Practice Test

2026 Practice Tests (digital)

2026 Practice Test D Math

Question 1 of 50

2026 Practice Test D Math, question 1

Question

What is the prime factorization of 3,575?

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

Show the answer

Correct answer: B

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

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

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

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

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

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

Question analysis

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

NYS standards: 6.NS.4 common factors and multiples
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Breaking a number into primes is grade 6 number work that relies on divisibility tests. The number here is larger than in most class examples, but the work is still a few divisions, lighter than typical grade 7 problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 2 of 50

2026 Practice Test D Math, question 2

Question

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

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

Show the answer

Correct answer: D

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

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

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

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

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

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

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

$x + 2y - 2x = z$

$2y - x = z$

Question analysis

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

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add and expand linear expressions with fraction coefficients, usually in one letter. Here three letters appear, one numerator is a difference, and the result is an expression rather than a number, which adds abstraction but no new skill.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 3 of 50

2026 Practice Test D Math, question 3

Question

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

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

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

What is the value of $x - y$ ?

A−7
B−5
C5
D7

Show the answer

Correct answer: B

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

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

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

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

$x + 16 = y + 11$

$x = y - 5$

$x - y = -5$

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

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

Question analysis

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

NYS standards: 7.EE.4 equations and inequalities from problems
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students write and solve equations with one unknown. This one has two unknowns that are never found separately; students must rearrange to get their difference, a less familiar goal, though the arithmetic is light.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The sums take seconds; the point is lost on the sign. Because set R's known numbers add to more than set S's, x has to be the smaller unknown, so x − y is negative, but subtracting the two known sums in the order they appear gives a positive 5. Choosing one pair of values that balances the sets, such as x = 1 and y = 6, settles the sign at once.

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

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 4 of 50

2026 Practice Test D Math, question 4

Question

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

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

Show the answer

Correct answer: C

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

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

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

Question analysis

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

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Substituting a value and dividing by a fraction are grade 6 skills, and here they take three short steps with friendly numbers. The stacked fraction asks for care, but the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 5 of 50

2026 Practice Test D Math, question 5

Question

In a school raffle, 1 out of every 25 students wins a prize. Each prize includes a randomly selected trading card from a set of 100 different cards, labeled C1 through C100, with each card equally likely to be chosen. If one student is selected at random, what is the probability that the student both wins a prize and receives Card C17?

Write an expression that represents the probability that a student wins the prize and receives Card C17.

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: 1/25 · 1/100

To calculate the joint probability, first determine the probability of each independent event.

The probability that a randomly selected student wins the prize is $\frac{1}{25}$ .

The probability that a randomly selected student will receive Card C17 is $\frac{1}{100}$ , because Card C17 is 1 card out of 100 labeled cards.

The probability that a student wins the prize and receives Card C17 is the product of those two probabilities, $\frac{1}{25} \cdot \frac{1}{100}$ .

Question analysis

In a school raffle, 1 out of every 25 students wins a prize, and each prize includes one of 100 equally likely trading cards. Students type an expression for the probability that a randomly chosen student both wins and gets one particular card. It tests the probability of two independent events, and combining the two chances correctly is the main hurdle.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.8a uniform probability models
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students find compound probabilities, usually by listing outcomes for coins or number cubes. These two stages have 25 and 100 outcomes, far too many pairs to list, so a rule must be used, and the answer is typed as an expression.
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
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 87 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 chance is stated almost directly, 1 out of 25 and 1 out of 100, so the question tests one rule: both events must happen, and the chances multiply. The word “and” pulls students toward adding, which gives a result larger than either chance, a sign that something is wrong. Others write only the card's chance and forget that the student must first win a prize.

Something you must already know moderate Working it out
The question never says that the chance of two independent events both happening is the product of their chances, so students must bring that rule to the 1/25 and the 1/100 themselves. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The answer is typed as an expression with no choices to compare, so a sum of the two chances, which looks just as finished, gets no warning on screen. (Enright, Morley & Sheehan, 2002)

Strategy: Write each event's chance as a fraction, then apply the rule for two independent events that must both happen, and check that the result is smaller than either chance alone.

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

1/25 + 1/100 Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Reads “both wins a prize and receives Card C17” as a cue to add, so sums the chances, though needing both events must be less likely than either.
1/100 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Writes only the chance of receiving Card C17 and leaves out that the student must first be one of the 1 in 25 who win a prize.
1/125 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Combines “1 out of 25” and “1 out of 100” by adding the totals, 25 + 100, a faulty rule that gives neither the sum nor the product.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance that something happens, from 0 to 1. The top counts the outcomes you want. The bottom counts all the outcomes that could happen. Use it when every outcome is equally likely, like one card picked from 100.
Probability that two independent events both happen: $P(A \text{ and } B) = P(A) \times P(B)$; A and B are two events. Independent means one does not change the chance of the other. P(A) is the chance of A, and P(B) is the chance of B. Multiply them to get the chance that both happen. The result is smaller than either chance, since both things must happen.

Sort out the question
Two things must both happen. First, the student wins a prize. Second, the card in that prize is Card C17. You need an expression for that chance. An expression is a math phrase made of numbers and signs, with no equal sign. You do not have to work it out to one number.
Find the chance of winning
The question says “1 out of every 25 students wins a prize.” One student is the part that wins, and 25 is the whole group. So:
$P(\text{win}) = \frac{1}{25}$
Find the chance of Card C17
Each prize has one card, picked at random from 100 cards. Every card is equally likely. Card C17 is 1 of those 100 cards:
$P(\text{C17}) = \frac{1}{100}$
Multiply the two chances
Winning a prize does not change which card is inside it. So the two events are independent. To find the chance that both happen, multiply:
$P(\text{win and C17}) = \frac{1}{25} \cdot \frac{1}{100}$
The dot · means times. Worked out, this is 1/2500.
Watch out: the word and does not mean add here. 1/25 + 1/100 is bigger than either chance, but needing both things must be less likely. Writing only 1/100 forgets that the student must win a prize first.
Enter the expression
On the test, press the fraction key, a small box over a small box. Type 1 on top and 25 on the bottom. Move the cursor out of the fraction with the → key. Press the dot key for times. Then use the fraction key again for 1 over 100. On this site, type 1/25 * 1/100 or 1/25 · 1/100. The value 1/2500 also counts.
So you would enter 1/25 · 1/100.

Check your answer

Picture 2,500 students. 1 in 25 wins, so about 100 students win a prize. Of those 100, about 1 gets Card C17. That is 1 student out of 2,500, which matches 1/25 · 1/100 = 1/2500. It is smaller than 1/25 and 1/100, as it should be.

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 D Math, page 78, question 5.

Question 6 of 50

2026 Practice Test D Math, question 6

Question

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

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

Show the answer

Correct answer: B

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

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

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

4 pounds is

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

5 pounds is

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

p pounds is

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 7 of 50

2026 Practice Test D Math, question 7

Question

A middle school library has a shelf with 40 books, each labeled with a different number from 201 to 240. One book is selected at random. What is the probability, as a percentage, of selecting book number 223?

Enter your answer in the space provided.

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

Show the answer

Correct answer: 2.5%

The probability that book number 223 will be selected at random from 40 books is 1 out of 40. To find the probability as a percentage, convert the fraction to a decimal.

The chance of selecting one specific book is $\frac{1}{40} = 0.025$ .

Convert the decimal to a percentage: $0.025 \times 100 = 2.5\%$

Question analysis

A library shelf holds 40 books numbered 201 to 240, and one is picked at random. Students find the probability of picking one particular book and type it as a percentage. It tests simple probability and converting a fraction to a percent; the book numbers play no part, and the decimal point needs care.

NYS standards: 6.SP.8a uniform probability models
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of one equally likely outcome is grade 6 content. The only further step is rewriting the fraction as a percent, which needs care with place value, so it is lighter than grade 7 compound probability.
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: An answer in an unexpected form moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The book numbers, 201 to 240, and the chosen number 223 play no part, since every book is equally likely and the shelf holds 40. The chance is 1 in 40, and the real demand is the form: changing it to a percent means a decimal with a zero in the tenths place, then a shift of two places. Students who know that 1/4 is 25% can check that 1/40 must be a tenth of that.

An answer in an unexpected form moderate Choosing the answer
Only the bold word percentage sets the form, since the empty box shows no % sign, so the probability 1/40 must be turned into 2.5, not entered as the fraction or as 0.025. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Dividing 1 by 40 gives 0.025, with a zero right after the decimal point, and losing that zero turns the percent into 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the books, write the chance of picking one particular book as a fraction, then turn the fraction into a decimal and multiply by 100 to get the percent.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

0.025 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the probability as a decimal, 1 ÷ 40 = 0.025, and enters it without multiplying by 100, though the box asks for a percentage.
25 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Loses the zero after the decimal point when dividing, taking 1 ÷ 40 as 0.25, as 1 ÷ 4 would be, and so enters 25.
1/40 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Enters the probability as the fraction 1/40, the form it was found in, overlooking the bold word percentage in the question.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance of an event. The top counts the outcomes you want. The bottom counts all the equally likely outcomes. Use it when each outcome has the same chance, like picking one book at random.
Changing a fraction to a percent: $\text{percent} = \frac{a}{b} \times 100$; a/b is the fraction. Percent means out of 100. Multiplying by 100 tells you how many out of 100 the fraction is. For example, 1/2 × 100 = 50, so 1/2 is 50%.

Sort out the question
The shelf has 40 books, numbered 201 to 240. One book is picked at random, so every book has the same chance. You need the chance of picking book 223, written as a percentage. Book 223 is on the shelf, since 223 is between 201 and 240.
Write the probability as a fraction
Only 1 book is number 223, and there are 40 books in all:
$P = \frac{1}{40}$
The numbers on the labels do not matter beyond this. They just give each book its own name.
Change the fraction to a decimal
Divide 1 by 40. It helps to start from 1/4 = 0.25. Since 40 is ten times 4, 1/40 is one tenth of 1/4:
$1 \div 40 = 0.025$
Watch out: there is a 0 right after the decimal point. Losing it gives 0.25 and then 25, which is wrong.
Change the decimal to a percent
Multiply by 100 to get a percent. That moves the decimal point two places to the right:
$0.025 \times 100 = 2.5$
So the chance is 2.5%.
Watch out: 0.025 is the probability as a decimal, and 1/40 is the same chance as a fraction. The question asks for a percentage, so neither one is the answer here.
Enter the number
The box shows no % sign. Type 2.5. Adding the sign, as 2.5%, also counts.
So you would enter 2.5.

Check your answer

2.5% means 2.5 out of every 100. Out of 40 books, that is 40 × 0.025 = 1 book, the one numbered 223. Also, one book out of 40 should be a small chance, and 2.5% is small.

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 D Math, page 80, question 7.

Question 8 of 50

2026 Practice Test D Math, question 8

Question

A square metal plate has a side length of 3 inches. A circular hole with a diameter of 2 inches is cut out from the center of the plate.

$A square, shaded gray, with a white circle cut out of its center. A side of the square is labeled 3 in., and the diameter of the circle is labeled 2 in. The shaded region is the part of the square outside the circle.$

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

Enter your answer in the space.

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

Show the answer

Correct answer: 5.86

To determine the area of a square, use $A = s^{2}$ .

Begin by finding the length of the side of the square, which is 3 inches. The area of the square is $(3\text{ in})^{2} = 9\text{ square inches}$ .

To determine the area of the circle, use $A = \pi r^{2}$ .

The radius of the circle (the hole) is one-half the diameter or $2\text{ in.} \div 2 = 1\text{ in.}$ , so the area of the entire circle is $\pi(1\text{ in.})^{2} \approx 3.14\text{ sq in.}$

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

$9\text{ sq in.} - \pi\text{ sq in.} =$

$9\text{ sq in.} - (3.14)\text{ sq in.} =$

$5.86\text{ sq in.}$

The shaded area to the nearest hundredth of a square inch $= 5.86$ sq in.

Question analysis

A 3-inch square metal plate has a circular hole 2 inches across cut from its center, and students find the area of the metal that is left, to the nearest hundredth of a square inch. It tests the areas of a circle and a square. Using the radius, not the diameter, and subtracting the right areas are the keys.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a circle's area is a grade 7 standard, and subtracting a hole from a shape is common grade 7 practice. The radius must come from the diameter and the result must be rounded, but the steps are routine.
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
Combines ideas minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 49 words.
Corrected from the DOE handbook: A wrong symbol in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every number is small, so the work is choosing the right pieces: the area of the square, the area of the hole, and the difference. A radius of 1 makes the circle's area simply π, which is easy to confuse with the circumference, 2π, when the formulas are half remembered. The region asked for is the gray metal, not the hole, and the final rounding goes up, since the third decimal place is 8.

Something you must already know moderate Working it out
Neither area formula is given, and the hole is labeled by its 2-inch diameter, so students must recall A = πr² and halve the diameter to get the radius 1 first. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
The shaded metal is a square with a circle removed, so two area formulas are combined by subtraction rather than either one used alone. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
The result is typed to the nearest hundredth with no choices to compare, so the long decimal 5.858… must be rounded up, not cut off after two places. (Enright, Morley & Sheehan, 2002)

Strategy: Halve the diameter to get the radius, find the area of the square and of the circle, subtract the circle from the square, and round only at the end.

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.
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.

The wrong answers, classified

2.72 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses the circumference, 2π × 1 ≈ 6.28, in place of the circle's area and subtracts it from the square's 9 square inches, getting about 2.72.
3.14 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the area of the circular hole, π × 1² ≈ 3.14 square inches, and enters it instead of the area of the shaded metal left around it.
5.85 Rounds too early (King et al., 2004)
Reveals a slip Expected pull: weak
Works the area out correctly as about 5.858 but cuts off the extra digits instead of rounding up to the nearest hundredth.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation gives the area of the hole as “π(1 in.)² ≅ 3.14 sq in.” The symbol ≅ means “is congruent to”, which an area and a number cannot be; the line means “is approximately equal to”, written ≈. The site now says “π(1 in.)² ≈ 3.14 sq in.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s^{2}$; A is the area, the amount of flat space inside. s is the length of one side. s² means s × s. It works because a square with side s holds s rows of s little unit squares.
Area of a circle: $A = \pi r^{2}$; A is the area. r is the radius, the distance from the center to the edge. It is half the diameter, the distance straight across through the center. π (pi) is a number a little more than 3, about 3.14. r² means r × r.

Sort out the question
The plate is a square with 3-inch sides. A circle with a 2-inch diameter is cut out of the middle. The diameter is the distance straight across the circle through its center. The shaded region is the metal that is left. You need its area to the nearest hundredth of a square inch.
Find the area of the square
Use the formula for the area of a square, with s = 3 inches:
$A = s^{2} = 3^{2} = 3 \times 3 = 9$
The whole plate, before the hole is cut, is 9 square inches.
Find the radius of the hole
The circle formula needs the radius, not the diameter. The figure labels the 2 in. distance across the hole. The radius is half of it:
$r = 2 \div 2 = 1$
The radius is 1 inch.
Find the area of the hole
Use the formula for the area of a circle, with r = 1 inch:
$A = \pi r^{2} = \pi \times 1^{2} = \pi \approx 3.14$
The sign ≈ means is about equal to. The hole is about 3.14 square inches.
Watch out: 2πr is the circumference, the distance around the circle, not its area. Using it gives 9 − 6.28 = 2.72. And 3.14 is the hole, not the metal, so do not stop there.
Subtract the hole from the square
The shaded metal is the square with the circle taken away:
$9 - \pi \approx 9 - 3.14 = 5.86$
The shaded region is about 5.86 square inches.
Round and enter the answer
With more digits of π, 3.1416, you get 9 − 3.1416 = 5.8584. The digit after the hundredths place is 8, which is 5 or more, so round up to 5.86. Chopping it to 5.85 is wrong. Type only the number, with no units.
So you would enter 5.86.

Check your answer

The hole is a bit more than a third of the plate, since 3.14 is a bit more than 9 ÷ 3 = 3. So the metal left should be a bit less than two thirds of 9, which is 6. The answer 5.86 is just under 6, so it makes sense.

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 D Math, page 81, question 8. This is the DOE’s version, before the corrections listed in the analysis.

Question 9 of 50

2026 Practice Test D Math, question 9

Question

Let m be a negative integer and k be a positive integer.

Which of the expressions below must result in a positive value?

Select the three 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 $-m + k$
B $-m \cdot k$
C $m - k$
D $-k \cdot m$
E $\frac{m}{k}$

Show the answer

Correct answer: A, B, and D

Each of the expressions must be considered:

$-m + k$ , where m is a negative integer and k is a positive integer: Correct

The “−” in front of m acts like a $-1$ multiplier that flips the sign of m, so $-m + k$ is positive when m is negative and k is positive.

$-m \cdot k$ , where m is a negative integer and k is a positive integer: Correct

The “−” in front of m acts like a $-1$ multiplier that flips the sign of m, so $-m \cdot k$ is positive when m is negative and k is positive.

$m - k$ , where m is a negative integer and k is a positive integer:

A negative number minus a positive number gives a negative number.

$-k \cdot m$ , where m is a negative integer and k is a positive integer: Correct

A positive times a negative is a negative. Then, taking the opposite of that makes it positive.

$\frac{m}{k}$ , where m is a negative integer and k is a positive integer:

A negative divided by a positive remains negative.

Therefore, $-m + k$ , $-m \cdot k$ , and $-k \cdot m$ must all result in a positive number.

Question analysis

Given that m is a negative integer and k a positive integer, students select the three expressions, out of five, that must have a positive value. It tests the sign rules for adding, subtracting, multiplying and dividing integers. Minus signs in front of letters, and letters in place of numbers, make each sign easy to misjudge, and all three must be found.

NYS standards: 7.NS.2a rational number multiplication
7.NS.1c rational number subtraction
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Sign rules for adding, subtracting, multiplying and dividing integers are core grade 7 content. Here they are applied to letters with known signs instead of numbers, across five expressions, and all three right ones must be chosen with no wrong one.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each expression needs only one sign rule, but the minus signs are placed to mislead. In −k · m the leading minus seems to make the result negative until the sign of m is counted, and in m − k two negatives seem to cancel, though subtracting makes the result more negative. Substituting one negative and one positive value into all five gives three positive results and two negative ones, the count the item asks for.

Variables instead of numbers moderate Understanding the question
Only the signs of m and k are known, so the sign of each expression must be reasoned from the rules or tested with values such as m = −2 and k = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
All five expressions mix the same two letters with minus signs, so none can be dismissed at a glance; each is judged on its own and exactly three must be picked. (Embretson & Wetzel, 1987)
Much to keep track of minor Working it out
The item scores only when all three positive expressions are chosen and both negative ones left out, so five separate verdicts must be kept and combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Substitute sample values, such as m = −2 and k = 3, into all five expressions and note each sign, then confirm with the sign rules that no other values could change it.

Traps

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

The wrong answers, classified

C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Sees two negatives, m and the minus before k, and applies “two negatives make a positive” to subtraction, though a negative minus a positive moves further below zero.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Left out by students who read the leading minus in −k · m as making the product negative, missing that m is negative too, so the product is positive.
E Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Sees no minus sign in m/k and takes it as positive, forgetting that m itself is negative, so the quotient of a negative and a positive is negative.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The opposite of a number: $-(-a) = a$; a is any positive number, so −a is negative. A minus sign in front of a number or letter means its opposite: the same distance from 0, on the other side. The opposite of a negative number is positive. So if m is negative, −m is positive.
Signs when you multiply or divide: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. Two negatives multiply to a positive. A negative and a positive multiply or divide to a negative. These rules tell you the sign of an answer without knowing the numbers.

Sort out the question
m is a negative integer, a whole number below 0, such as −2. k is a positive integer, such as 3. You must pick the three expressions that are always positive. Check all five, and test m = −2 and k = 3 as you go.
Understand the minus in front
A minus sign in front of a letter means the opposite of that number. Since m is negative, −m is positive. Since k is positive, −k is negative. With the test values, −m = 2 and −k = −3.
Check choices A and B
A: $-m + k$ is a positive plus a positive, so it is always positive. Test: 2 + 3 = 5.
B: $-m \cdot k$ is a positive times a positive, so it is always positive. The dot · means times. Test: 2 × 3 = 6.
Check choices C and E
C: $m - k$ starts below 0 and moves further down, so it is always negative. Test: −2 − 3 = −5.
E: $\frac{m}{k}$ is a negative divided by a positive, so it is always negative. Test: −2 ÷ 3 = −2/3.
Watch out: two negatives make a positive only when you multiply or divide them. Choice C subtracts, so it stays negative. Choice E shows no minus sign, but m itself is negative.
Check choice D
D: $-k \cdot m$ is a negative times a negative, so it is always positive. Test: −3 × (−2) = 6.
Watch out: the minus sign in front makes D look negative. But m is negative too, and the two negatives make a positive. Skipping D leaves you one answer short.
Choose the three answers
Choices C and E are always negative, so leave them out. The three positive ones are A, B and D. On the test, you check the box next to each one.
So the answers are A, B and D.

Check your answer

Try different values, such as m = −10 and k = 1. A: 10 + 1 = 11. B: 10 × 1 = 10. D: −1 × (−10) = 10. All three are positive again. C gives −11 and E gives −10, both negative.

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 D Math, page 82, question 9.

Question 10 of 50

2026 Practice Test D Math, question 10

Question

$2x + 11 > 3x + 9$

For what values of x is the above inequality true?

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

Show the answer

Correct answer: A

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

$2x + 11 > 3x + 9$

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

$-x + 11 > 9$

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

$-x > -2$

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

$-x > -2$

$x < 2$

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

$2x + 11 > 3x + 9$

$11 > x + 9$

$2 > x$

$x < 2$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 10 as shown on the DOE’s digital test, page 83 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 83, question 10. This is the DOE’s version, before the corrections listed in the analysis.

Question 11 of 50

2026 Practice Test D Math, question 11

Question

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

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

Show the answer

Correct answer: A

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

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

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

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

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

Question analysis

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

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are a grade 7 topic, and this is a standard scale problem. The one twist is that the scale is not stated: it must come from comparing the two perimeters before any side can be converted.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 12 of 50

2026 Practice Test D Math, question 12

Question

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

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

Show the answer

Correct answer: D

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

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

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

$x + 5 = 24$

$x = 19$

Question analysis

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

NYS standards: 7.G.4 circle area and circumference
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the fact that every radius has the same length, but spotting which segments are radii in a figure with extra markings is less routine than using a formula.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph moderate Understanding the question
The 12 labels the slanted segment SP, not the horizontal one, and the arrow for x runs from R past the center P to the foot of the vertical segment, so x is neither a radius nor a whole diameter. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
No obvious method minor Working it out
The right-angle mark and the triangle under S suggest the Pythagorean theorem, which is no help; the way in is seeing that RP and PT are radii as well. (Schoenfeld, 1985)

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

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Information you don't need: The question gives numbers or facts that play no part in the answer.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
P is the center of the circle, and R, S and T are on the circle. Segment SP is 12. Segment RT goes straight across through P. A short line from S meets RT at a right angle. It splits RT into x, from R to that point, and 5, from that point to T. You need x.
See that SP is a radius
SP runs from the center P to the point S on the circle. So SP is a radius, and the radius is 12. Every radius of a circle has the same length. RP and PT also run from the center to the circle, so RP = 12 and PT = 12.
Find the diameter RT
RT passes through the center, so it is a diameter. Use the formula with r = 12:
$\mathrm{RT} = 2r = 2 \times 12 = 24$
RT is 24 long. It is the radius RP plus the radius PT.
Solve for x
RT is also split into x and 5, so:
$x + 5 = 24$
Subtract 5 from both sides:
$x = 24 - 5 = 19$
Watch out: x runs from R past the center P, so it is longer than a radius. Reading it as 12, choice A, stops at P. Adding 12 + 5 = 17, choice C, puts the 12 in the wrong place.
Work from the numbers, not the picture
The figure is not drawn to scale. In it, x looks about three times as long as 5, which suggests 15, choice B. Trust the labels instead. The right triangle under S is not needed at all.
So the answer is D, 19.

Check your answer

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

More questions like this

The original DOE question

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

Question 13 of 50

2026 Practice Test D Math, question 13

Question

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

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

A43°
B44°
C46°
D47°

Show the answer

Correct answer: A

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

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

$5x - 5 = 90$

$5x = 95$

$x = 19$

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

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

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

The smallest angle is 43°.

Question analysis

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

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students write equations for unknown angles using complementary and supplementary angles. This also needs the angle sum of a triangle, which New York places in grade 8, plus substituting back and comparing, though each step is short.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Many steps minor
Something you must already know minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 50

2026 Practice Test D Math, question 14

Question

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

Show the answer

Correct answer: B

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

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

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

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

Question analysis

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

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking fractions of a quantity in stages is typical grade 7 work. The numbers divide evenly and the bold word flags the trap, so the four short steps match routine grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 15 of 50

2026 Practice Test D Math, question 15

Question

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

What is the value of the expression shown above?

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

Show the answer

Correct answer: C

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

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

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

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

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

Question analysis

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

NYS standards: 7.NS.1d properties for adding and subtracting
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding fractions with unlike denominators is taught before grade 7. Finding a common denominator for three related numbers and simplifying is short, routine work, lighter than typical grade 7 problems with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question checks whether students rename fractions before adding, and each wrong choice comes from a shortcut that skips that step or stops early. Seeing that 22 and 33 are multiples of 11 points to 66 as the common denominator. A size check helps too: the sum must be more than 1/11, its largest term, which rules out two of the wrong choices at once.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 16 of 50

2026 Practice Test D Math, question 16

Question

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

BREAD RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.5

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

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

$1 = 2p$

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

$0.5 = p$

The constant of proportionality is 0.5.

Question analysis

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

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from a table is a grade 7 standard, and here a single division in one row gives it. The demand is lighter than typical grade 7 work; the order of the ratio in the wording needs the most care.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 56 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 17 of 50

2026 Practice Test D Math, question 17

Question

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

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

Show the answer

Correct answer: D

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

$38 + 46 = 84$

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

$84 - 64 = 20$

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

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

Question analysis

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

NYS standards: 7.SP.8b sample spaces for compound events
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Overlapping groups are seldom taught directly in grade 7, though the arithmetic is whole-number addition and subtraction. Students must first find the overlap and then choose the right whole for the fraction, two decisions where errors are common.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 54 words.
Corrected from the DOE handbook: A misstatement in the explanation’s first step was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question hides its key number: the students in both groups appear only as the amount by which 38 + 46 overshoots 64. A sketch of two overlapping circles, with 20 in the middle, 18 in orchestra only and 26 in choir only, shows the whole group at once. After that, the remaining risk is the denominator, since the question asks for a fraction of all 64 students, not of one group.

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

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

Traps

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

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 17 as shown on the DOE’s digital test, page 90 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 90, question 17. This is the DOE’s version, before the corrections listed in the analysis.

Question 18 of 50

2026 Practice Test D Math, question 18

Question

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

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

Show the answer

Correct answer: C

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

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

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

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

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

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

Question analysis

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

NYS standards: 7.NS.2d fractions to decimals by long division
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding mixed numbers and writing a fraction as a repeating decimal are standard grade 7 skills. This combines them in three short steps; renaming a fraction and reading the bar notation are the only places that need care.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 13, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 19 of 50

2026 Practice Test D Math, question 19

Question

A store manager purchases a piece of furniture for $456.00. To determine the selling price, the manager increases the purchase cost by 125%. A customer buys the furniture and pays an additional 8% sales tax. How much does the customer pay for the furniture?

A$615.60
B$627.48
C$1,108.08
D$1,846.80

Show the answer

Correct answer: C

Given a purchase price of $456.00, the 125% increase in price is calculated as:

$456 \times 1.25 = 570$

The selling price equals the purchase price plus the increase, or:

$456 + 570 = 1{,}026$

The sales tax on the selling price is:

$1{,}026(0.08) = 82.08$

The customer pays the selling price plus the sales tax:

$1{,}026 + 82.08 = 1{,}108.08$

Question analysis

A store manager buys a piece of furniture for $456.00 and raises the price by 125% to set the selling price, and the customer also pays 8% sales tax. Students find the total the customer pays. It tests percent increase and sales tax in sequence, and reading an increase of more than 100% correctly is the main challenge.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students find markups and sales tax, usually with increases under 100%. An increase of more than 100% is unusual and easy to misread, and two percent changes must be applied in order, which goes beyond routine grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps moderate
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

A percent above 100 is unusual in grade 7 work, and it is the center of the question: the markup more than doubles the price. Students who treat “by 125%” as “125% of” still find a listed answer after adding tax, and so do students who add $125. A rough check helps: the selling price must be more than twice $456, about $1,000, and the tax adds less than a tenth of that.

Turning words into math moderate Understanding the question
“Increases the purchase cost by 125%” adds 125% of $456 to the $456 itself, a multiplier of 2.25; reading it as 125% of the cost leaves the original price out. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Four steps in a row, the increase, the selling price, the tax and the total, each depend on the one before, and two of them are easy to merge or skip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Taking 8% of $1,026 means multiplying by 0.08, and a decimal point one place off, 0.8, makes the tax ten times too large. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn each percent change into a multiplier, remembering that an increase by a percent is added to the original amount, apply the markup and then the tax, and estimate to check the size.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the increase, 125% of $456 = $570, and uses it as the selling price without adding back the $456 cost, then adds 8% tax.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 125% increase as $125, a fixed amount, so the selling price is $581, and then adds 8% tax correctly to reach $627.48.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds the $1,026 selling price correctly but writes 8% as 0.8, adding $820.80 in tax instead of $82.08.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $p\% \text{ of } n = \frac{p}{100} \times n$; n is the starting amount, and p is the percent. Percent means out of 100, so 125% is 125/100 = 1.25 and 8% is 8/100 = 0.08. Multiply n by that decimal to find the part.
Price after an increase: $\text{new price} = \text{old price} + \text{increase}$; The old price is what you start with. The increase is the amount added on. To increase by a percent, find that percent of the old price, then add it to the old price. Sales tax is added the same way.

Sort out the question
The manager pays $456.00 for the furniture. The selling price is that cost increased by 125%. The customer then pays 8% sales tax on top. Sales tax is an extra charge, a percent of the price. You need the total the customer pays.
Find the increase
Find 125% of $456. As a decimal, 125% is 1.25, which is 1 and a quarter. A quarter of 456 is 114:
$456 \times 1.25 = 456 + 114 = 570$
The increase is $570.
Find the selling price
Increases by means the increase is added to the cost:
$456 + 570 = 1{,}026$
The selling price is $1,026.
Watch out: $570 is only the increase. Using it as the price gives $615.60, choice A. And 125% is a percent, not $125. Adding $125 leads to $627.48, choice B.
Find the sales tax
As a decimal, 8% is 0.08. Find 8% of the selling price:
$1{,}026 \times 0.08 = 82.08$
The tax is $82.08.
Watch out: 8% is 0.08, not 0.8. Using 0.8 makes the tax ten times too big and gives $1,846.80, choice D.
Add the tax to the price
The customer pays the selling price plus the tax:
$1{,}026 + 82.08 = 1{,}108.08$
So the answer is C, $1,108.08.

Check your answer

Estimate. The price more than doubles: $456 plus $570 is about $1,000. Tax of 8% adds about $80. So the total should be a bit over $1,080, and only $1,108.08 is close. Or multiply in one go: 456 × 2.25 × 1.08 = 1,108.08.

More questions like this

The original DOE question

Question 19 as shown on the DOE’s digital test, page 92 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 92, question 19.

Question 20 of 50

2026 Practice Test D Math, question 20

Question

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

A−76
B−4
C4
D148

Show the answer

Correct answer: B

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

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

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

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

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

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

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

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

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

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

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

Question analysis

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

NYS standards: 7.NS.2a rational number multiplication
7.NS.2b rational number division
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Order of operations with negative numbers is standard grade 7 practice. This expression stacks a squared negative, two products and a division by a negative into one fraction, so it takes more steps than usual, each with a sign to track.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 6 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Nothing here is conceptually new; the question measures care with signs under the order of operations. The parentheses in (−4)² decide whether −9 multiplies 16 or −16, and the two products of −144 then nearly cancel the 304, leaving a small numerator whose sign is easy to lose. Writing each intermediate value with its sign, line by line, prevents most errors.

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

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 21 of 50

2026 Practice Test D Math, question 21

Question

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

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

Show the answer

Correct answer: C

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

$7.75 + 11.00 + 4.50 = 23.25$

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

$25.00 - 23.25 = 1.75$

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

Question analysis

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

NYS standards: 6.NS.3 multi-digit decimal operations
6.NS.5 positive and negative numbers
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 22 of 50

2026 Practice Test D Math, question 22

Question

Seth grows strawberries in his garden.

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

How many cups of strawberries did Seth have left?

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

Show the answer

Correct answer: B

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

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

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

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

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

$4\frac{1}{4}$

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

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

$4\frac{1}{4}$

Question analysis

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

NYS standards: 7.NS.3 four operations with rational numbers
7.EE.3 multi-step rational number problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding mixed numbers and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The bulleted list lays out the story day by day, so setting up the problem is easy; the points are lost in the fractions. The number of breakfast days, 3, sits right beside the 3/4 cup, and subtracting 3 cups instead of three quarters of a cup each day is the tempting shortcut. Writing every amount in fourths, 26/4 picked and 9/4 eaten, makes the last step a whole-number subtraction.

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

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

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 23 of 50

2026 Practice Test D Math, question 23

Question

Which situation involves quantities that combine to equal zero?

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

Show the answer

Correct answer: A

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

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

Question analysis

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

NYS standards: 7.NS.1a opposite quantities making 0
6.NS.5 positive and negative numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Describing situations where opposite quantities make zero is named in the grade 7 standards, and this is its most direct form: one reading of four everyday situations, easier than typical grade 7 work with signed numbers.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Turning words into math minor
Every choice must be checked minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 60 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The idea that a quantity and its opposite add to zero is familiar, and the right situation states it plainly. The wrong choices are built from equal amounts: two of them repeat a number but move in one direction, and the cave climb reaches zero partway before going 20 feet past it. Writing each situation as a sum of signed numbers takes a few seconds and settles it.

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

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

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 24 of 50

2026 Practice Test D Math, question 24

Question

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

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

Show the answer

Correct answer: B

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

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

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

Question analysis

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

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself, and this is a standard example: one fraction divided by another in two short steps, with the order of division the only real decision.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is a unit rate with fractions, and its wrong choices are the other ways to combine 1/8 and 3/4. Since 1 minute is only a third longer than 3/4 minute, the distance must be only a third more than 1/8 kilometer, which rules out the product and both large choices at once. Dividing with the units written in, kilometers over minutes, keeps the order right.

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

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

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 25 of 50

2026 Practice Test D Math, question 25

Question

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

Show the answer

Correct answer: B

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

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

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

Question analysis

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

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 50

2026 Practice Test D Math, question 26

Question

Jordan has a budget of $120 to rent a food truck. The rental company charges $18 per hour to rent the truck, plus a one-time cleaning fee of $30.

Write an inequality that represents this situation, in which x is the number of hours Jordan can rent the truck.

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: 30 + 18x ≤ 120

Let x represent the number of hours Jordan can rent the truck.

The hourly rental cost can be represented as $\$18 \cdot x$ , written as $18x$ .

The one-time cleaning fee of $30 is added to the hourly cost to obtain the total cost.

The sum of the cleaning fee and the hourly rental cost is represented as $30 + 18x$ .

As Jordan has $120 to spend, the total cost must be less than or equal to $120.

This is represented as $30 + 18x \leq 120$ or $120 \geq 30 + 18x$ .

Question analysis

Jordan has $120 to rent a food truck that costs $18 an hour plus a one-time $30 cleaning fee, and students type an inequality for x, the number of hours. It tests writing a two-step inequality from a word problem. Matching each dollar amount to its role and choosing the symbol that fits a spending limit are the tricky parts.

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 a two-step inequality from a word problem is the grade 7 standard itself. Nothing is offered to choose from: students type the whole inequality, so the numbers, the variable term and the symbol are all theirs to get 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: Turning words into math moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 57 words.
Corrected from the DOE handbook: A spelling 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 arithmetic is small, so the points ride on two decisions the keypad leaves open: which symbol, and whether to stop before solving. A budget means Jordan may spend all $120 but no more, so the total cost must stay at or under it, and a strict < leaves out the five-hour rental that costs exactly $120. Testing x = 5 and x = 6 in the typed inequality confirms both the symbol and the setup.

Turning words into math moderate Understanding the question
The stem never says “at most”: the limit comes only from “a budget of $120”, while “$18 per hour” must multiply x and the “one-time cleaning fee of $30” is added just once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The whole inequality is typed on a keypad that offers <, ≤, > and ≥ side by side, with no choices to compare, and the solved form x ≤ 5 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Work out what 1 hour and 2 hours would cost Jordan in all, see which amount repeats each hour, build the total for x hours, then compare it with the budget and test a few values.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

30 + 18x < 120 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Builds the cost 30 + 18x correctly but uses a strict less-than sign, which shuts out a five-hour rental costing exactly the $120 Jordan has.
30 + 18x ≥ 120 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Builds the cost correctly but points the symbol the wrong way, treating the $120 budget as an amount Jordan must spend at least instead of a limit.
x ≤ 5 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Writes and solves the inequality correctly, then enters its solution, x ≤ 5, though the item asks for the inequality showing the fee, the hourly charge and the budget.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “The onetime cleaning fee of $30 is added to the hourly cost …”. “Onetime” means “former”; a fee charged once is a “one-time” fee, as the question itself spells it. The site now says “one-time”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee and a rate: $\text{total cost} = \text{fee} + \text{rate} \times x$; The fee is a charge paid only once, so you add it once. The rate is the charge for each hour. x is the number of hours. You multiply the rate by x because it is paid again for every hour.
Staying within a budget: $\text{total cost} \leq \text{budget}$; The sign ≤ means is less than or equal to. A budget is the most money you can spend. You may spend all of it, or less, but never more. So the total cost must be less than or equal to the budget.

Sort out the question
You know three facts:
- Jordan can spend at most $120. That is the budget.
- The truck costs $18 for each hour.
- A cleaning fee of $30 is paid one time only.
x is the number of hours. You need an inequality. That is a math sentence that compares two amounts with a sign such as < or ≤. It must show all three facts.
Work out a few totals
Find the total cost for a few hours. Each hour adds one more $18:
Hours Hourly charges ($) Total cost ($)
1 $18 \times 1 = 18$ $30 + 18 = 48$
2 $18 \times 2 = 36$ $30 + 36 = 66$
x $18x$ $30 + 18x$
The $30 shows up once in every total. The $18 shows up once for each hour. So x hours cost 30 + 18x dollars. Here 18x means 18 times x.
Choose the right sign
The total cost must fit inside the budget. Jordan may spend exactly $120, or less, but not more. That is less than or equal to, the sign ≤. So:
$30 + 18x \leq 120$
Watch out: the sign < means less than, but not equal. $30 + 18x < 120$ leaves out spending exactly $120. The sign ≥ is backward too. It says Jordan must spend at least $120. Also keep each number in its place. $30x + 18$ charges $30 an hour.
Stop before you solve
You could solve the inequality. Subtract 30 from both sides to get $18x \leq 90$ . Then divide both sides by 18 to get $x \leq 5$ . So Jordan can rent the truck for at most 5 hours.
Watch out: the question asks for an inequality that models the situation. To model means to show the facts of the story in math. The solved form, x ≤ 5, is true, but it hides the $18, the $30 and the $120. So x ≤ 5 is not accepted. Enter the inequality before you solve it.
Enter it on the keypad
On the test, build the inequality on the keypad. Press 3, 0, +, 1, 8 and the x key. Then press the ≤ key, in the fourth row of keys. Then press 1, 2, 0. You need no dollar signs and no times sign. On this site, type it, with <= for ≤: 30 + 18x <= 120.
So you would enter 30 + 18x ≤ 120. The same inequality written the other way round, 120 ≥ 30 + 18x, is also right. The solved form, x ≤ 5, is not accepted, because it does not model the situation.

Hours	Hourly charges ($)	Total cost ($)
1	$18 \times 1 = 18$	$30 + 18 = 48$
2	$18 \times 2 = 36$	$30 + 36 = 66$
x	$18x$	$30 + 18x$

Check your answer

Test two numbers of hours. For 5 hours, $30 + 18 \times 5 = 120$ . Is 120 ≤ 120? Yes, so 5 hours fit the budget. For 6 hours, $30 + 18 \times 6 = 138$ . Is 138 ≤ 120? No, and $138 is over budget. The inequality gets both right.

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 D 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 D Math, question 27

Question

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

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

Show the answer

Correct answer: B

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

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

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

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

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

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

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

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3d converting measurement units
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multistep ratio problems with a unit conversion are typical grade 7 work. This one chains three short steps, a fraction of the mixture, pounds to ounces and a price per ounce, and the numbers are chosen to come out evenly.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 28 of 50

2026 Practice Test D Math, question 28

Question

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

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

Show the answer

Correct answer: C

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

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

Question analysis

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

NYS standards: 7.RP.2d points on proportional graphs
6.EE.9 dependent and independent variables
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on the graph of a proportional relationship means is named in the grade 7 standards. This is a direct case with no graph to read, though the stem introduces the two quantities in an order that needs care.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Reading a figure, table or graph minor
A very close wrong answer minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 76 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph minor Understanding the question
No graph is drawn, and the stem names y, the gallons, before x, the hours, so the pair (3, 24) has to be matched to its letters, not to the order the quantities were mentioned. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer minor Choosing the answer
The four statements reuse 3 and 24 in two frames, a total added “in” some hours and an amount “per hour”, so the right statement and its swapped twin differ only in which number counts gallons. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The graph links two amounts:
- x is the number of hours water has been added.
- y is the number of gallons added.
You need to say what the point (3, 24) means. No graph is drawn, so you work from the words.
Match each number to its letter
In an ordered pair, x always comes first and y second. So in (3, 24):
- x = 3, which means 3 hours.
- y = 24, which means 24 gallons.
Watch out: the question names gallons first and hours second. Do not match the numbers to the words in that order. That mistake reads 3 as gallons and 24 as hours, choice A.
Read the pair as a sentence
Put the two meanings together: after 3 hours, 24 gallons have been added in all. The 24 is a total for the whole 3 hours.
Watch out: 24 is not the amount for each hour. The rate per hour is $24 \div 3 = 8$ gallons. So choice D, 24 gallons per hour, is wrong. Choice B, 3 gallons per hour, is also not the rate.
So the answer is C, twenty-four gallons of water was added in 3 hours.

Check your answer

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

More questions like this

The original DOE question

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

Question 29 of 50

2026 Practice Test D Math, question 29

Question

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

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

Show the answer

Correct answer: B

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

1, 2, 3, 4

The possible outcomes of two spins are:

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

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

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

Question analysis

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

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a compound probability from a list or table of equally likely outcomes is standard grade 7 work. Two four-section spinners give a small sample space that is quick to write out, and the event needs one careful scan of the products.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 52 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is decided by the count of favorable outcomes, not by any calculation. Students who think of 4 only as 2 × 2 find one outcome, and those who think in factor pairs find two, which is not a listed choice and should send them back to the table. A 4-by-4 grid of products, red down the side and blue across, shows all three cells at once.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 30 of 50

2026 Practice Test D Math, question 30

Question

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

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

Show the answer

Correct answer: A

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

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

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

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

Question analysis

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

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is exactly the grade 7 standard. This one divides by a mixed number, and the cook's plan to enlarge the recipe adds wording without adding any numbers to use.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 31 of 50

2026 Practice Test D Math, question 31

Question

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

A80
B300
C400
D500

Show the answer

Correct answer: C

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

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

$2x = 800$

$x = 400$

Question analysis

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

NYS standards: 6.SP.1b samples and populations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a representative sample to reason about a population is grade 6 and 7 work. The twist is that the known count, the 100 red marbles added, belongs to the box rather than the sample, so the usual scaling runs in an unfamiliar direction.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 60 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is the logic of estimating a population by tagging, and the twist is that the known count belongs to the marbles that were added, not to the ones being estimated. Scaling the sample up by 50, so that its 2 red become the box’s 100, turns its 8 green into the estimate in one step. Students who scale by way of the whole box often stop at the box’s total, and those who treat 100 as the whole box scale the wrong amount.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 32 of 50

2026 Practice Test D Math, question 32

Question

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

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

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

Show the answer

Correct answer: B

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

Convert 72 cups to gallons:

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

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

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

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

Question analysis

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

NYS standards: 6.RP.3d converting measurement units
6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting units with given factors is grade 6 work. This adds a first step, the total number of cups, and a two-stage conversion through quarts, but the numbers are friendly and the path is routine for grade 7 students.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 32 words.
Corrected from the DOE handbook: A stray word at the end of the explanation was removed.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

Work backward. $4\frac{1}{2}$ gallons is $4\frac{1}{2} \times 4 = 18$ quarts. 18 quarts is $18 \times 4 = 72$ cups. Shared among 36 people, that is $72 \div 36 = 2$ cups each. It matches.

More questions like this

The original DOE question

Question 32 as shown on the DOE’s digital test, page 105 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 105, question 32. This is the DOE’s version, before the corrections listed in the analysis.

Question 33 of 50

2026 Practice Test D Math, question 33

Question

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

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

Show the answer

Correct answer: C

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

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

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

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding sales tax to a price is a standard grade 7 percent task. This one also subtracts a down payment at the end, three routine steps with round numbers, much like a typical classroom problem.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is the standard one for tax on a purchase, so the question checks order and completeness. Forgetting the tax or forgetting the down payment each produces an amount as reasonable-looking as the answer, and a careless 8 × 24 produces one just $300 below it. Multiplying the price by 1.08 in one step, which gives $25,920, is a second route that catches a slip in the tax.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 34 of 50

2026 Practice Test D Math, question 34

Question

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

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

Show the answer

Correct answer: C

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

Question analysis

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

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.8b probability models from observed frequencies
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Compound probability is grade 7 work, often done with a table of outcomes. Each chance here comes from one week's frequencies, and joining the morning and afternoon events in one step is about the usual grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 48 words.
Corrected from the DOE handbook: An inaccurate detail in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 34 as shown on the DOE’s digital test, page 107 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 107, question 34. This is the DOE’s version, before the corrections listed in the analysis.

Question 35 of 50

2026 Practice Test D Math, question 35

Question

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

Show the answer

Correct answer: C

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

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

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

AB1, AB2, AC1, AC2, AD1, AD2

BC1, BC2, BD1, BD2

CD1, CD2

There are 12 possible groups with only 1 male.

Question analysis

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

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students list outcomes with organized lists and tree diagrams, usually ordered ones. Counting unordered groups of three under an "only one" condition, built from two kinds of people, goes beyond that typical work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question hides a two-stage count in a hiring story. Students tend to get one stage right and forget the other: they count the pairs of women and stop, or they count ordered picks and so find every group twice. Naming the women A to D and the men 1 and 2, then listing the pairs of women alphabetically with each man attached, makes every group visible and checkable.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 36 of 50

2026 Practice Test D Math, question 36

Question

Leah has set a goal to read a total of 500 pages of a textbook. The table shows the book page numbers for the reading she did during three days.

DAILY READING LOG

Day	Pages Read
1	pages 45–90, inclusive
2	pages 100–135, inclusive
3	pages 145–190, inclusive

What percentage of her 500-page goal did Leah read on these three days? Round your answer to the nearest tenth.

Enter your answer in the space.

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

Show the answer

Correct answer: 25.6%

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

Day 1: $(90 - 45) + 1 = 46$ pages

Day 2: $(135 - 100) + 1 = 36$ pages

Day 3: $(190 - 145) + 1 = 46$ pages

Leah read a total of 128 pages, since $46 + 36 + 46 = 128$ .

To determine the percent, express 128 pages out of 500 pages as a fraction.

$\frac{128}{500} = \frac{128 \div 4}{500 \div 4} = \frac{32}{125}$

$32 \div 125 = 0.256 = 25.6\%$

Question analysis

Leah's daily reading log lists three ranges of textbook pages, each marked inclusive, and her goal is 500 pages. This percent question asks what percentage of the goal she read over the three days, rounded to the nearest tenth and typed in. Counting the pages in each inclusive range correctly is the tricky step before the percent.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Finding what percent one number is of another is grade 6 and 7 work. This adds three inclusive page ranges to count, which takes more care than subtracting, then a rounding step, all typed with no choices to check against.
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
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 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

Counting pages in a range is the whole difficulty; the percent at the end is routine once the total is right. Subtracting first from last feels natural, and here it even rewards the student with a round 25%, so nothing looks wrong. Checking the rule on a tiny range, pages 1 to 3, which is 3 pages and not 2, fixes the count, and the instruction to round to the nearest tenth hints that the answer is not a whole number.

Something you must already know moderate Working it out
Every row says inclusive, but nothing says that pages 45 to 90 hold 90 − 45 + 1 = 46 pages; subtracting alone loses one page on each of the three days. (Anderson & Krathwohl, 2001)
Many steps moderate Working it out
Three page counts, a sum, a division by 500 and a change to a percent come in sequence, and a one-page error in each count carries through to the end. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The answer is typed with no choices, and the miscounted total, 125 pages, gives a tidy 25% that looks more finished than the right value with its tenth. (Enright, Morley & Sheehan, 2002)

Strategy: Count each day's pages carefully, testing your method on a short range such as pages 1–3, add the three days, divide by 500, and write the result as a percent to the nearest tenth.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

25 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Subtracts the first page from the last for each day, 45 + 35 + 45 = 125 pages, forgetting that an inclusive range counts both end pages, so 125 ÷ 500 = 25%.
0.256 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Counts the 128 pages and divides by 500 correctly, then enters the decimal 0.256, one step short of multiplying by 100 to make a percent.
26 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds 25.6% correctly but rounds it to the nearest whole percent, overlooking the instruction to round to the nearest tenth.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range: $\text{pages} = \text{last page} - \text{first page} + 1$; Inclusive means the first and last pages are both read. Subtracting counts the steps from the first page to the last. That misses one page, so add 1. Test it: pages 1 to 3 are 3 pages, and 3 − 1 + 1 = 3.
Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you have, here the pages read. The whole is the amount you compare with, here the 500-page goal. Dividing gives a decimal. Multiplying by 100 turns it into a percent, a number out of 100.

Sort out the question
Leah’s goal is 500 pages. The log shows three days:
- Day 1: pages 45 to 90
- Day 2: pages 100 to 135
- Day 3: pages 145 to 190
Each range is inclusive. You need the percent of 500 pages she read, rounded to the nearest tenth.
Count the pages each day
Use the rule for an inclusive range: last page minus first page, plus 1.
$90 - 45 + 1 = 46$
$135 - 100 + 1 = 36$
$190 - 145 + 1 = 46$
Watch out: subtracting alone gives 45, 35 and 45 pages. That is 125 pages, which is exactly 25%. A round answer like that can hide the mistake.
Add the three days
Add the pages from all three days to get her total:
$46 + 36 + 46 = 128$
Leah read 128 pages in all. This is the part you compare with the goal.
Change to a percent
Divide the pages read by the goal, then multiply by 100:
$\frac{128}{500} \times 100 = 0.256 \times 100 = 25.6$
Leah read 25.6% of her goal. That already ends in tenths, so no rounding is needed.
Watch out: 0.256 is a decimal, not a percent. And 26 rounds to a whole number, not to the nearest tenth.
Enter the answer
The box has no % sign, but the question asks for a percentage, so the number in the box is the percent. Type it with its decimal point. Typing 25.6% also counts on this site. So you would enter 25.6.

Check your answer

Estimate. 10% of 500 pages is 50 pages. 128 pages is a bit more than 2 × 50 = 100, so the answer is between 20% and 30%. Then check exactly: $0.256 \times 500 = 128$ pages. It matches.

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 D Math, page 109, question 36.

Question 37 of 50

2026 Practice Test D Math, question 37

Question

What 3-digit number between 630 and 650 is divisible by both 5 and 8?

Enter your answer in the space.

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

Show the answer

Correct answer: 640

The final digit of the three-digit number is 0. Each of the digits 0 through 9, inclusive, must be considered.

To narrow the choices, first consider that the number between 630 and 650 must be divisible by both 5 and 8.

A number divisible by 5 must end in 0 or 5. Therefore, the only possibilities are 0 and 5.

If the number ended in 0, the whole number 640 must also be divisible by 8:

$640 \div 8 = 80 \text{ and } 640 \div 5 = 128$

Both result in a remainder of 0.
If the number ended in 5, the whole numbers 635 and 645 must also be divisible by 8:

$635 \div 8 = 79R3 \text{ and } 635 \div 5 = 127$

$645 \div 8 = 80R5 \text{ and } 645 \div 5 = 129$

Dividing 635 by 5 results in a remainder of 0, but dividing 635 by 8 results in a remainder of 3.

Dividing 645 by 5 results in a remainder of 0, but dividing 645 by 8 results in a remainder of 5.

Therefore, 640 is the only whole number between 630 and 650 that is divisible by both 5 and 8.

Question analysis

This number question asks for the 3-digit number between 630 and 650 that is divisible by both 5 and 8, typed into a box. Students use divisibility rules or common multiples to find it. The work is short, but the number has to pass both tests, so checking only one divisor is the easy slip.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Common multiples are grade 6 content. Divisibility by 5 limits the last digit, and only a few numbers lie between 630 and 650, so this is a quick check, shorter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Much to keep track of minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Few numbers need checking, so most students succeed; the risk is testing only one divisor. Students who start from multiples of 8 meet 632 first, and those who start from multiples of 5 can slip on 645 ÷ 8. Using the least common multiple, 40, shows at once that the number must be a multiple of 40, which leaves one candidate in the range.

Much to keep track of minor Working it out
Each candidate must pass two tests, divisible by 5 and by 8, and both ends of the range, 630 and 650, pass the first test but fail the second. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The shortcut that a multiple of 5 ends in 0 or 5 is assumed, and it cuts the search to 635, 640 and 645 before any division by 8. (Anderson & Krathwohl, 2001)

Strategy: Use divisibility by 5 to limit the last digit, then check only those few numbers in the range for divisibility by 8.

Traps

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

The wrong answers, classified

632 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Takes the first multiple of 8 after 630, 8 × 79 = 632, and stops without checking divisibility by 5, which a number ending in 2 fails.
645 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Picks a number ending in 5 so that 5 divides it, then slips on the check for 8, which leaves a remainder of 5.
40 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds the least common multiple of 5 and 8, 40, and enters it, one step before finding its multiple between 630 and 650.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisible by 5: $\text{last digit} = 0 \text{ or } 5$; Divisible means it divides evenly, with no remainder. A whole number is divisible by 5 only when its last digit is 0 or 5. That is because counting by 5s always lands on a number ending in 0 or 5.
Least common multiple: $\text{LCM}(5, 8) = 5 \times 8 = 40$; The least common multiple, or LCM, is the smallest number that both 5 and 8 divide evenly. 5 and 8 share no factor except 1, so their LCM is their product. Any number divisible by both is a multiple of 40.

Sort out the question
You need a 3-digit number between 630 and 650. It must be divisible by 5 and also by 8. Between means 630 and 650 themselves do not count.
Use the rule for 5
A number divisible by 5 ends in 0 or 5. Between 630 and 650, only three numbers do:
$635, \; 640, \; 645$
Only these need checking for 8.
Test each number for 8
Divide each one by 8 and look at the remainder, the amount left over:
- 635 ÷ 8 = 79, with 3 left over.
- 640 ÷ 8 = 80, with nothing left over.
- 645 ÷ 8 = 80, with 5 left over.
Only 640 divides evenly by 8.
Watch out: 645 passes the test for 5 but fails for 8. And 632, which is $8 \times 79$ , passes for 8 but ends in 2, so it fails for 5.
Enter the number
640 passes both tests. Type the whole number in the box.
Watch out: 40 is the smallest number divisible by 5 and 8, but it is not between 630 and 650.
So you would enter 640.

Check your answer

Use the LCM. A number divisible by both 5 and 8 must be a multiple of 40. $40 \times 16 = 640$ . The multiples next to it are 600 and 680, which are outside the range. So 640 is the only one.

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 D Math, page 110, question 37.

Question 38 of 50

2026 Practice Test D Math, question 38

Question

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

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

Show the answer

Correct answer: D

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

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

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

RR, RG, RW

GR, GG, GW

WR, WG, WW

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

RR, RG, RW, GG, GW, WW

Question analysis

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

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Representing the sample space of a compound event is a named grade 7 skill. Nothing is computed here; the demand is knowing what one outcome of drawing two marbles looks like and which pairs the bag allows, much as in grade 7 classwork.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 3, 71 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question tests a definition more than a skill, and three choices describe something real: the colors of one marble, the contents of the bag, and the pairs of different colors. The strongest pull comes from treating the colors as three single objects, which leaves out the pairs of one color although the bag holds three reds, three greens and two whites. Asking what Melissa could be holding after the draw settles it.

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

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Without replacement: In a second pick there is one fewer item, which changes the probability.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 39 of 50

2026 Practice Test D Math, question 39

Question

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

HOCKEY GOAL PROBABILITIES

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

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

A0.13
B0.21
C0.34
D0.50

Show the answer

Correct answer: C

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

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

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

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

$0.21 + 0.09 + 0.04 = 0.34.$

Question analysis

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

NYS standards: 6.SP.8 probability models for simple events
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math minor Understanding the question
The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.
Pick the right rows
Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:
- 3 goals: 0.21
- 4 goals: 0.09
- 5 goals: 0.04
Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.
Add the chances
A game can end with only one number of goals. So add the three chances:
$0.21 + 0.09 + 0.04 = 0.34$
Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.
So the answer is C, 0.34.

Check your answer

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

More questions like this

The original DOE question

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

Question 40 of 50

2026 Practice Test D Math, question 40

Question

A team is participating in a contest to collect canned goods for a food drive. The team has 53 points and must have a total of at least 100 points to win a prize. The team earns 5 points for each can of food a member collects. Which inequality can be used to find x, the number of cans of food the team needs to collect to win a prize?

A $5 + 53x \leq 100$
B $5 + 53x \geq 100$
C $53 + 5x \leq 100$
D $53 + 5x \geq 100$

Show the answer

Correct answer: D

Since each can is worth 5 points, the total number of points earned from all the cans, x, is 5x. The team started with 53 points. That is a constant value and doesn’t change. So the total number of points the team will earn is $53 + 5x$ . The team wants to earn at least (≥) 100 points, so they need $53 + 5x \geq 100$ total points.

Question analysis

A team has 53 points in a canned-food drive, earns 5 points for each can, and needs at least 100 points to win a prize. Students choose the inequality for x, the number of cans the team must collect. It tests writing an inequality from words, where the starting points and the points per can must each take the right place.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an inequality for a word problem with a starting amount and a rate is grade 7 work. The four choices pair two expressions with two symbols, so students make one decision about the terms and one about the phrase "at least".
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 71 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The choices form a two-by-two grid: which number goes with x, and which way the symbol points. Each decision can be made separately, so a student who settles both gets the item, while one mistake lands on a listed choice. Testing two numbers of cans, 5, which leaves the team at 78 points, and 10, which lifts it to 103, rules out every wrong inequality.

Turning words into math moderate Understanding the question
The 53 points the team “has” are a fixed start, the “5 points for each can” go with x, and “at least 100” calls for ≥, though “least” can sound like less. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The four choices are two expressions, 5 + 53x and 53 + 5x, each with ≤ and with ≥, so the right one and the strongest wrong one differ only in the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the team's total points after 2 cans and then after x cans, translate "at least 100" into a symbol, and compare the result with each choice.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Attaches x to the 53 points the team already has and also reverses “at least”, combining the mistakes behind the other two wrong choices.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Keeps the right symbol but multiplies the number of cans by 53 and adds 5, putting the starting points where the points per can belong.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Writes the points correctly as 53 + 5x but reads “at least 100” as at most 100, reversing the inequality.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total with a start and a rate: $\text{total} = \text{start} + \text{rate} \times x$; The start is the amount you have now, here 53 points. The rate is what each item adds, here 5 points for each can. x is the number of items. The start is added once, and the rate is multiplied by x.
At least: $\text{total} \geq \text{goal}$; The sign ≥ means is greater than or equal to. At least 100 means 100 or more. So the total must be greater than or equal to 100.

Sort out the question
You know:
- The team already has 53 points.
- It earns 5 points for each can.
- It needs at least 100 points to win.
x is the number of cans. You need the inequality that shows this.
Write the total points
Try 2 cans first. The team gets 5 points for each, so $53 + 5 \times 2 = 63$ points. For x cans, the total is:
$53 + 5x$
Watch out: the 53 points do not repeat for each can. $5 + 53x$ gives 53 points for each can, as in choices A and B.
Choose the sign
At least 100 means 100 or more, so use ≥:
$53 + 5x \geq 100$
Watch out: “least” sounds like “less,” but at least means the total can be 100 or higher. Using ≤ gives choice C.
So the answer is D, $53 + 5x \geq 100$ .

Check your answer

Test two numbers of cans. With 10 cans, $53 + 5 \times 10 = 103$ , and 103 ≥ 100 is true. The team wins. With 9 cans, $53 + 5 \times 9 = 98$ , and 98 ≥ 100 is false. The team is short. Choice D gets both right.

More questions like this

The original DOE question

Question 40 as shown on the DOE’s digital test, page 113 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 113, question 40.

Question 41 of 50

2026 Practice Test D Math, question 41

Question

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

Show the answer

Correct answer: B

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

Question analysis

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

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students find constants of proportionality in tables, graphs and equations. This one is a single division of miles by hours given in a sentence, the same unit rate work students do in grade 6.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One division answers the question, so it is among the easiest on the test. The wrong choices are the two numbers in the stem and one unrelated value, so a student who knows what the constant means is not tempted. Checking that the constant times 3 hours gives back the 90 miles confirms it.

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

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

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 42 of 50

2026 Practice Test D Math, question 42

Question

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

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

Show the answer

Correct answer: B

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

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

Then use division to find the unit rate.

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

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

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

Question analysis

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

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself. This is a short example with unit fractions, where scaling the time up to a whole hour is as quick as dividing.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 43 of 50

2026 Practice Test D Math, question 43

Question

Two student clubs are selling snacks as part of a fundraiser. The Art Club sells a 24-pack of granola bars for $12.00. The Science Club sells a 10-pack of granola bars for $6.50. To the nearest cent, what is the difference in cost per bar between the two clubs?

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

The cost for the Art Club’s bar is per bar.

The cost for the Science Club’s bar is per bar.

The difference between the cost of the bars is .

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: $0.50; blank 2: $0.65; blank 3: $0.15

To determine the cost per bar from the Art Club, divide the total cost by the total number of bars:

$\frac{\$12.00}{24} = \$0.50 \text{ per bar}$

To determine the cost per bar from the Science Club, divide the total cost by the total number of bars:

$\frac{\$6.50}{10} = \$0.65 \text{ per bar}$

To find the difference in price per bar, subtract the cost per bar for the Art Club from the cost per bar for the Science Club:

$\$0.65 \text{ per bar} - \$0.50 \text{ per bar} = \$0.15 \text{ per bar}$

Question analysis

Two clubs sell granola bars for a fundraiser: the Art Club a 24-pack for $12.00 and the Science Club a 10-pack for $6.50. Students complete three sentences with drop-down menus, giving each club's cost per bar and the difference between them. It tests unit prices with money, and all three menus must be right.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Comparing unit prices is grade 6 rate work. The menus split it into three short steps, two divisions of money amounts and one subtraction, which is lighter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Many steps minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 87 words.
Corrected from the DOE handbook: The order of the answers in the DOE’s key 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 menus guide the work, one sentence per step, and every option is a reasonable price, so the question rewards accurate division more than planning. No common misconception leads to the wrong options; they catch slips, such as misreading a pack size or subtracting carelessly. Checking that 24 bars at the first price cost $12.00 and 10 bars at the second cost $6.50 confirms both menus before the subtraction.

Many steps minor Working it out
Three menus must all be right to score: two divisions for the costs per bar, then a subtraction that uses both, so a slip in either division spoils the third menu. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Each division is with money, $12.00 ÷ 24 and $6.50 ÷ 10, so the decimal point must be placed carefully, though the menus offer only sensible amounts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide each pack's price by the number of bars in it, then subtract the smaller cost per bar from the larger; checking that 24 bars at your first price make $12.00 catches slips.

Traps

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

CBA Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Divides $12.00 by 20 instead of 24, misreading the size of the Art Club’s pack, so $0.60 a bar leaves a gap of only $0.05.
BBB An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds both costs per bar correctly, then slips in the subtraction, taking $0.65 − $0.50 as $0.10, the option beside the right one.

What was corrected from the DOE handbook, and why

Answer key: The DOE’s explanation opens with the key “($0.65; $0.50; $0.15)”, which puts $0.65 in the first drop-down and $0.50 in the second. Neither is possible: the first menu (the Art Club’s cost per bar) offers $0.40, $0.50, $0.60 and $0.75, and the second (the Science Club’s) offers $0.55, $0.65, $0.75 and $0.85. The explanation’s own working gives the Art Club $12.00 ÷ 24 = $0.50 and the Science Club $6.50 ÷ 10 = $0.65, so the key is $0.50, $0.65, $0.15. The site’s key was already in that order; the DOE’s key line is not shown on the site.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost of one item: $\text{cost per bar} = \text{price of the pack} \div \text{number of bars}$; The price of the pack is what you pay for all the bars together. Dividing shares that price equally among the bars. The result is the price of one bar, called the unit price.

Sort out the question
You know:
- The Art Club sells 24 bars for $12.00.
- The Science Club sells 10 bars for $6.50.
There are three menus. The first is the Art Club’s cost per bar. The second is the Science Club’s. The third is the difference, how much more one bar costs than the other.
Find the Art Club’s cost
Share $12.00 among 24 bars:
$12.00 \div 24 = 0.50$
Each bar costs $0.50. That makes sense: 24 half-dollars make $12. Pick $0.50 in the first menu.
Watch out: dividing by 20 instead of 24 gives $0.60, which is also in the menu.
Find the Science Club’s cost
Share $6.50 among 10 bars. Dividing by 10 moves the decimal point one place to the left:
$6.50 \div 10 = 0.65$
Each bar costs $0.65. Pick $0.65 in the second menu.
Find the difference
Subtract the smaller cost from the larger one:
$0.65 - 0.50 = 0.15$
The bars differ by $0.15. Pick $0.15 in the third menu.
Watch out: $0.10 sits right next to $0.15 in the menu. A slip in the subtraction lands there.
Fill in the three menus
On the test and on this site, you choose each answer from a drop-down menu.
So the menus read $0.50, $0.65 and $0.15.

Check your answer

Multiply back. $24 \times 0.50 = 12.00$ and $10 \times 0.65 = 6.50$ , so both costs per bar match the pack prices. And $0.50 + 0.15 = 0.65$ , so the difference fits too.

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 D Math, page 116, question 43. This is the DOE’s version, before the corrections listed in the analysis.

Question 44 of 50

2026 Practice Test D Math, question 44

Question

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

$p = 2.50 + 1.25c$

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

Show the answer

Correct answer: C

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

$22.50 = 2.50 + 1.25c$

$22.50 - 2.50 = 1.25c$

$20.00 \div 1.25 = c$

$16 = c$

Question analysis

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

NYS standards: 7.EE.4a two-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step equation of the form px + q = r in a money context is core grade 7 algebra. The equation is given, so the work is substituting the total and dividing by a decimal, a routine example.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The equation is supplied, so the question is a two-step solve with money. The tray’s $2.50 has to come off the total before dividing, and students who divide the whole $22.50 by $1.25 get a whole number that matches a choice, so nothing warns them. Substituting the chosen number back into p = 2.50 + 1.25c is a quick, decisive check.

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

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

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 45 of 50

2026 Practice Test D Math, question 45

Question

Marcus and Isla are saving money to buy concert tickets.

They decide to save the same amount each week.
They already have $8 saved.
After 3 weeks, the two have $26 saved in total.

How much money, in dollars, is being saved by Marcus and Isla each week?

Enter your answer in the space.

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

Show the answer

Correct answer: 6

Define the variable:

Let w represent the amount of money Marcus and Isla save per week.

Construct an equation:

Start with $8.

Marcus and Isla save w each week for 3 weeks, which can be expressed as $3 \cdot w$ or $3w$ .

The total amount saved can be represented by the expression $8 + 3w$ .

After 3 weeks, Marcus and Isla have $26. This can be represented by the equation $8 + 3w = 26$ .

Solve the equation:

$8 + 3w = 26$

$3w = 18$ (subtracting 8 from both sides)

$w = 6$ (dividing both sides by 3)

Marcus and Isla are saving $6 per week.

Question analysis

Marcus and Isla already have $8 toward concert tickets, save the same amount each week, and have $26 in all after 3 weeks. Students type how many dollars the two of them save each week. It tests writing and solving a two-step equation, keeping the money saved beforehand apart from the weekly savings.

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r from a word problem is the grade 7 standard. This one uses small whole numbers and two steps, a typical classroom example, with no choices to compare.
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
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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

Two inverse steps solve it, subtracting the starting $8 and dividing by the 3 weeks, and most mistakes drop one of them. The $18 saved over three weeks is the natural stopping place, and dividing $26 by 3 without removing the $8 gives an awkward value that should prompt a second look. Adding three weeks of the answer to $8 and getting $26 confirms it.

Turning words into math minor Understanding the question
The bullet points hold a start amount, “already have $8 saved”, and a total after three weeks, “$26 saved in total”, which must become 8 + 3w = 26. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The amount is typed with no choices, so stopping at the 3-week total of $18 leaves a whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Take away the amount saved before the weeks began, then share what is left equally among the weeks; check by adding three weeks of savings to the starting amount.

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

The wrong answers, classified

18 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Subtracts the $8 already saved from the $26 total and stops, giving the amount saved over all 3 weeks rather than in each week.
26/3 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $26 total by the 3 weeks, 26 ÷ 3 = 8 2/3, about $8.67, forgetting that $8 of it was saved before the weekly saving began.
3 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the weekly amount and then halves it for each person, though the question asks what Marcus and Isla save together each week.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total savings: $\text{total} = \text{start} + \text{weeks} \times \text{amount each week}$; The start is the money saved before the weeks begin. The amount each week is added once for every week, so you multiply it by the number of weeks. The total is all the money together.
Doing the same to both sides: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number (not zero when you divide). Doing the same thing to both sides keeps them equal. That is how you get the unknown by itself.

Sort out the question
You know:
- Marcus and Isla already have $8.
- They save the same amount each week.
- After 3 weeks they have $26 in all.
You need the amount the two of them save together each week.
Write an equation
Let w be the dollars saved each week. Three weeks add 3w dollars to the $8 they started with:
$8 + 3w = 26$
Subtract the starting money
Take 8 away from both sides:
$3w = 26 - 8 = 18$
They saved $18 over the 3 weeks.
Watch out: 18 is the amount for all 3 weeks, not for each week. Do not stop here.
Divide by the number of weeks
Divide both sides by 3:
$w = 18 \div 3 = 6$
They save $6 each week.
Watch out: dividing all $26 by 3 gives about 8.67. That forgets the $8 saved before. Halving 6 to get 3 is wrong too: the question asks what the two save together.
Enter the answer
Type only the number in the box, with no dollar sign and no units. So you would enter 6.

Check your answer

Count up week by week. Start at $8. After week 1: $8 + 6 = 14$ . After week 2: $14 + 6 = 20$ . After week 3: $20 + 6 = 26$ . That is the $26 in the question.

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 D Math, page 118, question 45.

Question 46 of 50

2026 Practice Test D Math, question 46

Question

A company charges $3.00 to ship a 4-pound package. The cost includes a base fee of $1.50 plus $0.50 for each pound over the first pound. The table below illustrates how the cost depends on the total weight of the package. Complete the table by matching each total weight to its correct total shipping cost.

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

SHIPPING COSTS

Total Weight (pounds)	Total Cost to Ship (dollars)
2
	2.50
6
	5.50

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: 2; blank 2: 3; blank 3: 4; blank 4: 9

Total Weight (pounds)	Total Cost to Ship (dollars)
2	2
3	2.50
6	4
9	5.50

Let x represent the weight of the package in pounds. The first pound is included in the base fee of $1.50, so $(x - 1)$ represents the number of pounds that cost $0.50 each.

The cost equation can be written as $\text{Cost} = \$1.50 + \$0.50\,(x - 1)$

The cost for a 2-pound package is $\$1.50 + \$0.50(2 - 1) = \$1.50 + \$0.50(1) = \$2.00$
The cost for a 6-pound package is $\$1.50 + \$0.50(6 - 1) = \$1.50 + \$0.50(5) = \$1.50 + \$2.50 = \$4.00$
The weight of a package that costs $2.50 to ship can be found by solving

$\$2.50 = \$1.50 + \$0.50(x - 1)$

$\$1.00 = \$0.50(x - 1)$

$2 = x - 1$

$x = 3 \text{ pounds}$
The weight of a package that costs $5.50 to ship can be found by solving

$\$5.50 = \$1.50 + \$0.50(x - 1)$

$\$4.00 = \$0.50(x - 1)$

$8 = x - 1$

$x = 9 \text{ pounds}$

To check:

Given that the total cost is $3.00 for the initial package

$\$3.00 = \$1.50 + \$0.50(x - 1)$

$\$1.50 = \$0.50(x - 1)$

$3 = x - 1$

$x = 4$ pounds fits the linear pattern of the table

Question analysis

A shipping company charges a base fee plus 50 cents for each pound over the first, and a 4-pound package costs $3.00. Students drag number tiles into a table to complete two costs and two weights, with some tiles left unused. It tests building and using a linear rule, and the wording about the first pound must be applied in every row.

NYS standards: 7.EE.4a two-step equations
6.EE.9 dependent and independent variables
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve word problems with equations like p(x + q) = r. This table needs that rule four times, twice run backward from a cost to a weight, and the phrase about the first pound must be applied every time.
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
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 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

The rule is stated in words and confirmed by the 4-pound example, so the work is applying it consistently in both directions. Charging every pound breaks the $3.00 example at once, since 4 pounds would cost $3.50, and models that ignore the wording leave costs or weights with no matching tile. A shortcut helps: each extra pound adds $0.50 to the $3.00, so the table can be filled by counting up or down from 4 pounds.

Turning words into math moderate Understanding the question
The phrase “$0.50 for each pound over the first pound” means the first pound is covered by the $1.50 fee, so a package of x pounds pays for only x − 1 of them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Four boxes must all be right, two found forward from a weight and two backward from a cost, and the backward rows need the first pound added back. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write the rule as the base fee plus $0.50 times the pounds after the first, check it against the 4-pound example, then use it forward for the costs and backward for the weights.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

BAEH Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Charges $0.50 for every pound, first included, so 2 pounds cost $2.50 and $2.50 ships 2 pounds; no tile fits its $4.50 or 8, so it settles for 4 and 9.
ACFH Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Triples the 2-pound cost for the 6-pound package, $2 × 3 = $6, treating the cost as proportional to weight though the base fee is paid only once.
ACBH Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Charges $0.50 for each of the 5 pounds over the first for the 6-pound package but stops there, leaving out the $1.50 base fee and entering 2.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Shipping cost: $C = 1.50 + 0.50 \times (w - 1)$; C is the total cost in dollars, and w is the weight in pounds. The $1.50 base fee is paid once, and it covers the first pound. Each pound after the first costs $0.50. There are w − 1 of those pounds.

Sort out the question
The base fee of $1.50 covers the first pound. Each pound over the first adds $0.50. A 4-pound package costs $3.00.
You fill four boxes: the costs for 2 and 6 pounds, and the weights that cost $2.50 and $5.50. Each tile is used once at most, and four tiles are not used.
Test the rule on 4 pounds
Check the rule with the example. A 4-pound package has 3 pounds over the first:
$C = 1.50 + 0.50 \times (4 - 1) = 1.50 + 1.50 = 3.00$
That is $3.00, so the rule fits.
Watch out: charging $0.50 for every pound, the first one too, gives $3.50 for 4 pounds. That does not match.
Find the two costs
For 2 pounds, 1 pound is over the first:
$1.50 + 0.50 \times 1 = 2.00$
For 6 pounds, 5 pounds are over the first:
$1.50 + 0.50 \times 5 = 1.50 + 2.50 = 4.00$
The costs are the tiles 2 and 4.
Watch out: 6 pounds is 3 times 2 pounds, but the cost is not 3 times $2. The base fee is paid once. Tripling gives the tile 6. Leaving out the base fee gives the tile 2.5.
Work back to the two weights
For $2.50, take off the base fee: $2.50 - 1.50 = 1.00$ . At $0.50 a pound, $1.00 pays for 2 extra pounds. Add the first pound: $2 + 1 = 3$ pounds.
For $5.50: $5.50 - 1.50 = 4.00$ . $4.00 pays for 8 extra pounds. Add the first pound: $8 + 1 = 9$ pounds.
Watch out: forgetting to add back the first pound gives 2 and 8.
Fill in the four boxes
From top to bottom, the boxes get 2, 3, 4 and 9. The tiles 2.5, 3.5, 6 and 7 are not used. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Each extra pound adds $0.50, so count from 4 pounds at $3.00. Going down: 3 pounds cost $2.50 and 2 pounds cost $2.00. Going up: 5 pounds $3.50, 6 pounds $4.00, 7 pounds $4.50, 8 pounds $5.00 and 9 pounds $5.50. All four boxes match.

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 D Math, page 119, question 46.

Question 47 of 50

2026 Practice Test D Math, question 47

Question

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

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

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

Show the answer

Correct answer: A

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

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

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

Question analysis

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

NYS standards: 7.G.3 slicing three-dimensional solids
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Describing the shape made by slicing a prism parallel to its base is exactly the grade 7 standard. This is its simplest case, a single slice of a labeled right prism, typical of a classroom example.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 2, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph minor Understanding the question
The prism is drawn standing on end, 6 inches tall on a 2-by-2 base, so the cut has to be pictured in a flat drawing where the tall front face is the most visible shape. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Students must know that a slice parallel to the base of a prism has the same shape and size as the base, which here is the square at the bottom. (Anderson & Krathwohl, 2001)

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

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Expects a slice to cut off a corner and make a triangle, as a cut through a pyramid might, not a cut parallel to this square base.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Describes a vertical slice, which runs down the 6-inch height and makes a 2-by-6 rectangle, mixing up horizontal and vertical.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Builds a triangle from the labels in the figure, one 2-inch and two 6-inch sides, a shape no horizontal cut of this prism can make.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The figure is a right rectangular prism, a box shape whose faces are all rectangles. It stands on end. The two edges along the bottom are labeled 2 in. The tall edge is labeled 6 in.
You need the shape you get from a horizontal slice. Horizontal means flat, like the floor or a tabletop.
Find the base
The base is the face the prism stands on. The two bottom edges you can see are both labeled 2 in. The right-angle mark on the top face shows its corners are square. The top and the bottom match. So the base is a square with 2-inch sides.
Picture the slice
A horizontal slice cuts straight across the prism, parallel to the base. The walls of the prism go straight up, so the slice is a copy of the base. It is a square with 2-inch sides, at whatever height you cut.
The 6-inch height does not matter. It only tells how tall the stack of squares is.
Rule out the other shapes
Watch out: a 2-inch by 6-inch rectangle, choice C, comes from a vertical slice. That cut goes straight down, through the 6-inch height. A triangle, choice B or D, needs a slanted cut through a corner.
So the answer is A, a square that has 2-inch sides.

Check your answer

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

More questions like this

The original DOE question

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

Question 48 of 50

2026 Practice Test D Math, question 48

Question

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

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

Show the answer

Correct answer: C

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

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

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

Sides:

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

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

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4 equations and inequalities from problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 ratio problems usually supply numbers to work with. This one gives only a relationship between two perimeters, so students must bring in the number of sides of each shape themselves and keep the ratio in the order asked.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question turns a statement about perimeters into a ratio of sides, and the natural guess, that the sides keep the perimeters’ 2 to 1, ignores that the square shares its perimeter among 4 sides and the triangle among 3. Choosing a triangle perimeter such as 12 makes the comparison concrete. The order of the ratio still has to be checked at the end, because its reverse is listed too.

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

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 49 of 50

2026 Practice Test D Math, question 49

Question

Anise has a piece of cloth measuring 45 inches by 75 inches. She wants to cut it into squares of equal size, without any cloth left over. What is the greatest possible side length, in inches, of one of those squares?

Show the answer

Correct answer: C

Given that each piece will be a square, the length and width will be the same. The greatest possible length can be determined by finding the greatest common factor of the length and width of the original piece of cloth.

Begin by finding the prime factorization of the length (75 in.) and width (45 in.).

$45 = 3 \times 3 \times 5$

$75 = 3 \times 5 \times 5$

The common factors can be seen as $3 \times 5$ .

The greatest possible side length, in inches, will be $3 \times 5 = 15$ .

Question analysis

Anise wants to cut a 45-inch by 75-inch piece of cloth into equal squares with nothing left over. This factors question asks for the greatest possible side length of each square, in inches. Students must see that the side has to divide both lengths evenly, then find the largest number that does.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Greatest common factors of two numbers up to 100 are grade 6 content. The cloth setting asks students to recognize that this is what is needed, but the work is one short factor search, below typical grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the side is seen as a common factor, the work is short, since 45 and 75 share both 3 and 5. Students who spot one shared factor and stop reach a listed choice, and those who think only of the largest square that fits use the full 45-inch width. Checking that 45 and 75 both divide by the chosen side, and that no larger number does, closes it.

Turning words into math minor Understanding the question
The words “squares of equal size, without any cloth left over” mean the side must divide both 45 and 75 exactly; nothing names the greatest common factor. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding the greatest common factor of 45 and 75 calls for prime factorization or a list of factors, a grade 6 skill the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Write each length as a product of primes, multiply the primes they share, and check that both 45 and 75 divide evenly by the result.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds a factor that 45 and 75 share, 3, and stops without combining it with the 5 they also share.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Spots the common factor 5, since both lengths end in 5, and stops before multiplying it by the 3 they also share.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Uses the cloth’s full width, 45 inches, as the side, the largest square that fits, but 45 does not divide 75, leaving a 45-by-30 strip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor: $\text{GCF} = \text{product of the prime factors both numbers share}$; A factor divides a number evenly. The greatest common factor, or GCF, is the biggest number that divides both. Write each number as a product of primes, numbers whose only factors are 1 and themselves. Then multiply the primes they share.

Sort out the question
The cloth is 45 inches by 75 inches. It is cut into equal squares, with no cloth left over. You need the greatest side the squares can have.
No cloth left over means the side must fit evenly into 45 and into 75. So the side is a common factor of both. The greatest one is the GCF.
Break each number into primes
Write each length as a product of prime numbers:
$45 = 3 \times 3 \times 5$
$75 = 3 \times 5 \times 5$
The two lists share one 3 and one 5.
Multiply the shared primes
Multiply the primes the two numbers share:
$\text{GCF} = 3 \times 5 = 15$
The squares can be 15 inches on a side.
Watch out: 3 and 5 are each common factors, choices A and B. But they are not the greatest. Use every prime the numbers share.
Rule out the full width
Watch out: a 45-inch square, choice D, fits the width. But 75 ÷ 45 = 1 with 30 left over. That leaves a strip of cloth 45 inches by 30 inches.
So the answer is C, 15.

Check your answer

Divide each length by 15: $45 \div 15 = 3$ and $75 \div 15 = 5$ . So the cloth cuts into 3 rows of 5 squares, with nothing left. Both divisions come out even.

More questions like this

The original DOE question

Question 49 as shown on the DOE’s digital test, page 122 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 122, question 49.

Question 50 of 50

2026 Practice Test D Math, question 50

Question

This table shows the number of birds that came to a bird feeder on Monday and Tuesday.

BIRDS AT BIRD FEEDER

Day	Number of Birds
Monday	30
Tuesday	24

What was the percent decrease in the number of birds from Monday to Tuesday?

A6%
B20%
C24%
D25%

Show the answer

Correct answer: B

To determine the percent decrease from Monday to Tuesday, first calculate the decrease in the number of birds at the feeder.

$30 - 24 = 6$

The difference in the number of birds at the feeder divided by the number of birds on Monday will yield the percent decrease from Monday to Tuesday.

$\frac{6}{30} = 0.20$

0.20 represents 20%.

There was a 20% decrease from Monday to Tuesday.

Question analysis

A table shows that 30 birds came to a bird feeder on Monday and 24 on Tuesday. This percent change question asks for the percent decrease from Monday to Tuesday. Students find the change and compare it with one day's count, and choosing the right day as the base decides the result.

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Percent decrease is core grade 7 work, and this is a plain example: one subtraction and one division with friendly numbers. The only real demand is measuring the change against the right day's count.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The subtraction is trivial, so the question is about the base. Both counts sit in the table, and dividing by the later one gives an equally neat percent that is easy to accept. Saying “a decrease from Monday” and putting Monday’s 30 under the change avoids it, and checking that 20% of 30 is 6 confirms the result.

A very close wrong answer minor Choosing the answer
Dividing the drop of 6 by Monday’s 30 gives 20% and by Tuesday’s 24 gives 25%, two round, listed percents only 5 points apart, so only the choice of base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that a percent decrease is the change divided by the starting amount, here Monday’s count, a rule the question does not state. (Anderson & Krathwohl, 2001)

Strategy: Subtract to find how many fewer birds came on Tuesday, divide by Monday's count, the starting amount, and change the fraction to a percent.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the decrease of 6 birds and reads it as 6 percent, as if the count had dropped from 100 rather than from 30.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Reads Tuesday’s 24 birds as a percent, taking a number straight from the table without working out the change at all.
D Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the decrease of 6 by Tuesday’s 24 birds instead of Monday’s 30, measuring the change from the later day: 6/24 = 25%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent decrease: $\text{percent decrease} = \frac{\text{old} - \text{new}}{\text{old}} \times 100$; Old is the starting amount, here Monday’s count. New is the later amount, Tuesday’s count. The top is how much it dropped. You divide by the starting amount because a change is measured from where you started.

Sort out the question
The table shows 30 birds on Monday and 24 birds on Tuesday. You need the percent decrease from Monday to Tuesday. From Monday means Monday is the starting amount.
Find the drop
Subtract Tuesday’s count from Monday’s:
$30 - 24 = 6$
There were 6 fewer birds.
Watch out: 6 is a number of birds, not a percent. Calling it 6% leads to choice A.
Compare the drop with Monday
Use the formula. Divide the drop by Monday’s count, then multiply by 100:
$\frac{6}{30} \times 100 = 0.20 \times 100 = 20$
The number of birds went down by 20%.
Watch out: dividing by Tuesday’s 24 gives $\frac{6}{24} = 0.25$ , or 25%, choice D. Measure the change from the starting day. Choice C, 24%, just copies a number from the table.
So the answer is B, 20%.

Check your answer

Take 20% of Monday’s count: $0.20 \times 30 = 6$ . Then $30 - 6 = 24$ , which is Tuesday’s count. The answer checks out.

More questions like this

The original DOE question

Question 50 as shown on the DOE’s digital test, page 123 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 123, question 50.