Practice Test

2026 Practice Tests (digital)

2026 Practice Test C Math

Question 1 of 50

2026 Practice Test C Math, question 1

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 2 of 50

2026 Practice Test C Math, question 2

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 3 of 50

2026 Practice Test C Math, question 3

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 4 of 50

2026 Practice Test C Math, question 4

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 5 of 50

2026 Practice Test C Math, question 5

Question

Maya read n pages for her English homework. Trevor read three times as many pages as Maya. Together, Maya and Trevor read 12 more pages than Ali read. Write an expression that shows how many pages Ali read, in terms of n.

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

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

Show the answer

Correct answer: 4n – 12

Let n represent the number of pages that Maya read.

Trevor read three times as many pages as Maya: Trevor’s pages $= 3n$

To find how many pages they read together, add Maya’s and Trevor’s pages: $n + 3n = 4n$

This means that together, they read $4n$ pages.

Ali read 12 fewer pages than Maya and Trevor read together. So 12 should be subtracted from the total:

Ali’s pages $= 4n - 12$

Question analysis

Maya reads n pages, Trevor reads three times as many, and together they read 12 more pages than Ali. Students type an expression in n for the number of pages Ali read. It tests writing and simplifying expressions from words, and the comparison with Ali has to be translated with care.

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 chains three relationships, and the last compares the pair with Ali rather than describing Ali directly. Typing the answer with no choices to compare adds risk.
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
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 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

Building the pair's total, n + 3n = 4n, is quick; the question turns on the direction of the last comparison. The pair read more than Ali, so Ali's count is the smaller one, yet the word more pulls toward adding 12. Testing n = 10 settles it: Maya and Trevor read 40 pages together, so Ali must have read 28, not 52.

Turning words into math moderate Understanding the question
“Together, Maya and Trevor read 12 more pages than Ali read” compares the pair with Ali, so Ali read 12 fewer than 4n, though the word more invites adding 12. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
Every count is written in terms of n, so there is no page total to check an expression against unless the student picks a value for n and tests it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The expression is typed into a box with no choices to compare, so an entry such as 4n + 12 gets no warning that the comparison was read backward. (Enright, Morley & Sheehan, 2002)

Strategy: Write Trevor's pages and the pair's total in terms of n, then test your expression with a value such as n = 10 to check that the pair really read 12 more than Ali.

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.
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.
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

4n + 12 Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Follows the word more and adds 12 to the pair's 4n, though the sentence says the pair read 12 more than Ali, so Ali read fewer.
3n – 12 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Takes Trevor's 3n as the pair's total, leaving out Maya's own n pages, then subtracts the 12 correctly.
12 – 4n Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Sees that 12 and 4n must be subtracted but writes the subtraction in the wrong order, 12 − 4n, which is negative for most values of n.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding like terms: $n + 3n = 4n$; n is the number of pages Maya read. Terms with the same letter are called like terms. Add them by adding the numbers in front: 1 n plus 3 n is 4 n. It works because 1 group of n pages and 3 more groups make 4 groups.
Turning a more-than fact around: $\text{if } T = A + 12 \text{, then } A = T - 12$; T is the pages Maya and Trevor read together. A is the pages Ali read. The pair read 12 more than Ali, so Ali read 12 fewer than the pair. Subtracting 12 undoes adding 12.

Sort out the question
Maya read n pages. n is a variable, a letter that stands for a number you don’t know. Trevor read 3 times as many pages as Maya. Together they read 12 more pages than Ali. You need Ali’s pages, written with n.
Write Trevor’s pages
Three times as many as n is 3 times n:
$\text{Trevor} = 3n$
A number written next to a letter means multiply. So 3n means 3 × n pages.
Add Maya’s and Trevor’s pages
Together means add. Maya read 1 group of n, and Trevor read 3 groups:
$n + 3n = 4n$
Together they read 4n pages.
Watch out: do not leave Maya out. Using only Trevor’s 3n gives 3n − 12, which is wrong.
Decide who read more
The pair read 12 more than Ali. So Ali read 12 fewer than the pair. Take 12 away from their total:
$\text{Ali} = 4n - 12$
Watch out: the word more tempts you to add 12. But 4n + 12 would mean Ali read more than the pair. And 12 − 4n subtracts in the wrong order.
Enter the expression
On the test, the keypad has an n key. Press 4, then n. Then press the minus key beside the plus key, then 1 and 2. You don’t need a times sign. On this site, type 4n-12.
So you would enter 4n − 12. Any expression equal to it, such as −12 + 4n, is also right.

Check your answer

Pick a number for n, such as 10. Maya read 10 pages and Trevor read 30, so together they read 40. Your expression gives Ali $4 \times 10 - 12 = 28$ pages. Is 40 twelve more than 28? Yes, $28 + 12 = 40$ . So the expression works.

More questions like this

The original DOE question

Question 5 as shown on the DOE’s digital test, page 78 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 78, question 5.

Question 6 of 50

2026 Practice Test C Math, question 6

Question

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

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

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 7 of 50

2026 Practice Test C Math, question 7

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 8 of 50

2026 Practice Test C Math, question 8

Question

Jordan is helping plan snacks for a school dance. He knows that 0.75 of the students voted for pizza, and he wants to record the same amount using a fraction.

Which fractions are equivalent to the decimal 0.75?

Select the four correct answers.

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

A $\frac{3}{4}$
B $\frac{15}{20}$
C $\frac{75}{100}$
D $\frac{2}{5}$
E $\frac{6}{8}$
F $\frac{8}{10}$

Show the answer

Correct answer: A, B, C, and E

Check each fraction to determine whether it is equal to 0.75 when simplified or converted to a decimal.

$\frac{3}{4} \rightarrow 3 \div 4 = 0.75$
This is the most common fraction equal to 0.75.

$\frac{15}{20} \rightarrow 15 \div 20 = 0.75$
Simplifies to 3/4, because 15 and 20 can both be divided by 5.

$\frac{75}{100} \rightarrow 75 \div 100 = 0.75$
This is a direct match to 0.75 when written as a decimal.

$\frac{2}{5} \rightarrow 2 \div 5 = 0.4$ , which is less than 0.75.

$\frac{6}{8} \rightarrow 6 \div 8 = 0.75$
Simplifies to $\frac{3}{4}$ because both 6 and 8 can be divided by 2.

$\frac{8}{10} \rightarrow 8 \div 10 = 0.8$ , which is more than 0.75.

Question analysis

Jordan wants to record 0.75 of the students' pizza votes as a fraction, and students select the four fractions, out of six, that equal 0.75. It tests converting between decimals and fractions and recognizing equivalent fractions. Each choice has to be checked by its value, not by how its numbers look.

NYS standards: 7.NS.2d fractions to decimals by long division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Matching fractions to a decimal such as 0.75 builds on grade 4 to 6 work, and the denominators here divide easily. The added demand is the format: four of the six must be chosen, and one wrong or missing pick loses the item.
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: Every choice must be checked moderate
A very close wrong answer minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Three of the four right fractions, 3/4, 75/100 and 6/8, are familiar forms of 0.75, so the item comes down to the fourth. Students who stop after the familiar ones must decide between 15/20, which needs dividing by 5 to recognize, and 8/10, which sits close to 0.75 on a number line. Dividing each numerator by its denominator, or scaling 3/4 up by 5, settles it.

Every choice must be checked moderate Choosing the answer
Six fractions with six different denominators have to be tested one by one, by dividing or by simplifying, since 15/20 and 8/10 cannot be judged by how they look. (Embretson & Wetzel, 1987)
A very close wrong answer minor Choosing the answer
8/10 is 0.8, only 0.05 above 0.75, and it follows 6/8 by adding 2 to the top and the bottom, so it looks like one more member of the family. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
Four of the six must be chosen and all four must be right for the item to count, so each verdict has to be kept while the remaining fractions are tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Turn each choice into a decimal by dividing its numerator by its denominator, or reduce it to lowest terms, and keep only the ones whose value is exactly the decimal given.

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

D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No usual working links 2/5 to 0.75, since 2 ÷ 5 = 0.4, so it mostly catches students filling the fourth slot by guessing.
F Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Builds 8/10 from 6/8 by adding 2 to the top and the bottom, as if equal additions kept a fraction's value, or accepts 0.8 as close enough.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing a fraction to a decimal: $\frac{a}{b} = a \div b$; a is the numerator, the top. b is the denominator, the bottom. A fraction bar means divide. So dividing the top by the bottom gives the fraction’s value as a decimal.
Equivalent fractions: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top and b is the bottom. n is a number that goes evenly into both. Dividing the top and the bottom by the same number gives an equal fraction. Equivalent fractions name the same amount in different ways.

Sort out the question
Six fractions are given, A to F. You need the four that equal the decimal 0.75. Equivalent means equal in value. Test every choice, because some fractions look alike but are not equal.
Write 0.75 as a fraction
0.75 means 75 hundredths:
$0.75 = \frac{75}{100}$
So choice C works. Now simplify it. 25 goes into both 75 and 100:
$\frac{75 \div 25}{100 \div 25} = \frac{3}{4}$
So choice A, 3/4, works too.
Test choices B and E
Simplify each one to see if it becomes 3/4:
- B: 5 goes into 15 and 20. $\frac{15 \div 5}{20 \div 5} = \frac{3}{4}$ . It works.
- E: 2 goes into 6 and 8. $\frac{6 \div 2}{8 \div 2} = \frac{3}{4}$ . It works.
Test choices D and F
Divide the top by the bottom:
- D: $2 \div 5 = 0.4$ . That is less than 0.75.
- F: $8 \div 10 = 0.8$ . That is more than 0.75.
Watch out: F looks like 6/8 with 2 added to the top and the bottom. Adding the same number to both changes the value. Only multiplying or dividing both keeps it.
Name the four answers
A, B, C and E all equal 0.75. On the test, you check the box beside each of them. On this site, you choose all four, then check your answer.
So the answers are A, B, C and E.

Check your answer

Turn each answer into a decimal. $3 \div 4$ , $15 \div 20$ , $75 \div 100$ and $6 \div 8$ all give 0.75. Exactly four choices passed, and the question asks for four. So none were missed.

More questions like this

The original DOE question

Question 8 as shown on the DOE’s digital test, page 81 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 81, question 8.

Question 9 of 50

2026 Practice Test C Math, question 9

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 10 of 50

2026 Practice Test C Math, question 10

Question

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

In the figure above, $n \parallel p$ . 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 11 of 50

2026 Practice Test C Math, question 11

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 12 of 50

2026 Practice Test C Math, question 12

Question

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

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

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

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

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

Point

Show the answer

Correct answer: the point 0.2

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

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

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

$\frac{2}{5} \cdot \frac{1}{2} = \frac{2}{10} = 0.2$

Question analysis

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

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.6 probability from 0 to 1
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the chance of two independent events together is grade 7 compound probability, here with one spinner and one coin. Counting sections from the picture and placing the result on a number line add two small steps.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 59 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph moderate Understanding the question
The two sections labeled 1, upper left and right, are not side by side, and the number line labels only 0, 1/2 and 1, so its nine points must be read as tenths. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Nothing says that the chance of two independent events both happening is the product of their chances, 2/5 × 1/2, rather than their sum. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

0.4 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the spinner's chance of landing on 1, two sections out of five, and plots 2/5 without multiplying by the coin's chance of tails.
0.9 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds the two chances, 2/5 + 1/2 = 9/10, instead of multiplying them, though both events must happen together.
0.1 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Counts only one section labeled 1, so multiplies 1/5 by 1/2 and plots 1/10, missing the second 1 on the spinner.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The spinner has 5 congruent sections. Congruent means the same size and shape, so each is equally likely. The coin has a heads side and a tails side. You need the chance that the spinner lands on 1 and the coin lands on tails.
Find the spinner’s chance
Two sections show 1: the upper left one and the one on the right. So 2 of the 5 equal sections give a 1:
$P(1) = \frac{2}{5}$
Watch out: the two 1s are not next to each other. Counting only one gives 1/5 here, and 0.1 at the end.
Find the coin’s chance
A fair coin is equally likely to land on either side. Tails is 1 of its 2 sides:
$P(\text{tails}) = \frac{1}{2}$
Multiply the two chances
The spin and the flip do not affect each other, and both must happen. So multiply:
$\frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = 0.2$
Watch out: stopping at 2/5 gives 0.4. Adding gives $\frac{2}{5} + \frac{1}{2} = \frac{9}{10}$ , or 0.9. Both are wrong.
Select the point
On the test, the number line labels only 0, 1/2 and 1. Between 0 and 1 are 9 targets, one at each tenth. 0.2 is the second target to the right of 0. On this site, choose 0.2 from the Point menu.
So the answer is the point at 0.2.

Check your answer

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

More questions like this

The original DOE question

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

Question 13 of 50

2026 Practice Test C Math, question 13

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 14 of 50

2026 Practice Test C Math, question 14

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 15 of 50

2026 Practice Test C Math, question 15

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.

Fill-in question: work out the answer and type 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 16 of 50

2026 Practice Test C Math, question 16

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 17 of 50

2026 Practice Test C Math, question 17

Question

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

What is the distance between the midpoints of $\overline{MN}$ 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 18 of 50

2026 Practice Test C Math, question 18

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 19 of 50

2026 Practice Test C Math, question 19

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 20 of 50

2026 Practice Test C Math, question 20

Question

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

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

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

Select one answer in each row.

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

Show the answer

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

$A = 25\pi$ : Circle A

$C = 5\pi$ : Circle C

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

Circle A:

$A = \pi r^{2}$

$A = 25\pi$

$r = 5$

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

Circle C:

$C = \pi \cdot d$

$C = 5\pi$

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

Question analysis

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

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying the area and circumference formulas is the grade 7 circle standard. This gives a radius for one circle and, for the other, a segment known only through congruence, and both rows of the table must be right.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know moderate
Much to keep track of minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph moderate Understanding the question
Circle C's size is shown only by segment DE, drawn through center C with no label, and only the stem's sentence that DE is congruent to AB makes it a diameter of 5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know moderate Working it out
Neither A = πr² nor C = πd is given, and one circle is measured by a radius and the other by a diameter, so each formula needs the right length. (Anderson & Krathwohl, 2001)
Much to keep track of minor Working it out
Four values are in play, areas of 25π and 6.25π and circumferences of 10π and 5π, and both rows must be matched correctly for the item to count. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

1:B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Squares the diameter instead of the radius, so Circle C's area seems to be π × 5² = 25π while Circle A's becomes 100π.
1:C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses 2πr for the area, so Circle A seems to have an area of 10π and Circle C one of 5π, and no circle seems to have 25π.
2:A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses π times the radius for the circumference, so Circle A's radius of 5 seems to give a circumference of 5π.
2:C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Overlooks that DE is congruent to AB, so Circle C has no known size, and with Circle A's circumference at 10π picks Neither.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Circle A has its center at A. Its radius, AB, is 5. Circle C has its center at C. Segment DE goes through C, so it is a diameter. Congruent means the same length, so DE is 5 too.
Careful: in the table, A means area and C means circumference. Circle A and Circle C are just names.
Find each radius and diameter
Circle A: the radius is 5, so the diameter is $2 \times 5 = 10$ .
Circle C: the diameter is 5, so the radius is $5 \div 2 = 2.5$ .
Watch out: without the sentence about DE, Circle C seems to have no size. Then you might pick Neither for the second row.
Check the area row
Use the area formula for each circle. For Circle A:
$A = \pi \times 5^{2} = 25\pi$
For Circle C:
$A = \pi \times 2.5^{2} = 6.25\pi$
Only Circle A has an area of 25π.
Watch out: squaring Circle C’s diameter instead of its radius gives 25π by mistake.
Check the circumference row
Use the circumference formula for each circle. For Circle A:
$C = \pi \times 10 = 10\pi$
For Circle C:
$C = \pi \times 5 = 5\pi$
Only Circle C has a circumference of 5π.
Watch out: π times the radius is not the circumference. It would give Circle A 5π.
Fill in the table
Choose one button in each row. In the row for A = 25π, choose the Circle A column. In the row for C = 5π, choose the Circle C column. Neither stays empty.
So the answers are A = 25π: Circle A and C = 5π: Circle C.

Check your answer

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

More questions like this

The original DOE question

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

Question 21 of 50

2026 Practice Test C Math, question 21

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 22 of 50

2026 Practice Test C Math, question 22

Question

During a hiking trip, a group of students recorded the elevation (in feet) at several points along the trail:

$-120\text{ ft},\ -80\text{ ft},\ 0\text{ ft},\ 45\text{ ft},\ -30\text{ ft},\ 90\text{ ft}$

What is the difference between the highest and lowest elevations recorded on the trail?

Enter your answer in the space.

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

Show the answer

Correct answer: 210

Determine the difference between the highest and lowest elevations:

Determine the highest elevation:

Of −120, −80, 0, 45, −30, and 90, the greatest value is 90 because its location is farthest to the right on a number line.

Highest elevation $= 90\text{ ft}$

Determine the lowest elevation:

Of −120, −80, 0, 45, −30, and 90, the lowest value is −120 because its location is farthest to the left on a number line.

Lowest elevation $= -120\text{ ft}$

Determine the difference between the two elevations:

$90\text{ ft} - (-120)\text{ ft} = 90\text{ ft} + 120\text{ ft} = 210\text{ ft}$

Question analysis

A group of students on a hiking trip records six elevations in feet, some below zero and some above, and finds the difference between the highest and the lowest, typing the result. It tests ordering positive and negative numbers and subtracting integers. Spotting the lowest reading among several negatives and subtracting a negative number correctly are where errors creep in.

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 Finding the gap between a positive and a negative value by subtracting is core grade 7 work with integers. Picking the extremes from six readings is quick, so the demand is the subtraction of a negative number.
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, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Picking the highest and lowest of six readings is quick, though −120 must be recognized as lower than −80 and −30. The points are lost in the subtraction: the span runs 120 feet up to zero and 90 more above it, and students who treat the readings as plain sizes find only 30. Picturing the two readings on a vertical number line shows the whole span.

Error-prone arithmetic moderate Working it out
The difference is 90 − (−120), and subtracting a negative means adding 120, so the part of the trail below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The difference is typed with no choices, so 30 or −30, the results of mishandling the negative, gets no warning that the two readings lie on opposite sides of zero. (Enright, Morley & Sheehan, 2002)

Strategy: Pick the greatest and the least elevation by their places on a number line, then subtract the least from the greatest, turning the subtraction of a negative into an addition.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
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).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

30 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 120 − 90 = 30, as if both readings were on the same side of zero, leaving out the stretch below zero.
-30 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Combines the two extremes by adding, 90 + (−120) = −30, the net of the readings, instead of subtracting the lowest from the highest.
-210 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −120 − 90 = −210, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between highest and lowest: $\text{difference} = \text{highest} - \text{lowest}$; The difference between the highest and lowest values is how far apart they are. Subtract the lowest from the highest, so the answer is never negative.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding a positive. On a number line, it means moving to the right.

Sort out the question
Elevation is height above or below a starting level, here 0 feet. Negative elevations are below 0. The six readings are −120, −80, 0, 45, −30 and 90 feet. You need the difference between the highest and the lowest.
Find the highest and the lowest
Picture the readings on a number line. The highest is farthest right: 90 feet. The lowest is farthest left: −120 feet.
For negative numbers, the bigger the number after the minus sign, the lower it is. So −120 is lower than −80.
Subtract the lowest from the highest
Use the difference formula. Subtracting a negative is the same as adding:
$90 - (-120) = 90 + 120 = 210$
So the two readings are 210 feet apart.
Watch out: $120 - 90 = 30$ treats both readings as if they were above 0. Adding them, $90 + (-120) = -30$ , is wrong too.
Enter the answer
A difference like this is a distance, so it is positive. Subtracting in the other order gives −210, which is wrong. Type only the number, with no units.
So you would enter 210.

Check your answer

Count the distance in two parts. From −120 feet up to 0 is 120 feet. From 0 up to 90 is 90 more feet. Together, $120 + 90 = 210$ feet. That matches.

More questions like this

The original DOE question

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

Question 23 of 50

2026 Practice Test C Math, question 23

Question

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

On parallelogram PQRS above, the length of $\overline{QT}$ 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 24 of 50

2026 Practice Test C Math, question 24

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 25 of 50

2026 Practice Test C Math, question 25

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 26 of 50

2026 Practice Test C Math, question 26

Question

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

Which straight line is the graph of a proportional relationship?

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 27 of 50

2026 Practice Test C Math, question 27

Question

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

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

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

Show the answer

Correct answer: C

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

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

Question analysis

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

NYS standards: 7.RP.2d points on proportional graphs
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on a proportional graph means in its situation is a named grade 7 skill. The graph is clear and starts at the origin, but each statement has to be read closely against the axis labels.
Estimated difficulty: Straightforward: two steps or one careful reading.
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 28 of 50

2026 Practice Test C Math, question 28

Question

If $\frac{z+3w}{4} = 5w$ , what is the value of z in terms of w?

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

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

z =

Show the answer

Correct answer: z = 17w

The goal is to isolate the variable z by using inverse operations.

First, multiply both sides by 4:

$\frac{z+3w}{4} = 5w$

$z + 3w = 20w$

Then subtract $3w$ from both sides:

$z = 17w$

Question analysis

This algebra question gives an equation in which z plus 3w, all divided by 4, equals 5w, and students type z in terms of w after a printed "z =". It tests solving a literal equation with inverse operations. The 4 divides the whole sum, so undoing the steps in the right order, with the right signs, is the tricky part.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve two-step equations with numbers, such as (x + 3)/4 = 5. Solving for one letter in terms of another is grade 8 work, and the typed answer gives no choices 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
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The equation takes only two inverse operations, but their order is the trap: the 4 divides the whole sum z + 3w, so it has to be cleared first. Students who subtract 3w from 5w straight away, or who move 3w to the right without changing its sign, reach a tidy multiple of w that looks as finished as the right one. Substituting w = 1, which needs z = 17 to make (z + 3)/4 equal 5, checks any entry.

Variables instead of numbers moderate Understanding the question
With w on both sides and no numbers to substitute, the result has to stay a multiple of w, and a slip such as 23w gives no sign that it is wrong. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The answer is typed after a printed “z =”, with no choices to compare, so the entry must be a single term in w, not an equation or a number. (Enright, Morley & Sheehan, 2002)

Strategy: Treat w like a known number: undo the division by 4 first, multiplying the whole of each side, then undo the added 3w, keeping the terms in w together.

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

8w Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Undoes the + 3w before the ÷ 4, subtracting 3w from 5w as if only z were divided by 4, so z/4 = 2w and z = 8w.
23w A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Multiplies correctly to get z + 3w = 20w, then moves 3w to the right side without changing its sign, so z = 20w + 3w = 23w.
-1.75w Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 5w by 4 instead of multiplying, undoing a division with another division, so z + 3w = 1.25w and z = 1.25w − 3w = −1.75w.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same thing to both sides: $\text{if } A = B, \text{ then } 4A = 4B \text{ and } A - C = B - C$; A and B are the two sides of an equation, so they are equal. C is any amount. An equation is like a balanced scale. If you multiply both sides by the same number, or take the same amount from both, it stays balanced. That is how you get one letter alone.

Sort out the question
The equation is $\frac{z + 3w}{4} = 5w$ . You need z in terms of w. That means getting z alone on one side. The other side will use only w and numbers.
Treat w like a number you do not know yet. The answer box already shows z =, so you type only what comes after it.
Undo the division first
The fraction bar means the whole top, z + 3w, is divided by 4. So undo that first. Multiply both sides by 4, because multiplying undoes dividing:
$\frac{z + 3w}{4} \times 4 = 5w \times 4$
$z + 3w = 20w$
Watch out: do not subtract 3w first. The 3w sits inside the fraction, so it is divided by 4 too. Subtracting it from 5w first leads to the wrong answer 8w.
Undo the adding
Now 3w is added to z. Subtract 3w from both sides:
$z + 3w - 3w = 20w - 3w$
$z = 17w$
Think of w as a thing, like apples. 20 apples take away 3 apples leaves 17 apples.
Watch out: moving 3w to the other side without changing its sign gives $20w + 3w = 23w$ . That is wrong.
Enter it after the printed z =
On the test, the box already shows z =. Press 1, then 7, then the w key. You do not need a times sign between 17 and w. Do not type z = again. On this site, type 17w in the box.
So you would enter 17w, which makes z = 17w.

Check your answer

Pick a number for w, such as w = 1. Then z = 17. Put both into the first equation. The left side is $\frac{17 + 3}{4} = \frac{20}{4} = 5$ . The right side is $5 \times 1 = 5$ . The two sides match, so 17w works.

More questions like this

The original DOE question

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

Question 29 of 50

2026 Practice Test C Math, question 29

Question

A smoothie shop uses a consistent ratio of fruit to yogurt in its recipes. The table shows the proportional relationship between the number of cups of fruit (x) and the number of cups of yogurt (y) in different smoothie batches.

SMOOTHIE RECIPE

Cups of Fruit x	Cups of Yogurt y
6	1.5
8	2.0
10	2.5
12	3.0

What is the constant of proportionality of the number of cups of yogurt to the number of cups of fruit?

Enter your answer in the space.

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

Show the answer

Correct answer: 0.25

To find p, the constant of proportionality of the number of cups of yogurt, y, to the number of cups of fruit, x, use the equation $y = px$ .

The table shows that there are 1.5 cups of yogurt for 6 cups of fruit.

$1.5 = 6p$

$\frac{1.5}{6} = \frac{6p}{6}$

$\frac{1}{4} = 0.25 = p$

The constant of proportionality is 0.25.

Question analysis

A smoothie shop's recipe table lists cups of fruit and cups of yogurt for four batches in a proportional relationship. Students type the constant of proportionality of yogurt to fruit. It tests finding a unit rate from a table, and since fruit is listed first while the question names yogurt first, the ratio is easy to take upside down.

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: At grade 7 Finding the constant of proportionality in a table is a core grade 7 skill, and these values are friendly decimals. Typing the answer with no choices, and taking the ratio in the order the question gives, are the added demands.
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: 1 step
Reading level: Question and choices: about grade 6, 75 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 answer with one division, and the values divide evenly, so the points are lost on direction. Fruit fills the first column and comes first in the stem’s own ratio, which invites dividing it by the yogurt. The table also climbs by a steady half cup of yogurt, a number that looks like a constant but ignores the 2 cups of fruit added each time. Checking that 0.25 times each fruit value gives its yogurt value settles it.

Turning words into math moderate Understanding the question
The opening sentence speaks of a ratio of fruit to yogurt, but the question asks for the constant of yogurt to fruit, so the division is y ÷ x, against the table’s column order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed into a box with no choices, so 4, the ratio taken the other way round, is a believable entry that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Check which quantity the question names first, divide that column by the other in any row, and confirm the quotient is the same in a second row.

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

4 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides cups of fruit by cups of yogurt, 6 ÷ 1.5 = 4, following the stem’s “ratio of fruit to yogurt” and turning the requested constant upside down.
0.5 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the steady rise of 0.5 cup of yogurt from row to row as the constant, ignoring that the fruit rises by 2 cups each time.
1.5 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives 1.5, the first yogurt value in the table, as if the constant were an amount in one batch rather than a ratio of two amounts.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; k is the constant of proportionality. y and x are two matching amounts from the same row of the table. In a proportional relationship, y = kx, so y is always k times x. That is why dividing y by x gives the same k in every row.

Sort out the question
The table lists cups of fruit (x) and cups of yogurt (y) for four batches. You need the constant of proportionality of yogurt to fruit.
That is the number of cups of yogurt for each 1 cup of fruit. It is also the number you multiply x by to get y.
Divide yogurt by fruit
Use the first row: 6 cups of fruit and 1.5 cups of yogurt. Yogurt goes on top, because the question says yogurt to fruit:
$k = \frac{y}{x} = \frac{1.5}{6} = 0.25$
So there is 0.25 cup, or one quarter of a cup, of yogurt for each cup of fruit.
Watch out: the first sentence says fruit to yogurt, but the question asks the other way. Dividing $6 \div 1.5 = 4$ turns the answer upside down.
Check two more rows
The constant must be the same in every row. Test two more:
$\frac{2.0}{8} = 0.25 \qquad \frac{3.0}{12} = 0.25$
They match. Watch out: the yogurt goes up by 0.5 cup from row to row. That 0.5 is not the constant. The fruit goes up by 2 cups at the same time, and $0.5 \div 2 = 0.25$ .
Enter the number
Type only the number in the box, with no units. You can type 0.25 or .25. On this site, 1/4 counts too.
So you would enter 0.25.

Check your answer

Multiply each amount of fruit by 0.25. $6 \times 0.25 = 1.5$ , $8 \times 0.25 = 2$ , $10 \times 0.25 = 2.5$ and $12 \times 0.25 = 3$ . Every result matches the yogurt column. Also, each batch has less yogurt than fruit, so the constant should be less than 1.

More questions like this

The original DOE question

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

Question 30 of 50

2026 Practice Test C Math, question 30

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 31 of 50

2026 Practice Test C Math, question 31

Question

NUMBER OF VISITORS
TO BROOKSIDE ZOO

	Male	Female
Children	360	340
Adults	210	290

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

A30%
B28%
C24%
D17%

Show the answer

Correct answer: A

First, determine the total number of visitors.

$360 + 340 + 210 + 290 = 1200$

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 32 of 50

2026 Practice Test C Math, question 32

Question

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 15

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

Maximum value $= 84^{\circ}\text{ F}$ , 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 33 of 50

2026 Practice Test C Math, question 33

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 34 of 50

2026 Practice Test C Math, question 34

Question

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

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

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

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 35 of 50

2026 Practice Test C Math, question 35

Question

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

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

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

a =

Show the answer

Correct answer: a = 10b

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

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

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

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

$4a - 2a = 28b - 8b$

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

$2a = 20b$

$a = 10b$

Question analysis

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

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve equations like 4(x + 2) = 28 with numbers. This one has letters on both sides, needs the distributive property and like terms, and asks for a in terms of b, which is grade 8 algebra.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Many steps moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
Both letters appear on both sides, and the answer must stay a multiple of b, so there is no number to check an entry against unless students invent one. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps moderate Working it out
Four moves run in a chain, distributing the 4, gathering the a terms, gathering the b terms and dividing by 2, and an error at any one carries into the typed result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The answer is typed after a printed “a =”, with no choices, so 20b, the line before dividing by 2, can be entered without anything on screen questioning it. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

13b A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies only the a by 4, writing 4a + 2b = 28b + 2a, so 2a = 26b and a = 13b.
18b A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Distributes correctly but moves 8b to the right without changing its sign, so 2a = 28b + 8b = 36b and a = 18b.
20b Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Reaches 2a = 20b correctly and enters 20b, stopping before dividing both sides by 2 to get a alone.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The equation is $4(a + 2b) = 28b + 2a$ . You need a in terms of b. That means a alone on one side, with only b and numbers on the other.
The answer box already shows a =, so you type only what comes after it.
Multiply out the parentheses
Use the distributive property. The 4 multiplies both terms inside:
$4 \times a + 4 \times 2b = 28b + 2a$
$4a + 8b = 28b + 2a$
Watch out: multiplying only the a by 4 gives 4a + 2b. That leads to the wrong answer 13b.
Gather the a terms
Like terms are terms with the same letter, such as 4a and 2a. Put the a terms on the left. Subtract 2a from both sides:
$4a - 2a + 8b = 28b$
$2a + 8b = 28b$
Gather the b terms
Now subtract 8b from both sides:
$2a = 28b - 8b$
$2a = 20b$
Watch out: moving 8b without changing its sign gives $28b + 8b = 36b$ . That leads to 18b, which is wrong.
Divide both sides by 2
2a means 2 times a. Divide both sides by 2 to get a alone:
$a = \frac{20b}{2} = 10b$
Watch out: 20b is equal to 2a, not a. Do not stop there.
Enter it after the printed a =
On the test, the box already shows a =. Press 1, then 0, then the b key. You do not need a times sign. Do not type a = again. On this site, type 10b in the box.
So you would enter 10b, which makes a = 10b.

Check your answer

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

More questions like this

The original DOE question

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

Question 36 of 50

2026 Practice Test C Math, question 36

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 37 of 50

2026 Practice Test C Math, question 37

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 38 of 50

2026 Practice Test C Math, question 38

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 39 of 50

2026 Practice Test C Math, question 39

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 40 of 50

2026 Practice Test C Math, question 40

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 41 of 50

2026 Practice Test C Math, question 41

Question

Elijah is at the school bookstore. He wants to buy a water bottle for $12.50 and some notebooks that cost $4.75 each. He has at most $35.00 to spend. Write an inequality that represents the number of notebooks, n, Elijah can purchase.

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

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

Show the answer

Correct answer: 12.50 + 4.75n ≤ 35.00

The question uses the phrase “at most,” which means “less than or equal to” (≤).

Since notebooks cost $4.75 each and Elijah will buy n notebooks, the total cost just for the notebooks is found by multiplying $4.75 by n, or $4.75n$ .

Combining the cost of a water bottle ($12.50) with the cost of n notebooks, the inequality is $12.50 + 4.75n \leq 35.00$ .

This may also be written as $4.75n + 12.50 \leq 35.00$ .

Question analysis

Elijah has at most $35.00 to buy one water bottle and some notebooks at the school bookstore, with a price given for each. Students type an inequality for the number of notebooks, n, he can buy. It tests writing an inequality from a word problem. The two prices play different roles, and the phrase "at most" decides the inequality symbol.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an inequality for a budget with a one-time cost and a cost per item is typical grade 7 work. The answer is typed, so both the symbol for "at most" and the role of each price must be right.
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: 2 steps
Reading level: Question and choices: about grade 5, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The prices are given in the order they are used, so the expression for the total is the easier half. The points are lost on the symbol: “at most” has to be heard as no more than, with equality allowed, and the keypad offers all four inequality signs side by side. Pricing one and two notebooks with the bottle, $17.25 and $22.00, shows which amount repeats and confirms the total.

Turning words into math moderate Understanding the question
The $12.50 bottle is bought once and the $4.75 is paid for each of n notebooks, and “at most $35.00” has to become ≤, which allows a total of exactly $35.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The whole inequality, with n, its symbol and both prices, is built on a keypad offering ≤, <, ≥ and >, so a wrong symbol looks as complete as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Price a few cases by hand, such as 1 and 2 notebooks with the water bottle, to see which amount repeats, then decide whether a total of exactly $35.00 is allowed.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

12.50 + 4.75n ≥ 35.00 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Builds the total correctly but reads “at most” as at least, writing ≥, which would mean Elijah must spend $35.00 or more.
12.50 + 4.75n < 35.00 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Builds the total correctly but uses a strict less-than sign, leaving out a total of exactly $35.00, which “at most” allows.
17.25n ≤ 35.00 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the bottle’s price to each notebook’s, $12.50 + $4.75 = $17.25, and multiplies by n, as if a bottle came with every notebook.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a one-time cost: $\text{total} = \text{one-time cost} + (\text{price each}) \times n$; The one-time cost is paid once, no matter what. The price each is paid once for every item. n is the number of items. You multiply the price each by n because you pay it n times.
At most: $\text{total} \leq \text{limit}$; The sign ≤ means less than or equal to. The total can be under the limit or exactly equal to it, but never over it. Use it for words like at most and no more than.

Sort out the question
Elijah buys one water bottle for $12.50. He also buys n notebooks at $4.75 each. He can spend at most $35.00.
You need an inequality for n. An inequality compares two amounts with a sign such as ≤ instead of =.
Price a few cases
Work out the cost for a few numbers of notebooks:
Notebooks Total ($)
1 $12.50 + 4.75 = 17.25$
2 $12.50 + 9.50 = 22.00$
3 $12.50 + 14.25 = 26.75$
The $12.50 shows up once. The $4.75 shows up once for each notebook.
Write the total cost
Use the total cost formula. The bottle is the one-time cost, and each notebook costs $4.75:
$12.50 + 4.75n$
Here 4.75n means 4.75 times n.
Watch out: do not add the two prices first. $17.25n$ charges for a new bottle with every notebook.
Choose the sign
At most $35.00 means the total can be less than $35.00 or exactly $35.00. It cannot be more. So use ≤:
$12.50 + 4.75n \leq 35.00$
Watch out: ≥ means at least, the opposite. And < leaves out a total of exactly $35.00, which Elijah is allowed to spend.
Enter it on the keypad
Type the whole inequality. Nothing is printed before the box. On the test, press 1, 2, ., 5, 0, +, 4, ., 7, 5, the n key, the ≤ key, then 3, 5, ., 0, 0. The ≤ key is in the fourth row. Skip the dollar signs. On this site, type <= for ≤.
So you would enter 12.50 + 4.75n ≤ 35.00. An equal form, such as 4.75n + 12.50 ≤ 35.00, counts too.

Notebooks	Total ($)
1	$12.50 + 4.75 = 17.25$
2	$12.50 + 9.50 = 22.00$
3	$12.50 + 14.25 = 26.75$

Check your answer

Test some numbers. For 4 notebooks, $12.50 + 4.75 \times 4 = 31.50$ , which is not over 35.00. For 5 notebooks, $12.50 + 4.75 \times 5 = 36.25$ , which is too much. So Elijah can buy up to 4 notebooks. That makes sense for a budget of $35.00.

More questions like this

The original DOE question

Question 41 as shown on the DOE’s digital test, page 114 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 114, question 41.

Question 42 of 50

2026 Practice Test C Math, question 42

Question

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

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 7,500

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

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

$V_c = 20\text{ cm} \times 20\text{ cm} \times 20\text{ cm} = 8{,}000\text{ cm}^3$

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

$V_p = 5\text{ cm} \times 5\text{ cm} \times 20\text{ cm} = 500\text{ cm}^3$

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

$8{,}000\text{ cm}^3 - 500\text{ cm}^3 = 7{,}500\text{ cm}^3$

Question analysis

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

NYS standards: 7.G.6 area, surface area, and volume
6.G.2 volume with fractional edge lengths
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the volume of a solid made from rectangular prisms, here by taking one away, is named in the grade 7 standards. This is a direct case with whole numbers, two volumes and one subtraction.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Reading a figure, table or graph minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Something you must already know minor Working it out
The volume of a rectangular prism, length times width times height, is not given, and it is needed twice, once for the 20 cm cube and once for the hole. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The figure labels only the 5 cm opening and the cube’s 20 cm edge; the hole’s 20 cm height comes from the words, while the drawing shows it reaching from top to bottom. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
An answer in an unexpected form minor Choosing the answer
The volume is typed with no choices to compare, so the hole’s 500 cubic centimeters or the cube’s 8,000 can be entered without any warning. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

500 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the volume of the hole, 5 × 5 × 20 = 500 cubic centimeters, and gives the clay removed instead of the clay that is left.
7,875 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Treats the hole as a 5 cm cube, 5 × 5 × 5 = 125, though it runs the full 20 cm height, so 8,000 − 125 = 7,875.
7,975 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Subtracts only the area of the square opening, 5 × 5 = 25, from the cube’s 8,000, mixing a face’s area with the hole’s volume.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Tariq’s block starts as a cube 20 cm on each side. He removes a hole shaped like a rectangular prism, a box shape with flat rectangle sides. Its base is 5 cm by 5 cm, and its height is 20 cm. The figure shows it running from the top face to the bottom face.
You need the volume of the clay that is left, in cubic centimeters.
Find the volume of the cube
Use the volume formula. The length, width and height are all 20 cm:
$V = 20 \times 20 \times 20 = 8{,}000$
The full cube holds 8,000 cubic cm of clay.
Find the volume of the hole
The hole is a prism too. Its length and width are 5 cm, and its height is 20 cm:
$V = 5 \times 5 \times 20 = 500$
So 500 cubic cm of clay are removed.
Watch out: the hole is 20 cm tall, not 5 cm. Treating it as a small 5 cm cube gives 125 and the wrong answer 7,875. And 5 × 5 = 25 is only the area of the opening, not a volume. Subtracting it gives 7,975, also wrong.
Subtract the hole from the cube
The clay left is the cube minus the hole:
$8{,}000 - 500 = 7{,}500$
The block has 7,500 cubic cm of clay left.
Watch out: 500 is the clay taken out, not the clay that is left.
Enter the number
Type only the number, with no units. You can type 7500 or 7,500.
So you would enter 7,500.

Check your answer

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

More questions like this

The original DOE question

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

Question 43 of 50

2026 Practice Test C Math, question 43

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 44 of 50

2026 Practice Test C Math, question 44

Question

A smartphone screen is designed to be 72 millimeters wide. The quality control team allows for a variation of $\pm 4\%$ in the width. Select the minimum and maximum screen widths that fall within the acceptable range of screen widths, in millimeters.

Select the correct answer from each drop-down to complete the sentence.

The acceptable screen width is from to millimeters.

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

Show the answer

Correct answer: blank 1: 69.12; blank 2: 74.88

A variation of $\pm 4\%$ in the width means that the width of the smartphone can be from 4% less than the standard of 72 mm to 4% greater than 72 mm.

The least allowed measure is

$72 - 0.04(72) = 72 - 2.88 = 69.12\text{ mm}$

The greatest allowed measure is

$72 + 0.04(72) = 72 + 2.88 = 74.88\text{ mm}$

Therefore, the allowable range is from 69.12 mm to 74.88 mm in width.

Question analysis

A smartphone screen is designed to be 72 millimeters wide, and quality control allows a variation of plus or minus 4%. Students choose the minimum and maximum acceptable widths from two drop-down menus. It tests finding a percent of a quantity, and the 4% has to be turned into millimeters before it is added or subtracted.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a percent of a number and adding or subtracting it is grade 7 percent work. The ± sign, the decimal result and two linked menus, which must both be right, add a little to the usual demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
A very close wrong answer minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 59 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards turning the percent into millimeters before doing anything else. Students who treat ±4% as ±4 millimeters find the two whole-number options at either end of the menus, 68.00 and 76.00, which look like a natural pair. Those who compute 2.88 correctly can still lose the item on the maximum by rounding to 3 and picking 75.00. Multiplying 72 by 0.96 and by 1.04 is a quick double-check.

Error-prone arithmetic moderate Working it out
Four percent of 72 is 0.04 × 72 = 2.88, a decimal product, and taking it from 72 means borrowing across the decimal point to reach 69.12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “a variation of ±4% in the width” has to become 4% of 72 millimeters, a length of 2.88 mm, before anything is added or subtracted. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The second menu lists 75.00 just 0.12 above 74.88, so a rounded estimate of 72 + 3 lands on the wrong option, and estimating cannot separate the two. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
Both menus must be right to score, so the minimum and the maximum each need their own subtraction or addition after the shared percent is found. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find 4% of the 72-millimeter design width first, then subtract that amount for the minimum and add it for the maximum, before looking at the menus.

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.
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

AD Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reads ±4% as ±4 millimeters, so the widths run from 72 − 4 = 68.00 to 72 + 4 = 76.00.
BC Rounds too early (King et al., 2004)
Reveals a slip Expected pull: moderate
Finds the 69.12 minimum, then estimates the maximum by rounding the 2.88 change up to 3, so 72 + 3 = 75.00 is picked over 74.88.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The percent is how many out of 100. Dividing it by 100 turns it into a decimal, so 4% becomes 0.04. The whole is the amount you take the percent of. The part is the amount that the percent stands for.

Sort out the question
The screen is designed to be 72 millimeters (mm) wide. The sign ±4% means plus or minus 4%. So a screen may be up to 4% narrower or up to 4% wider than 72 mm.
You need the minimum, the smallest allowed width, and the maximum, the largest. You pick each one from a drop-down menu.
Find 4% of 72 mm
Use the percent formula. 4% is $\frac{4}{100} = 0.04$ :
$0.04 \times 72 = 2.88$
So the width may change by up to 2.88 mm either way.
Watch out: 4% is not 4 mm. Taking 4 away and adding 4 gives 68.00 and 76.00. Both are in the menus, and both are wrong.
Find the minimum width
Take the change away from the design width:
$72 - 2.88 = 69.12$
The narrowest allowed screen is 69.12 mm wide.
Find the maximum width
Add the change to the design width:
$72 + 2.88 = 74.88$
The widest allowed screen is 74.88 mm wide.
Watch out: rounding 2.88 up to 3 gives 75.00. That is in the second menu too, but it is too wide.
Choose from each menu
Open the first drop-down menu and choose 69.12. Open the second one and choose 74.88.
So the sentence reads: the acceptable screen width is from 69.12 to 74.88 millimeters.

Check your answer

Use multipliers. A screen 4% narrower keeps 96% of the width: $0.96 \times 72 = 69.12$ . A screen 4% wider is 104% of it: $1.04 \times 72 = 74.88$ . Both match. Also, both ends are the same 2.88 mm away from 72.

More questions like this

The original DOE question

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

Question 45 of 50

2026 Practice Test C Math, question 45

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 46 of 50

2026 Practice Test C Math, question 46

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 47 of 50

2026 Practice Test C Math, question 47

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 48 of 50

2026 Practice Test C Math, question 48

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 49 of 50

2026 Practice Test C Math, question 49

Question

The box plot illustrates the number of hours eighth-grade students spend on screens during a typical week. What is the difference between the range and the interquartile range (IQR) of the screen time data?

$A box plot titled NUMBER OF SCREEN HOURS FOR 8th GRADE STUDENTS above a number line labeled Hours, from 5 to 20 with a labeled tick at every 1. The left whisker starts at 5, the box runs from 8 to 15 with the median line at 12, and the right whisker ends at 20.$

Move the correct answer to each box. Not all answers will be used.

The range of screen time hours is hours.

The IQR is hours.

The difference between the range and the IQR is hours.

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. Each tile can be used once. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 15; blank 2: 7; blank 3: 8

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

Maximum value $= 20\text{ hr}$ , Minimum value $= 5\text{ hr}$

$20\text{ hr} - 5\text{ hr} = 15\text{ hr}$

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 $= 15\text{ hr}$ , Lower Quartile Value $= 8\text{ hr}$

$15\text{ hr} - 8\text{ hr} = 7\text{ hr}$

Determine the difference between the range (15 hr) and the interquartile range (7 hr):

$15\text{ hr} - 7\text{ hr} = 8\text{ hr}$

Question analysis

A box plot shows how many hours a week eighth-grade students spend on screens, on a number line marked every hour. Students move number tiles to complete three sentences: the range, the interquartile range, and the difference between them. Each tile can be used only once, and keeping the range and the interquartile range apart is the tricky part.

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Reading a box plot and finding its range and interquartile range are grade 7 statistics skills. The values sit on labeled whole-number ticks and the three sentences lead through the steps, a typical grade 7 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 minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 69 words.
Corrected from the DOE handbook: A missing comma in the explanation 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

The plot is drawn on whole-hour ticks, and the three sentences lay out the steps, so this is a direct check of two definitions. The main risk is taking a value off the plot, such as the lower quartile of 8, for a spread between two values. Because the four tiles allow only one use each, a student who confuses the measures can still fill every box, which hides the error.

Something you must already know minor Working it out
The sentences name the range and the IQR but never define them; students must know the range spans the whiskers and the IQR spans only the box. (Anderson & Krathwohl, 2001)
Many steps minor Working it out
Three boxes must all be right, the third depends on the first two, and each tile can be used once, so one mix-up spreads to a second box. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Read the five key values off the plot, subtract the least from the greatest for the range, subtract the lower quartile from the upper for the IQR, then subtract those results.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

DCB Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the lower quartile, 8, as the interquartile range, reading a value off the box instead of its width, so the difference becomes 15 − 8 = 7.
BDC Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Swaps the two measures, giving the box’s width, 7, as the range and the whiskers’ spread, 15, as the IQR, while the difference stays 8.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation runs two values together: “Maximum value = 20 hr Minimum value = 5 hr”. Its next line is “Upper Quartile Value = 15 hr, Lower Quartile Value = 8 hr”, and Question 32’s explanation has “Maximum value = 84° F, Minimum value = 39° F”. The site adds the comma.

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
The box plot shows the hours of screen time for eighth graders in a week. You need three numbers: the range, the IQR, and the difference between them.
There are four tiles: 5, 7, 8 and 15. Each tile is used at most once, and one tile is left over.
Read the five values
The number line has a tick at every hour, and every value sits on a tick:
- The left whisker starts at 5. This is the minimum.
- The box starts at 8. This is the lower quartile, the middle of the lower half of the data.
- The line inside the box is at 12. This is the median, the middle value.
- The box ends at 15. This is the upper quartile, the middle of the upper half.
- The right whisker ends at 20. This is the maximum.
Find the range
Use the range formula with the two ends of the whiskers:
$\text{range} = 20 - 5 = 15$
The screen times spread over 15 hours, from the least to the most.
Find the IQR
Use the IQR formula with the two edges of the box:
$\text{IQR} = 15 - 8 = 7$
The middle half of the students spend between 8 and 15 hours, a spread of 7 hours.
Watch out: the IQR is a width, not a point on the plot. The tile 8 is the lower quartile. Putting it in the IQR box also makes the last box 7, which is wrong.
Find the difference
Subtract the IQR from the range:
$15 - 7 = 8$
Watch out: do not swap the two measures. The range uses the whiskers’ ends, and the IQR uses only the box.
Fill in the three boxes
On the test, you drag a tile into each box. On this site, you choose each tile from its box’s menu. The range box gets 15, the IQR box gets 7, and the difference box gets 8. The tile 5 is not used.
So the boxes get 15, 7 and 8.

Check your answer

The difference is just the two whiskers added together. The left whisker is $8 - 5 = 3$ hours long. The right whisker is $20 - 15 = 5$ hours long. $3 + 5 = 8$ , which matches the third box.

More questions like this

The original DOE question

Question 49 as shown on the DOE’s digital test, page 122 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 122, question 49. This is the DOE’s version, before the corrections listed in the analysis.

Question 50 of 50

2026 Practice Test C Math, question 50

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.