this relationship, where b represents the number of 2-point baskets Marcus makes.

$2b + 8 \geq 25$

What is the number of 2-point baskets Marcus needs to make if he wants to score at least his minimum total points for a game and make his mean number of free throws?

A $b \leq 8$
B $b \geq 9$
C $b \geq 16$
D $b \leq 25$

Show the answer

Correct answer: B

Solve the inequality $2b + 8 \geq 25$ .

$2b + 8 - 8 \geq 25 - 8$

$2b \geq 17$

$\frac{2b}{2} \geq \frac{17}{2}$

$b \geq 8.5$

Since the number of baskets made must be a whole number, the fewest 2-point baskets, b, he must make is 9.

Question analysis

In basketball, Marcus makes a mean of 8 one-point free throws a game and wants at least 25 points, shown by the inequality 2b + 8 ≥ 25. Students find how many 2-point baskets he needs. It tests solving a two-step inequality and interpreting the solution in context, since a player cannot make part of a basket.

NYS standards: 7.EE.4b two-step inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step inequality from a word problem is a grade 7 standard. Here the inequality is already written, so the extra demand is interpreting its solution as a count of baskets, typical of grade 7 tasks.
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 minor
Complex sentences minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 92 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 inequality is handed over, so the solving is two routine moves; the judgment comes after. Dividing 17 by 2 leaves a half, and since a basket cannot be split, the boundary has to move up to 9: eight baskets give 2(8) + 8 = 24 points, one short of 25. Students who flip the symbol while dividing, or who move the 8 without changing its sign, still find a choice to match.

An answer in an unexpected form minor Choosing the answer
Solving gives b ≥ 8.5, which no choice shows; it has to be recognized as b ≥ 9, because Marcus cannot make half a basket. (Enright, Morley & Sheehan, 2002)
Complex sentences minor Understanding the question
The 33-word question restates the given inequality as “at least his minimum total points” and “make his mean number of free throws”, wording harder than the math it describes. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Undo the added 8 and then the 2 to solve for b, then ask which whole numbers of baskets actually reach 25 points, testing the boundary value in the original inequality.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
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 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reverses the inequality when dividing by 2, as if dividing always flipped it, so b ≤ 8.5, and whole baskets give b ≤ 8.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves the 8 to the right without changing its sign, so 2b ≥ 33 and b ≥ 16.5, then drops the half to match b ≥ 16.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 25-point total from the inequality as a limit on baskets and writes at most, a number copied from the question rather than solved for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality: $\text{if } A \geq B, \text{ then } A - C \geq B - C \text{ and } \frac{A}{2} \geq \frac{B}{2}$; A and B are the two sides. The sign ≥ means greater than or equal to. Taking the same amount C from both sides keeps the sign the same. Dividing both sides by a positive number, like 2, also keeps it the same. Only a negative number flips it.

Sort out the question
A free throw is worth 1 point, and each other basket is worth 2 points. Marcus makes 8 free throws, which give 8 points. The letter b is the number of 2-point baskets, so they give 2b points.
He wants at least 25 points in all. That is the inequality $2b + 8 \geq 25$ . An inequality compares two amounts. The sign ≥ means greater than or equal to. You need the values of b that work.
Subtract 8 from both sides
Undo the added 8 first:
$2b + 8 - 8 \geq 25 - 8$
$2b \geq 17$
Watch out: adding 8 instead gives $2b \geq 33$ . That leads to choice C, which is wrong.
Divide both sides by 2
2b means 2 times b. Divide both sides by 2. Since 2 is positive, the sign stays the same:
$\frac{2b}{2} \geq \frac{17}{2}$
$b \geq 8.5$
Watch out: do not flip the sign here. Flipping it gives $b \leq 8.5$ , which leads to choice A.
Think about whole baskets
Marcus cannot make half a basket. So test the whole numbers near 8.5:
$2 \times 8 + 8 = 24$
$2 \times 9 + 8 = 26$
Eight baskets give 24 points, one short of 25. Nine baskets give 26 points, which is enough. So he needs 9 or more baskets.
Give the answer
The number of baskets is 9 or more, written $b \geq 9$ . Choice D, $b \leq 25$ , just copies the 25 from the points.
So the answer is B, $b \geq 9$ .

Check your answer

Try a bigger number, such as b = 12. Then $2 \times 12 + 8 = 32$ points, which is at least 25. Any number of baskets above 9 also works, and that is what ≥ says. Below 9, the total stays at 24 points or fewer.

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 C Math, page 113, question 40.

Question 50 of 57

2022 Test A Math, question 50

Question

Nadia paid $1.00 to mail a letter. The cost to mail a letter is $0.55 for the first ounce of weight plus $0.15 for each additional ounce. What was the weight, in ounces, of the letter Nadia mailed?

A1.4
B1.5
C3
D4

Show the answer

Correct answer: D

Let x represent the number of ounces Nadia’s letter weighed. The first ounce costs $0.55, so $(x - 1)$ represents the number of ounces that cost $0.15 each.

Given that the total cost is $1.00, the equation that describes the situation can be written and solved as shown.

$0.55 + 0.15(x - 1) = 1.00$

$0.55 + 0.15x - 0.15 = 1.00$

$0.15x = 1.00 - 0.55 + 0.15$

$0.15x = 0.60$

$\frac{0.15x}{0.15} = \frac{0.60}{0.15}$

$x = 4$

So the letter Nadia mailed weighed 4 ounces.

Question analysis

Nadia pays $1.00 to mail a letter, and postage costs $0.55 for the first ounce plus $0.15 for each additional ounce. Students find the weight of the letter in ounces. It tests writing and solving a two-step equation from a pricing rule in which the two prices apply to different ounces.

NYS standards: 7.EE.4a two-step equations
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve word problems that lead to equations such as px + q = r. This one takes that form once each price is matched to the right ounces, a step that makes it a little more demanding than routine practice.
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
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 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

Pricing the letter ounce by ounce, $0.55, $0.70, $0.85, $1.00, reaches the total at the fourth ounce without any algebra. The equation route works too, but the first ounce is easy to lose: dividing the $0.45 left by $0.15 counts only the ounces after the first. The two decimal choices come from merging the two prices or from treating dollars as ounces.

Turning words into math moderate Understanding the question
“$0.55 for the first ounce of weight plus $0.15 for each additional ounce” makes the cost 0.55 + 0.15(x − 1), so the $0.15 covers every ounce except the first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
The money is in decimals, $1.00 − $0.55 = $0.45 and then $0.45 ÷ $0.15 = 3, and dividing by a decimal invites a misplaced point. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the cost of a letter weighing x ounces, $0.55 plus $0.15 for each ounce after the first, set it equal to $1.00 and solve, then check by pricing the letter ounce by ounce.

Traps

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.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Merges the two prices into $0.70 for every ounce and divides $1.00 by $0.70, about 1.43, which rounds to 1.4.
B Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Counts the first ounce, then treats the $0.45 left over as 0.45 ounce, getting 1.45 ounces, which rounds to 1.5.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 3 additional ounces that the $0.45 left over pays for, 0.45 ÷ 0.15 = 3, and stops before adding the first ounce.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost to mail a letter: $C = 0.55 + 0.15 \times (w - 1)$; C is the cost in dollars. w is the weight in ounces. The first ounce costs $0.55. Each ounce after the first costs $0.15, and there are w − 1 of them. Subtracting 1 leaves out the first ounce, which is already paid for.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
Nadia paid $1.00 to mail a letter. The first ounce costs $0.55. Each additional ounce, meaning each ounce after the first, costs $0.15. You need the weight of the letter in ounces.
Write the equation
Let w be the weight in ounces. The first ounce is 1 of the w ounces. So the extra ounces number w − 1. The cost is $1.00.
$0.55 + 0.15 \times (w - 1) = 1.00$
Take off the first ounce’s cost
Subtract 0.55 from both sides. This keeps the equation balanced.
$0.15 \times (w - 1) = 1.00 - 0.55 = 0.45$
So $0.45 pays for the extra ounces.
Find the extra ounces
Divide both sides by 0.15. Multiply both numbers by 100 to make them whole.
$w - 1 = 0.45 \div 0.15 = 45 \div 15 = 3$
There are 3 extra ounces after the first one.
Add back the first ounce
$w = 3 + 1 = 4$
The letter weighed 4 ounces. Watch out: 3 counts only the ounces after the first. It is choice C. Also, don’t join $0.55 and $0.15 into one price. $1.00 \div 0.70$ is about 1.4, choice A. So the answer is D, 4 ounces.

Check your answer

Price the letter one ounce at a time. 1 ounce: $0.55. 2 ounces: $0.70. 3 ounces: $0.85. 4 ounces: $1.00. The cost reaches $1.00 at 4 ounces. That matches.

More questions like this

The original DOE question

Question 107 as printed on page 95 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 95, question 107. Open page 95 of the handbook (PDF)

Question 51 of 57

2022 Test A Math, question 51

Question

COST OF ORANGES

Number of Oranges	Cost
4	$1.80
6	$2.70
9	$4.05

The table shows the cost of oranges. What is the unit rate for the cost per orange?

A$0.45
B$0.90
C$1.35
D$1.80

Show the answer

Correct answer: A

The unit rate (cost per orange) can be determined by dividing the cost of x oranges by x.

$\frac{1.80}{4} = \frac{2.70}{6} = \frac{4.05}{9} = 0.45$

The unit rate is $0.45.

Question analysis

A table shows what 4, 6 and 9 oranges cost, in dollars and cents. Students find the unit rate, the cost of a single orange. It tests identifying a unit rate in a proportional table, and the tricky part is keeping straight which column holds the cost and which holds the number of oranges.

NYS standards: 7.RP.2b constant of proportionality
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a unit rate in a table belongs to grade 7 proportional relationships, but here it takes one division with money in a clearly proportional table, so the task is easier 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: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 3, 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

Any row gives the rate in one division, and the three rows agree, so the question is easy once unit rate is read as the cost of one orange. The traps are differences between rows, which look like a change per orange but cover two or three oranges, and the first cost in the table, which is for four. Dividing a second row confirms the result.

Reading a figure, table or graph minor Understanding the question
The table lists 4, 6 and 9 oranges but never a single orange, so the cost of one has to be computed from a row, such as $1.80 ÷ 4. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The term unit rate is not defined; students must know it means the cost of one orange, a row's cost divided by its number of oranges. (Anderson & Krathwohl, 2001)

Strategy: Pick any row and divide its cost by its number of oranges; doing the same with a second row confirms that the rate stays the same.

Traps

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

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Subtracts the first two costs, $2.70 − $1.80 = $0.90, the price of 2 more oranges, without dividing by the 2 oranges it covers.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Subtracts the last two costs, $4.05 − $2.70 = $1.35, the price of 3 more oranges, rather than the price of one.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes $1.80, the first cost in the table, which is what 4 oranges cost together, not one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{cost}}{\text{number of oranges}}$; The unit rate is the cost of one item, here one orange. The cost is what a group of oranges costs together. Divide by the number of oranges in that group. Any row of the table gives the same rate, because each orange costs the same.

Sort out the question
The table gives the cost of 4, 6 and 9 oranges. It never shows the cost of one orange. A unit rate is the amount for one unit. Here that means the cost of one orange. You need to find it.
Divide one row
Use the first row: 4 oranges cost $1.80.
$\text{unit rate} = \frac{\text{cost}}{\text{number of oranges}}$
$\text{unit rate} = \frac{1.80}{4} = 0.45$
One orange costs $0.45. Watch out: $1.80 is the cost of 4 oranges, not one. Don’t pick choice D.
Check with the other rows
$\frac{2.70}{6} = 0.45 \qquad \frac{4.05}{9} = 0.45$
Every row gives $0.45, so the rate stays the same. Watch out: $2.70 - 1.80 = 0.90$ is the cost of 2 more oranges, choice B. $4.05 - 2.70 = 1.35$ is the cost of 3 more, choice C. Neither is for one orange. So the answer is A, $0.45.

Check your answer

Multiply back: $0.45 \times 4 = 1.80$ and $0.45 \times 9 = 4.05$ . Both match the table. Even a difference works once you divide: $0.90 \div 2 = 0.45$ .

More questions like this

The original DOE question

Question 108 as printed on page 95 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 95, question 108. Open page 95 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 A Math, question 52

Question

A softball player bought bottles of water for her team to have at practice. The total number of bottles she bought, b, is proportional to n, the number of players at practice. Which equation represents the relationship between b and n? The constant of proportionality is p.

A $b = pn$
B $b = p + n$
C $b + n = p$
D $b + n + p = 1$

Show the answer

Correct answer: A

In this context, the constant of proportionality, p, represents the relationship between b, the number of bottles of water, and n, the number of players on the team.

The number of bottles of water is the product of the constant of proportionality and the number of players.

The equation that represents this relationship is $b = pn$ .

Question analysis

A softball player buys bottles of water for practice, and the number of bottles, b, is proportional to the number of players, n, with constant of proportionality p. Students choose the equation that represents the relationship. It tests writing an equation for a proportional relationship using letters only, so the meaning of proportional has to be clear.

NYS standards: 7.RP.2c proportional relationship equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students write equations for proportional relationships from tables, graphs and descriptions, usually with a number to find. Here the constant is already given as a letter, so only the general form must be recalled, a short task.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly remember
What makes it hard: Something you must already know minor
Variables instead of numbers minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 51 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 recall question dressed in a short story: the softball team adds nothing but names for the letters. The one idea needed is that proportional means multiplied by a constant, so doubling the players doubles the bottles. Students who think of a constant as something added, or as a fixed total, are drawn to the two sums that use the same letters.

Something you must already know minor Working it out
The general form of a proportional relationship, one amount equal to the constant times the other, has to be recalled; the stem names p but never says it multiplies. (Anderson & Krathwohl, 2001)
Variables instead of numbers minor Understanding the question
No numbers are given, only b, n and p, so a choice cannot be checked by computing unless students make up a case, such as 2 bottles per player for 10 players. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Recall how two proportional quantities are linked through the constant of proportionality, then test each choice with sample numbers: if the players double, the bottles should double too.

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

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats proportional as adding a fixed amount, so the bottles would be the players plus p, rather than the players multiplied by a constant.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes the constant of proportionality as the total of bottles and players, confusing a constant sum with a constant ratio.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Puts all three letters into one sum equal to 1, an equation that uses the right letters but matches no idea of proportion, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx$; y and x are two amounts that change together. k is the constant of proportionality, the number you always multiply x by to get y. Because it is a multiplication, doubling x doubles y.

Sort out the question
The letter b is the number of bottles, and n is the number of players. The bottles are proportional to the players. The constant of proportionality is p.
You need the equation that links b and n. There are no numbers, only letters.
Recall what proportional means
Proportional means one amount is always a fixed number times the other. For example, say there are 2 bottles for each player. Then 10 players need 20 bottles, and 20 players need 40 bottles. Doubling the players doubles the bottles.
The fixed number, here 2, is the constant of proportionality.
Write it with these letters
Start with the rule $y = kx$ . The bottles take the place of y, the players take the place of x, and p is the constant:
$b = pn$
Here pn means p times n.
Check the other choices
Watch out: choice B, $b = p + n$ , adds instead of multiplying. With p = 2, 10 players give 12 bottles. Then 20 players give 22, which is not double.
Choice C makes p a total of bottles and players, not a multiplier. Choice D makes all three letters add up to 1, which does not fit at all.
Give the answer
Only $b = pn$ multiplies the players by a constant. So it is the only choice where doubling the players doubles the bottles.
So the answer is A, $b = pn$ .

Check your answer

Try p = 3 bottles per player. With 5 players, $b = 3 \times 5 = 15$ . With 10 players, $b = 3 \times 10 = 30$ . Twice the players, twice the bottles. Also, $15 \div 5 = 3$ , so b ÷ n gives back the constant.

More questions like this

The original DOE question

Question 43 as shown on the DOE’s digital test, page 116 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 116, question 43.

Question 53 of 57

2022 Test A Math, question 53

Question

Malik earns $10 per hour at his job. He wants to change to a job that will pay $12 per hour. What will be the percent increase in Malik’s hourly pay if he makes this job change?

A2%
B20%
C83%
D120%

Show the answer

Correct answer: B

Subtract to determine the amount of increase in Malik’s hourly pay if he changes jobs.

$12 - 10 = 2$

Divide the increase amount by the original hourly pay rate.

$2 \div 10 = 0.2$

Multiply the quotient by 100 to express the decimal amount as a percentage.

$0.2 \times 100 = 20$

Question analysis

Malik earns $10 an hour and wants to switch to a job that pays $12 an hour. Students find the percent increase in his hourly pay. It tests finding a percent change. The numbers are easy, so the challenge is knowing which amounts to compare and how to turn that comparison into a percent.

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Percent increase is grade 7 content, and this case uses whole-dollar wages that make the arithmetic easy. The demand is comparing the change with the right amount rather than reporting the change itself or a ratio of the wages.
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: Something you must already know minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 2, 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

The numbers could hardly be friendlier, a $2 raise on $10, so the question checks what to divide by what. Dividing $12 by $10 gives 1.2, the new pay as a percent of the old, and stopping there is tempting because the division feels like the whole job. Reading the $2 itself as 2% is the other quick error, since dollars of change are not percent of change.

Something you must already know minor Working it out
The rule for percent change, the change divided by the starting amount, is not given, and each wrong choice comes from a different comparison of the $10 and the $12. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
“Percent increase in Malik’s hourly pay” has to be read as the $2 rise measured against the $10 he earns now, not as the new wage compared with the old. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many dollars the hourly pay goes up, then divide that change by the pay Malik earns now and write the result as a percent.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the $2 increase and reads it as 2%, as if a change in dollars were a percent, without dividing it by the original $10.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the old pay by the new, 10 ÷ 12 ≈ 0.83, turning the ratio of the wages upside down and reporting 83%.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the new pay by the old, 12 ÷ 10 = 1.2, which gives the new wage as a percent of the old, not the increase.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; The original is the amount before the change. The new amount is the amount after it. The top is how much it went up. Dividing by the original compares the change with where you started. Multiplying by 100 makes it a percent.

Sort out the question
Malik earns $10 an hour now. The new job pays $12 an hour. You need the percent increase.
That means: how much does his pay go up, compared with what he earns now? So $10 is the original amount, and $12 is the new amount.
Find the increase in dollars
Subtract the old pay from the new pay:
$12 - 10 = 2$
His pay goes up by $2 an hour.
Watch out: $2 is an amount of money, not a percent. Calling it 2% gives choice A.
Compare the increase with the original
Use the percent increase formula. Divide the $2 by the original $10, then multiply by 100:
$\frac{2}{10} \times 100 = 0.2 \times 100 = 20$
So the pay goes up by 20%.
Watch out: $12 \div 10 = 1.2$ is 120%, choice D. That is the new pay as a percent of the old, not the increase. And $10 \div 12$ is about 83%, choice C, which is upside down.
Give the answer
The $2 raise is 20% of Malik’s current pay.
So the answer is B, 20%.

Check your answer

Work forward. 20% of $10 is $0.20 \times 10 = 2$ dollars. Add that to $10, and you get $12, the new pay. It works. Note that 120% is the new pay compared with the old pay, which is the same as a 20% increase.

More questions like this

The original DOE question

Question 45 as shown on the DOE’s digital test, page 118 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 118, question 45.

Question 54 of 57

2022 Test A Math, question 54

Question

A rain gauge shows that 10 centimeters of rain fell yesterday. If 1 inch is equal to approximately 2.5 centimeters, about how many inches of rain fell yesterday?

A1
B2.5
C4
D10

Show the answer

Correct answer: C

Set up a proportion to solve:

$\frac{1\text{ in.}}{2.5\text{ cm}} = \frac{x\text{ in.}}{10\text{ cm}}$

$2.5x = 10$

$x = 10 \div 2.5$

$x = 4$

10 centimeters is approximately 4 inches.

Question analysis

A rain gauge shows that 10 centimeters of rain fell, and 1 inch is about 2.5 centimeters. Students find about how many inches of rain fell. It tests converting between metric and customary units with a given conversion fact, and the main risk is converting in the wrong direction or reusing a number from the question.

NYS standards: 6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting units with a given conversion fact is grade 6 content, and this needs a single division with friendly numbers, lighter than grade 7 work with rates and percents.
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
Solution: 1 step
Reading level: Question and choices: about grade 10, 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

A single division answers the question, and its numbers are friendly: four lengths of 2.5 centimeters make 10. Every wrong choice is simply a number from the stem, so the only real risk is reaching for a given value instead of converting. Inches are longer than centimeters, so the rainfall in inches has to be less than 10 but more than the 1 inch that 2.5 centimeters make.

Turning words into math minor Understanding the question
The fact “1 inch is equal to approximately 2.5 centimeters” has to be turned into a division, how many 2.5-centimeter lengths fit into 10, rather than a multiplication. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Ask how many groups of 2.5 centimeters fit into the 10 centimeters of rain, since each group of 2.5 centimeters makes one inch.

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.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks the 1 from the conversion fact, 1 inch for 2.5 centimeters, reading the given fact as the answer instead of applying it to 10 centimeters.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 2.5, the number of centimeters in one inch, straight from the conversion fact rather than finding how many inches 10 centimeters make.
D Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Keeps the 10 from the rain gauge, as if centimeters and inches were the same size, so no conversion is done at all.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Converting centimeters to inches: $\text{inches} = \text{centimeters} \div 2.5$; About 2.5 centimeters make 1 inch. So each group of 2.5 centimeters is 1 inch. Dividing the centimeters by 2.5 counts how many of those groups fit. That count is the number of inches.

Sort out the question
The rain gauge shows 10 centimeters (cm) of rain. You know that 1 inch is about 2.5 cm. You need the rain in inches.
Decide whether to multiply or divide
An inch is longer than a centimeter. So it takes fewer inches than centimeters to measure the same rain. The answer must be less than 10. That tells you to divide, not multiply.
Watch out: choice D, 10, keeps the same number, as if inches and centimeters were the same size.
Divide by 2.5
Use the conversion formula:
$10 \div 2.5 = 4$
You can see it by adding: $2.5 + 2.5 + 2.5 + 2.5 = 10$ . Four groups of 2.5 cm fit into 10 cm. So 10 cm is about 4 inches.
Give the answer
Watch out: choices A and B, 1 and 2.5, are just the numbers in the conversion fact. They do not use the 10 cm at all.
About 4 inches of rain fell. So the answer is C, 4.

Check your answer

Change back. Each inch is 2.5 cm, so 4 inches is $4 \times 2.5 = 10$ cm. That is what the gauge showed. Also, 4 inches is less than 10, as it should be, because inches are longer.

More questions like this

The original DOE question

Question 46 as shown on the DOE’s digital test, page 119 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 119, question 46.

Question 55 of 57

2022 Test A Math, question 55

Question

The height of a plant, in millimeters, is proportional to the number of days the plant has grown. The plant has grown 20 millimeters in 8 days. What is the constant of proportionality, in millimeters per day?

A0.4
B2.5
C12
D20

Show the answer

Correct answer: B

The constant of proportionality is determined by dividing the number of millimeters by the number of days:

$\frac{20\text{ mm}}{8\text{ days}} = \frac{20}{8} = 2.5\text{ mm/day}$

Question analysis

A plant's height in millimeters is proportional to the number of days it has grown, and it grew 20 millimeters in 8 days. Students find the constant of proportionality in millimeters per day. It tests finding a unit rate, and the order of the division decides whether the rate comes out in the units asked for.

NYS standards: 7.RP.2b constant of proportionality
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from one pair of values is a basic grade 7 skill. With one division and the units named in the question, it asks less than most 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: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One division gives the constant, and the question itself names the order by asking for millimeters per day. The tempting slip is dividing the smaller number by the larger, 8 by 20, out of habit, which gives a rate in days per millimeter. A plant that grows 20 millimeters in 8 days grows more than 2 millimeters a day, a quick sense check on any result.

Something you must already know minor Working it out
Students must know that the constant is the height divided by the days; the units in the question, millimeters per day, name the order of the division. (Anderson & Krathwohl, 2001)

Strategy: Use the units in the question, millimeters per day, as the instruction: divide the millimeters grown by the number of days.

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 partial knowledge Expected pull: strong
Divides the days by the millimeters, 8 ÷ 20 = 0.4, which gives days per millimeter, the rate turned upside down.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts the days from the millimeters, 20 − 8 = 12, combining the two numbers without thinking about a rate.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the 20 millimeters the plant grew over all 8 days, the total growth, rather than the growth for each single day.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; Here y is the height in millimeters, and x is the number of days. k is the growth in one day, in millimeters per day. In a proportional relationship, y = kx, so dividing y by x gives k.

Sort out the question
The plant’s height is proportional to the number of days it grows. It grew 20 millimeters (mm) in 8 days.
You need the constant of proportionality in millimeters per day. Per day means for each day. So you need how much the plant grows in one day.
Divide millimeters by days
The units tell you the order: millimeters per day means millimeters divided by days. Use the formula:
$k = \frac{20}{8} = \frac{5}{2} = 2.5$
So the plant grows 2.5 mm each day.
Avoid the tempting slips
Watch out: $8 \div 20 = 0.4$ , choice A, is days per millimeter. That is the rate upside down.
Choice C, 12, comes from $20 - 8$ . Subtracting does not make a rate. Choice D, 20, is the growth over all 8 days, not one day.
Give the answer
The constant of proportionality is 2.5 mm per day.
So the answer is B, 2.5.

Check your answer

Multiply back. At 2.5 mm a day for 8 days, the plant grows $2.5 \times 8 = 20$ mm. That matches. Also, 2 mm a day would give only 16 mm, and 3 mm a day would give 24 mm. So the rate must be between 2 and 3.

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 C Math, page 120, question 47.

Question 56 of 57

2022 Test A Math, question 56

Question

Which table shows a proportional relationship?

A
x y
0 0
4 $\frac{3}{2}$
8 3
B
x y
0 1
4 $\frac{7}{2}$
8 6
C
x y
0 −1
4 $-\frac{1}{2}$
8 0
D
x y
0 0
4 12
8 16

x	y
0	0
4	$\frac{3}{2}$
8	3

x	y
0	1
4	$\frac{7}{2}$
8	6

x	y
0	−1
4	$-\frac{1}{2}$
8	0

x	y
0	0
4	12
8	16

Show the answer

Correct answer: A

The first table is the only one with a constant of proportionality, $\frac{3}{8}$ .

In each ordered pair, $\frac{3}{8}x = y$ .

$\frac{3}{8}(0) = 0$

$\frac{3}{8}(4) = \frac{12}{8} = \frac{3}{2}$

$\frac{3}{8}(8) = \frac{24}{8} = 3$

Question analysis

Four tables each give y for x = 0, 4 and 8, with some values written as fractions or negatives. Students choose the table that shows a proportional relationship. It tests recognizing proportional relationships from a table, which means checking every row against one rule rather than judging by the pattern.

NYS standards: 7.RP.2a testing for proportionality
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Deciding from a table whether a relationship is proportional is a core grade 7 skill. The fractions and negative values in the tables make each ratio a careful check rather than a glance.
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: Every choice must be checked moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 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 quick tests decide proportionality, and the tables are built so that each test alone lets a wrong table through. The last table starts at (0, 0) like the right one, while the middle two grow by equal steps like the right one. Only checking that y ÷ x is the same in every row with x not 0 separates the one table whose ratio stays at 3/8.

Every choice must be checked moderate Choosing the answer
Each table has to be tested row by row, and two of them start at (0, 0), so the opening row settles nothing; only y ÷ x in the other rows separates them. (Embretson & Wetzel, 1987)
Error-prone arithmetic minor Working it out
The y-values include 3/2, 7/2 and −1/2, so the ratios mean dividing fractions by 4, such as (3/2) ÷ 4 = 3/8, and comparing them with 3 ÷ 8. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: For each table, divide y by x in every row where x is not zero, and check that the quotient never changes and that y is 0 when x is 0.

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.
Checks only one number: Tries one value in each choice and takes the first that works, when that value happens to work for a wrong choice too; a second value would rule it out.

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: moderate
Sees y rise by the same 5/2 each time x rises by 4 and takes the steady change as proportional, missing that y is 1 when x is 0.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Spots a 0 in the y column and a steady rise of 1/2, but that 0 belongs to x = 8, while at x = 0 the table gives −1.
D Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: strong
Checks only the row (0, 0), which every proportional table has, or only 12 ÷ 4 = 3, without testing 16 ÷ 8 = 2 in the last row.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx$; y and x are the two amounts in each row. k is the constant of proportionality, the same number in every row. Two tests follow from it. When x is 0, y must be 0. And y ÷ x must equal k in every row where x is not 0.

Sort out the question
Each table gives y for x = 0, 4 and 8. You need the one that shows a proportional relationship. That means y is always the same number times x.
Use two tests. First, y must be 0 when x is 0. Second, y ÷ x must be the same in the other rows.
Test the row where x is 0
Look at the top row of each table:
- Table A: y = 0. It passes.
- Table B: y = 1. It fails.
- Table C: y = −1. It fails.
- Table D: y = 0. It passes.
Watch out: tables B and C go up by equal steps. Steady steps are not enough, because they do not start at 0. Table C does have a 0, but it is next to x = 8.
Test the ratio in table A
Divide y by x. Dividing by 4 is the same as multiplying by $\frac{1}{4}$ :
$\frac{3}{2} \div 4 = \frac{3}{2} \times \frac{1}{4} = \frac{3}{8}$
$3 \div 8 = \frac{3}{8}$
Both rows give $\frac{3}{8}$ . Table A passes both tests.
Test the ratio in table D
Divide y by x in the same way:
$12 \div 4 = 3 \qquad 16 \div 8 = 2$
The ratios are different, so table D is not proportional.
Watch out: table D starts at (0, 0) like a proportional table. If you check only that row, or only the row with 4, you will pick choice D.
Give the answer
Only table A starts at (0, 0) and keeps y ÷ x the same.
So the answer is A, the table with y = 0, $\frac{3}{2}$ and 3.

Check your answer

Use the rule for table A, $y = \frac{3}{8}x$ . When x = 4, $y = \frac{3}{8} \times 4 = \frac{12}{8} = \frac{3}{2}$ . When x = 8, $y = \frac{3}{8} \times 8 = 3$ . Both match. Doubling x from 4 to 8 also doubles y.

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 C Math, page 121, question 48.

Question 57 of 57

2022 Test A Math, question 57

Question

Terry wants to pour cement around the edge of the circular patio in her backyard. The patio has a radius of 5 feet. What is the distance, in feet, around the edge of the patio? Use 3.14 for $\pi$ .

A15.7
B31.4
C49.3
D78.5

Show the answer

Correct answer: B

To find the distance around a circle if the radius is known, use the formula for circumference, $C = 2\pi r$ .

$C = 2(3.14)(5) = 31.4$

Question analysis

Terry wants to pour cement around the edge of a circular patio with a radius of 5 feet. Students find the distance around the edge, using 3.14 for π. It tests the circumference of a circle, and because the formulas for a circle's area and circumference both use the radius, they are easy to mix up.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying a circle formula is a grade 7 standard, and this is a one-formula problem with a given value for π. The only demand beyond routine is recalling the right formula, which the test does not print.
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: Something you must already know moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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

One formula and one multiplication finish the problem once the right formula is chosen. The cement in the story pulls toward area, and the area, 78.5, is listed, as is half the circumference, from multiplying 3.14 by the radius alone. Remembering that the circumference is π times the diameter, here 10 feet, and that the answer is in feet, not square feet, settles it.

Something you must already know moderate Working it out
The circumference formula is not printed, and the area formula uses the same radius and π, so recalling the wrong one gives a number that looks just as finished. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
Pouring cement suggests covering a surface, but the question asks for “the distance, in feet, around the edge”, which is the circumference, not the area. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Decide which circle measurement the distance around the edge is, recall its formula, and substitute the radius and 3.14, doubling the radius wherever the formula calls for the diameter.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
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 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies π by the radius, 3.14 × 5 = 15.7, using the radius where C = πd needs the diameter, so it finds half the distance around.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Starts from the area formula and squares π instead of the radius, 3.14 × 3.14 × 5 ≈ 49.3, mixing the wrong formula with a slip in squaring.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Finds the patio’s area, 3.14 × 5² = 78.5 square feet, using the area formula instead of the circumference for the distance around.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \pi r$; C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, about 3.14. 2r is the diameter, the distance all the way across. The distance around is always about 3.14 times that.
Area of a circle: $A = \pi r^2$; A is the area, the space inside the circle, in square units. r² means r times r. It is here so you can tell the two formulas apart. This question does not use it.

Sort out the question
The patio is a circle with a radius of 5 feet. The radius is the distance from the center to the edge. Terry wants cement around the edge.
You need the distance around the edge, in feet. The distance around a circle is called its circumference. Use 3.14 for π.
Choose the right formula
The distance around is a length, so use the circumference formula, $C = 2\pi r$ .
Watch out: pouring cement may make you think of area. The area, $3.14 \times 5 \times 5 = 78.5$ square feet, is choice D. That is the space inside the patio, not the distance around it.
Put in the numbers
Put 3.14 in for π and 5 in for r. Multiply 2 by 5 first, because that is easy:
$C = 2 \times 3.14 \times 5 = 10 \times 3.14 = 31.4$
The distance around the patio is 31.4 feet.
Watch out: $3.14 \times 5 = 15.7$ , choice A, forgets the 2. It goes only halfway around.
Give the answer
The distance around the edge of the patio is 31.4 feet.
So the answer is B, 31.4.

Check your answer

Estimate. The diameter is 10 feet. The distance around a circle is a little more than 3 times its diameter. $3 \times 10 = 30$ , so about 31 feet makes sense. Choice D, 78.5, is far too big, and it is in square feet.

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 C Math, page 123, question 50.