SHSAT Sunset

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Math

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Question 1 of 120

2026 Practice Test A Math, question 8

Question

Liam and Kevin each collect stamps. Liam has twice as many stamps as Kevin. Liam gives 8 of his stamps to Kevin. Liam now has 12 more stamps than Kevin.

How many stamps did Liam start with?

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

An equation that relates the number of stamps originally in Liam’s and Kevin’s collections is L=L = KK.

After Liam gives 8 stamps to Kevin, the equation becomes L−8=L - 8 = .

By solving the equation, it can be determined that Liam started with stamps.

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: 2; blank 2: K + 20; blank 3: 56

Liam originally has twice as many stamps as Kevin, so the equation is L=2KL = 2K.

After Liam gives 8 stamps to Kevin, the equation becomes L−8=(K+8)+12L - 8 = (K + 8) + 12, which simplifies to L−8=K+20L - 8 = K + 20.

Substitute L=2KL = 2K into the equation: 2K−8=K+202K - 8 = K + 20.

Solve for K:

2K−K=20+82K - K = 20 + 8

K=28K = 28

Substitute to find L: L=2K=2×28=56L = 2K = 2 \times 28 = 56.

So, Liam started with 56 stamps.

Question analysis

This word problem about two stamp collections is answered with three drop-down menus: an equation for the starting counts, an equation for after Liam gives Kevin 8 stamps, and the number Liam started with. Students turn "twice as many" and "12 more" into algebra and solve. Tracking both boys' counts before and after the gift is the tricky part, and every menu must be right.

NYS standards
  • 8.EE.8c systems in real-world problems
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve one equation in one unknown. The menus lay out the equations, but the problem still ties two unknowns together before and after a gift, a grade 8 system, and all three menus must be right.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 5, 86 words.
Students, last 30 days
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 do much of the setting up, since each sentence already shows the shape of its equation. The second menu is the hard one: the 8 stamps leave Liam and join Kevin before the gap of 12 is measured. Solving gives Kevin's 28 first, and K + 20 then equals 48, Liam's count after the gift, which waits on the last menu beside his starting 56. Testing each starting count against the story is a quick check on the last menu.

  1. Turning words into math moderate Understanding the question

    The gift changes both counts, so "Liam now has 12 more stamps than Kevin" compares Liam's L − 8 with Kevin's K + 8, and the second menu's K + 20 holds a 20 that appears nowhere in the stem. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of moderate Working it out

    Each boy has a count before and after the gift, four numbers in all, and all three menus must be right; the last asks for Liam's start, not the 48 he has after giving 8 away. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let K be Kevin's starting count, write each boy's count after the gift in terms of K, and set Liam's 12 above Kevin's; testing each number in the last menu is a fast check.

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.
  • 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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Twice as many as
L=2KL = 2K
L is the number of stamps Liam starts with, and K is the number Kevin starts with. Twice as many means 2 times as many, so Liam’s count is 2 times Kevin’s. Because L and 2K are equal, you can put 2K in place of L in any equation.
Doing the same to both sides
if a=b, then a+c=b+c\text{if } a = b \text{, then } a + c = b + c
a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. Add the same thing to both sides, or take the same thing away from both, and the scale stays level. This is how you move a term to the other side.

  1. Sort out the question

    Let L be Liam’s stamps at the start and K be Kevin’s. You know three facts:

    • Liam has twice as many stamps as Kevin.
    • Liam gives 8 of his stamps to Kevin.
    • After that, Liam has 12 more stamps than Kevin.

    The first menu finishes the starting equation. The second finishes the equation after the gift. The third is Liam’s count at the start.

  2. Fill in the first menu

    Twice as many means 2 times as many. Liam has twice as many as Kevin, so:

    L=2KL = 2K

    Pick 2 in the first menu.

    Watch out: the 0.5 in the menu turns the fact around. L=0.5KL = 0.5K would give Liam half as many as Kevin. Liam has more, so his count must be the bigger one.

  3. Write both counts after the gift

    The gift changes both counts. Liam loses 8 stamps, and Kevin gains those same 8.

    • Liam now has L−8L - 8 stamps.
    • Kevin now has K+8K + 8 stamps.

    Liam has 12 more than Kevin. So Liam’s count equals Kevin’s count plus 12:

    L−8=(K+8)+12=K+20L - 8 = (K + 8) + 12 = K + 20

    Pick K + 20 in the second menu.

    Watch out: the 12 goes on Kevin’s count, not Liam’s. Picking L + 12 adds it to the wrong boy.

  4. Solve for K

    Now use the first equation. Since L=2KL = 2K, put 2K in place of L:

    2K−8=K+202K - 8 = K + 20

    Subtract K from both sides. Then add 8 to both sides. Each move keeps the scale level.

    K−8=20K - 8 = 20

    K=28K = 28

    Kevin started with 28 stamps.

  5. Find Liam’s starting count

    Liam started with twice as many as Kevin:

    L=2K=2×28=56L = 2K = 2 \times 28 = 56

    Liam started with 56 stamps.

    Watch out: K+20=48K + 20 = 48 is Liam’s count after the gift. The menu offers 48, but the question asks about the start.

    So the menus read 2, K + 20 and 56.

Check your answer

At the start, Liam has 56 stamps and Kevin has 28, and 56 is twice 28. After the gift, Liam has 48 and Kevin has 36. 48−36=1248 - 36 = 12, so Liam has 12 more. Every fact fits.

More questions like this

The original DOE question
Question 8 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 A Math, page 79, question 8.

Question 2 of 120

2026 Practice Test A Math, question 10

Question

Three expressions are given:

−2(2.5n+4)-2(2.5n + 4)

5−n5 - n

3(2n+7)3(2n + 7)

Which of these are equivalent to the sum of these three expressions?

Select the two 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.

  • A5−n+6n+21−5n−85 - n + 6n + 21 - 5n - 8
  • B5−n+6n+21−5n+85 - n + 6n + 21 - 5n + 8
  • C5−n+5n+21−5n−85 - n + 5n + 21 - 5n - 8
  • D18−n18 - n
  • E18
  • F34
Show the answer

Correct answer: A and E

Add the three expressions: 5−n+3(2n+7)+(−2)(2.5n+4)5 - n + 3(2n + 7) + (-2)(2.5n + 4).

Distribute 3 and −2: 5−n+6n+21−5n−85 - n + 6n + 21 - 5n - 8.

Combine the variable terms and combine the constants: (−n+6n−5n)+(5+21−8)=18(-n + 6n - 5n) + (5 + 21 - 8) = 18.

Question analysis

This algebra question gives three expressions, one with a decimal coefficient and a negative factor, and asks which two of six choices are equivalent to their sum. It tests distributing, combining like terms and recognizing equivalent expressions. Signs from the negative factor are easy to lose, and several choices are built from exactly those slips.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.4 identifying equivalent expressions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Expanding and combining linear expressions with a negative factor and a decimal coefficient is standard grade 7 work. Choosing the two equivalent ones out of six, where each common slip appears as a choice, takes more care than one simplification.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 8, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The six choices form three consistent pairs, each an expansion with its simplified total: the right pair, a pair built on a sign slip with −2 × 4 that totals 34, and a pair built on 3 × 2n read as 5n that totals 18 − n. A student who makes either slip finds two choices that agree with each other and seem to confirm the work. The question rewards one careful simplification and the confidence to accept a sum with no n left.

  1. Error-prone arithmetic moderate Working it out

    The first expression puts a negative factor in front of a decimal term: −2 times 2.5n is −5n, and the −2 must reach the 4 as well, giving −8, not +8. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The first two expansions are identical except for their last sign, − 8 against + 8, and the third differs from the first only in 5n for 6n, so only a term-by-term check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    Two distributions come first, then the n terms and the constants are collected separately, and six choices are compared with the result; both right ones must be picked to score. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    One right answer is the expansion before any terms are combined and the other is the plain number 18, which has to be accepted as equal to a sum in which every expression held n. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute each factor across its parentheses, keeping the sign with every product, then combine the n terms and the constants separately and test each of the six choices against the result.

Traps

  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
  • 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.
  • Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number in front of the parentheses, and b and c are the terms inside. You multiply a by every term inside, not only the first one. If a is negative, its minus sign goes into every product. It works because a groups of b + c is a groups of b plus a groups of c.
Multiplying signed numbers
(−a)×b=−(a×b)(-a) \times b = -(a \times b)
a and b are positive numbers, so −a is negative. A negative number times a positive number is negative. Use it whenever a minus sign sits in front of the parentheses: −2 times 4 is −8, not 8.
Combining like terms
an+bn=(a+b)nan + bn = (a + b)n
Like terms have the same letter part, such as 6n and −5n. a and b are their coefficients, the numbers in front of n. Add the coefficients and keep the n. Constants, the plain numbers, combine only with other constants.

  1. Sort out the question

    An expression is a math phrase made of numbers, letters and signs, with no equal sign. The sum of the three expressions means you add all of them:

    −2(2.5n+4)+(5−n)+3(2n+7)-2(2.5n + 4) + (5 - n) + 3(2n + 7)

    You need the two choices that are equivalent to this sum. Equivalent means equal for every value of n. Some choices show the sum before it is simplified. Others show the result.

  2. Distribute the 3

    To distribute means to multiply the number outside the parentheses by each term inside.

    3(2n+7)=3×2n+3×7=6n+213(2n + 7) = 3 \times 2n + 3 \times 7 = 6n + 21

    Watch out: 3 times 2n is 6n. Adding 3 and 2 instead gives 5n. That slip leads to choice C, and then to 18 − n, choice D.

  3. Distribute the −2

    Now multiply −2 by both terms. A negative times a positive is negative, so both products are negative.

    −2(2.5n+4)=−2×2.5n+(−2)×4=−5n−8-2(2.5n + 4) = -2 \times 2.5n + (-2) \times 4 = -5n - 8

    Watch out: the −2 multiplies the 4 too. Writing +8 instead of −8 leads to choice B. That sum works out to 34, choice F.

  4. Write the whole sum

    Put the three pieces together, starting with 5 − n:

    5−n+6n+21−5n−85 - n + 6n + 21 - 5n - 8

    This matches choice A exactly, term by term. So A is one of the answers.

  5. Combine like terms

    Gather the n terms first. Remember that −n means −1 times n.

    −n+6n−5n=(−1+6−5)n=0n=0-n + 6n - 5n = (-1 + 6 - 5)n = 0n = 0

    Then gather the constants:

    5+21−8=185 + 21 - 8 = 18

    The n terms cancel out, so the whole sum is just 18.

    Watch out: a sum with no n left is fine. Keeping an extra −n leads to choice D.

  6. Pick the two answers

    Choice A is the sum before you combine like terms. Choice E, 18, is the sum after you combine them. Both equal the sum for every value of n. So the answers are A and E.

Check your answer

Try a value, such as n = 2, in each expression. −2(2.5×2+4)=−2×9=−18-2(2.5 \times 2 + 4) = -2 \times 9 = -18. 5−2=35 - 2 = 3. 3(2×2+7)=3×11=333(2 \times 2 + 7) = 3 \times 11 = 33. The sum is −18+3+33=18-18 + 3 + 33 = 18, which matches E. Choice D would give 18 − 2 = 16 instead.

More questions like this

The original DOE question
Question 10 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 A Math, page 81, question 10.

Question 3 of 120

2026 Practice Test A Math, question 17

Question

Consider this expression:

|(−8)−(−3)+6|−|6−11||(-8) - (-3) + 6| - |6 - 11|

Create an equivalent expression.

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

|| − ||

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: 1; blank 2: −5

To create an expression equivalent to |(−8)−(−3)+6|−|6−11||(-8) - (-3) + 6| - |6 - 11|:

Step 1: Simplify the first absolute value term.

(−8)−(−3)=−8+3=−5(-8) - (-3) = -8 + 3 = -5

−5+6=1-5 + 6 = 1

Simplify the second absolute value term.

6−11=−56 - 11 = -5

Therefore the equivalent expression is

|1|−|−5||1| - |-5|.

Question analysis

This absolute value question shows an expression with two pairs of bars and asks students to build an equivalent one by placing number tiles in two boxes. It tests subtracting negative integers and understanding what absolute value bars do. Not every tile is used, and several tiles match the results of common sign mistakes.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simplifying signed integers inside absolute value bars, including subtracting a negative, is core grade 7 work. This asks only for the values inside the bars, not the final number, but one sign slip leads straight to a wrong tile.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every move here involves a sign, and the tiles are built from the usual misreadings of absolute value: 11 for making each number positive before combining, 17 for erasing every minus sign. A slip on − (−3) gives −5 for the first pair, but the second pair needs the only −5 tile, so that slip leaves a box with no tile to fit. Students who simplify each inside completely, and leave the bars untouched, finish quickly.

  1. Error-prone arithmetic moderate Working it out

    Inside the first bars, subtracting −3 must become adding 3, so −8 + 3 = −5, and adding 6 then crosses zero to reach 1; reading − (−3) as − 3 gives −5 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Two pairs of bars are simplified separately, and both boxes need the right tile, kept in the original order, before the answer is right. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    Each box takes what is left inside a pair of bars, not its absolute value, so the second box needs the tile −5 rather than a 5, and the expression's value, −4, is never entered. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −3 as adding 3, then simplify the inside of each pair of bars separately, left to right; each box replaces the inside of one pair, in the same order.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a is any number, and b is a positive number, so −b is negative. Subtracting a negative number is the same as adding its opposite, the same number with the other sign. Think of money: taking away a debt of $3 leaves you $3 richer.
Absolute value
|a|=the distance from a to 0|a| = \text{the distance from } a \text{ to } 0
The bars around a number stand for its absolute value. a is any number. The absolute value is how far a is from 0 on a number line, so it is never negative. Both |5| and |−5| equal 5. Always simplify inside the bars first.

  1. Sort out the question

    The expression has two pairs of absolute value bars. The new expression keeps both pairs, with a box inside each one:

    |□|−|□||\square| - |\square|

    Each box gets the number you reach when you simplify inside that pair of bars. The bars stay, so you do not take the absolute value yet. Some tiles are not used.

  2. Simplify inside the first bars

    Inside the first bars is (−8)−(−3)+6(-8) - (-3) + 6. Subtracting −3 is the same as adding 3. Work from left to right:

    (−8)−(−3)=−8+3=−5(-8) - (-3) = -8 + 3 = -5

    −5+6=1-5 + 6 = 1

    The first box gets the tile 1.

    Watch out: the bars do not make each number positive before you combine. Changing −8 and −3 to 8 and 3 first gives 8 − 3 + 6 = 11. Dropping every minus sign gives 8 + 3 + 6 = 17. Both are tiles, and both are wrong.

  3. Simplify inside the second bars

    Inside the second bars is 6−116 - 11. You take a bigger number from a smaller one, so the result is negative:

    6−11=−56 - 11 = -5

    The second box gets the tile −5.

    Watch out: do not change −5 to 5. The bars are still there in the new expression, so the box holds what is inside them.

  4. Keep the boxes in order

    The first bars came first, so 1 goes in the first box and −5 in the second. The new expression has the same value as the old one:

    |1|−|−5|=1−5=−4|1| - |-5| = 1 - 5 = -4

    Swapping the tiles gives |−5|−|1|=5−1=4|-5| - |1| = 5 - 1 = 4, which is not equal. So the boxes get 1 and −5, making |1| − |−5|.

    On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Use a number line for the first pair. Start at −8 and move 3 to the right to reach −5. Then move 6 more to the right to reach 1. For the second pair, start at 6 and move 11 to the left. You land on −5. Both boxes check out.

More questions like this

The original DOE question
Question 17 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 A Math, page 88, question 17.

Question 4 of 120

2026 Practice Test A Math, question 27

Question

Complete the steps to show one way to solve the equation 24=−0.5(x+8)24 = -0.5(x + 8) for x.

Move the correct answer to each box. Each answer may be used more than once. Not all answers will be used.

24=−0.5(x+8)24 = -0.5(x + 8)

24=−0.5x+24 = -0.5x +

=−0.5x= -0.5x

=x= x

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. A tile can be used more than 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: −4; blank 2: 28; blank 3: −56

Solution:

Answer for second equation: −4

Answer for third equation: 28

Answer for fourth equation: −56

24=−0.5(x+8)24 = -0.5(x + 8)

Distribute the −0.5 on the right side of the equation: 24=−0.5x−424 = -0.5x - 4.

Add 4 to both sides of the equation to eliminate the −4 on the right side: 28=−0.5x28 = -0.5x.

Divide both sides by −0.5: −56=x-56 = x.

Question analysis

Students complete a step-by-step solution of the equation 24 = −0.5(x + 8) by dragging number tiles into three boxes: one after distributing, one after collecting the constants, and one for x. It tests solving a two-step linear equation with a negative decimal. Tiles may be reused, and many come in positive and negative pairs, so every sign counts.

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving equations of the form p(x + q) = r with rational numbers is named in the grade 7 standards. Here the steps are laid out, but the negative decimal factor forces a sign decision at every one of the three boxes.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The layout settles the method, so the question comes down to one sign at each box. The printed plus before the first box nudges students toward the positive 4, and the pool then supplies every number that follows from it, down to −40. Students who distribute correctly can still lose the item at the last box by dividing 28 by 0.5. Substituting −56 back, −0.5 × (−48) = 24, confirms the final box.

  1. Error-prone arithmetic moderate Working it out

    Two of the three boxes turn on a negative decimal: −0.5 times 8 must give −4, and dividing 28 by −0.5 means multiplying by −2, which doubles the size and makes it negative. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    The pool pairs −4 with 4, −40 with 40 and −56 with 56, so a box filled with the right size and the wrong sign looks just as finished; estimating cannot separate −56 from 56. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    The three boxes form a chain and all three must be right, so a slip in the first carries through; the pool even holds the whole wrong chain 4, 20 and −40. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The second line prints a plus sign before its box, so the distributed −0.5x − 4 must be written as −0.5x + (−4), with the negative tile in the box. (Enright, Morley & Sheehan, 2002)

Strategy: Multiply −0.5 by both x and 8, undo the constant term by doing the opposite operation to both sides, then divide by −0.5, which is the same as multiplying by −2.

Traps

  • 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.)
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number in front of the parentheses, and b and c are the terms inside. Multiply a by each term inside. If a is negative, the minus sign goes into every product. It works because a groups of b + c is the same as a groups of b plus a groups of c.
Signs when you multiply or divide
(−)×(+)=(−)(−)×(−)=(+)(-) \times (+) = (-) \qquad (-) \times (-) = (+)
A negative number times a positive number gives a negative result. A negative times a negative gives a positive result. The same rules hold when you divide. Use them at every step where a negative number appears.
Doing the same to both sides
if a=b, then a+c=b+c and ac=bc\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). An equation is like a balanced scale. Doing the same thing to both sides keeps it level. That is how you get x by itself.

  1. Sort out the question

    You solve 24=−0.5(x+8)24 = -0.5(x + 8) by filling three boxes:

    • Box 1 comes after −0.5x+-0.5x +, once the parentheses are gone.
    • Box 2 is the number equal to −0.5x-0.5x.
    • Box 3 is the value of x.

    A tile can be used more than once, and some tiles are not used.

  2. Distribute the −0.5

    Multiply −0.5 by each term inside the parentheses. A negative times a positive is negative:

    −0.5×x=−0.5x-0.5 \times x = -0.5x

    −0.5×8=−4-0.5 \times 8 = -4

    So 24=−0.5x+(−4)24 = -0.5x + (-4). The line already shows a plus sign, so box 1 gets −4.

    Watch out: the −0.5 multiplies the 8 too. Writing 4 instead starts a wrong chain: 4, then 20, then −40. Every number in that chain is a tile.

  3. Add 4 to both sides

    You want −0.5x alone on one side, and the −4 is in the way. Adding 4 undoes subtracting 4, so add 4 to both sides:

    24+4=−0.5x−4+424 + 4 = -0.5x - 4 + 4

    28=−0.5x28 = -0.5x

    Box 2 gets 28.

    Watch out: taking 4 away from 24 instead gives 20, another tile.

  4. Divide both sides by −0.5

    Now x is multiplied by −0.5. Dividing undoes multiplying, so divide both sides by −0.5:

    28÷(−0.5)=−5628 \div (-0.5) = -56

    Dividing by 0.5 is the same as multiplying by 2, since every whole holds two halves. A positive divided by a negative is negative. So x=−56x = -56.

    Watch out: dropping the minus sign gives 56, which is also a tile.

  5. Name the three tiles

    From top to bottom, the finished steps read:

    24=−0.5x+(−4)24 = -0.5x + (-4)

    28=−0.5x28 = -0.5x

    −56=x-56 = x

    So the boxes get −4, 28 and −56, in that order. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Put x=−56x = -56 back into the first equation. −56+8=−48-56 + 8 = -48. Then −0.5×(−48)=24-0.5 \times (-48) = 24, since a negative times a negative is positive. That gives 24, the left side, so −56 is right.

More questions like this

The original DOE question
Question 27 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 A Math, page 98, question 27.

Question 5 of 120

2026 Practice Test A Math, question 32

Question

What is the value of x in the equation 6x=9x+136x = 9x + 13?

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: x = -13/3

Solution:

Start with 6x=9x+136x = 9x + 13.

Subtract 9x9x from both sides to move all terms involving x to one side.

6x−9x=136x - 9x = 13

Combine like terms.

−3x=13-3x = 13

Divide both sides by −3 to solve for x.

x=−133x = -\frac{13}{3}

x=−413x = -4\frac{1}{3}

Question analysis

This equation question asks for the value of x when 6x equals 9x plus 13, typed into a box after "x =". Students solve a linear equation with the variable on both sides by gathering the x terms and dividing. With no answer choices to check against, keeping the sign of each moved term is the main challenge.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.NS.2b rational number division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 The variable appears on both sides, a grade 8 topic, yet collecting it takes one subtraction and one division, much like a grade 7 two-step equation. The typed answer, with no choices to compare, adds some risk.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two moves solve the equation, so the method is not in doubt; the points are lost in the result. Subtracting 6x instead, which gives 0 = 3x + 13, works as well but moves the sign onto the 13, and either route ends in a division that must keep the negative. The answer is not a whole number, which unsettles students who expect one, and a switch to decimals leads to a rounded entry that is not exact.

  1. Error-prone arithmetic moderate Working it out

    Collecting the x terms gives 6x − 9x = −3x, a negative coefficient, and 13 ÷ (−3) is a negative fraction that does not come out whole, so both the sign and the fraction must survive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The answer is typed after the printed x =, with no choices to compare, and its decimal form, −4.333…, repeats forever, so only a fraction or mixed number is exact. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract 9x from both sides so that x appears once, divide both sides by the coefficient left on x, and check by substituting the result into both sides of the original equation.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a=b, then a−c=b−c and ac=bc\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 or term (not zero when you divide). Taking the same thing away from both sides, or dividing both by the same number, keeps the sides equal. That is how you get x by itself.
Dividing by a negative number
a−b=−ab\frac{a}{-b} = -\frac{a}{b}
a and b are positive numbers, so −b is negative. A positive number divided by a negative number is negative. You can write the minus sign in front of the whole fraction.
Changing a fraction to a mixed number
ab=qrb\frac{a}{b} = q\frac{r}{b}
Use this when the top of a fraction is bigger than the bottom. Divide a by b. q is the number of whole times b fits into a. r is the remainder, what is left over. The remainder goes on top of the leftover fraction.

  1. Sort out the question

    You need the value of x that makes 6x=9x+136x = 9x + 13 true. The letter x appears on both sides, so first gather the x terms on one side. There are no choices to check against, so work carefully.

  2. Move the x terms to one side

    Subtract 9x from both sides. This clears the x term off the right side:

    6x−9x=9x+13−9x6x - 9x = 9x + 13 - 9x

    −3x=13-3x = 13

    Since 6 − 9 = −3, the left side is −3 times x.

  3. Divide both sides by −3

    x is multiplied by −3, so divide both sides by −3:

    x=13−3=−133x = \frac{13}{-3} = -\frac{13}{3}

    A positive divided by a negative is negative, so x is negative.

    Watch out: dropping the minus sign gives 13/3, which is wrong. Also, the number you divide by goes on the bottom. Writing −3/13 turns the fraction upside down.

  4. Write it as a mixed number

    A mixed number is a whole number with a fraction, like 4 1/3. Divide 13 by 3. The 3 fits 4 times, with 1 left over.

    133=413\frac{13}{3} = 4\frac{1}{3}

    So x=−413x = -4\frac{1}{3}.

    Watch out: as a decimal, −13/3 is −4.333…, and the 3s never end. Rounding to −4.33 is close, but it is not exact. Keep the fraction.

  5. Enter the answer

    The box already shows x =, so you enter only the number. On the test, press the minus key first. Then press the fraction key, a small box over a small box. Type 13 in the top box and 3 in the bottom box. On this site, type −13/3 or −4 1/3.

    So the finished answer reads x = −13/3. You type only the −13/3, which is the same as −4 1/3.

Check your answer

Put x=−133x = -\frac{13}{3} into both sides. The left side is 6×(−133)=−266 \times \left(-\frac{13}{3}\right) = -26. The right side is 9×(−133)+13=−39+13=−269 \times \left(-\frac{13}{3}\right) + 13 = -39 + 13 = -26. Both sides equal −26, so the answer is right.

More questions like this

The original DOE question
Question 32 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 A Math, page 103, question 32.

Question 6 of 120

2026 Practice Test A Math, question 37

Question

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

Select all the correct answers.

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

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

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

Correct answer: B, E, and F

Solution:

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

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

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

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

Check all seven points:

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

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

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

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

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

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

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

Question analysis

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

NYS standards
  • 7.RP.2b constant of proportionality
  • 7.RP.2c proportional relationship equations
  • 6.RP.3a tables of equivalent ratios
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a unit rate and deciding which points fit a proportional relationship are grade 7 skills, and the arithmetic here is easy. What adds demand is testing all seven points with no count given, since one miss or extra pick loses the item.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Every choice must be checked moderate Choosing the answer

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

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

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

  3. A very close wrong answer minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the price of one notebook

    Split the $12 equally among the 4 notebooks:

    12÷4=312 \div 4 = 3

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

    y=3xy = 3x

  3. Test each point

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

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

  4. Watch for look-alike points

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

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

  5. Name the points

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

Check your answer

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

More questions like this

The original DOE question
Question 37 as shown on the DOE’s digital test, page 108 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 108, question 37.

Question 7 of 120

2026 Practice Test A Math, question 43

Question

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

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

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

-5-4-3-2-1012345

Show the answer

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

Solution:

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

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

−3≤2x−5-3 \leq 2x - 5

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

Divide both sides by 2: 1≤x1 \leq x

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

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

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality and graphing its solution set is a named grade 7 skill. This one puts the variable on the right of the symbol and asks students to build the graph themselves rather than pick it from four.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

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

  2. Many steps minor Working it out

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

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

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

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

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Add 5 to both sides

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

    −3+5≤2x−5+5-3 + 5 \leq 2x - 5 + 5

    2≤2x2 \leq 2x

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

  3. Divide both sides by 2

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

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

    1≤x1 \leq x

  4. Read it with x first

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

    x≥1x \geq 1

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

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

  5. Choose a closed dot

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

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

  6. Draw the ray

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

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

Check your answer

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

More questions like this

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

Question 8 of 120

2026 Practice Test A Math, question 50

Question

Let b=3b = 3, c=4c = 4, and d=5d = 5. Determine the value of cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}.

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

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: 7/5

Solution:

75\frac{7}{5}

Substitute b=3b = 3, c=4c = 4, and d=5d = 5 in cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}: 4543+45\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5}.

Then simplify the complex fraction by multiplying the complex fraction by a form of 1, (3434)\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right): 4543+45=4543(3434)+45\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5} = \frac{\frac{4}{5}}{\frac{4}{3}}\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right) + \frac{4}{5}.

Simplify after multiplying by a form of 1: (45)(34)(43)(34)+45=351+45\frac{\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)}{\left(\frac{4}{3}\right)\left(\frac{3}{4}\right)} + \frac{4}{5} = \frac{\frac{3}{5}}{1} + \frac{4}{5}.

Then simplify again and add the two fractions: 351+45=35+45=75\frac{\frac{3}{5}}{1} + \frac{4}{5} = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}.

Question analysis

This substitution question gives values for b, c and d and asks for the value of an expression made of a fraction divided by a fraction, plus another fraction. Students substitute, simplify the complex fraction and add, then type the result. Dividing fractions in the stacked form is the main hurdle, and the item wants a fraction, not a decimal.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.NS.1 dividing fractions by fractions
  • 7.NS.2c properties for multiplying and dividing
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students evaluate expressions and divide and add fractions. A fraction stacked over another fraction looks unfamiliar and there are no choices to check, but with these values the work is three short steps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the values are in, the work is one fraction division and one addition, but the division is printed as a fraction over a fraction, in small type. Multiplying the two fractions, or flipping the top one, gives a sum in fifteenths that looks as finished as the right one. Noticing that the 4s cancel, so the stacked part is simply 3/5, turns the last step into adding fifths.

  1. Error-prone arithmetic moderate Working it out

    Dividing 4/5 by 4/3 means multiplying by 3/4, giving 3/5, which is then added to 4/5; multiplying straight across or flipping the top fraction instead both lead to fifteenths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    Nothing on the page says that a fraction over a fraction is a division, or that dividing by 4/3 means multiplying by its reciprocal, and the stacked form is rare in grade 7 work. (Anderson & Krathwohl, 2001)

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

    The value is typed into a fraction box with no choices to check against, and the item asks for a fraction, so the equal decimal 1.4 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Substitute the values, rewrite the stacked fraction as the top fraction times the reciprocal of the bottom one, where the 4s cancel, then add the last fraction and leave the result as a fraction.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • 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.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions
ab÷cd=ab×dc\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. It works because a fraction times its reciprocal is 1, so the reciprocal undoes the fraction, just as dividing does.
Adding fractions with the same bottom
ad+bd=a+bd\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}
d is the denominator, the bottom number, which tells the size of each piece. a and b are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

  1. Sort out the question

    You know b=3b = 3, c=4c = 4 and d=5d = 5. You need the value of:

    cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}

    The first part is a fraction on top of a fraction. A fraction bar means divide, so this part is cd÷cb\frac{c}{d} \div \frac{c}{b}. The answer must be typed as a fraction.

  2. Put in the numbers

    Replace each letter with its value. Look closely: the bottom fraction has b under the c, not d.

    cd=45cb=43\frac{c}{d} = \frac{4}{5} \qquad \frac{c}{b} = \frac{4}{3}

    So the expression becomes:

    45÷43+45\frac{4}{5} \div \frac{4}{3} + \frac{4}{5}

  3. Divide the two fractions

    Use the rule for dividing fractions. The reciprocal of 4/3 is 3/4. To multiply fractions, multiply the tops and multiply the bottoms:

    45÷43=45×34=1220=35\frac{4}{5} \div \frac{4}{3} = \frac{4}{5} \times \frac{3}{4} = \frac{12}{20} = \frac{3}{5}

    To simplify 12/20, divide its top and bottom by 4.

    Watch out: flip only the bottom fraction. Multiplying without flipping gives 16/15 and a total of 28/15. Flipping the top one gives 5/3 and a total of 37/15.

  4. Add the last fraction

    Both fractions are now fifths, so add the tops and keep the bottom:

    35+45=3+45=75\frac{3}{5} + \frac{4}{5} = \frac{3 + 4}{5} = \frac{7}{5}

    The value is 7/5.

  5. Enter it as a fraction

    The item asks for a fraction. 7/5 equals the decimal 1.4, but 1.4 is not accepted. On the test, the answer box already holds a fraction template, a small box over a small box. Type 7 in the top box. Then click the bottom box and type 5. On this site, type 7/5.

    So you would enter 7/5.

Check your answer

Check with decimals. 45=0.8\frac{4}{5} = 0.8, and 4/3 is about 1.33. Then 0.8÷1.330.8 \div 1.33 is about 0.6, which is 3/5. Last, 0.6+0.8=1.40.6 + 0.8 = 1.4, and 1.4 is the same as 7/5.

More questions like this

The original DOE question
Question 50 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 A Math, page 121, question 50.

Question 9 of 120

2026 Practice Test B Math, question 2

Question

The list shows the results of a survey. A survey was taken to see whether students like basketball, soccer, both, or neither. These were the results:

  • 25 students like basketball.
  • 19 students like soccer.
  • 7 students like both basketball and soccer.
  • 10 students don’t like either basketball or soccer.

Use the information provided to determine how many students were surveyed and how many like only soccer.

Move the correct answer to each box.

There were students surveyed, and students like only soccer.

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: 47; blank 2: 12

There were 47 students surveyed, and 12 students like only soccer.

To determine how many students were surveyed, start with the total number of students who like basketball and soccer separately and subtract the number of students who were counted twice.

There were 25 students who like basketball, 19 students who like soccer, and 7 students who like both basketball and soccer.

So, the total number of students who like basketball or soccer is (25+19)−7=37(25 + 19) - 7 = 37.

Then, add the students who don’t like either basketball or soccer: 10 students don’t like either sport.

Total students surveyed: 37+10=4737 + 10 = 47

To determine how many students like only soccer, subtract the number of students who like both basketball and soccer from the total number of students who like soccer.

Students who like only soccer: 19−7=1219 - 7 = 12

Question analysis

This survey problem lists how many students like basketball, soccer, both sports or neither, and students drag number tiles into two boxes: how many were surveyed and how many like only soccer. It is a Venn diagram counting question. Keeping track of the students who like both sports, who appear in two of the listed counts, is the tricky part.

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 Overlapping groups like these are seldom taught on their own in grade 7, though the arithmetic is simple whole-number addition and subtraction. Two results come from the same four facts, and both boxes must be right to score.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 78 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the work is keeping track of one overlap that touches both answers. Students often remove the 7 for one box and not the other, and the pool holds 54 and 19, the tiles those half-finished methods produce. A sketch of two overlapping circles, with 7 in the middle, 18 and 12 beside it and 10 outside, turns both boxes into reading off the picture.

  1. Much to keep track of moderate Working it out

    The 7 students who like both sports sit inside the 25 and inside the 19, so they come out once for the total and again for only soccer, and both boxes must be right to score. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    The list never says that the 25 basketball fans already include the 7 who like both sports, and only the word “only” in the question separates the 12 from all 19 soccer fans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Sketch two overlapping circles, write the students who like both sports in the middle, fill each outer part by subtraction, then add every region and the students who like neither.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap
total=B+S−both+neither\text{total} = B + S - \text{both} + \text{neither}
B is the number who like basketball, and S is the number who like soccer. The students who like both are inside B and inside S, so B + S counts them twice. Subtract them once to fix that. Then add the students who like neither sport, since they are in neither group.
Only one group
only soccer=S−both\text{only soccer} = S - \text{both}
S is everyone who likes soccer, and both is the number who like both sports. Only soccer means the soccer fans who do not also like basketball. Take the both group out of the soccer group to find them.

  1. Sort out the question

    The survey gives you four facts:

    • 25 students like basketball.
    • 19 students like soccer.
    • 7 students like both sports.
    • 10 students like neither sport.

    You need two numbers. The first box is how many students were surveyed in all. The second box is how many like only soccer, which means soccer but not basketball. Each tile can be used once.

  2. See who is counted twice

    The 7 students who like both sports are part of the 25 basketball fans. They are also part of the 19 soccer fans. So if you add 25 and 19, those 7 students get counted two times.

    A Venn diagram shows this. It is a picture of two circles that overlap. The 7 go in the middle, where the circles share space.

  3. Count who likes at least one sport

    Use the overlap rule. Add the two groups, then take away the 7 who were counted twice:

    25+19−7=44−7=3725 + 19 - 7 = 44 - 7 = 37

    So 37 students like basketball, soccer or both.

  4. Add the students who like neither

    The 10 students who like neither sport are not in either group. They were still surveyed, so add them:

    37+10=4737 + 10 = 47

    So 47 students were surveyed.

    Watch out: adding 25 + 19 + 10 = 54 counts the 7 students twice. The tile 54 is waiting for that mistake.

  5. Find who likes only soccer

    The 19 soccer fans include the 7 who like basketball too. Take those 7 away:

    19−7=1219 - 7 = 12

    So 12 students like only soccer.

    Watch out: the word only matters. The tile 19 counts every soccer fan, not just the ones who like only soccer.

  6. Fill in both boxes

    On the test, you drag the tile 47 into the first box and the tile 12 into the second box. On this site, you choose each tile from its box’s menu.

    So the boxes get 47 and 12: there were 47 students surveyed, and 12 students like only soccer.

Check your answer

Split the students into four groups that do not overlap. Only basketball is 25−7=1825 - 7 = 18. Both sports is 7. Only soccer is 12. Neither is 10. Add the four groups: 18+7+12+10=4718 + 7 + 12 + 10 = 47. That matches the first box.

More questions like this

The original DOE question
Question 2 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 B Math, page 74, question 2.

Question 10 of 120

2026 Practice Test B Math, question 3

Question

At a diner, 18 dozen buns were used yesterday for 150 lunch customers. Today, there were 250 lunch customers. Assuming that the bun usage rate is the same for both days, which proportions can be used to determine how many dozen buns are needed for 250 lunch customers?

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.

  • A18150=x250\frac{18}{150} = \frac{x}{250}
  • B18250=x150\frac{18}{250} = \frac{x}{150}
  • C15018=250x\frac{150}{18} = \frac{250}{x}
  • Dx18=250150\frac{x}{18} = \frac{250}{150}
  • Ex150=18250\frac{x}{150} = \frac{18}{250}
Show the answer

Correct answer: A, C, and D

The three correct answers are:

18150=x250\frac{18}{150} = \frac{x}{250}

15018=250x\frac{150}{18} = \frac{250}{x}

x18=250150\frac{x}{18} = \frac{250}{150}

Set up a proportion to compare the number of buns per customer. These are all equivalent:

18 dozen buns150 customers=x dozen buns250 customers\frac{18\text{ dozen buns}}{150\text{ customers}} = \frac{x\text{ dozen buns}}{250\text{ customers}}

To check for equivalency, solve each proportion for x.

Each of the three correct proportions simplifies to 150x=18(250)150x = 18(250).

Then solve for x to determine how many dozen buns are needed for 250 lunch customers: 150x=4,500150x = 4{,}500

x=4,500150x = \frac{4{,}500}{150}

x=30x = 30

Question analysis

A diner used 18 dozen buns for 150 lunch customers yesterday, and students select the three proportions, out of five, that can find how many dozen buns 250 customers need at the same rate. It tests setting up proportions and recognizing equivalent ones. All five equations use the same numbers, so each has to be checked for which quantities it pairs.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Setting up a proportion from a rate is everyday grade 7 work. Here nothing is solved: five setups built from the same numbers must each be judged, some arranged in unfamiliar ways, and all three usable ones found.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 54 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Nothing needs solving here; the question is which pairings keep the usage rate fixed. Students who check only for buns over customers on both sides find exactly three such setups, two of them wrong, and pass over the two right ones written another way. The wrong pair gives 10.8 dozen, fewer buns for more customers, which a quick sense check would catch, and cross-multiplying every setup settles all five.

  1. Every choice must be checked moderate Choosing the answer

    All five setups are built from 18, 150, 250 and x, so none can be dismissed on sight; each has to be cross-multiplied and compared with 150x = 18 × 250. (Embretson & Wetzel, 1987)

  2. A very close wrong answer moderate Choosing the answer

    The first two setups differ only in where 150 and 250 sit, 18/150 = x/250 against 18/250 = x/150, and both show buns over customers on each side, so a units check passes both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    One right setup puts customers over dozens of buns, and another sets today’s x over yesterday’s 18 equal to 250 customers over 150, forms that look wrong to students who learned only rate equals rate. (Enright, Morley & Sheehan, 2002)

  4. Much to keep track of minor Working it out

    Three of the five must be picked and all three must be right to score, so the verdict on each setup has to be kept until the last one is tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write one proportion you trust by matching both units and days, cross-multiply it, then cross-multiply each choice and keep only those that give the same equation.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A rate
rate=dozens of bunscustomers\text{rate} = \frac{\text{dozens of buns}}{\text{customers}}
The rate is how many dozen buns are used for each customer. Yesterday it was 18 dozen over 150 customers. Today it is x dozen over 250 customers. The question says the rate is the same both days, so those two fractions are equal.
Cross-multiplying a proportion
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
A proportion says two fractions are equal. a and c are the tops, and b and d are the bottoms. Multiply each top by the other bottom, and the two products are equal. This turns any proportion into a plain equation, so you can compare setups that look different.

  1. Sort out the question

    Yesterday the diner used 18 dozen buns for 150 customers. Today it has 250 customers. Let x be the number of dozen buns needed today. The bun usage rate is the same both days.

    A proportion is two equal fractions. You must pick the three proportions that give the right value of x.

  2. Write one proportion you trust

    Put each day’s buns over that same day’s customers. The rates are equal, so the fractions are equal:

    18150=x250\frac{18}{150} = \frac{x}{250}

    This is choice A. Cross-multiply it:

    150×x=18×250150 \times x = 18 \times 250

    150x=4,500150x = 4{,}500

    Any choice that leads to 150x = 4,500 is a right one.

  3. Solve for x

    Divide both sides by 150 to get x by itself:

    x=4,500÷150=30x = 4{,}500 \div 150 = 30

    So the diner needs 30 dozen buns today. A wrong setup gives a different number.

  4. Test choices C and D

    Cross-multiply each one.

    • C: 15018=250x\frac{150}{18} = \frac{250}{x} gives 150×x=18×250150 \times x = 18 \times 250. That is the same equation, so C works. It is choice A with both fractions flipped.
    • D: x18=250150\frac{x}{18} = \frac{250}{150} gives x×150=18×250x \times 150 = 18 \times 250. The same equation again, so D works too.

    D compares today with yesterday. Today’s buns over yesterday’s buns equals today’s customers over yesterday’s customers.

    Watch out: D looks odd because it puts buns over buns. Do not skip it. It still keeps the rate the same.

  5. Test choices B and E

    • B: 18250=x150\frac{18}{250} = \frac{x}{150} gives 250×x=18×150250 \times x = 18 \times 150, so 250x = 2,700.
    • E: x150=18250\frac{x}{150} = \frac{18}{250} gives x×250=150×18x \times 250 = 150 \times 18, the same 250x = 2,700.

    Both give x=2,700÷250=10.8x = 2{,}700 \div 250 = 10.8. That is fewer buns for more customers, which makes no sense.

    Watch out: B and E show buns over customers, so they look right. But each one mixes up the days. B puts yesterday’s 18 dozen over today’s 250 customers. E puts today’s buns over yesterday’s 150 customers.

  6. Name the three answers

    Choices A, C and D all give 150x = 4,500, so x = 30. On the test, you check the box next to each one. So the answers are A, C and D.

Check your answer

Find the buns a second way. Today has 250150=53\frac{250}{150} = \frac{5}{3} times as many customers as yesterday. So it needs 5/3 times as many buns: 18×53=3018 \times \frac{5}{3} = 30 dozen. That matches the 30 from choices A, C and D.

More questions like this

The original DOE question
Question 3 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 B Math, page 75, question 3.

Question 11 of 120

2026 Practice Test B Math, question 4

Question

Maria is playing a game in which players earn points called minks, zogs, and barts.

1 mink=3 zogs1\text{ mink} = 3\text{ zogs}

1 mink=14 bart1\text{ mink} = \frac{1}{4}\text{ bart}

Maria has 60 zogs and 24 barts. She wants to exchange her zogs and barts for minks.

How many minks will Maria receive from the exchange?

Enter your answer in the space.

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

Show the answer

Correct answer: 116

Step 1: Convert the zogs to minks.
Since 1 mink=3 zogs1\text{ mink} = 3\text{ zogs}, divide the number of zogs Maria has by 3: 60÷3=2060 \div 3 = 20.

So, 60 zogs=20 minks60\text{ zogs} = 20\text{ minks}.

Step 2: Convert the barts to minks.
Since 1 mink=14 bart1\text{ mink} = \frac{1}{4}\text{ bart}, multiply the number of barts by 4 (because 1 bart is equal to 4 minks): 24×4=9624 \times 4 = 96.

So, 24 barts=96 minks24\text{ barts} = 96\text{ minks}.

Step 3: Add the mink conversions.
Add the results from Steps 1 and 2 to find the total number of minks Maria will receive: 20+96=11620 + 96 = 116.

Question analysis

In a game with made-up points called minks, zogs and barts, Maria trades 60 zogs and 24 barts for minks, and students type how many minks she receives. It tests unit conversion with invented units and a fraction. One rate says a mink equals a fraction of a bart, so deciding whether to multiply or divide is the tricky part.

NYS standards
  • 6.RP.3d converting measurement units
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Converting units with a given rate is grade 6 ratio reasoning. Invented units remove any real-world sense check, and the two conversions run in opposite directions, one through a fraction, which matches the care grade 7 problems need.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 2, 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

Each step is one multiplication, division or addition; the risk is direction. The two rates point opposite ways, three zogs to a mink but four minks to a bart, and the fraction in the second rate invites dividing by 4. Asking of each unit whether one of it is worth more or less than a mink fixes the direction before any arithmetic, and it also shows why the 60 and the 24 can never simply be added.

  1. Turning words into math moderate Understanding the question

    Both rates put the mink on the left, so “1 mink = 1/4 bart” has to be turned around into 1 bart = 4 minks, and the invented units give no sense of which is worth more. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Two conversions run in opposite directions, dividing the zogs by 3 and multiplying the barts by 4, and their results are added, so a slip in either step changes the total. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The total is typed into a box with no choices to compare, so a conversion done the wrong way still produces a believable whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: For each kind of point, decide whether one of it is worth more or less than a mink, then multiply or divide to turn it into minks, and add the two results.

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).
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Converting with a rate
minks=number of points×minks for one point\text{minks} = \text{number of points} \times \text{minks for one point}
Number of points is how many zogs or barts Maria has. Minks for one point is what one zog, or one bart, is worth in minks. Find that worth first for each kind of point. Then multiply, because every point of the same kind trades for the same number of minks.

  1. Sort out the question

    Maria trades two kinds of points for minks. You know:

    • 3 zogs trade for 1 mink.
    • 1 mink is worth 14\frac{1}{4} of a bart.
    • Maria has 60 zogs and 24 barts.

    You need the total number of minks. Zogs and barts are different units, like dimes and dollars. So you cannot just add them.

    Watch out: 60 + 24 = 84 adds zogs to barts. That is not a number of minks.

  2. Find what one zog is worth

    It takes 3 zogs to make 1 mink. So one zog is only part of a mink. One zog is worth 13\frac{1}{3} of a mink.

    A zog is smaller than a mink. So Maria will get fewer minks than she has zogs.

  3. Turn the bart rule around

    The rule says 1 mink is 14\frac{1}{4} of a bart. So a mink is a quarter of a bart. It takes 4 quarters to make a whole:

    4×14=14 \times \frac{1}{4} = 1

    So 4 minks make 1 bart. One bart is worth 4 minks. A bart is bigger than a mink.

    Watch out: the fraction tempts you to divide by 4. Then 24 barts become only 6 minks, and the total is 26.

  4. Change the zogs into minks

    Use the rate. Each of the 60 zogs is worth 1/3 of a mink:

    60×13=60÷3=2060 \times \frac{1}{3} = 60 \div 3 = 20

    Multiplying by 1/3 is the same as dividing by 3. So 60 zogs trade for 20 minks.

    Watch out: multiplying 60 by 3 gives 180. That treats a zog as bigger than a mink, and the total becomes 276.

  5. Change the barts into minks

    Each of the 24 barts is worth 4 minks:

    24×4=9624 \times 4 = 96

    So 24 barts trade for 96 minks.

  6. Add the minks

    Both amounts are now in minks, so you can add them:

    20+96=11620 + 96 = 116

    Maria receives 116 minks. Type only the number in the box, with no units. So you would enter 116.

Check your answer

Work backward. 20 minks are worth 20×3=6020 \times 3 = 60 zogs. 96 minks are worth 96×14=2496 \times \frac{1}{4} = 24 barts. Both match what Maria started with. Also, a bart is worth more than a mink, so 24 barts should give many more than 24 minks. They do.

More questions like this

The original DOE question
Question 4 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 B Math, page 76, question 4.

Question 12 of 120

2026 Practice Test B Math, question 13

Question

A basket contains 3 apples, 4 oranges, 5 bananas, and 8 pears.

If one piece of fruit is randomly pulled from the basket, what is the probability that it will be a pear?

Select the place on the number line to plot the point.

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

00.10.20.30.40.50.60.70.80.91

Show the answer

Correct answer: the point 0.4

The total number of fruits is:

3(apples)+4(oranges)+5(bananas)+8(pears)=20 fruits3(\text{apples}) + 4(\text{oranges}) + 5(\text{bananas}) + 8(\text{pears}) = 20\text{ fruits}

The probability of picking a pear is:

8 (pears)20 (total)=820 or 0.4\frac{8\ (\text{pears})}{20\ (\text{total})} = \frac{8}{20} \text{ or } 0.4

Question analysis

A basket holds apples, oranges, bananas and pears, and students find the probability that a piece of fruit pulled at random is a pear, then plot it as a point on a number line from 0 to 1. It tests simple probability as a part of a whole. With no choices to pick from, the fraction has to be turned into tenths to find its place.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of a simple event from counts is grade 6 work in New York. This one adds four whole numbers, forms one fraction and plots it among marked tenths, a shorter and more routine task than grade 7 probability.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
3 steps
Reading level
Question and choices: about grade 6, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Adding the four counts to 20 and writing 8 over 20 is routine, so most students succeed. The losses come from the number line: its decimal labels make it easy to read the 8 pears as 0.8, and a part-to-part fraction such as 8/12 falls between marks and has to be rounded, which should prompt a second look. Simplifying 8/20 to 2/5 and then to 4/10 makes the mark plain.

  1. Something you must already know minor Working it out

    The question does not say that a probability is the favorable count over the whole count, here the 8 pears over all 20 pieces of fruit rather than over the other 12. (Anderson & Krathwohl, 2001)

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

    The number line is labeled only in tenths from 0 to 1, so 8/20 has to be rewritten as a decimal before the right mark can be chosen. (Enright, Morley & Sheehan, 2002)

Strategy: Add up all the fruit for the total, write the number of pears over that total, rewrite the fraction in tenths, and select that mark on the line.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event
P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}
P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want, here pears. The bottom counts every possible result, here every piece of fruit. It works when each result is equally likely, like grabbing fruit without looking.
Equal fractions
ab=a÷nb÷n\frac{a}{b} = \frac{a \div n}{b \div n}
a is the top of a fraction, and b is the bottom. n is any number except 0. Dividing the top and the bottom by the same number gives an equal fraction. Use it to get 10 on the bottom, which is easy to write as a decimal.

  1. Sort out the question

    The basket has 3 apples, 4 oranges, 5 bananas and 8 pears. You pull out one piece without looking. You need the probability that it is a pear. Then you plot that number as a point. The number line runs from 0 to 1, with a mark at every tenth.

  2. Count all the fruit

    The bottom of the fraction is every piece of fruit you could pull out. Add all four kinds:

    3+4+5+8=203 + 4 + 5 + 8 = 20

    There are 20 pieces of fruit in all.

  3. Write the probability as a fraction

    Use the probability formula. There are 8 pears out of 20 pieces of fruit:

    P=820P = \frac{8}{20}

    Watch out: the bottom is all the fruit, not just the other fruit. 812\frac{8}{12} compares the pears with the 12 fruits that are not pears. It is about 0.67, and it leads you to the mark at 0.7. And 12 over 20 is the chance of not getting a pear.

  4. Change it to tenths

    The number line is marked in tenths. So rewrite the fraction with 10 on the bottom. Divide the top and the bottom by 2:

    820=8÷220÷2=410=0.4\frac{8}{20} = \frac{8 \div 2}{20 \div 2} = \frac{4}{10} = 0.4

    Four tenths is written 0.4.

    Watch out: the 8 pears do not make 0.8. That skips dividing by the 20 pieces of fruit.

  5. Plot the point

    Start at 0 and count four marks to the right: 0.1, 0.2, 0.3, 0.4. On the test, you click the mark labeled 0.4. On this site, you choose 0.4 from the menu.

    So the answer is the point at 0.4.

Check your answer

The chance of not getting a pear is 1220=0.6\frac{12}{20} = 0.6. A pear and not a pear cover every result, so the two must add to 1: 0.4+0.6=10.4 + 0.6 = 1. Also, pears are less than half of the fruit, so the point should be left of 0.5. It is.

More questions like this

The original DOE question
Question 13 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 B Math, page 85, question 13.

Question 13 of 120

2026 Practice Test B Math, question 16

Question

Which tables represent a proportional relationship between the number of books purchased and the total cost?

Select the two 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
    Number of Books Purchased10152025
    Total Cost ($)406080100
  • B
    Number of Books Purchased10152025
    Total Cost ($)70110150250
  • C
    Number of Books Purchased10152025
    Total Cost ($)80120160200
  • D
    Number of Books Purchased10152025
    Total Cost ($)90130170210
  • E
    Number of Books Purchased10152025
    Total Cost ($)100100100100
Show the answer

Correct answer: A and C

The two tables that represent a proportional relationship are tables A and C.

In table A, the total costs divided by the numbers of books for each set of quantities are:

4010=4; 6015=4; 8020=4; 10025=4\frac{40}{10} = 4;\ \frac{60}{15} = 4;\ \frac{80}{20} = 4;\ \frac{100}{25} = 4

These ratios are equivalent; therefore, these quantities are in a proportional relationship.

In table C, the total costs divided by the numbers of books for each set of quantities are:

8010=8; 12015=8; 16020=8; 20025=8\frac{80}{10} = 8;\ \frac{120}{15} = 8;\ \frac{160}{20} = 8;\ \frac{200}{25} = 8

These ratios are equivalent; therefore, these quantities are in a proportional relationship.

Question analysis

Five tables list the total cost of buying 10, 15, 20 and 25 books, and students select the two tables that show a proportional relationship between books and cost. It tests recognizing proportional relationships from a constant unit rate. A steady rise in cost is not enough on its own, so each pair of values has to be checked.

NYS standards
  • 7.RP.2a testing for proportionality
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Deciding from a table whether two quantities are proportional is a core grade 7 skill, and the divisions here are easy. The demand lies in checking every column of five tables rather than judging by how the costs change.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The divisions are easy, so the question tests what proportional means. Three tables grow by the same amount for every 5 books, and the one with a $10 head start looks just as regular as the right one it copies; only its cost per book, drifting from $9 down to $8.40, gives it away. The flat $100 table and the one that jumps to $250 catch fewer students, and dividing the first and last column of each table is a quick, reliable check.

  1. A very close wrong answer moderate Choosing the answer

    The table running $90, $130, $170, $210 is the $80-to-$200 table with $10 added to every cost; both climb $40 for each 5 books, and only the cost per book separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Every choice must be checked minor Choosing the answer

    Five tables of four columns each have to be tested by dividing cost by books, and both right tables must be picked to score, so no table can be judged from one column. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    Students must know that proportional means the same cost per book in every column, a line through zero, not merely a cost that grows by the same amount each time. (Anderson & Krathwohl, 2001)

Strategy: Divide the total cost by the number of books in every column of each table; a table is proportional only when that cost per book stays the same all the way across.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost per book
cost per book=total cost÷number of books\text{cost per book} = \text{total cost} \div \text{number of books}
This is the unit rate, the price of one book. Divide the total cost in dollars by the number of books bought. If every book has the same price, this number comes out the same in every column.
A proportional relationship
y=kxy = kx
x is the number of books, and y is the total cost in dollars. k is the cost per book, a number that never changes. In a proportional relationship, y is always k times x. So 0 books cost $0, and twice as many books cost twice as much.

  1. Sort out the question

    Each table shows the total cost of 10, 15, 20 and 25 books. Proportional means the cost per book is the same in every column. In other words, each book has one fixed price. You must pick the two tables where that is true.

  2. Test the first table

    Divide each total cost by its number of books. For table A:

    40÷10=460÷15=480÷20=4100÷25=440 \div 10 = 4 \qquad 60 \div 15 = 4 \qquad 80 \div 20 = 4 \qquad 100 \div 25 = 4

    Every column gives $4 per book. So table A is proportional.

  3. Test the other four tables

    Do the same division for every table. The numbers are dollars per book:

    Table10 books15 books20 books25 booksSame?
    A4444Yes
    B7about 7.337.5010No
    C8888Yes
    D9about 8.678.508.40No
    E10about 6.6754No

    Only tables A and C keep the same price all the way across.

  4. Watch for steady growth

    Watch out: table D climbs $40 for every 5 more books, just like table C. That steady climb looks proportional, but it is not enough. D starts $10 higher than C, so its price per book slides from $9 down to $8.40.

    Watch out: table E never changes. But $100 for 10 books is $10 each, and $100 for 25 books is $4 each. A total that stays the same is not a price that stays the same.

  5. Name the two answers

    Table A charges $4 per book and table C charges $8 per book, in every column. On the test, you check the box next to each of those tables. So the answers are A and C.

Check your answer

In a proportional table, twice the books cost twice as much. In table A, 10 books cost $40 and 20 books cost $80, which is double. In table C, $80 doubles to $160. In table D, $90 would have to double to $180, but the table shows $170. So D fails.

More questions like this

The original DOE question
Question 16 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 B Math, page 88, question 16.

Question 14 of 120

2026 Practice Test B Math, question 17

Question

How many positive integers satisfy the inequality x+7<23x + 7 < 23?

Enter your answer in the space.

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

Show the answer

Correct answer: 15

Solve the inequality for x.

x+7<23x + 7 < 23

x<16x < 16

There are 15 positive integers that are less than 16.

Question analysis

This inequality question asks how many positive integers make x + 7 < 23 true, and the count is typed into a box. Students solve a one-step inequality, then count its whole-number solutions. The strict less-than sign and the word positive decide where the count starts and stops, so both ends need checking.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 inequality problems usually take two steps, such as 2x − 5 > 9. This one needs a single subtraction; the care goes into counting the whole-number solutions and deciding whether the boundary itself counts.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 12, 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

One subtraction solves the inequality, so the points are lost in the count that follows. The boundary 16 is the last number the solving produces and is easy to type in, or to count up to; checking that 16 + 7 is 23, which is not less than 23, rules it out. Listing 1 through 15, the positive integers below the boundary, settles the count.

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

    Solving gives x < 16, but the box asks how many positive integers make the inequality true, a count that still has to be made after the solving is done. (Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Two conditions set the ends of the count at once: the word positive starts it at 1 rather than 0, and the strict < stops it below 16. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Undo the addition to find the boundary for x, then count the positive whole numbers below it, checking the largest one and the boundary in the original inequality.

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).
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a<b, then a−c<b−c\text{if } a < b \text{, then } a - c < b - c
a and b are the two sides of an inequality, and c is any number. Taking the same number away from both sides keeps the inequality true. Think of two people who each walk down the same number of stairs: the lower one stays lower. Use it to get x by itself.

  1. Sort out the question

    An inequality compares two amounts. The sign < means is less than. A positive integer is a whole number bigger than 0: 1, 2, 3, and so on. You need to count how many of them make x+7<23x + 7 < 23 true. The answer is a count, not a value of x.

  2. Get x by itself

    The 7 is added to x. Subtracting 7 undoes adding 7, so subtract 7 from both sides:

    x+7−7<23−7x + 7 - 7 < 23 - 7

    x<16x < 16

    So x must be less than 16.

    Watch out: adding 7 to 23 instead gives x < 30, and a count of 29.

  3. Check the boundary number

    The number 16 is the boundary, where the solutions stop. Does 16 itself work? Try it:

    16+7=2316 + 7 = 23

    But 23 is not less than 23. It is equal. So 16 does not count. The largest number that works is 15, since 15+7=2215 + 7 = 22, and 22 is less than 23.

    Watch out: entering 16 counts the boundary, which fails.

  4. Count the positive integers

    The numbers that work start at 1, the smallest positive integer. Zero is not positive. They stop at 15:

    1,2,3,…,14,151, 2, 3, \ldots, 14, 15

    Counting from 1 up to 15 gives 15 numbers.

  5. Enter the count

    There are 15 positive integers that make the inequality true. Type the count in the box as a plain whole number. So you would enter 15.

Check your answer

Test both ends. For 1: 1+7=81 + 7 = 8, and 8 is less than 23, so 1 works. For 15: 22 is less than 23, so 15 works. For 16: 23 is not less than 23, so 16 fails. That leaves exactly the numbers 1 through 15.

More questions like this

The original DOE question
Question 17 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 B Math, page 89, question 17.

Question 15 of 120

2026 Practice Test B Math, question 24

Question

A chemical decays in such a way that the amount left at the end of each week is 20% less than the amount at the beginning of that same week. What percentage of the original amount is left after two weeks?

  • A40%
  • B60%
  • C64%
  • D80%
Show the answer

Correct answer: C

Let x be the original amount of the chemical. It loses 20% after each week, which means 80% of the chemical remains at the end of each week. End of first week: 0.80x0.80x

At the end of the second week, 80% of the amount left at the end of the first week remains.

End of second week: 0.80(0.80x)=0.64x0.80(0.80x) = 0.64x or 64%

Question analysis

A chemical loses 20% of the amount present at the start of each week, and students find what percentage of the original amount is left after two weeks. It tests applying a percent decrease more than once. The question turns on which amount each week's loss is measured against.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 A typical grade 7 percent task applies one increase or decrease to one amount. This one applies a weekly decrease twice in a row, an early form of exponential decay that goes beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
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

Two equal decreases in a row feel like one decrease of twice the size, and that is the trap: 20% and 20% suggest 40% gone. The stem states the rule, each loss measured from that week’s start, in one quick phrase. Working week by week from 100 shows the second loss is 16, not 20, which separates the answer from the choice four points away, while stopping after one week or reporting the loss costs the point more simply.

  1. A very close wrong answer major Choosing the answer

    The choices 60% and 64% sit four percentage points apart, so a rough estimate cannot choose between them; only seeing that the second 20% is taken from 80, not 100, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math moderate Understanding the question

    The phrase “the amount at the beginning of that same week” sets each week’s base, so the second week’s 20% is measured from what was left after the first week. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Variables instead of numbers minor Understanding the question

    No starting amount is given, so students must pick one, such as 100 grams, or work with a percent of a percent, 0.8 × 0.8. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Use a starting amount of 100 so that percents become units, find what remains after the first week, then apply the same 20% loss to that remainder.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of an amount
part=p100×amount\text{part} = \frac{p}{100} \times \text{amount}
p is the percent, and amount is the whole you take the percent of. Percent means out of 100, so 20% is 20/100. Multiply to find the part. Be sure the amount is the right whole. Here, each week uses the amount at the start of that week.
Percent left after a loss
percent left=100%−percent lost\text{percent left} = 100\% - \text{percent lost}
The whole amount is 100%. If 20% goes away, the rest stays. So 80% of each week’s starting amount is left at the end of that week. Keeping 80% is the same as multiplying by 0.8.

  1. Sort out the question

    Each week the chemical loses 20% of what it has. The key words are “the amount at the beginning of that same week.” So each week’s 20% comes from that week’s starting amount. No amount is given. You need the percent of the original amount that is left after two weeks.

  2. Pick an easy starting amount

    Pretend the chemical starts at 100 grams. With 100, the number of grams left is the same as the percent left. Any starting amount gives the same percent, so choose the easiest one.

  3. Work out the first week

    Use the percent formula. The first week starts with 100 grams:

    20100×100=20\frac{20}{100} \times 100 = 20

    The chemical loses 20 grams, so 100−20=80100 - 20 = 80 grams are left. That is 80% after one week.

    Watch out: 80% is only the first week. Stopping here gives choice D.

  4. Work out the second week

    The second week starts with 80 grams, not 100. So take 20% of 80:

    20100×80=16\frac{20}{100} \times 80 = 16

    This time the chemical loses only 16 grams, so 80−16=6480 - 16 = 64 grams are left.

    Watch out: taking away 20 grams again uses the original amount. That gives 60%, choice B. Choice A, 40%, adds the two losses. But the question asks what is left, and only 36% was lost.

  5. Read the answer as a percent

    WeekStart (grams)Loss (grams)Left (grams)
    11002080
    2801664

    Another way: each week keeps 80%, or 0.8, of its start. Two weeks gives 0.8×0.8=0.640.8 \times 0.8 = 0.64, which is 64%.

    64 grams out of the original 100 grams is 64%. So the answer is C, 64%.

Check your answer

Think about the size of the second loss. The second week starts with less, so it loses less than 20 grams. That means more than 60% is left. It does lose something, so less than 80% is left. Of the four choices, only 64% is between 60% and 80%.

More questions like this

The original DOE question
Question 24 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 B Math, page 96, question 24.

Question 16 of 120

2026 Practice Test B Math, question 28

Question

A circle has a radius of 5 inches. What is the area of the circle in terms of π\pi?

Use π\pi to express your answer.

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: 25π

Solution:

25π25\pi

The area of a circle is its radius squared times π\pi.

A=πr2A = \pi r^2

r=5r = 5

A=π(5)2A = \pi(5)^2

A=25πA = 25\pi

Question analysis

This circle question gives a radius of 5 inches and asks for the area in terms of π, typed into a box rather than chosen. Students recall and apply the formula for the area of a circle. The formula is not printed on the test, and the area and circumference formulas are easy to mix up.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the area of a circle from its radius is the grade 7 circle standard applied directly, in one routine calculation. Typing the answer in terms of π, with no choices to compare, adds a little risk.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is one formula and one square, so the question comes down to recall and to the form of the answer. The area and circumference formulas both pair π with the radius, and 2πr gives a tidy value with π in it that looks just as right. The other losses are about form: replacing π with 3.14, which the instructions rule out twice, or dropping π after squaring 5.

  1. Something you must already know moderate Working it out

    The formula A = πr² is not printed with the question, and the circumference formula 2πr uses the same r and π, so recalling the wrong one gives a value that looks just as finished. (Anderson & Krathwohl, 2001)

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

    The answer is built on a keypad with a π key and must keep π as a symbol; a decimal worked out with 3.14 is not accepted, and no choices show the expected form. (Enright, Morley & Sheehan, 2002)

Strategy: Recall the area formula rather than the circumference formula, square the radius before multiplying, and leave π as a symbol instead of replacing it with 3.14.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle
A=πr2A = \pi r^2
A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. r² means r times r. Square the radius first, then multiply by π. The area is in square units.
Circumference of a circle
C=2πrC = 2 \pi r
C is the circumference, the distance around the circle. r is the radius, and π is the same number as above. This formula measures a length, not the space inside. It is here so you can tell the two formulas apart.

  1. Sort out the question

    The circle has a radius of 5 inches. The radius is the distance from the center to the edge. You need the area, the space inside the circle. In terms of π means π stays in the answer as a symbol. Do not change it to 3.14.

  2. Choose the area formula

    The area formula is A=πr2A = \pi r^2. It uses the radius squared.

    Watch out: the circumference formula, C=2πrC = 2\pi r, also uses π and the radius. It gives 2×π×5=10π2 \times \pi \times 5 = 10\pi, the distance around the circle in inches. That is not the area.

  3. Square the radius

    Squaring a number means multiplying it by itself:

    r2=52=5×5=25r^2 = 5^2 = 5 \times 5 = 25

    Watch out: squaring is not doubling. 5×2=105 \times 2 = 10 would lead to 10π again.

  4. Multiply by π

    Put 25 into the formula in place of r²:

    A=π×25=25πA = \pi \times 25 = 25\pi

    The area is 25π square inches. Write the number first and then π, the way you write 3x for 3 times x.

    Watch out: leave π as π. Using 3.14 gives 78.5, which is not accepted. Dropping π leaves 25, which is not the area.

  5. Enter it with the π key

    On the test, the keypad has a π key in its bottom left corner. Press 2, then 5, then the π key. You do not need a times sign. Enter only the answer, with no A = and no units. On this site, type 25π or 25pi.

    So you would enter 25π.

Check your answer

Estimate. π is a little more than 3, so 25π is a little more than 75 square inches. A square drawn around the circle is 10 inches on each side, so its area is 100 square inches. The circle fits inside that square, so its area must be less than 100. It is.

More questions like this

The original DOE question
Question 28 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 B Math, page 100, question 28.

Question 17 of 120

2026 Practice Test B Math, question 31

Question

A pizza shop offers a choice of 3 sizes (small, medium, and large) and 7 different toppings. Different pizzas can be created by changing the size and the choice of toppings. If Cody wants to order a pizza with exactly 2 different toppings, how many different pizzas can he create?

  • A6
  • B21
  • C63
  • D126
Show the answer

Correct answer: C

Figure out how many different topping pairs are possible. Use 1, 2, 3, 4, 5, 6, 7 to represent the toppings, and create a list of possible pairs:

1,2; 1,3; 1,4; 1,5; 1,6; 1,7
2,3; 2,4; 2,5; 2,6; 2,7
3,4; 3,5; 3,6; 3,7
4,5; 4,6; 4,7
5,6; 5,7
6,7

So there are 21 different topping combinations for one pizza.

Since there are 3 pizza sizes, multiply the total number of combinations by 3 to get the total number of different pizzas Cody can create: 3×21=633 \times 21 = 63.

Question analysis

Cody orders a pizza with exactly 2 of 7 toppings in one of 3 sizes, and students count how many different pizzas he could create. It tests counting combinations, where the order of the toppings does not matter, combined with a second choice. The challenge is to count every possible pizza exactly once.

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 In grade 7, compound events are usually ordered pairs, such as a coin with a number cube. Here an unordered pair of toppings is chosen from seven and then matched with three sizes, a combination count beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 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 question packs two counts into one sentence, each with its own trap. Order does not matter among the toppings, while the size multiplies every pair, and students tend to remember one of these and forget the other. Listing pairs systematically, each topping matched only with later ones, keeps the first count honest, and the stem’s “changing the size and the choice of toppings” is the reminder that size counts too.

  1. Combines ideas moderate Working it out

    Two counts are joined: the unordered pairs of toppings from 7, then the 3 sizes each pair can come in, and a mistake in either count changes the total. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Students must know that a pizza with mushroom and onion is the same as one with onion and mushroom, so the ordered count 7 × 6 has to be halved. (Anderson & Krathwohl, 2001)

Strategy: Number the toppings 1 to 7, list each pair once by matching every topping only with higher-numbered ones, count the list, then bring in the choice of size.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The counting principle
total=m×n\text{total} = m \times n
m is the number of ways to make one choice, and n is the number of ways to make a second choice. Every first choice can go with every second choice, so you multiply. Here, every pair of toppings can come in every size.
Pairs when order does not matter
pairs=n×(n−1)2\text{pairs} = \frac{n \times (n - 1)}{2}
n is the number of items to choose from. There are n ways to pick a first item and n − 1 ways to pick a different second one. That counts each pair twice, once in each order. Dividing by 2 counts each pair once.

  1. Sort out the question

    Each of Cody’s pizzas has one size and exactly 2 different toppings. There are 3 sizes and 7 toppings. Pizzas differ by “changing the size and the choice of toppings.” So a small and a large with the same toppings are two different pizzas. You need to count every pizza once.

  2. List the topping pairs

    Order does not matter: onion with peppers is the same pizza as peppers with onion. So number the toppings 1 to 7, and pair each one only with bigger numbers. Then no pair shows up twice.

    ToppingPairs withPairs
    12, 3, 4, 5, 6, 76
    23, 4, 5, 6, 75
    34, 5, 6, 74
    45, 6, 73
    56, 72
    671

    6+5+4+3+2+1=216 + 5 + 4 + 3 + 2 + 1 = 21

    There are 21 different pairs of toppings.

  3. Check the pairs with the formula

    Use the pairs formula with 7 toppings:

    7×62=422=21\frac{7 \times 6}{2} = \frac{42}{2} = 21

    Watch out: 7 × 6 = 42 counts every pair twice, once as 1 then 2 and once as 2 then 1. Keeping 42 leads to 42×3=12642 \times 3 = 126, choice D.

  4. Bring in the sizes

    Each of the 21 topping pairs can come in any of the 3 sizes. Use the counting principle:

    21×3=6321 \times 3 = 63

    Watch out: stopping at 21 leaves out the sizes, which gives choice B. Choice A, 6, multiplies the 3 sizes by the 2 toppings on one pizza. That does not count pizzas.

  5. Give the answer

    There are 21 topping pairs, and each one comes in 3 sizes. Cody can create 63 different pizzas. So the answer is C, 63.

Check your answer

Count one size at a time. A small pizza can have any of the 21 pairs, so there are 21 small pizzas. There are also 21 medium and 21 large ones: 21+21+21=6321 + 21 + 21 = 63. That is more than the 21 pairs but fewer than 126, so it is reasonable.

More questions like this

The original DOE question
Question 31 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 B Math, page 103, question 31.

Question 18 of 120

2026 Practice Test B Math, question 46

Question

Amiya subscribes to an online streaming service. She pays a onetime setup fee of $50, plus $20 per month for service.

Write an equation that can be used to determine the total amount Amiya will pay, y, for x months of service.

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

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: y = 20x + 50

Solution: y=20x+50y = 20x + 50

An equation of the form px+q=rpx + q = r can be used to represent the total amount Amiya will pay.

Using the information from the scenario, the onetime setup fee of $50 is the constant term (q), and the $20 per month of service is the coefficient (p) of the number of months (x).

So, the equation that can be used to determine the total amount Amiya will pay, y, for x months of service is y=20x+50y = 20x + 50.

Question analysis

Amiya pays a one-time setup fee plus a monthly charge for a streaming service, and students type an equation for her total cost y after x months. It tests writing a linear equation from a word problem with a starting amount and a rate. The two dollar amounts play different roles, so which one goes with x has to be thought through.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an equation for a fixed fee plus a steady monthly charge is a familiar grade 7 task, and the numbers are whole. Typing the full equation, with both variables and no choices to compare, makes it less forgiving.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are small and whole, so the work is in the roles. The $50 is paid once and the $20 every month, and a student who places them in the order they appear writes the fee as the monthly rate; others leave the fee out entirely. Checking the equation at x = 1 and x = 2, where Amiya has paid $70 and then $90, shows at once whether each amount is in the right place.

  1. Turning words into math moderate Understanding the question

    Only the words “onetime setup fee” and “per month” tell which amount is multiplied by x and which is added once, and the $50 comes first in the sentence. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The whole equation, with y, an equals sign and x, is built on a keypad with no choices, so a reversed or incomplete equation looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Work out Amiya's total for 1 month and for 2 months by hand, see which amount is added again each month, and build an equation that gives both totals.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee and a rate
y=(rate)x+(fee)y = (\text{rate})x + (\text{fee})
y is the total cost in dollars, and x is the number of months. The rate is the amount paid every month. You multiply it by x because it is paid once for each month. The fee is paid only once, so you add it once, no matter how many months.

  1. Sort out the question

    Amiya pays two kinds of charges:

    • a setup fee of $50, paid one time only;
    • a charge of $20 for every month of service.

    y is her total cost in dollars, and x is the number of months. You need an equation that links them. The answer is an equation, not a number.

  2. Work out a few totals

    Find the total for a few months by hand. Each month adds one more $20:

    MonthsMonthly charges ($)Total ($)
    11×20=201 \times 20 = 2050+20=7050 + 20 = 70
    22×20=402 \times 20 = 4050+40=9050 + 40 = 90
    33×20=603 \times 20 = 6050+60=11050 + 60 = 110

    The $50 shows up once in every total. The $20 shows up once for each month.

  3. Match each number to its job

    Use the fee-and-rate model, y=(rate)x+(fee)y = (\text{rate})x + (\text{fee}). The rate is the amount that repeats each month, $20. It multiplies the months, so it becomes 20x. The fee, $50, is paid once, so it is added once.

    Watch out: the $50 comes first in the question, but it is not the rate. y=50x+20y = 50x + 20 charges $50 a month. y=70xy = 70x charges the fee every month. y=20xy = 20x forgets the fee.

  4. Write the equation

    Put the rate and the fee into the model:

    y=20x+50y = 20x + 50

    Here 20x means 20 times x. An equal form is fine too, such as y=50+20xy = 50 + 20x, because you can add in either order.

  5. Enter it on the keypad

    Type the whole equation, including y =. Without it, you have an expression, not an equation. On the test, the keypad has x, y and = keys. Press y, =, 2, 0, x, +, 5, 0. You do not need a times sign or dollar signs. On this site, type it the same way.

    So you would enter y = 20x + 50.

Check your answer

Test the equation with the totals you found. For 1 month, y=20×1+50=70y = 20 \times 1 + 50 = 70, and Amiya pays $50 + $20 = $70. For 2 months, y=20×2+50=90y = 20 \times 2 + 50 = 90. Both match. For 0 months, y = 50, just the setup fee.

More questions like this

The original DOE question
Question 46 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 B Math, page 118, question 46.

Question 19 of 120

2026 Practice Test C Math, question 1

Question

What is the prime factorization of 756?

  • A22⋅32⋅212^{2} \cdot 3^{2} \cdot 21
  • B22⋅3⋅7⋅92^{2} \cdot 3 \cdot 7 \cdot 9
  • C22⋅7⋅272^{2} \cdot 7 \cdot 27
  • D22⋅33⋅72^{2} \cdot 3^{3} \cdot 7
Show the answer

Correct answer: D

Determine the prime factors of 756 using a factor tree:

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

The prime factors are 2, 2, 3, 3, 3, and 7, or 2⋅2⋅3⋅3⋅3⋅72 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7, which can be expressed using exponents as 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

Question analysis

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

NYS standards
  • 6.NS.4 common factors and multiples
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Factor trees and exponent notation are grade 6 skills, and 756 splits quickly by small divisors. What adds a little is that every choice multiplies out to 756, so checking the product alone settles nothing.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 11, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    2² · 3² · 21 shares the right answer's 2² and differs only in 3² · 21 against 3³ · 7, and both equal 756, so neither an estimate nor a product check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Students must know that a prime factorization may hold only primes, and must spot that 21, 9 and 27 are composite even though none of them is even. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You need the prime factorization of 756. That means writing 756 as a product of prime numbers only. A product is the answer to a multiplication. A prime number has exactly two factors, 1 and itself. Examples are 2, 3, 5 and 7.

    All four choices multiply to 756. So multiplying a choice back does not pick the answer. You must check that every factor is prime.

  2. Divide by 2 while you can

    A factor tree splits a number into two factors, again and again. Start with the smallest prime, 2. 756 is even, so 2 goes into it:

    756=2×378756 = 2 \times 378

    378 is even too:

    378=2×189378 = 2 \times 189

    189 is odd, so 2 does not go into it. You have pulled out two 2s.

  3. Divide by 3 while you can

    Add the digits of 189: 1+8+9=181 + 8 + 9 = 18. A number can be divided by 3 when its digits add up to a number that 3 goes into. 3 goes into 18, so it goes into 189:

    189=3×63189 = 3 \times 63

    63=3×2163 = 3 \times 21

    21=3×721 = 3 \times 7

    7 is prime, so the tree stops. You have pulled out three 3s and one 7.

  4. Write the factors with exponents

    Collect every prime you pulled out:

    756=2×2×3×3×3×7756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7

    Use the exponent rule. Two 2s make 222^{2}, and three 3s make 333^{3}. The choices use a raised dot, ·, for times. So 756 is 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

  5. Choose the answer

    Watch out: each wrong choice is a factor tree that stopped too soon. A ends in 21, which is 3 × 7. B has 9, which is 3 × 3. C has 27, which is 3 × 3 × 3. Being odd does not make a number prime.

    Only D uses primes alone. So the answer is D, 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

Check your answer

Multiply your answer back: 22×33=4×27=1082^{2} \times 3^{3} = 4 \times 27 = 108, and 108×7=756108 \times 7 = 756. Then look at each factor. 2, 3 and 7 are all prime. D passes both tests.

More questions like this

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

Question 20 of 120

2026 Practice Test C Math, question 2

Question

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

  • A13\frac{1}{3}
  • B12\frac{1}{2}
  • C2
  • D3
Show the answer

Correct answer: B

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

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

Subtract 4 from both sides:

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

Multiply both sides of the equation by m:

8=16m8 = 16m

Solve for m:

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

Simplify:

(8(÷8)16(÷8))=(12)\left(\frac{8(\div 8)}{16(\div 8)}\right) = \left(\frac{1}{2}\right)

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Undoing an addition and then a division is typical two-step grade 7 equation work. The unknown sitting in a denominator is less common than px + q = r, which makes the last step easier to get backward.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    The unknown is the divisor, so 8 ÷ m = 16 ends in m = 8 ÷ 16, a quotient less than 1; the familiar order, 16 ÷ 8, gives a whole number instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    Students must see that (8 ÷ m) is the fraction 8/m and that freeing m from a denominator means multiplying both sides by it, a move grade 7 equations rarely need. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The equation is (8÷m)+4=20(8 \div m) + 4 = 20. You need the value of m. Here m is the number you divide 8 by. In words: 8 divided by some number, plus 4, makes 20.

  2. Undo the plus 4

    A fraction bar means divide, so 8÷m8 \div m is the same as 8m\frac{8}{m}. Subtract 4 from both sides to undo the plus 4:

    8m+4−4=20−4\frac{8}{m} + 4 - 4 = 20 - 4

    8m=16\frac{8}{m} = 16

    Now 8 divided by m equals 16.

    Watch out: adding 4 instead gives 24 on the right. That leads to choice A, 1/3, or to D, 3.

  3. Get m out of the bottom

    m is on the bottom of the fraction. Multiply both sides by m to free it:

    8=16×m8 = 16 \times m

    Now divide both sides by 16:

    m=816=8÷816÷8=12m = \frac{8}{16} = \frac{8 \div 8}{16 \div 8} = \frac{1}{2}

    8 goes into both 8 and 16, so 8/16 simplifies to 1/2.

  4. Choose the answer

    Watch out: it is tempting to divide 16 by 8 and get 2, choice C. But ask yourself: 8 divided by what number makes 16? To get an answer bigger than 8, you must divide by a number less than 1.

    So the answer is B, 1/2.

Check your answer

Put 1/2 back into the equation. How many halves are in 8? There are 16, so 8÷12=168 \div \frac{1}{2} = 16. Then 16+4=2016 + 4 = 20. That matches the equation, so m = 1/2 is right.

More questions like this

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

Question 21 of 120

2026 Practice Test C Math, question 3

Question

6x=x−1,6806x = x - 1{,}680

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

  • A336
  • B240
  • C−240
  • D−336
Show the answer

Correct answer: D

Simplify 6x=x−1,6806x = x - 1{,}680. First, subtract x from both sides of the equation. Then, divide both sides by 5.

6x−x=x−x−1,6806x - x = x - x - 1{,}680

5x=−1,6805x = -1{,}680

5x5=−1,6805\frac{5x}{5} = \frac{-1{,}680}{5}

x=−336x = -336

Question analysis

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

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.NS.2b rational number division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 An unknown on both sides of the equals sign is a grade 8 topic, yet gathering the x terms here takes one subtraction and one division, the same work as a two-step grade 7 equation.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 12 words.
Corrected from the DOE handbook
  • A line of the explanation’s working, printed without its fraction bars and equals sign, was restored.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    Subtracting x from both sides leaves 5x = −1,680, and 1,680 ÷ 5 = 336 has to be worked out by hand while the negative sign is carried through to the result. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The choices are two sizes, 336 and 240, each offered with both signs, so 336 and −336 are written the same except for the minus; only tracking the sign of 1,680 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The equation is 6x=x−1,6806x = x - 1{,}680. You need the value of x. Here 6x means 6 times x. The unknown x is on both sides. So first gather it on one side.

  2. Subtract x from both sides

    Take x away from both sides. On the right, x − x is 0, so only −1,680 is left:

    6x−x=x−x−1,6806x - x = x - x - 1{,}680

    5x=−1,6805x = -1{,}680

    Six x’s take away one x leaves five x’s.

    Watch out: moving x to the left without changing its sign gives 7x. Dividing by 7 then leads to choice C, −240, or B, 240.

  3. Divide both sides by 5

    5x means 5 times x. Divide both sides by 5 to undo it:

    5x5=−1,6805\frac{5x}{5} = \frac{-1{,}680}{5}

    Find the size first. Split 1,680 into 1,500 + 180:

    1,500÷5+180÷5=300+36=3361{,}500 \div 5 + 180 \div 5 = 300 + 36 = 336

    A negative divided by a positive is negative:

    x=−336x = -336

  4. Choose the answer

    Watch out: the minus sign is easy to drop. Positive 336 is choice A. Carry the sign of −1,680 all the way to the end.

    So the answer is D, −336.

Check your answer

Put −336 into both sides. The left side is 6×(−336)=−2,0166 \times (-336) = -2{,}016. The right side is −336−1,680=−2,016-336 - 1{,}680 = -2{,}016. The two sides match, so x = −336 is right.

More questions like this

The original DOE question
Question 3 as shown on the DOE’s digital test, page 76 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 76, question 3. This is the DOE’s version, before the corrections listed in the analysis.

Question 22 of 120

2026 Practice Test C Math, question 4

Question

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

  • A$2,025
  • B$9,900
  • C$11,100
  • D$13,500
Show the answer

Correct answer: C

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

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

275×36=9,900275 \times 36 = 9{,}900

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

9,900+1,200=11,1009{,}900 + 1{,}200 = 11{,}100

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

Question analysis

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

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 A total made of one payment up front plus equal monthly payments is a short whole-number calculation, simpler than grade 7 word problems with percents or negatives. The only conversion is from years to months.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 29 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The situation is familiar and the method direct, so this is one of the easier questions on the test. Points are lost in two places: turning 3 years into 36 payments, and remembering the money paid up front after the long multiplication. Each slip, like paying the down payment once a year, leads to a listed total, so checking that the answer holds exactly one $1,200 and 36 payments of $275 is worth the seconds.

  1. Much to keep track of minor Working it out

    The $1,200 down payment is paid once while the $275 repeats every month, so it has to be set aside while 36 payments are totaled and then added at the end. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Error-prone arithmetic minor Working it out

    The product 275 × 36 = 9,900 is a two-digit multiplication done by hand; splitting it as 275 × 4 = 1,100 and then 1,100 × 9 makes it quick and safe. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The person pays $1,200 once, at the start. This is the down payment. Then the person pays $275 every month for 3 years. You need the total amount paid for the car.

    The payments come once a month, but the time is given in years. So change the years into months first.

  2. Change 3 years into months

    Use the months formula:

    months=12×3=36\text{months} = 12 \times 3 = 36

    So the person makes 36 monthly payments.

    Watch out: multiplying $275 by 3 counts only 3 payments. That leads to choice A, $2,025.

  3. Add up the monthly payments

    Multiply $275 by 36. A quick way is to split 36 into 4 × 9:

    275×4=1,100275 \times 4 = 1{,}100

    1,100×9=9,9001{,}100 \times 9 = 9{,}900

    So the 36 payments add up to $9,900.

    Watch out: stopping here gives choice B, $9,900. The down payment is still missing.

  4. Add the down payment once

    Use the total-cost formula:

    total=1,200+9,900=11,100\text{total} = 1{,}200 + 9{,}900 = 11{,}100

    The person pays $11,100 in all. Adding $1,200 for every year counts it three times. That gives choice D, $13,500.

    So the answer is C, $11,100.

Check your answer

Estimate. $275 is a little less than $300, and 36×300=10,80036 \times 300 = 10{,}800. Add the $1,200: 10,800+1,200=12,00010{,}800 + 1{,}200 = 12{,}000. The real total should be a little less than $12,000. $11,100 fits.

More questions like this

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

Question 23 of 120

2026 Practice Test C Math, question 5

Question

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

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

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

Show the answer

Correct answer: 4n – 12

Let n represent the number of pages that Maya read.

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

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

This means that together, they read 4n4n pages.

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

Ali’s pages =4n−12= 4n - 12

Question analysis

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

NYS standards
  • 6.EE.6 variables representing unknown numbers
  • 6.EE.2a writing algebraic expressions
  • 6.EE.3 generating equivalent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing an expression from words is grade 6 content, but this chains three relationships, and the last compares the pair with Ali rather than describing Ali directly. Typing the answer with no choices to compare adds risk.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 52 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “Together, Maya and Trevor read 12 more pages than Ali read” compares the pair with Ali, so Ali read 12 fewer than 4n, though the word more invites adding 12. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    Every count is written in terms of n, so there is no page total to check an expression against unless the student picks a value for n and tests it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The expression is typed into a box with no choices to compare, so an entry such as 4n + 12 gets no warning that the comparison was read backward. (Enright, Morley & Sheehan, 2002)

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

Traps

  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Maya read n pages. n is a variable, a letter that stands for a number you don’t know. Trevor read 3 times as many pages as Maya. Together they read 12 more pages than Ali. You need Ali’s pages, written with n.

  2. Write Trevor’s pages

    Three times as many as n is 3 times n:

    Trevor=3n\text{Trevor} = 3n

    A number written next to a letter means multiply. So 3n means 3 × n pages.

  3. Add Maya’s and Trevor’s pages

    Together means add. Maya read 1 group of n, and Trevor read 3 groups:

    n+3n=4nn + 3n = 4n

    Together they read 4n pages.

    Watch out: do not leave Maya out. Using only Trevor’s 3n gives 3n − 12, which is wrong.

  4. Decide who read more

    The pair read 12 more than Ali. So Ali read 12 fewer than the pair. Take 12 away from their total:

    Ali=4n−12\text{Ali} = 4n - 12

    Watch out: the word more tempts you to add 12. But 4n + 12 would mean Ali read more than the pair. And 12 − 4n subtracts in the wrong order.

  5. Enter the expression

    On the test, the keypad has an n key. Press 4, then n. Then press the minus key beside the plus key, then 1 and 2. You don’t need a times sign. On this site, type 4n-12.

    So you would enter 4n − 12. Any expression equal to it, such as −12 + 4n, is also right.

Check your answer

Pick a number for n, such as 10. Maya read 10 pages and Trevor read 30, so together they read 40. Your expression gives Ali 4×10−12=284 \times 10 - 12 = 28 pages. Is 40 twelve more than 28? Yes, 28+12=4028 + 12 = 40. So the expression works.

More questions like this

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

Question 24 of 120

2026 Practice Test C Math, question 6

Question

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

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

  • A36
  • B18
  • C6
  • D5
Show the answer

Correct answer: A

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

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

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

Solve for x:

5x=1805x = 180

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

x=36x = 36

Question analysis

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

NYS standards
  • 7.G.5 angle relationships and equations
  • 8.G.5 angle sums, transversals, angle-angle similarity
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 The angle fact needed, about consecutive angles of a parallelogram, rests on grade 8 parallel-line reasoning. Once it is known, the work is one short equation, like a typical grade 7 problem about angle sums.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know moderate Working it out

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write an equation for the angles

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

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

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

  3. Solve for x

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

    5x=1805x = 180

    Divide both sides by 5 to get x alone:

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

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

  4. Choose the answer

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

    So the answer is A, 36.

Check your answer

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

More questions like this

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

Question 25 of 120

2026 Practice Test C Math, question 7

Question

−12>6x-12 > 6x

For what values of x is the above inequality true?

  • Ax>−6x > -6
  • Bx<−6x < -6
  • Cx<−2x < -2
  • Dx>−2x > -2
Show the answer

Correct answer: C

Divide to solve for x:

(−12)6>6x6\frac{(-12)}{6} > \frac{6x}{6}

−2>x-2 > x

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

x<−2x < -2

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 inequalities usually take two steps, such as 3x + 4 > 19. This one needs a single division, and the care goes into reading the result correctly when the variable ends up on the right of the symbol.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 10, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    x < −2 and x > −2 share the boundary −2 and differ only in direction, so finding the right number settles nothing until the direction is read correctly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Students must know that the symbol flips only when both sides are multiplied or divided by a negative number, and that −2 > x says the same as x < −2. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The inequality is −12>6x-12 > 6x. The symbol > means is greater than. So it says −12 is greater than 6 times x. You need every value of x that makes this true.

  2. Divide both sides by 6

    To get x alone, divide both sides by 6:

    −126>6x6\frac{-12}{6} > \frac{6x}{6}

    −2>x-2 > x

    You divided by 6, which is positive. So the symbol does not flip.

    Watch out: −12 is negative, but that does not matter. Only dividing by a negative number flips the symbol.

  3. Put x first

    −2>x-2 > x says −2 is greater than x. That means x is less than −2. Swap the sides and turn the symbol around:

    x<−2x < -2

    Watch out: keeping > when you swap the sides gives choice D, x>−2x > -2. That is the opposite meaning.

  4. Choose the answer

    Choices A and B use −6. That comes from adding 6 to −12 instead of dividing by 6. The right boundary is −2, and x is below it.

    So the answer is C, x<−2x < -2.

Check your answer

Test a number below −2, such as −3. Then 6×(−3)=−186 \times (-3) = -18, and −12>−18-12 > -18 is true. Now test 0, which is above −2. Then 6×0=06 \times 0 = 0, and −12>0-12 > 0 is false. So only numbers below −2 work.

More questions like this

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

Question 26 of 120

2026 Practice Test C Math, question 8

Question

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

Which fractions are equivalent to the decimal 0.75?

Select the four correct answers.

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

  • A34\frac{3}{4}
  • B1520\frac{15}{20}
  • C75100\frac{75}{100}
  • D25\frac{2}{5}
  • E68\frac{6}{8}
  • F810\frac{8}{10}
Show the answer

Correct answer: A, B, C, and E

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

34→3÷4=0.75\frac{3}{4} \rightarrow 3 \div 4 = 0.75
This is the most common fraction equal to 0.75.

1520→15÷20=0.75\frac{15}{20} \rightarrow 15 \div 20 = 0.75
Simplifies to 3/4, because 15 and 20 can both be divided by 5.

75100→75÷100=0.75\frac{75}{100} \rightarrow 75 \div 100 = 0.75
This is a direct match to 0.75 when written as a decimal.

25→2÷5=0.4\frac{2}{5} \rightarrow 2 \div 5 = 0.4, which is less than 0.75.

68→6÷8=0.75\frac{6}{8} \rightarrow 6 \div 8 = 0.75
Simplifies to 34\frac{3}{4} because both 6 and 8 can be divided by 2.

810→8÷10=0.8\frac{8}{10} \rightarrow 8 \div 10 = 0.8, which is more than 0.75.

Question analysis

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

NYS standards
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Matching fractions to a decimal such as 0.75 builds on grade 4 to 6 work, and the denominators here divide easily. The added demand is the format: four of the six must be chosen, and one wrong or missing pick loses the item.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Every choice must be checked moderate Choosing the answer

    Six fractions with six different denominators have to be tested one by one, by dividing or by simplifying, since 15/20 and 8/10 cannot be judged by how they look. (Embretson & Wetzel, 1987)

  2. A very close wrong answer minor Choosing the answer

    8/10 is 0.8, only 0.05 above 0.75, and it follows 6/8 by adding 2 to the top and the bottom, so it looks like one more member of the family. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Much to keep track of minor Working it out

    Four of the six must be chosen and all four must be right for the item to count, so each verdict has to be kept while the remaining fractions are tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Six fractions are given, A to F. You need the four that equal the decimal 0.75. Equivalent means equal in value. Test every choice, because some fractions look alike but are not equal.

  2. Write 0.75 as a fraction

    0.75 means 75 hundredths:

    0.75=751000.75 = \frac{75}{100}

    So choice C works. Now simplify it. 25 goes into both 75 and 100:

    75÷25100÷25=34\frac{75 \div 25}{100 \div 25} = \frac{3}{4}

    So choice A, 3/4, works too.

  3. Test choices B and E

    Simplify each one to see if it becomes 3/4:

    • B: 5 goes into 15 and 20. 15÷520÷5=34\frac{15 \div 5}{20 \div 5} = \frac{3}{4}. It works.
    • E: 2 goes into 6 and 8. 6÷28÷2=34\frac{6 \div 2}{8 \div 2} = \frac{3}{4}. It works.

  4. Test choices D and F

    Divide the top by the bottom:

    • D: 2÷5=0.42 \div 5 = 0.4. That is less than 0.75.
    • F: 8÷10=0.88 \div 10 = 0.8. That is more than 0.75.

    Watch out: F looks like 6/8 with 2 added to the top and the bottom. Adding the same number to both changes the value. Only multiplying or dividing both keeps it.

  5. Name the four answers

    A, B, C and E all equal 0.75. On the test, you check the box beside each of them. On this site, you choose all four, then check your answer.

    So the answers are A, B, C and E.

Check your answer

Turn each answer into a decimal. 3÷43 \div 4, 15÷2015 \div 20, 75÷10075 \div 100 and 6÷86 \div 8 all give 0.75. Exactly four choices passed, and the question asks for four. So none were missed.

More questions like this

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

Question 27 of 120

2026 Practice Test C Math, question 9

Question

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

  • A10
  • B15
  • C24
  • D30
Show the answer

Correct answer: C

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

15 miles1 hour=15 miles60 minutes=1 mile4 minutes\frac{15\text{ miles}}{1\text{ hour}} = \frac{15\text{ miles}}{60\text{ minutes}} = \frac{1\text{ mile}}{4\text{ minutes}}, or 14\frac{1}{4} mile per minute.

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

The distance to school is 6 miles, so it will take 6×4=246 \times 4 = 24 minutes to ride to school.

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Distance, speed and time with whole numbers is grade 6 rate work. The only extra demand is an answer in minutes when the speed is per hour, a conversion grade 7 students make routinely.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

    Nothing on the page says that time is distance divided by speed or that an hour has 60 minutes, and the bold word minutes requires both. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    6 ÷ 15 is less than 1, 0.4 hour, and turning it into minutes means 0.4 × 60; the cleaner route, 60 ÷ 15 = 4 minutes a mile, avoids decimals. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Maria rides at a constant speed of 15 miles per hour. Constant means it never changes. School is 6 miles away. You need the time in minutes, not hours. You must know that one hour has 60 minutes.

  2. Find the minutes for one mile

    15 miles per hour means Maria rides 15 miles in 60 minutes. Use the minutes-per-mile formula:

    60÷15=460 \div 15 = 4

    So each mile takes Maria 4 minutes.

  3. Find the minutes for 6 miles

    Each of the 6 miles takes 4 minutes:

    6×4=246 \times 4 = 24

    So the ride takes 24 minutes.

    Watch out: 15 is the speed, not a time, so choice B is a trap. And 60÷6=1060 \div 6 = 10, choice A, ignores the speed.

  4. Choose the answer

    Another way is the time formula. 6÷15=0.46 \div 15 = 0.4, so the ride takes 0.4 of an hour. In minutes, that is 0.4×60=240.4 \times 60 = 24.

    Watch out: rounding 0.4 hour up to half an hour gives choice D, 30.

    So the answer is C, 24.

Check your answer

Check with the speed. 24 minutes is 2460=25\frac{24}{60} = \frac{2}{5} of an hour. In a full hour Maria rides 15 miles. So in 2/5 of an hour she rides 25×15=6\frac{2}{5} \times 15 = 6 miles. That is exactly the distance to school.

More questions like this

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

Question 28 of 120

2026 Practice Test C Math, question 10

Question

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

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

  • A17
  • B117
  • C163
  • D173
Show the answer

Correct answer: C

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

180∘=73∘+90∘+x180^{\circ} = 73^{\circ} + 90^{\circ} + x

180∘=163∘+x180^{\circ} = 163^{\circ} + x

17∘=x17^{\circ} = x

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

s=180∘−17∘s = 180^{\circ} - 17^{\circ}

s=163∘s = 163^{\circ}

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

Question analysis

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

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle problems use supplementary, vertical and adjacent angles. This one also needs the angle sum of a triangle and the angles formed by parallel lines and a transversal, grade 8 facts, chained across a busy figure.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 3, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

    The 73° sits between line p and the vertical segment, not between p and m, and r is the obtuse angle above m on the left of line n, far from every marked angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Three facts chain together: the triangle's angles sum to 180°, the angles along straight line m sum to 180°, and corresponding angles at the parallel lines n and p are equal. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Something you must already know minor Working it out

    The angle pairs formed by a line crossing two parallel lines are a grade 8 topic, and the question uses them without naming them. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Lines n and p are parallel, and line m crosses both. Near line p, a straight-up segment meets line m. The small square shows a right angle, 90°. Line p, line m and that segment make a small triangle. Its top angle is 73°.

    You need r, the wide angle where line n crosses line m.

  2. Find the triangle’s third angle

    The third angle is where line p crosses line m. Call it x. Use the triangle rule:

    73+90+x=18073 + 90 + x = 180

    x=180−163=17x = 180 - 163 = 17

    So that angle is 17°.

    Watch out: this is not r. Stopping here gives choice A, 17.

  3. Use the straight line m

    The 17° angle sits above line m, on the right of line p. Beside it, on the left of line p, is a wide angle. Call it s. The two angles fill the straight line m, so they add to 180°:

    s=180−17=163s = 180 - 17 = 163

    So s is 163°.

    Watch out: borrow carefully. 180 − 17 is 163, not 173, choice D.

  4. Carry the angle to line n

    Angle s is above line m and to the left of line p. Angle r is above line m and to the left of line n. They are in the same position, so they are corresponding angles. Since n and p are parallel:

    r=s=163r = s = 163

    So the answer is C, 163.

Check your answer

Look at the figure. Angle r is wider than a right angle, so it must be more than 90°. That rules out 17. Also, the narrow angle beside r, on the right of line n, matches the 17° angle at line p. And 163+17=180163 + 17 = 180, as it should on a straight line.

More questions like this

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

Question 29 of 120

2026 Practice Test C Math, question 11

Question

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

  • A198
  • B612
  • C620
  • D632
Show the answer

Correct answer: B

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

5×576=2,8805 \times 576 = 2{,}880

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

6×378=2,2686 \times 378 = 2{,}268

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

2,880−2,268=6122{,}880 - 2{,}268 = 612

Question analysis

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

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Whole-number multiplication and subtraction in a three-step word problem is easier than grade 7 work with fractions, decimals or negatives. Because the products are large, a slip or a rounded estimate easily gives another choice.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Allison has 5 albums with 576 stamps in each. She moves all her stamps into 6 new albums. Each new album holds 378 stamps. Some stamps will not fit. You need the number left over.

  2. Count all her stamps

    Use the equal-groups formula. Split 576 into 500 + 76 to make it easy:

    5×576=5×500+5×765 \times 576 = 5 \times 500 + 5 \times 76

    2,500+380=2,8802{,}500 + 380 = 2{,}880

    Allison has 2,880 stamps.

  3. Count what the new albums hold

    Do the same for the 6 new albums. Split 378 into 300 + 78:

    6×378=6×300+6×786 \times 378 = 6 \times 300 + 6 \times 78

    1,800+468=2,2681{,}800 + 468 = 2{,}268

    The new albums hold 2,268 stamps in all.

  4. Subtract to find the leftovers

    The stamps left over are the ones she has minus the ones that fit:

    2,880−2,268=6122{,}880 - 2{,}268 = 612

    Watch out: 576−378=198576 - 378 = 198, choice A, compares just one old album with one new one. Rounding to 580 and 380 first gives 620, choice C. Rounding only 576 gives 632, choice D. Use the exact numbers.

    So the answer is B, 612.

Check your answer

Add the leftovers to the stamps that fit: 2,268+612=2,8802{,}268 + 612 = 2{,}880. That is exactly the number Allison started with. So no stamps were lost or counted twice.

More questions like this

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

Question 30 of 120

2026 Practice Test C Math, question 12

Question

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

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

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

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

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

00.10.20.30.40.50.60.70.80.91

Show the answer

Correct answer: the point 0.2

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

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

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Something you must already know minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the spinner’s chance

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

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

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

  3. Find the coin’s chance

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

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

  4. Multiply the two chances

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

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

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

  5. Select the point

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

    So the answer is the point at 0.2.

Check your answer

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

More questions like this

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

Question 31 of 120

2026 Practice Test C Math, question 13

Question

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

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

What is the result?

  • A13.5
  • B135
  • C135.2
  • D135.3
Show the answer

Correct answer: D

Begin by multiplying 135.28 by 10:

135.28×10=1,352.8135.28 \times 10 = 1{,}352.8

Add 0.5:

1,352.8+0.5=1,353.31{,}352.8 + 0.5 = 1{,}353.3

Drop the digits after the decimal point, so 1,353.31{,}353.3 becomes 1,3531{,}353.

Divide by 10:

1,353÷10=135.31{,}353 \div 10 = 135.3

Question analysis

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

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Multiplying and dividing a decimal by 10 and adding 0.5 are grade 5 and 6 skills. The four steps are spelled out, so the demand is following them exactly and in order, not choosing a method.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 3, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The steps spell out how to round 135.28 to the nearest tenth, but nothing says so, and they have to be followed exactly. The decisive moment is step 2, where adding 0.5 pushes 1,352.8 past 1,353; skip it and the digits are simply cut, not rounded. Students who recognize the routine can check the result directly, since 135.28 lies closer to 135.3 than to 135.2.

  1. A very close wrong answer moderate Choosing the answer

    135.3 and 135.2 differ only in the tenths digit, and 135 shares their whole-number part, so a rough sense of the result's size rules out nothing. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Many steps minor Working it out

    Four listed steps move the decimal point twice, and doing them in another order, or skipping one, still produces a believable number near 135. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    Adding 0.5 to 1,352.8 carries into the ones place, since 8 tenths and 5 tenths make 1.3, which turns 1,352 into 1,353 before the decimals are dropped. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You start with 135.28 and do four steps, in order. Write down the number after each step. Doing them in your head, or in another order, leads to a wrong choice. You need the number at the end.

  2. Multiply by 10

    Step 1. Multiplying by 10 moves the decimal point one place to the right:

    135.28×10=1,352.8135.28 \times 10 = 1{,}352.8

    The digits stay the same. Only the decimal point moves.

  3. Add 0.5

    Step 2. Line up the decimal points. 8 tenths plus 5 tenths is 13 tenths. That is 1 whole and 3 tenths, so carry the 1 into the ones place:

    1,352.8+0.5=1,353.31{,}352.8 + 0.5 = 1{,}353.3

    Watch out: skipping this step, or forgetting to carry, keeps 1,352. That leads to choice C, 135.2.

  4. Drop the digits after the point

    Step 3. Cut off the digits after the decimal point. Do not round. So 1,353.3 becomes 1,353.

    Watch out: do step 3 before step 4. Dividing first gives 135.33. Dropping its decimals then gives 135, choice B.

  5. Divide by 10

    Step 4. Dividing by 10 moves the decimal point one place to the left:

    1,353÷10=135.31{,}353 \div 10 = 135.3

    Skipping step 1 gives 13.5, choice A. So the answer is D, 135.3.

Check your answer

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

More questions like this

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

Question 32 of 120

2026 Practice Test C Math, question 14

Question

If x=−2x = -2, what is the value of |x+x2+x3|−|x+2x+3x||x + x^{2} + x^{3}| - |x + 2x + 3x|?

  • A−18
  • B−6
  • C−2
  • D2
Show the answer

Correct answer: B

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

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

Evaluate the terms involving exponents:

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

Evaluate the terms involving multiplication:

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

Simplify the expression:

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

|−6|−|−12|\lvert -6 \rvert - \lvert -12 \rvert

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

6−12=(−6)6 - 12 = (-6)

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    With x = −2 the square is 4 but the cube is −8, so the first bars hold −2 + 4 − 8 = −6, and the sign changes from each term to the next. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Six terms are substituted, two pairs of bars are simplified and made positive, and only then is 12 subtracted from 6, four stages that must come in a fixed order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You need the value of |x+x2+x3|−|x+2x+3x||x + x^{2} + x^{3}| - |x + 2x + 3x| when x = −2. The bars | | mean absolute value. Work out each pair of bars on its own. Subtract only at the end.

  2. Find the powers of −2

    Put −2 in for x. Keep each negative sign:

    (−2)2=(−2)×(−2)=4(-2)^{2} = (-2) \times (-2) = 4

    (−2)3=4×(−2)=−8(-2)^{3} = 4 \times (-2) = -8

    The square is positive, but the cube is negative.

    Watch out: taking (−2)3(-2)^{3} as +8 makes the first bars hold 10. That leads to choice C, −2.

  3. Work out the first bars

    Add the three terms inside the first pair of bars:

    |(−2)+4+(−8)|=|−6|=6|(-2) + 4 + (-8)| = |-6| = 6

    −2 + 4 is 2, and 2 + (−8) is −6. The distance from −6 to 0 is 6.

  4. Work out the second bars

    Here 2x means 2 times x, so it is −4. And 3x is −6:

    |(−2)+(−4)+(−6)|=|−12|=12|(-2) + (-4) + (-6)| = |-12| = 12

    All three terms are negative, so they add to −12. Its absolute value is 12.

  5. Subtract the second value

    Now subtract the second value from the first:

    6−12=−66 - 12 = -6

    Watch out: the bars make each part positive, but the difference can still be negative. Keeping −6 inside the first bars gives −6 − 12 = −18, choice A.

    So the answer is B, −6.

Check your answer

Check the second bars a faster way. x+2x+3x=6xx + 2x + 3x = 6x, and 6×(−2)=−126 \times (-2) = -12. That matches. Also, the second value, 12, is bigger than the first, 6. So the difference must be negative, as −6 is.

More questions like this

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

Question 33 of 120

2026 Practice Test C Math, question 15

Question

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

Enter your answer in the space.

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

Show the answer

Correct answer: 3

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

5.5 oz×6=33 oz5.5\text{ oz} \times 6 = 33\text{ oz}

So, 33 oz of orange juice costs $1.79.

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

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

Cross multiply to solve for x in the proportion:

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

33x=$229.1233x = \$229.12

x≈$6.94x \approx \$6.94

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

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

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

Round the difference to the nearest dollar.

$2.95 rounds to $3.

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3d converting measurement units
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Unit prices are grade 6 and 7 work, but this chains five steps with decimals and needs a fact the question does not give, the number of fluid ounces in a gallon. Typing a rounded difference, with no choices, adds risk.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
5 steps
Reading level
Question and choices: about grade 5, 60 words.
Corrected from the DOE handbook
  • A rounded value in the explanation is now marked as approximate (≈ instead of =).
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The calculation is long, and its key fact is missing from the page. Students who know the 128 fluid ounces still face a division by 33 that does not come out evenly, which tempts them to round the price per ounce to 5 cents and lose about 50 cents a gallon. Keeping the exact quotient to the end, or scaling 33 ounces up by about 3.9, keeps the difference near $2.95, which rounds to 3.

  1. Something you must already know moderate Working it out

    The question never says that a gallon holds 128 fluid ounces, yet the whole comparison depends on it; the 6-pack holds only 6 × 5.5 = 33 ounces. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Five moves follow one another: the ounces in a pack, the scale-up to 128 ounces, the division by 33, the subtraction of $3.99 and the rounding to a whole dollar. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    1.79 × 128 = 229.12 and 229.12 ÷ 33 are long decimal calculations without a calculator, and a slip in either moves the price of a gallon. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The rounded difference is typed with no choices, so an unrounded $2.95, or a figure rounded too soon, looks just as finished as the whole-dollar answer. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Milk costs $3.99 for 1 gallon. Orange juice comes in a 6-pack of 5.5-fluid-ounce bottles for $1.79. You need how much more a gallon of juice would cost than a gallon of milk. Round to the nearest dollar.

    The question doesn’t say how big a gallon is. You must know this fact: 1 gallon is 128 fluid ounces.

  2. Find the ounces in a 6-pack

    Each of the 6 bottles holds 5.5 fluid ounces:

    6×5.5=336 \times 5.5 = 33

    So $1.79 buys 33 fluid ounces of juice.

  3. Set up a proportion

    Let x be the cost in dollars of a gallon, 128 fluid ounces, of juice. The price per ounce stays the same. So ounces over dollars is the same for the pack and the gallon:

    331.79=128x\frac{33}{1.79} = \frac{128}{x}

    Cross-multiply:

    33×x=1.79×128=229.1233 \times x = 1.79 \times 128 = 229.12

  4. Find the cost of a gallon

    Divide both sides by 33:

    x=229.12÷33≈6.94x = 229.12 \div 33 \approx 6.94

    The sign ≈ means about equal. The exact value is 6.943…, so a gallon of juice costs about $6.94.

    Watch out: don’t round too early. One ounce costs about $0.054. Rounding that to $0.05 makes a gallon $6.40, and your answer becomes 2.

  5. Subtract the milk’s price

    Find how much more the juice costs:

    6.94−3.99=2.956.94 - 3.99 = 2.95

    A gallon of juice costs $2.95 more than a gallon of milk.

    Watch out: entering 7, the juice’s price, forgets to subtract the milk.

  6. Round and enter it

    The question asks for the nearest dollar. $2.95 is between $2 and $3. Its cents, 95, are 50 or more, so it rounds up to $3. Do not enter 2.95.

    Type only the number in the box. So you would enter 3.

Check your answer

Estimate. 128 ounces is almost 4 packs, since 4×33=1324 \times 33 = 132. Four packs cost 4×1.79=7.164 \times 1.79 = 7.16 dollars, so a gallon costs a bit less than that. Milk costs about $4. About $7 minus $4 is $3, which matches.

More questions like this

The original DOE question
Question 15 as shown on the DOE’s digital test, page 88 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 88, question 15. This is the DOE’s version, before the corrections listed in the analysis.

Question 34 of 120

2026 Practice Test C Math, question 16

Question

3+(−3)3−(−3)=\frac{3+(-3)}{3-(-3)} =

  • A−1
  • B0
  • C16\frac{1}{6}
  • D1
Show the answer

Correct answer: B

Simplify the expression.

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

3+(−3)3−(−3)=3−33+3=06\frac{3+(-3)}{3-(-3)} = \frac{3-3}{3+3} = \frac{0}{6}

The zero property of division states that zero divided by any nonzero number equals zero, so 06=0\frac{0}{6} = 0.

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding and subtracting a negative integer and dividing are grade 7 basics, and here the numbers are small. It is a short routine calculation, easier than typical grade 7 work, though the two signs invite a quick misreading.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 2 words.
Corrected from the DOE handbook
  • A grammatical error in the explanation was corrected.
  • A rule stated too broadly in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The top of the fraction is 3+(−3)3 + (-3), and the bottom is 3−(−3)3 - (-3). A fraction bar means divide. Work out the top and the bottom separately first. Then divide the top by the bottom.

  2. Work out the top

    Adding −3 is the same as subtracting 3:

    3+(−3)=3−3=03 + (-3) = 3 - 3 = 0

    A number plus its opposite is always 0.

    Watch out: do not add the 3s and keep the minus sign. That makes the top −6 and leads to choice A, −1.

  3. Work out the bottom

    Subtracting −3 is the same as adding 3:

    3−(−3)=3+3=63 - (-3) = 3 + 3 = 6

    Watch out: the top and bottom look alike, but they are not. Making both of them 6 gives 6/6 = 1, choice D.

  4. Divide the top by the bottom

    Now the fraction is 0 over 6. Zero divided by any number other than zero is zero:

    3+(−3)3−(−3)=06=0\frac{3+(-3)}{3-(-3)} = \frac{0}{6} = 0

    A top of 0 does not make the fraction 1/6, choice C. So the answer is B, 0.

Check your answer

Check by multiplying. A division is right when the answer times the bottom gives the top. Here 0×6=00 \times 6 = 0, which is the top. So 0/6 really is 0.

More questions like this

The original DOE question
Question 16 as shown on the DOE’s digital test, page 89 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 89, question 16. This is the DOE’s version, before the corrections listed in the analysis.

Question 35 of 120

2026 Practice Test C Math, question 17

Question

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

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

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

Correct answer: A

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

7−1=67 - 1 = 6

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

6÷2=36 \div 2 = 3

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

1+3=41 + 3 = 4

7−3=47 - 3 = 4

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

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

6−3=36 - 3 = 3

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

3÷2=1.53 \div 2 = 1.5

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

3+1.5=4.53 + 1.5 = 4.5

6−1.5=4.56 - 1.5 = 4.5

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

Determine the distance between the two midpoints by subtracting:

4.5−4=0.54.5 - 4 = 0.5

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Many steps minor Working it out

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the midpoint of MN

    Use the midpoint formula with the ends 1 and 7:

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

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

  3. Find the midpoint of RS

    Use the same formula with the ends 3 and 6:

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

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

  4. Subtract the two midpoints

    Use the distance formula:

    4.5−4=0.54.5 - 4 = 0.5

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

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

  5. Choose the answer

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

Check your answer

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

More questions like this

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

Question 36 of 120

2026 Practice Test C Math, question 18

Question

I. s−ts - t
II. st2st^{2}
III. sts^{t}

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

  • AII only
  • BI and II only
  • CII and III only
  • DI, II, and III
Show the answer

Correct answer: D

Each of the three expressions must be considered:

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

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

Expression II: st2st^{2}, where s is a positive integer and t is a negative integer:

Squaring a negative integer results in a positive integer, so s(−t)2s(-t)^{2} is equivalent to finding the value of st2st^{2}, which must result in a positive number.

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know moderate Working it out

    Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.

  2. Try one pair of numbers

    Try s = 2 and t = −3 in each expression:

    • I: 2−(−3)=2+3=52 - (-3) = 2 + 3 = 5
    • II: 2×(−3)2=2×9=182 \times (-3)^{2} = 2 \times 9 = 18
    • III: 2−3=123=182^{-3} = \frac{1}{2^{3}} = \frac{1}{8}

    All three are positive this time. Now see why that always happens.

  3. Explain expression I

    Use the rule for subtracting:

    s−t=s+(−t)s - t = s + (-t)

    t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.

    Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.

  4. Explain expression II

    t2t^{2} means t × t. A negative times a negative is positive, so t2t^{2} is always positive. Then s×t2s \times t^{2} is a positive times a positive. That is always positive too.

  5. Explain expression III

    Use the rule for a negative exponent:

    st=1s−ts^{t} = \frac{1}{s^{-t}}

    −t is a positive integer, so s−ts^{-t} is a positive integer. So sts^{t} is 1 over a positive number. That is a positive fraction.

    Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.

  6. Choose the answer

    Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.

    So the answer is D, I, II, and III.

Check your answer

Try the values closest to 0: s = 1 and t = −1. I gives 1−(−1)=21 - (-1) = 2. II gives 1×(−1)2=11 \times (-1)^{2} = 1. III gives 1−1=11=11^{-1} = \frac{1}{1} = 1. All three are still positive.

More questions like this

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

Question 37 of 120

2026 Practice Test C Math, question 19

Question

What is the value of (−34)(−25)\left(-\frac{3}{4}\right)\left(-\frac{2}{5}\right)?

  • A−720-\frac{7}{20}
  • B−310-\frac{3}{10}
  • C310\frac{3}{10}
  • D720\frac{7}{20}
Show the answer

Correct answer: C

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

(−3)×(−2)4×5=620\frac{(-3) \times (-2)}{4 \times 5} = \frac{6}{20}

Simplify the result:

6÷220÷2=310\frac{6 \div 2}{20 \div 2} = \frac{3}{10}

Question analysis

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

NYS standards
  • 7.NS.2a rational number multiplication
  • 7.NS.2c properties for multiplying and dividing
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multiplying two negative fractions is a standard grade 7 exercise, done here in one product and one simplification. It matches typical classroom work, with the sign rule and lowest terms as the only points of care.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 10, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer minor Choosing the answer

    3/10 and −3/10 are identical except for the sign, so the size alone does not separate them; only the rule for multiplying two negatives does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Two negative fractions are multiplied, 3 × 2 over 4 × 5, and the product 6/20 has to be reduced to 3/10 before it matches a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The two fractions sit side by side in parentheses, with nothing between them. That means multiply. Both fractions are negative. You need their product, the answer to the multiplication.

  2. Decide the sign first

    A negative times a negative is positive. So the answer will be positive. That rules out choices A and B.

    Watch out: choice B, −3/10, has the right size but the wrong sign. The two minus signs cancel.

  3. Multiply tops and bottoms

    Now multiply the sizes. Multiply the tops, then the bottoms:

    34×25=3×24×5=620\frac{3}{4} \times \frac{2}{5} = \frac{3 \times 2}{4 \times 5} = \frac{6}{20}

    Watch out: do not find a common denominator. That is for adding and subtracting. Subtracting the fractions gives 7/20 or −7/20, choices D and A.

  4. Simplify the fraction

    2 goes into both 6 and 20. Divide the top and the bottom by 2:

    6÷220÷2=310\frac{6 \div 2}{20 \div 2} = \frac{3}{10}

    The product is positive 3/10. So the answer is C, 3/10.

Check your answer

Use decimals. 3/4 is 0.75 and 2/5 is 0.4. Then 0.75×0.4=0.30.75 \times 0.4 = 0.3, which is 3/10. Two negatives give a positive, so the product is positive 3/10.

More questions like this

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

Question 38 of 120

2026 Practice Test C Math, question 20

Question

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

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

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

Select one answer in each row.

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

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

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

A=25πA = 25\pi: Circle A

C=5πC = 5\pi: Circle C

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

Circle A:

A=πr2A = \pi r^{2}

A=25πA = 25\pi

r=5r = 5

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

Circle C:

C=π⋅dC = \pi \cdot d

C=5πC = 5\pi

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

Question analysis

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

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying the area and circumference formulas is the grade 7 circle standard. This gives a radius for one circle and, for the other, a segment known only through congruence, and both rows of the table must be right.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 5, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

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

  2. Something you must already know moderate Working it out

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

  3. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Find each radius and diameter

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

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

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

  3. Check the area row

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

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

    For Circle C:

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

    Only Circle A has an area of 25π.

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

  4. Check the circumference row

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

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

    For Circle C:

    C=π×5=5πC = \pi \times 5 = 5\pi

    Only Circle C has a circumference of 5π.

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

  5. Fill in the table

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

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

Check your answer

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

More questions like this

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

Question 39 of 120

2026 Practice Test C Math, question 21

Question

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

  • A2
  • B3
  • C18
  • D38
Show the answer

Correct answer: A

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

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

ao=cw\frac{a}{o} = \frac{c}{w}

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

a6=1648\frac{a}{6} = \frac{16}{48}

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

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

Solve for the value of a:

48a=9648a = 96

a=2a = 2

Question analysis

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

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Equal ratios with whole numbers are grade 6 work, and the numbers divide evenly. Grade 7 proportion problems are usually longer; here the only care is keeping apples with oranges in the same order as cherries with walnuts.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    The sentence pairs apples with oranges and cherries with walnuts in the same order, so a/6 = 16/48; reading either pair the other way turns one ratio upside down. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Four kinds of fruit are named and three counts given, and the unknown apples must be matched with the smaller count, the cherries, not the larger, the walnuts. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    A ratio compares two amounts. The ratio of apples to oranges equals the ratio of cherries to walnuts. You know there are 6 oranges, 16 cherries and 48 walnuts. You need the number of apples.

  2. Write the proportion

    Put the first thing of each pair on top. Apples go with oranges, and cherries go with walnuts. Let a be the number of apples:

    a6=1648\frac{a}{6} = \frac{16}{48}

    Watch out: keep the order the same. Writing 48 over 16 flips the second ratio. That gives 18, choice C.

  3. Simplify the known ratio

    16 goes into both 16 and 48. Divide the top and the bottom by 16:

    16÷1648÷16=13\frac{16 \div 16}{48 \div 16} = \frac{1}{3}

    So there is 1 cherry for every 3 walnuts. There must also be 1 apple for every 3 oranges.

  4. Find the apples

    Now the proportion is:

    a6=13\frac{a}{6} = \frac{1}{3}

    The bottom 6 is 3 × 2, so the top is 1 × 2. That makes a=2a = 2.

    Watch out: 48÷16=348 \div 16 = 3 is choice B, but that is not the number of apples. And 38 comes from using differences instead of ratios.

    So the answer is A, 2.

Check your answer

Compare the ratios. Apples to oranges is 2 to 6. Cherries to walnuts is 16 to 48. In both, the second number is 3 times the first: 2×3=62 \times 3 = 6 and 16×3=4816 \times 3 = 48. So the ratios are equal.

More questions like this

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

Question 40 of 120

2026 Practice Test C Math, question 22

Question

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

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 210

Determine the difference between the highest and lowest elevations:

Determine the highest elevation:

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

Highest elevation =90 ft= 90\text{ ft}

Determine the lowest elevation:

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

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

Determine the difference between the two elevations:

90 ft−(−120) ft=90 ft+120 ft=210 ft90\text{ ft} - (-120)\text{ ft} = 90\text{ ft} + 120\text{ ft} = 210\text{ ft}

Question analysis

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

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.7b order statements in context
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the gap between a positive and a negative value by subtracting is core grade 7 work with integers. Picking the extremes from six readings is quick, so the demand is the subtraction of a negative number.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    The difference is 90 − (−120), and subtracting a negative means adding 120, so the part of the trail below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The difference is typed with no choices, so 30 or −30, the results of mishandling the negative, gets no warning that the two readings lie on opposite sides of zero. (Enright, Morley & Sheehan, 2002)

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

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Elevation is height above or below a starting level, here 0 feet. Negative elevations are below 0. The six readings are −120, −80, 0, 45, −30 and 90 feet. You need the difference between the highest and the lowest.

  2. Find the highest and the lowest

    Picture the readings on a number line. The highest is farthest right: 90 feet. The lowest is farthest left: −120 feet.

    For negative numbers, the bigger the number after the minus sign, the lower it is. So −120 is lower than −80.

  3. Subtract the lowest from the highest

    Use the difference formula. Subtracting a negative is the same as adding:

    90−(−120)=90+120=21090 - (-120) = 90 + 120 = 210

    So the two readings are 210 feet apart.

    Watch out: 120−90=30120 - 90 = 30 treats both readings as if they were above 0. Adding them, 90+(−120)=−3090 + (-120) = -30, is wrong too.

  4. Enter the answer

    A difference like this is a distance, so it is positive. Subtracting in the other order gives −210, which is wrong. Type only the number, with no units.

    So you would enter 210.

Check your answer

Count the distance in two parts. From −120 feet up to 0 is 120 feet. From 0 up to 90 is 90 more feet. Together, 120+90=210120 + 90 = 210 feet. That matches.

More questions like this

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

Question 41 of 120

2026 Practice Test C Math, question 23

Question

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

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

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

Correct answer: C

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

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Something you must already know minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Pick the base and the height

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

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

  3. Multiply the base by the height

    Use the area formula:

    A=34×18A = 34 \times 18

    Break it up to multiply:

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

    The area is 612 square centimeters.

  4. Choose the answer

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

    So the answer is C, 612 sq cm.

Check your answer

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

More questions like this

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

Question 42 of 120

2026 Practice Test C Math, question 24

Question

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

  • A36
  • B63
  • C84
  • D108
Show the answer

Correct answer: B

The ratio of paintings to drawings to photographs is 3:5:43 : 5 : 4.

First, determine the number of actual paintings and photographs.

Since the number of actual drawings is 45, each ratio is multiplied by 9, since 5×9=455 \times 9 = 45.

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

Then, add the actual number of paintings and photographs.

27+36=6327 + 36 = 63

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scaling a ratio from one known part is grade 6 and 7 work. A three-part ratio, and a question about two of the three groups combined, make it a three-step task typical of grade 7 ratio problems.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Much to keep track of moderate Working it out

    The ratio has three parts, the known count belongs to the middle part, drawings, and the question wants the first and third parts combined, so three roles must be tracked. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    The word respectively assigns 3, 5 and 4 to paintings, drawings and photographs in that order, and the question then names two of them in the other order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The ratio of paintings to drawings to photographs is 3:5:4. Respectively means in that order: 3 for paintings, 5 for drawings and 4 for photographs. There were 45 drawings. You need the number of photographs and paintings together.

  2. Find the value of one part

    Drawings are 5 parts, and there were 45 drawings:

    45÷5=945 \div 5 = 9

    So each part of the ratio stands for 9 works.

  3. Find each group

    Multiply each group’s parts by 9:

    TypePartsActual number
    Paintings33×9=273 \times 9 = 27
    Drawings55×9=455 \times 9 = 45
    Photographs44×9=364 \times 9 = 36

    The drawings row gives 45, which matches the question.

  4. Add only the two groups asked for

    Add the paintings and the photographs:

    27+36=6327 + 36 = 63

    Watch out: 36 alone, choice A, leaves out the paintings. Adding the 45 drawings too gives 108, choice D, the whole show.

    So the answer is B, 63.

Check your answer

Paintings and photographs are 3+4=73 + 4 = 7 parts. Each part is 9 works, so 7×9=637 \times 9 = 63. That matches. Also, 63 to 45 is the same as 7 to 5, just as the ratio says.

More questions like this

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

Question 43 of 120

2026 Practice Test C Math, question 25

Question

For each 38\frac{3}{8} foot of curtain width, 56\frac{5}{6} yard of fabric is needed to make ruffles. How many yards of fabric are needed per foot of curtain width?

  • A516\frac{5}{16} yd
  • B2924\frac{29}{24} yd
  • C209\frac{20}{9} yd
  • D72\frac{7}{2} yd
Show the answer

Correct answer: C

It takes 56\frac{5}{6} yard of fabric to make 38\frac{3}{8} foot of curtain width.

To find the ratio of 1 foot of curtain width to 38\frac{3}{8} foot of curtain width (1:38)\left(1:\frac{3}{8}\right), divide.

1÷38=1×83=831 \div \frac{3}{8} = 1 \times \frac{8}{3} = \frac{8}{3}

So the amount of fabric needed for 1 foot of curtain width is 83\frac{8}{3} times the amount needed for 38\frac{3}{8} foot of curtain width.

83×56=4018\frac{8}{3} \times \frac{5}{6} = \frac{40}{18}

40÷218÷2=209\frac{40 \div 2}{18 \div 2} = \frac{20}{9}

The amount of fabric needed for 1 foot of curtain width is 209\frac{20}{9} yards.

Question analysis

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

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Computing a unit rate from a ratio of two fractions is the grade 7 rate standard itself. Two units, feet and yards, and an unfamiliar sewing setting make it harder to see which fraction divides which.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “For each 3/8 foot of curtain width, 5/6 yard of fabric” gives the rate as a pair of fractions, and “per foot” means dividing the yards by the feet. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Dividing by 3/8 means multiplying by 8/3, and the product 40/18 has to be reduced to 20/9, an improper fraction rather than a whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    For every 3/8 foot of curtain width, you need 5/6 yard of fabric. You need the yards of fabric for 1 whole foot. That is a unit rate, an amount for exactly 1 foot.

    3/8 foot is less than a whole foot. So a whole foot needs more than 5/6 yard.

  2. Set up the division

    Per foot means divide the yards by the feet:

    56÷38\frac{5}{6} \div \frac{3}{8}

    Watch out: multiplying 5/6 by 3/8 gives 5/16, choice A. That is less fabric for more curtain, which makes no sense. Adding them gives 29/24, choice B, but you can’t add yards and feet.

  3. Multiply by the reciprocal

    Flip 3/8 to get 8/3, and multiply:

    56×83=5×86×3=4018\frac{5}{6} \times \frac{8}{3} = \frac{5 \times 8}{6 \times 3} = \frac{40}{18}

    The 8/3 says a whole foot is 8/3 times as long as 3/8 foot. So it needs 8/3 times as much fabric.

    Watch out: adding 8/3 and 5/6 gives 7/2, choice D. Multiply them instead.

  4. Simplify the fraction

    2 goes into both 40 and 18:

    40÷218÷2=209\frac{40 \div 2}{18 \div 2} = \frac{20}{9}

    So 1 foot of curtain width needs 20/9 yards of fabric, a little more than 2 yards. So the answer is C, 20/9 yd.

Check your answer

Multiply back. 3/8 foot should need 3/8 of the fabric for a whole foot: 209×38=6072=56\frac{20}{9} \times \frac{3}{8} = \frac{60}{72} = \frac{5}{6} yard. That is the amount the question gives. Also, 20/9 is more than 5/6, as it should be.

More questions like this

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

Question 44 of 120

2026 Practice Test C Math, question 26

Question

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

Which straight line is the graph of a proportional relationship?

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

Correct answer: D

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

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

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

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

Question analysis

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

NYS standards
  • 7.RP.2a testing for proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Deciding from a graph whether a relationship is proportional is a named grade 7 skill, and this is a typical case: four lines on a clean grid, each checked against the same rule, with no calculation needed.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 9, 14 words.
Corrected from the DOE handbook
  • Coordinates and an equation in the explanation were corrected to match the graph.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

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

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

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Recall what a proportional graph looks like

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

    y=k×0=0y = k \times 0 = 0

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

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

  3. Check where each line starts

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

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

  4. Test the ratio on line TR

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

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

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

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

  5. Give the answer

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

    So the answer is D, line TR.

Check your answer

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

More questions like this

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

Question 45 of 120

2026 Practice Test C Math, question 27

Question

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

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

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

Correct answer: C

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Every choice must be checked minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

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

  2. Find the cost of one pound

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

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

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

  3. Test the point in choice C

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

    10×0.15=1.510 \times 0.15 = 1.5

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

  4. Check the other statements

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

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

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

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

  5. Give the answer

    Only choice C gives the right meaning to both numbers.

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

Check your answer

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

More questions like this

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

Question 46 of 120

2026 Practice Test C Math, question 28

Question

If z+3w4=5w\frac{z+3w}{4} = 5w, what is the value of z in terms of w?

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

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

Show the answer

Correct answer: z = 17w

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

First, multiply both sides by 4:

z+3w4=5w\frac{z+3w}{4} = 5w

z+3w=20wz + 3w = 20w

Then subtract 3w3w from both sides:

z=17wz = 17w

Question analysis

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

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.4a two-step equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations with numbers, such as (x + 3)/4 = 5. Solving for one letter in terms of another is grade 8 work, and the typed answer gives no choices to check against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 4, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    With w on both sides and no numbers to substitute, the result has to stay a multiple of w, and a slip such as 23w gives no sign that it is wrong. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The answer is typed after a printed “z =”, with no choices to compare, so the entry must be a single term in w, not an equation or a number. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The equation is z+3w4=5w\frac{z + 3w}{4} = 5w. You need z in terms of w. That means getting z alone on one side. The other side will use only w and numbers.

    Treat w like a number you do not know yet. The answer box already shows z =, so you type only what comes after it.

  2. Undo the division first

    The fraction bar means the whole top, z + 3w, is divided by 4. So undo that first. Multiply both sides by 4, because multiplying undoes dividing:

    z+3w4×4=5w×4\frac{z + 3w}{4} \times 4 = 5w \times 4

    z+3w=20wz + 3w = 20w

    Watch out: do not subtract 3w first. The 3w sits inside the fraction, so it is divided by 4 too. Subtracting it from 5w first leads to the wrong answer 8w.

  3. Undo the adding

    Now 3w is added to z. Subtract 3w from both sides:

    z+3w−3w=20w−3wz + 3w - 3w = 20w - 3w

    z=17wz = 17w

    Think of w as a thing, like apples. 20 apples take away 3 apples leaves 17 apples.

    Watch out: moving 3w to the other side without changing its sign gives 20w+3w=23w20w + 3w = 23w. That is wrong.

  4. Enter it after the printed z =

    On the test, the box already shows z =. Press 1, then 7, then the w key. You do not need a times sign between 17 and w. Do not type z = again. On this site, type 17w in the box.

    So you would enter 17w, which makes z = 17w.

Check your answer

Pick a number for w, such as w = 1. Then z = 17. Put both into the first equation. The left side is 17+34=204=5\frac{17 + 3}{4} = \frac{20}{4} = 5. The right side is 5×1=55 \times 1 = 5. The two sides match, so 17w works.

More questions like this

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

Question 47 of 120

2026 Practice Test C Math, question 29

Question

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

SMOOTHIE RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.25

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

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

1.5=6p1.5 = 6p

1.56=6p6\frac{1.5}{6} = \frac{6p}{6}

14=0.25=p\frac{1}{4} = 0.25 = p

The constant of proportionality is 0.25.

Question analysis

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

NYS standards
  • 7.RP.2b constant of proportionality
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the constant of proportionality in a table is a core grade 7 skill, and these values are friendly decimals. Typing the answer with no choices, and taking the ratio in the order the question gives, are the added demands.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
1 step
Reading level
Question and choices: about grade 6, 75 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Any row gives the answer with one division, and the values divide evenly, so the points are lost on direction. Fruit fills the first column and comes first in the stem’s own ratio, which invites dividing it by the yogurt. The table also climbs by a steady half cup of yogurt, a number that looks like a constant but ignores the 2 cups of fruit added each time. Checking that 0.25 times each fruit value gives its yogurt value settles it.

  1. Turning words into math moderate Understanding the question

    The opening sentence speaks of a ratio of fruit to yogurt, but the question asks for the constant of yogurt to fruit, so the division is y ÷ x, against the table’s column order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The constant is typed into a box with no choices, so 4, the ratio taken the other way round, is a believable entry that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

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

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The table lists cups of fruit (x) and cups of yogurt (y) for four batches. You need the constant of proportionality of yogurt to fruit.

    That is the number of cups of yogurt for each 1 cup of fruit. It is also the number you multiply x by to get y.

  2. Divide yogurt by fruit

    Use the first row: 6 cups of fruit and 1.5 cups of yogurt. Yogurt goes on top, because the question says yogurt to fruit:

    k=yx=1.56=0.25k = \frac{y}{x} = \frac{1.5}{6} = 0.25

    So there is 0.25 cup, or one quarter of a cup, of yogurt for each cup of fruit.

    Watch out: the first sentence says fruit to yogurt, but the question asks the other way. Dividing 6÷1.5=46 \div 1.5 = 4 turns the answer upside down.

  3. Check two more rows

    The constant must be the same in every row. Test two more:

    2.08=0.253.012=0.25\frac{2.0}{8} = 0.25 \qquad \frac{3.0}{12} = 0.25

    They match. Watch out: the yogurt goes up by 0.5 cup from row to row. That 0.5 is not the constant. The fruit goes up by 2 cups at the same time, and 0.5÷2=0.250.5 \div 2 = 0.25.

  4. Enter the number

    Type only the number in the box, with no units. You can type 0.25 or .25. On this site, 1/4 counts too.

    So you would enter 0.25.

Check your answer

Multiply each amount of fruit by 0.25. 6×0.25=1.56 \times 0.25 = 1.5, 8×0.25=28 \times 0.25 = 2, 10×0.25=2.510 \times 0.25 = 2.5 and 12×0.25=312 \times 0.25 = 3. Every result matches the yogurt column. Also, each batch has less yogurt than fruit, so the constant should be less than 1.

More questions like this

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

Question 48 of 120

2026 Practice Test C Math, question 30

Question

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

  • A0.955%
  • B1.91%
  • C9.55%
  • D19.1%
Show the answer

Correct answer: C

The fraction 1912000\frac{191}{2000} represents the 191 high school seniors out of 2,000 enrolled in the Honor Society.

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

1912000=0.0955=9.55%\frac{191}{2000} = 0.0955 = 9.55\%

Question analysis

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

NYS standards
  • 6.RP.3c percent as a rate per 100
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Writing a part of a whole as a percent is grade 6 work, done here with one division and one conversion. The awkward numbers, 191 out of 2,000, make placing the decimal point the only real demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 24 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is a single fraction, 191 over 2,000, so the question tests decimal places. The four choices share the digits 955 or 191 and differ only in where the point falls, which means a slip of one place still finds a choice waiting. A rough check settles it: 191 is a little under 200, and 200 is 10% of 2,000, so the share must be a little under 10%.

  1. Error-prone arithmetic moderate Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    There are 2,000 seniors in all. Of them, 191 are in the Honor Society. You need the percent of all 2,000 seniors who are in it.

    So 191 is the part, and 2,000 is the whole.

  2. Write the share as a fraction

    Put the part over the whole:

    partwhole=1912,000\frac{\text{part}}{\text{whole}} = \frac{191}{2{,}000}

    This fraction says 191 out of every 2,000 seniors. Next, turn it into a decimal by dividing.

  3. Divide to get a decimal

    Dividing by 2,000 is the same as dividing by 2 and then by 1,000:

    191÷2=95.5191 \div 2 = 95.5

    95.5÷1,000=0.095595.5 \div 1{,}000 = 0.0955

    Dividing by 1,000 moves the decimal point 3 places to the left. So the share is 0.0955.

  4. Change the decimal to a percent

    Multiply by 100. That moves the decimal point 2 places to the right:

    0.0955×100=9.550.0955 \times 100 = 9.55

    So the share is 9.55%.

    Watch out: moving the point only 1 place gives 0.955%, choice A. Dividing by 1,000 instead of 2,000 gives 19.1%, choice D. Dividing 191 by 100 gives 1.91%, choice B.

  5. Give the answer

    The 191 Honor Society members are 9.55% of all 2,000 seniors. That is a little less than one tenth of them.

    So the answer is C, 9.55%.

Check your answer

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

More questions like this

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

Question 49 of 120

2026 Practice Test C Math, question 31

Question

NUMBER OF VISITORS
TO BROOKSIDE ZOO

Male Female
Children 360 340
Adults 210 290

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

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

Correct answer: A

First, determine the total number of visitors.

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

The largest group is 360 (Male Children).

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Many steps minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

    MaleFemale
    Children360340
    Adults210290

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

  2. Add up all the visitors

    Add all four groups to get the whole:

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

    So 1,200 people visited the zoo that day.

  3. Find the largest group

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

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

  4. Find the percent

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

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

    Now multiply by 100 to get a percent:

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

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

  5. Give the answer

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

    So the answer is A, 30%.

Check your answer

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

More questions like this

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

Question 50 of 120

2026 Practice Test C Math, question 32

Question

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 15

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

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

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

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Many steps minor Working it out

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

  3. Something you must already know minor Working it out

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

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

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Read the values between the ticks

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

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

  3. Find the range

    Use the range formula:

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

    The data spread over 45 degrees.

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

  4. Find the IQR

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

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

    The middle half of the data spreads over 30 degrees.

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

  5. Subtract the IQR from the range

    The question asks for the difference between the two:

    45−30=1545 - 30 = 15

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

    So you would enter 15.

Check your answer

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

More questions like this

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

Question 51 of 120

2026 Practice Test C Math, question 33

Question

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

  • A6
  • B15
  • C18
  • D30
Show the answer

Correct answer: A

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

18÷6120÷6=320\frac{18 \div 6}{120 \div 6} = \frac{3}{20}

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

320(40)=12020=6\frac{3}{20}(40) = \frac{120}{20} = 6

Question analysis

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

NYS standards
  • 6.SP.7 long-run relative frequency
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Predicting from repeated trials is grade 6 probability, and grade 7 students make similar estimates from samples. The twist is scaling down, from 120 draws to a bag of 40 tiles, rather than up to a larger group.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 61 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    The 18 red draws come from 120 draws with the tile put back each time, so they give a share, 3/20, to take of the 40 tiles, not a count of red tiles. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Students must know that the share of red in repeated draws estimates the share of red tiles in the bag; the stem never says to multiply 40 by that share. (Anderson & Krathwohl, 2001)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The bag has 40 tiles, each red, green or blue. A tile is drawn, its color is written down, and it goes back in the bag. This happens 120 times, and red comes up 18 times.

    You need the most likely number of red tiles in the bag.

  2. See why 18 is not the answer

    Each tile goes back in the bag after it is drawn. So the same red tile can be drawn again and again. That means 18 counts draws, not tiles.

    Watch out: choice C, 18, mixes up the number of red draws with the number of red tiles.

  3. Find the share of red draws

    Red came up 18 times out of 120. Simplify by dividing the top and bottom by 6:

    18120=320\frac{18}{120} = \frac{3}{20}

    As a decimal, that is 0.15, or 15%.

    Watch out: 15 is a percent here, not a number of tiles. Stopping now gives choice B.

  4. Take that share of the bag

    Use the same share of the 40 tiles in the bag:

    320×40=12020=6\frac{3}{20} \times 40 = \frac{120}{20} = 6

    So about 6 of the 40 tiles are red.

  5. Give the answer

    The best prediction is 6 red tiles. The other 34 tiles are probably green or blue.

    So the answer is A, 6.

Check your answer

There were 120 draws for 40 tiles. That is 3 draws for every tile. So divide the red draws by 3: 18÷3=618 \div 3 = 6. Same answer. Choice D, 30 red tiles, would be three quarters of the bag. Then red would come up far more than 18 times.

More questions like this

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

Question 52 of 120

2026 Practice Test C Math, question 34

Question

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

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

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

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

Correct answer: B

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

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. List every path in the tree

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

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

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

  3. Mark the ones that fit

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

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

  4. Find the probability

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

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

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

Check your answer

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

More questions like this

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

Question 53 of 120

2026 Practice Test C Math, question 35

Question

If 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a, what is the value of a in terms of b?

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

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

Show the answer

Correct answer: a = 10b

To find the value of a in terms of b, solve 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a for a. Begin by distributing 4 on the left side of the equation:

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

4a+8b=28b+2a4a + 8b = 28b + 2a

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

4a−2a=28b−8b4a - 2a = 28b - 8b

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

2a=20b2a = 20b

a=10ba = 10b

Question analysis

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

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve equations like 4(x + 2) = 28 with numbers. This one has letters on both sides, needs the distributive property and like terms, and asks for a in terms of b, which is grade 8 algebra.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 4, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    Both letters appear on both sides, and the answer must stay a multiple of b, so there is no number to check an entry against unless students invent one. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Many steps moderate Working it out

    Four moves run in a chain, distributing the 4, gathering the a terms, gathering the b terms and dividing by 2, and an error at any one carries into the typed result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The answer is typed after a printed “a =”, with no choices, so 20b, the line before dividing by 2, can be entered without anything on screen questioning it. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
k(x+y)=kx+kyk(x + y) = kx + ky
k is the number outside the parentheses. x and y are the terms inside. The number outside multiplies every term inside, not just the first one. It works because 4 groups of (x + y) hold 4 x’s and 4 y’s.
Doing the same thing to both sides
if A=B, then A−C=B−C and A2=B2\text{if } A = B, \text{ then } A - C = B - C \text{ and } \frac{A}{2} = \frac{B}{2}
A and B are the two equal sides of an equation. C is any amount. Taking the same amount from both sides, or dividing both by the same number, keeps them equal, like a balanced scale.

  1. Sort out the question

    The equation is 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a. You need a in terms of b. That means a alone on one side, with only b and numbers on the other.

    The answer box already shows a =, so you type only what comes after it.

  2. Multiply out the parentheses

    Use the distributive property. The 4 multiplies both terms inside:

    4×a+4×2b=28b+2a4 \times a + 4 \times 2b = 28b + 2a

    4a+8b=28b+2a4a + 8b = 28b + 2a

    Watch out: multiplying only the a by 4 gives 4a + 2b. That leads to the wrong answer 13b.

  3. Gather the a terms

    Like terms are terms with the same letter, such as 4a and 2a. Put the a terms on the left. Subtract 2a from both sides:

    4a−2a+8b=28b4a - 2a + 8b = 28b

    2a+8b=28b2a + 8b = 28b

  4. Gather the b terms

    Now subtract 8b from both sides:

    2a=28b−8b2a = 28b - 8b

    2a=20b2a = 20b

    Watch out: moving 8b without changing its sign gives 28b+8b=36b28b + 8b = 36b. That leads to 18b, which is wrong.

  5. Divide both sides by 2

    2a means 2 times a. Divide both sides by 2 to get a alone:

    a=20b2=10ba = \frac{20b}{2} = 10b

    Watch out: 20b is equal to 2a, not a. Do not stop there.

  6. Enter it after the printed a =

    On the test, the box already shows a =. Press 1, then 0, then the b key. You do not need a times sign. Do not type a = again. On this site, type 10b in the box.

    So you would enter 10b, which makes a = 10b.

Check your answer

Pick b = 1, so a = 10. The left side is 4(10+2)=4×12=484(10 + 2) = 4 \times 12 = 48. The right side is 28+2×10=4828 + 2 \times 10 = 48. The sides match, so 10b works.

More questions like this

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

Question 54 of 120

2026 Practice Test C Math, question 36

Question

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

  • A110\frac{1}{10}
  • B14\frac{1}{4}
  • C313\frac{3}{13}
  • D13\frac{1}{3}
Show the answer

Correct answer: B

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

52−12=4052 - 12 = 40

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

13−3=1013 - 3 = 10

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

10 hearts remaining40 total cards remaining=1040=14\frac{10\text{ hearts remaining}}{40\text{ total cards remaining}} = \frac{10}{40} = \frac{1}{4}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a simple probability is grade 6 content. This one adds a step many grade 7 problems skip: the cards in Sara's hand must be taken out of both the hearts and the whole deck first.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The careful route takes two subtractions and a fraction: 10 hearts among 40 cards. The numbers are kind, because Sara’s hand is a quarter hearts, just like the full deck, so a student who ignores her hand and divides 13 by 52 also reaches 1/4. The wrong choices come from describing her hand instead of the deck, or from comparing hearts with the cards that are not hearts.

  1. Much to keep track of minor Working it out

    Four counts are in play, 52 cards, 13 hearts, Sara’s 12 cards and her 3 hearts, and both the hearts and the total must be reduced before the fraction is formed. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    “The other cards remain in the deck” and “the remainder of the deck” point to the 40 cards Sara is not holding, not to the full 52 or to her hand. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work out how many cards are left in the deck and how many of them are hearts, then write the hearts left over the cards left and simplify.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    A full deck has 52 cards, and 13 of them are hearts. Sara holds 12 cards, and 3 of those are hearts. The rest of the cards stay in the deck.

    You need the chance that a card drawn from the rest of the deck is a heart. Her hand changes both counts.

  2. Count the cards left

    Take Sara’s 12 cards out of the deck:

    52−12=4052 - 12 = 40

    So 40 cards are left in the deck.

  3. Count the hearts left

    Sara holds 3 of the 13 hearts. Take those out:

    13−3=1013 - 3 = 10

    So 10 hearts are left in the deck.

  4. Find the probability

    Use the probability formula. Put the hearts left over all the cards left, then simplify:

    P=1040=14P = \frac{10}{40} = \frac{1}{4}

    Watch out: comparing the 10 hearts with the 30 cards that are not hearts gives 13\frac{1}{3}, choice D. The bottom must be all 40 cards. Choice C, 313\frac{3}{13}, describes Sara’s hand, not the deck. Choice A, 110\frac{1}{10}, uses the 10 hearts but forgets the 40 cards.

  5. Give the answer

    The chance of drawing a heart from the rest of the deck is 1 in 4.

    So the answer is B, 14\frac{1}{4}.

Check your answer

10 is one quarter of 40, because 4×10=404 \times 10 = 40. Here is another check. Sara’s hand is 3 hearts out of 12, which is also one quarter. Taking a hand that is one quarter hearts out of a deck that is one quarter hearts leaves one quarter hearts.

More questions like this

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

Question 55 of 120

2026 Practice Test C Math, question 37

Question

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

PICKING CANDY FROM
A BOWL

Flavor Probability
Butterscotch 0.57
Lemon 0.095
Maple 0.125
Strawberry 0.21

Which flavor is most likely to be randomly picked?

  • Abutterscotch
  • Blemon
  • Cmaple
  • Dstrawberry
Show the answer

Correct answer: A

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

Question analysis

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

NYS standards
  • 6.SP.6 probability from 0 to 1
  • 6.NS.7 ordering and absolute value
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 9, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

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

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

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.

    The greater the probability, the more likely the pick. So you need the greatest of the four decimals.

  2. Line up the decimals

    The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:

    FlavorProbability
    Butterscotch0.570
    Lemon0.095
    Maple0.125
    Strawberry0.210

  3. Compare the tenths digits

    Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.

    Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.

  4. Give the answer

    Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.

    So the answer is A, butterscotch.

Check your answer

0.57 is more than 0.5, which is one half. So more than half the candies are butterscotch. The other three flavors add up to 0.095+0.125+0.21=0.430.095 + 0.125 + 0.21 = 0.43, less than half. So butterscotch must be the most likely.

More questions like this

The original DOE question
Question 37 as shown on the DOE’s digital test, page 110 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 110, question 37.

Question 56 of 120

2026 Practice Test C Math, question 38

Question

This formula can be used to determine p, the number of pages that will be in a book that contains w words, with an average of r words per page, and f extra pages in the front for the title and publishing information.

p=1rw+fp = \frac{1}{r}w + f

If the formula is solved for w, it can be used to find the number of words in a book with p pages. What will the formula be if it is solved for w?

  • Aw=r(p−f)w = r(p - f)
  • Bw=1rp+fw = \frac{1}{r}p + f
  • Cw=f+rpw = f + rp
  • Dw=r+p+fw = r + p + f
Show the answer

Correct answer: A

Apply the properties of equality to solve for w.

p=1rw+fp = \frac{1}{r}w + f

Subtract f from both sides of the equation.

p−f=1rwp - f = \frac{1}{r}w

Multiply both sides by r.

r(p−f)=r(1rw)r(p - f) = r\left(\frac{1}{r}w\right)

r(p−f)=wr(p - f) = w

w=r(p−f)w = r(p - f)

Question analysis

A formula gives the number of pages p in a book from its number of words w, the average words per page r, and the extra front pages f. Students choose the formula solved for w. It tests rearranging a formula with inverse operations, and with only letters there are no numbers to check, so the order of the steps matters.

NYS standards
  • 8.EE.7 linear equations in one variable
  • 7.EE.4a two-step equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations such as 5 = x/4 + 2 with numbers. Rearranging a formula with four letters to solve for a different one is grade 8 work, with no numbers to check against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 82 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The formula is short, but solving it for w means undoing two operations in reverse order, with letters only. The tempting shortcut multiplies p by r and leaves the f untouched, which yields a formula that looks as finished as the right one. Swapping p and w without undoing anything is the other quick trap. Substituting a small book, 5,000 words at 250 a page with 4 front pages, shows which formula gives back 5,000.

  1. Variables instead of numbers moderate Understanding the question

    Four letters and no numbers mean each choice has to be judged by its structure, unless students pick values such as r = 250, w = 5,000 and f = 4 to test it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Many steps minor Working it out

    Two inverse operations must come in order, subtracting f before multiplying by r; multiplying first means the f would have to be multiplied by r as well. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Treat it like a two-step equation with numbers: undo what was done to w last, then the step before, applying each operation to the whole of the other side.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying by the reciprocal
r×1r=1r \times \frac{1}{r} = 1
r is any number except 0. The reciprocal of r is 1/r, the number flipped upside down. A number times its reciprocal is always 1. So multiplying by r undoes the 1/r in front of w and leaves w by itself.

  1. Sort out the question

    The formula is p=1rw+fp = \frac{1}{r}w + f. Here p is the number of pages, w the number of words, r the words on each page, and f the extra front pages.

    You need the formula solved for w. That means w alone on one side of the equal sign.

  2. See what happens to w

    Read the formula like a recipe. Start with w. First it is multiplied by 1r\frac{1}{r}, which is the same as dividing by r. Then f is added.

    To get w back, undo the steps in reverse order. Take away f first. Then undo the 1r\frac{1}{r}.

  3. Subtract f from both sides

    Taking f from both sides keeps them equal:

    p−f=1rwp - f = \frac{1}{r}w

    Now the only thing left to undo is the 1r\frac{1}{r}.

  4. Multiply both sides by r

    Use the reciprocal rule. Multiply both whole sides by r:

    r(p−f)=r×1rw=wr(p - f) = r \times \frac{1}{r}w = w

    So w=r(p−f)w = r(p - f).

    Watch out: w=f+rpw = f + rp, choice C, multiplies only p by r. The f must be taken away before you multiply. Or, if you multiply first, every term gets multiplied by r.

  5. Give the answer

    Choice B only trades the places of p and w, and it undoes nothing. Choice D just adds the letters.

    So the answer is A, w=r(p−f)w = r(p - f).

Check your answer

Try a book with 5,000 words, 250 words a page and 4 front pages. The first formula gives p=5,000÷250+4=24p = 5{,}000 \div 250 + 4 = 24 pages. Choice A gives back the words: 250×(24−4)=5,000250 \times (24 - 4) = 5{,}000. Choice C gives 4+250×24=6,0044 + 250 \times 24 = 6{,}004, which is wrong.

More questions like this

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

Question 57 of 120

2026 Practice Test C Math, question 39

Question

A shipment contains 170 small boxes of medical supplies. In a random sample of 20 of these boxes, 8 have damage and the others are undamaged. Based on this sample, what is the best prediction of the number of undamaged boxes in the shipment, not including the boxes in the random sample?

  • A60
  • B68
  • C90
  • D102
Show the answer

Correct answer: C

Determine the fraction of undamaged boxes in the sample, given 8 of the 20 boxes are damaged.

20−820=1220=35 or 0.6\frac{20-8}{20} = \frac{12}{20} = \frac{3}{5}\text{ or }0.6

Subtract the number of boxes in the sample from 170.

170−20=150170 - 20 = 150

Basing the prediction of undamaged boxes on the sample, multiply the number of boxes not in the sample by the fraction of undamaged boxes in the sample.

150×0.6=90150 \times 0.6 = 90

Question analysis

A shipment holds 170 boxes of medical supplies, and in a random sample of 20, 8 are damaged. Students predict how many undamaged boxes are in the shipment, leaving out the boxes in the sample. It tests using a sample to make a prediction, and the question's two conditions each lead to a different tempting wrong choice.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.SP.1b samples and populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Predicting from a random sample is grade 7 proportional reasoning. This version adds two conditions, undamaged rather than damaged and the shipment without the sample, so it takes three steps and a close reading.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 49 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The proportion itself is routine; the question is built around two conditions that each change the numbers. The stem reports the damaged boxes, but the question asks about undamaged ones, and its bold not removes the 20 sampled boxes from the shipment. Each wrong choice matches one way of mishandling the two conditions, so checking the result against both words, undamaged and not, is what the question rewards.

  1. Much to keep track of moderate Working it out

    Two conditions point away from the obvious numbers: the question wants undamaged boxes while the stem counts the 8 damaged, and it leaves out the 20 sampled, so 150 boxes, not 170. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps minor Working it out

    Three steps follow in order, the undamaged share 12/20, the 150 boxes outside the sample, and their product, and each wrong turn lands on a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the fraction of the sample that is undamaged, count the boxes that were not sampled, and multiply; then check that the answer describes the group of boxes asked about.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from a sample
prediction=part of the samplesample size×group size\text{prediction} = \frac{\text{part of the sample}}{\text{sample size}} \times \text{group size}
A sample is a small group chosen at random from a bigger group. The fraction is the share of the sample with some feature. A random sample is likely to look like the bigger group. So use the same share of the group you are predicting for.

  1. Sort out the question

    The shipment has 170 boxes. A random sample of 20 boxes was checked, and 8 of them were damaged.

    You need to predict the undamaged boxes in the shipment, not counting the 20 in the sample. Two words matter: undamaged and not.

  2. Find the undamaged share

    The sample has 20−8=1220 - 8 = 12 undamaged boxes. Write that as a share of the sample, and simplify:

    1220=35=0.6\frac{12}{20} = \frac{3}{5} = 0.6

    So 0.6, or 60%, of the sample is undamaged.

    Watch out: 820=0.4\frac{8}{20} = 0.4 is the damaged share. Using it leads to choice A or B.

  3. Count the boxes outside the sample

    The question leaves out the 20 boxes in the sample. Take them away from the shipment:

    170−20=150170 - 20 = 150

    So the prediction is about 150 boxes.

  4. Make the prediction

    Use the same undamaged share for the 150 boxes:

    0.6×150=900.6 \times 150 = 90

    About 90 of those boxes should be undamaged.

    Watch out: 0.6×170=1020.6 \times 170 = 102, choice D, forgets the bold not. It counts the sample boxes again.

  5. Give the answer

    The best prediction is 90 undamaged boxes outside the sample.

    So the answer is C, 90.

Check your answer

Predict the damaged boxes outside the sample too: 0.4×150=600.4 \times 150 = 60. Then 90+60=15090 + 60 = 150, every box outside the sample. Also, 12 out of 20 is a bit more than half. And 90 is a bit more than half of 150.

More questions like this

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

Question 58 of 120

2026 Practice Test C Math, question 40

Question

In basketball, a player can earn 1 point for making a free throw and 2 points for making another basket. Marcus makes a mean of 8 free throws per game, and he scores a total of at least 25 points per game. The inequality below shows this relationship, where b represents the number of 2-point baskets Marcus makes.

2b+8≥252b + 8 \geq 25

What is the number of 2-point baskets Marcus needs to make if he wants to score at least his minimum total points for a game and make his mean number of free throws?

  • Ab≤8b \leq 8
  • Bb≥9b \geq 9
  • Cb≥16b \geq 16
  • Db≤25b \leq 25
Show the answer

Correct answer: B

Solve the inequality 2b+8≥252b + 8 \geq 25.

2b+8−8≥25−82b + 8 - 8 \geq 25 - 8

2b≥172b \geq 17

2b2≥172\frac{2b}{2} \geq \frac{17}{2}

b≥8.5b \geq 8.5

Since the number of baskets made must be a whole number, the fewest 2-point baskets, b, he must make is 9.

Question analysis

In basketball, Marcus makes a mean of 8 one-point free throws a game and wants at least 25 points, shown by the inequality 2b + 8 ≥ 25. Students find how many 2-point baskets he needs. It tests solving a two-step inequality and interpreting the solution in context, since a player cannot make part of a basket.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality from a word problem is a grade 7 standard. Here the inequality is already written, so the extra demand is interpreting its solution as a count of baskets, typical of grade 7 tasks.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 92 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The inequality is handed over, so the solving is two routine moves; the judgment comes after. Dividing 17 by 2 leaves a half, and since a basket cannot be split, the boundary has to move up to 9: eight baskets give 2(8) + 8 = 24 points, one short of 25. Students who flip the symbol while dividing, or who move the 8 without changing its sign, still find a choice to match.

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

    Solving gives b ≥ 8.5, which no choice shows; it has to be recognized as b ≥ 9, because Marcus cannot make half a basket. (Enright, Morley & Sheehan, 2002)

  2. Complex sentences minor Understanding the question

    The 33-word question restates the given inequality as “at least his minimum total points” and “make his mean number of free throws”, wording harder than the math it describes. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Undo the added 8 and then the 2 to solve for b, then ask which whole numbers of baskets actually reach 25 points, testing the boundary value in the original inequality.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality
if A≥B, then A−C≥B−C and A2≥B2\text{if } A \geq B, \text{ then } A - C \geq B - C \text{ and } \frac{A}{2} \geq \frac{B}{2}
A and B are the two sides. The sign ≥ means greater than or equal to. Taking the same amount C from both sides keeps the sign the same. Dividing both sides by a positive number, like 2, also keeps it the same. Only a negative number flips it.

  1. Sort out the question

    A free throw is worth 1 point, and each other basket is worth 2 points. Marcus makes 8 free throws, which give 8 points. The letter b is the number of 2-point baskets, so they give 2b points.

    He wants at least 25 points in all. That is the inequality 2b+8≥252b + 8 \geq 25. An inequality compares two amounts. The sign ≥ means greater than or equal to. You need the values of b that work.

  2. Subtract 8 from both sides

    Undo the added 8 first:

    2b+8−8≥25−82b + 8 - 8 \geq 25 - 8

    2b≥172b \geq 17

    Watch out: adding 8 instead gives 2b≥332b \geq 33. That leads to choice C, which is wrong.

  3. Divide both sides by 2

    2b means 2 times b. Divide both sides by 2. Since 2 is positive, the sign stays the same:

    2b2≥172\frac{2b}{2} \geq \frac{17}{2}

    b≥8.5b \geq 8.5

    Watch out: do not flip the sign here. Flipping it gives b≤8.5b \leq 8.5, which leads to choice A.

  4. Think about whole baskets

    Marcus cannot make half a basket. So test the whole numbers near 8.5:

    2×8+8=242 \times 8 + 8 = 24

    2×9+8=262 \times 9 + 8 = 26

    Eight baskets give 24 points, one short of 25. Nine baskets give 26 points, which is enough. So he needs 9 or more baskets.

  5. Give the answer

    The number of baskets is 9 or more, written b≥9b \geq 9. Choice D, b≤25b \leq 25, just copies the 25 from the points.

    So the answer is B, b≥9b \geq 9.

Check your answer

Try a bigger number, such as b = 12. Then 2×12+8=322 \times 12 + 8 = 32 points, which is at least 25. Any number of baskets above 9 also works, and that is what ≥ says. Below 9, the total stays at 24 points or fewer.

More questions like this

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

Question 59 of 120

2026 Practice Test C Math, question 41

Question

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

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

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

Show the answer

Correct answer: 12.50 + 4.75n ≤ 35.00

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

Since notebooks cost $4.75 each and Elijah will buy n notebooks, the total cost just for the notebooks is found by multiplying $4.75 by n, or 4.75n4.75n.

Combining the cost of a water bottle ($12.50) with the cost of n notebooks, the inequality is 12.50+4.75n≤35.0012.50 + 4.75n \leq 35.00.

This may also be written as 4.75n+12.50≤35.004.75n + 12.50 \leq 35.00.

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an inequality for a budget with a one-time cost and a cost per item is typical grade 7 work. The answer is typed, so both the symbol for "at most" and the role of each price must be right.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 50 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    The $12.50 bottle is bought once and the $4.75 is paid for each of n notebooks, and “at most $35.00” has to become ≤, which allows a total of exactly $35.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The whole inequality, with n, its symbol and both prices, is built on a keypad offering ≤, <, ≥ and >, so a wrong symbol looks as complete as the right one. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Elijah buys one water bottle for $12.50. He also buys n notebooks at $4.75 each. He can spend at most $35.00.

    You need an inequality for n. An inequality compares two amounts with a sign such as ≤ instead of =.

  2. Price a few cases

    Work out the cost for a few numbers of notebooks:

    NotebooksTotal ($)
    112.50+4.75=17.2512.50 + 4.75 = 17.25
    212.50+9.50=22.0012.50 + 9.50 = 22.00
    312.50+14.25=26.7512.50 + 14.25 = 26.75

    The $12.50 shows up once. The $4.75 shows up once for each notebook.

  3. Write the total cost

    Use the total cost formula. The bottle is the one-time cost, and each notebook costs $4.75:

    12.50+4.75n12.50 + 4.75n

    Here 4.75n means 4.75 times n.

    Watch out: do not add the two prices first. 17.25n17.25n charges for a new bottle with every notebook.

  4. Choose the sign

    At most $35.00 means the total can be less than $35.00 or exactly $35.00. It cannot be more. So use ≤:

    12.50+4.75n≤35.0012.50 + 4.75n \leq 35.00

    Watch out: ≥ means at least, the opposite. And < leaves out a total of exactly $35.00, which Elijah is allowed to spend.

  5. Enter it on the keypad

    Type the whole inequality. Nothing is printed before the box. On the test, press 1, 2, ., 5, 0, +, 4, ., 7, 5, the n key, the ≤ key, then 3, 5, ., 0, 0. The ≤ key is in the fourth row. Skip the dollar signs. On this site, type <= for ≤.

    So you would enter 12.50 + 4.75n ≤ 35.00. An equal form, such as 4.75n + 12.50 ≤ 35.00, counts too.

Check your answer

Test some numbers. For 4 notebooks, 12.50+4.75×4=31.5012.50 + 4.75 \times 4 = 31.50, which is not over 35.00. For 5 notebooks, 12.50+4.75×5=36.2512.50 + 4.75 \times 5 = 36.25, which is too much. So Elijah can buy up to 4 notebooks. That makes sense for a budget of $35.00.

More questions like this

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

Question 60 of 120

2026 Practice Test C Math, question 42

Question

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

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 7,500

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

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

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Something you must already know minor Working it out

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

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

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Find the volume of the cube

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

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

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

  3. Find the volume of the hole

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

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

    So 500 cubic cm of clay are removed.

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

  4. Subtract the hole from the cube

    The clay left is the cube minus the hole:

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

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

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

  5. Enter the number

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

    So you would enter 7,500.

Check your answer

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

More questions like this

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

Question 61 of 120

2026 Practice Test C Math, question 43

Question

A softball player bought bottles of water for her team to have at practice. The total number of bottles she bought, b, is proportional to n, the number of players at practice. Which equation represents the relationship between b and n? The constant of proportionality is p.

  • Ab=pnb = pn
  • Bb=p+nb = p + n
  • Cb+n=pb + n = p
  • Db+n+p=1b + n + p = 1
Show the answer

Correct answer: A

In this context, the constant of proportionality, p, represents the relationship between b, the number of bottles of water, and n, the number of players on the team.

The number of bottles of water is the product of the constant of proportionality and the number of players.

The equation that represents this relationship is b=pnb = pn.

Question analysis

A softball player buys bottles of water for practice, and the number of bottles, b, is proportional to the number of players, n, with constant of proportionality p. Students choose the equation that represents the relationship. It tests writing an equation for a proportional relationship using letters only, so the meaning of proportional has to be clear.

NYS standards
  • 7.RP.2c proportional relationship equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 students write equations for proportional relationships from tables, graphs and descriptions, usually with a number to find. Here the constant is already given as a letter, so only the general form must be recalled, a short task.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 7, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a recall question dressed in a short story: the softball team adds nothing but names for the letters. The one idea needed is that proportional means multiplied by a constant, so doubling the players doubles the bottles. Students who think of a constant as something added, or as a fixed total, are drawn to the two sums that use the same letters.

  1. Something you must already know minor Working it out

    The general form of a proportional relationship, one amount equal to the constant times the other, has to be recalled; the stem names p but never says it multiplies. (Anderson & Krathwohl, 2001)

  2. Variables instead of numbers minor Understanding the question

    No numbers are given, only b, n and p, so a choice cannot be checked by computing unless students make up a case, such as 2 bottles per player for 10 players. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Recall how two proportional quantities are linked through the constant of proportionality, then test each choice with sample numbers: if the players double, the bottles should double too.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship
y=kxy = kx
y and x are two amounts that change together. k is the constant of proportionality, the number you always multiply x by to get y. Because it is a multiplication, doubling x doubles y.

  1. Sort out the question

    The letter b is the number of bottles, and n is the number of players. The bottles are proportional to the players. The constant of proportionality is p.

    You need the equation that links b and n. There are no numbers, only letters.

  2. Recall what proportional means

    Proportional means one amount is always a fixed number times the other. For example, say there are 2 bottles for each player. Then 10 players need 20 bottles, and 20 players need 40 bottles. Doubling the players doubles the bottles.

    The fixed number, here 2, is the constant of proportionality.

  3. Write it with these letters

    Start with the rule y=kxy = kx. The bottles take the place of y, the players take the place of x, and p is the constant:

    b=pnb = pn

    Here pn means p times n.

  4. Check the other choices

    Watch out: choice B, b=p+nb = p + n, adds instead of multiplying. With p = 2, 10 players give 12 bottles. Then 20 players give 22, which is not double.

    Choice C makes p a total of bottles and players, not a multiplier. Choice D makes all three letters add up to 1, which does not fit at all.

  5. Give the answer

    Only b=pnb = pn multiplies the players by a constant. So it is the only choice where doubling the players doubles the bottles.

    So the answer is A, b=pnb = pn.

Check your answer

Try p = 3 bottles per player. With 5 players, b=3×5=15b = 3 \times 5 = 15. With 10 players, b=3×10=30b = 3 \times 10 = 30. Twice the players, twice the bottles. Also, 15÷5=315 \div 5 = 3, so b ÷ n gives back the constant.

More questions like this

The original DOE question
Question 43 as shown on the DOE’s digital test, page 116 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 116, question 43.

Question 62 of 120

2026 Practice Test C Math, question 44

Question

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

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

The acceptable screen width is from to millimeters.

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

Show the answer

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

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

The least allowed measure is

72−0.04(72)=72−2.88=69.12 mm72 - 0.04(72) = 72 - 2.88 = 69.12\text{ mm}

The greatest allowed measure is

72+0.04(72)=72+2.88=74.88 mm72 + 0.04(72) = 72 + 2.88 = 74.88\text{ mm}

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

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a percent of a number and adding or subtracting it is grade 7 percent work. The ± sign, the decimal result and two linked menus, which must both be right, add a little to the usual demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 59 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    Four percent of 72 is 0.04 × 72 = 2.88, a decimal product, and taking it from 72 means borrowing across the decimal point to reach 69.12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “a variation of ±4% in the width” has to become 4% of 72 millimeters, a length of 2.88 mm, before anything is added or subtracted. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    The second menu lists 75.00 just 0.12 above 74.88, so a rounded estimate of 72 + 3 lands on the wrong option, and estimating cannot separate the two. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Many steps minor Working it out

    Both menus must be right to score, so the minimum and the maximum each need their own subtraction or addition after the shared percent is found. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The percent is how many out of 100. Dividing it by 100 turns it into a decimal, so 4% becomes 0.04. The whole is the amount you take the percent of. The part is the amount that the percent stands for.

  1. Sort out the question

    The screen is designed to be 72 millimeters (mm) wide. The sign ±4% means plus or minus 4%. So a screen may be up to 4% narrower or up to 4% wider than 72 mm.

    You need the minimum, the smallest allowed width, and the maximum, the largest. You pick each one from a drop-down menu.

  2. Find 4% of 72 mm

    Use the percent formula. 4% is 4100=0.04\frac{4}{100} = 0.04:

    0.04×72=2.880.04 \times 72 = 2.88

    So the width may change by up to 2.88 mm either way.

    Watch out: 4% is not 4 mm. Taking 4 away and adding 4 gives 68.00 and 76.00. Both are in the menus, and both are wrong.

  3. Find the minimum width

    Take the change away from the design width:

    72−2.88=69.1272 - 2.88 = 69.12

    The narrowest allowed screen is 69.12 mm wide.

  4. Find the maximum width

    Add the change to the design width:

    72+2.88=74.8872 + 2.88 = 74.88

    The widest allowed screen is 74.88 mm wide.

    Watch out: rounding 2.88 up to 3 gives 75.00. That is in the second menu too, but it is too wide.

  5. Choose from each menu

    Open the first drop-down menu and choose 69.12. Open the second one and choose 74.88.

    So the sentence reads: the acceptable screen width is from 69.12 to 74.88 millimeters.

Check your answer

Use multipliers. A screen 4% narrower keeps 96% of the width: 0.96×72=69.120.96 \times 72 = 69.12. A screen 4% wider is 104% of it: 1.04×72=74.881.04 \times 72 = 74.88. Both match. Also, both ends are the same 2.88 mm away from 72.

More questions like this

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

Question 63 of 120

2026 Practice Test C Math, question 45

Question

Malik earns $10 per hour at his job. He wants to change to a job that will pay $12 per hour. What will be the percent increase in Malik’s hourly pay if he makes this job change?

  • A2%
  • B20%
  • C83%
  • D120%
Show the answer

Correct answer: B

Subtract to determine the amount of increase in Malik’s hourly pay if he changes jobs.

12−10=212 - 10 = 2

Divide the increase amount by the original hourly pay rate.

2÷10=0.22 \div 10 = 0.2

Multiply the quotient by 100 to express the decimal amount as a percentage.

0.2×100=200.2 \times 100 = 20

Question analysis

Malik earns $10 an hour and wants to switch to a job that pays $12 an hour. Students find the percent increase in his hourly pay. It tests finding a percent change. The numbers are easy, so the challenge is knowing which amounts to compare and how to turn that comparison into a percent.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent increase is grade 7 content, and this case uses whole-dollar wages that make the arithmetic easy. The demand is comparing the change with the right amount rather than reporting the change itself or a ratio of the wages.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 2, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers could hardly be friendlier, a $2 raise on $10, so the question checks what to divide by what. Dividing $12 by $10 gives 1.2, the new pay as a percent of the old, and stopping there is tempting because the division feels like the whole job. Reading the $2 itself as 2% is the other quick error, since dollars of change are not percent of change.

  1. Something you must already know minor Working it out

    The rule for percent change, the change divided by the starting amount, is not given, and each wrong choice comes from a different comparison of the $10 and the $12. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    “Percent increase in Malik’s hourly pay” has to be read as the $2 rise measured against the $10 he earns now, not as the new wage compared with the old. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many dollars the hourly pay goes up, then divide that change by the pay Malik earns now and write the result as a percent.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
percent increase=new−originaloriginal×100\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100
The original is the amount before the change. The new amount is the amount after it. The top is how much it went up. Dividing by the original compares the change with where you started. Multiplying by 100 makes it a percent.

  1. Sort out the question

    Malik earns $10 an hour now. The new job pays $12 an hour. You need the percent increase.

    That means: how much does his pay go up, compared with what he earns now? So $10 is the original amount, and $12 is the new amount.

  2. Find the increase in dollars

    Subtract the old pay from the new pay:

    12−10=212 - 10 = 2

    His pay goes up by $2 an hour.

    Watch out: $2 is an amount of money, not a percent. Calling it 2% gives choice A.

  3. Compare the increase with the original

    Use the percent increase formula. Divide the $2 by the original $10, then multiply by 100:

    210×100=0.2×100=20\frac{2}{10} \times 100 = 0.2 \times 100 = 20

    So the pay goes up by 20%.

    Watch out: 12÷10=1.212 \div 10 = 1.2 is 120%, choice D. That is the new pay as a percent of the old, not the increase. And 10÷1210 \div 12 is about 83%, choice C, which is upside down.

  4. Give the answer

    The $2 raise is 20% of Malik’s current pay.

    So the answer is B, 20%.

Check your answer

Work forward. 20% of $10 is 0.20×10=20.20 \times 10 = 2 dollars. Add that to $10, and you get $12, the new pay. It works. Note that 120% is the new pay compared with the old pay, which is the same as a 20% increase.

More questions like this

The original DOE question
Question 45 as shown on the DOE’s digital test, page 118 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 118, question 45.

Question 64 of 120

2026 Practice Test C Math, question 46

Question

A rain gauge shows that 10 centimeters of rain fell yesterday. If 1 inch is equal to approximately 2.5 centimeters, about how many inches of rain fell yesterday?

  • A1
  • B2.5
  • C4
  • D10
Show the answer

Correct answer: C

Set up a proportion to solve:

1 in.2.5 cm=x in.10 cm\frac{1\text{ in.}}{2.5\text{ cm}} = \frac{x\text{ in.}}{10\text{ cm}}

2.5x=102.5x = 10

x=10÷2.5x = 10 \div 2.5

x=4x = 4

10 centimeters is approximately 4 inches.

Question analysis

A rain gauge shows that 10 centimeters of rain fell, and 1 inch is about 2.5 centimeters. Students find about how many inches of rain fell. It tests converting between metric and customary units with a given conversion fact, and the main risk is converting in the wrong direction or reusing a number from the question.

NYS standards
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting units with a given conversion fact is grade 6 content, and this needs a single division with friendly numbers, lighter than grade 7 work with rates and percents.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 10, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A single division answers the question, and its numbers are friendly: four lengths of 2.5 centimeters make 10. Every wrong choice is simply a number from the stem, so the only real risk is reaching for a given value instead of converting. Inches are longer than centimeters, so the rainfall in inches has to be less than 10 but more than the 1 inch that 2.5 centimeters make.

  1. Turning words into math minor Understanding the question

    The fact “1 inch is equal to approximately 2.5 centimeters” has to be turned into a division, how many 2.5-centimeter lengths fit into 10, rather than a multiplication. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Ask how many groups of 2.5 centimeters fit into the 10 centimeters of rain, since each group of 2.5 centimeters makes one inch.

Traps

  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Converting centimeters to inches
inches=centimeters÷2.5\text{inches} = \text{centimeters} \div 2.5
About 2.5 centimeters make 1 inch. So each group of 2.5 centimeters is 1 inch. Dividing the centimeters by 2.5 counts how many of those groups fit. That count is the number of inches.

  1. Sort out the question

    The rain gauge shows 10 centimeters (cm) of rain. You know that 1 inch is about 2.5 cm. You need the rain in inches.

  2. Decide whether to multiply or divide

    An inch is longer than a centimeter. So it takes fewer inches than centimeters to measure the same rain. The answer must be less than 10. That tells you to divide, not multiply.

    Watch out: choice D, 10, keeps the same number, as if inches and centimeters were the same size.

  3. Divide by 2.5

    Use the conversion formula:

    10÷2.5=410 \div 2.5 = 4

    You can see it by adding: 2.5+2.5+2.5+2.5=102.5 + 2.5 + 2.5 + 2.5 = 10. Four groups of 2.5 cm fit into 10 cm. So 10 cm is about 4 inches.

  4. Give the answer

    Watch out: choices A and B, 1 and 2.5, are just the numbers in the conversion fact. They do not use the 10 cm at all.

    About 4 inches of rain fell. So the answer is C, 4.

Check your answer

Change back. Each inch is 2.5 cm, so 4 inches is 4×2.5=104 \times 2.5 = 10 cm. That is what the gauge showed. Also, 4 inches is less than 10, as it should be, because inches are longer.

More questions like this

The original DOE question
Question 46 as shown on the DOE’s digital test, page 119 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 119, question 46.

Question 65 of 120

2026 Practice Test C Math, question 47

Question

The height of a plant, in millimeters, is proportional to the number of days the plant has grown. The plant has grown 20 millimeters in 8 days. What is the constant of proportionality, in millimeters per day?

  • A0.4
  • B2.5
  • C12
  • D20
Show the answer

Correct answer: B

The constant of proportionality is determined by dividing the number of millimeters by the number of days:

20 mm8 days=208=2.5 mm/day\frac{20\text{ mm}}{8\text{ days}} = \frac{20}{8} = 2.5\text{ mm/day}

Question analysis

A plant's height in millimeters is proportional to the number of days it has grown, and it grew 20 millimeters in 8 days. Students find the constant of proportionality in millimeters per day. It tests finding a unit rate, and the order of the division decides whether the rate comes out in the units asked for.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a constant of proportionality from one pair of values is a basic grade 7 skill. With one division and the units named in the question, it asks less than most grade 7 problems.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 7, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One division gives the constant, and the question itself names the order by asking for millimeters per day. The tempting slip is dividing the smaller number by the larger, 8 by 20, out of habit, which gives a rate in days per millimeter. A plant that grows 20 millimeters in 8 days grows more than 2 millimeters a day, a quick sense check on any result.

  1. Something you must already know minor Working it out

    Students must know that the constant is the height divided by the days; the units in the question, millimeters per day, name the order of the division. (Anderson & Krathwohl, 2001)

Strategy: Use the units in the question, millimeters per day, as the instruction: divide the millimeters grown by the number of days.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality
k=yxk = \frac{y}{x}
Here y is the height in millimeters, and x is the number of days. k is the growth in one day, in millimeters per day. In a proportional relationship, y = kx, so dividing y by x gives k.

  1. Sort out the question

    The plant’s height is proportional to the number of days it grows. It grew 20 millimeters (mm) in 8 days.

    You need the constant of proportionality in millimeters per day. Per day means for each day. So you need how much the plant grows in one day.

  2. Divide millimeters by days

    The units tell you the order: millimeters per day means millimeters divided by days. Use the formula:

    k=208=52=2.5k = \frac{20}{8} = \frac{5}{2} = 2.5

    So the plant grows 2.5 mm each day.

  3. Avoid the tempting slips

    Watch out: 8÷20=0.48 \div 20 = 0.4, choice A, is days per millimeter. That is the rate upside down.

    Choice C, 12, comes from 20−820 - 8. Subtracting does not make a rate. Choice D, 20, is the growth over all 8 days, not one day.

  4. Give the answer

    The constant of proportionality is 2.5 mm per day.

    So the answer is B, 2.5.

Check your answer

Multiply back. At 2.5 mm a day for 8 days, the plant grows 2.5×8=202.5 \times 8 = 20 mm. That matches. Also, 2 mm a day would give only 16 mm, and 3 mm a day would give 24 mm. So the rate must be between 2 and 3.

More questions like this

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

Question 66 of 120

2026 Practice Test C Math, question 48

Question

Which table shows a proportional relationship?

  • A
    xy
    00
    432\frac{3}{2}
    83
  • B
    xy
    01
    472\frac{7}{2}
    86
  • C
    xy
    0−1
    4−12-\frac{1}{2}
    80
  • D
    xy
    00
    412
    816
Show the answer

Correct answer: A

The first table is the only one with a constant of proportionality, 38\frac{3}{8}.

In each ordered pair, 38x=y\frac{3}{8}x = y.

38(0)=0\frac{3}{8}(0) = 0

38(4)=128=32\frac{3}{8}(4) = \frac{12}{8} = \frac{3}{2}

38(8)=248=3\frac{3}{8}(8) = \frac{24}{8} = 3

Question analysis

Four tables each give y for x = 0, 4 and 8, with some values written as fractions or negatives. Students choose the table that shows a proportional relationship. It tests recognizing proportional relationships from a table, which means checking every row against one rule rather than judging by the pattern.

NYS standards
  • 7.RP.2a testing for proportionality
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Deciding from a table whether a relationship is proportional is a core grade 7 skill. The fractions and negative values in the tables make each ratio a careful check rather than a glance.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 9, 13 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two quick tests decide proportionality, and the tables are built so that each test alone lets a wrong table through. The last table starts at (0, 0) like the right one, while the middle two grow by equal steps like the right one. Only checking that y ÷ x is the same in every row with x not 0 separates the one table whose ratio stays at 3/8.

  1. Every choice must be checked moderate Choosing the answer

    Each table has to be tested row by row, and two of them start at (0, 0), so the opening row settles nothing; only y ÷ x in the other rows separates them. (Embretson & Wetzel, 1987)

  2. Error-prone arithmetic minor Working it out

    The y-values include 3/2, 7/2 and −1/2, so the ratios mean dividing fractions by 4, such as (3/2) ÷ 4 = 3/8, and comparing them with 3 ÷ 8. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: For each table, divide y by x in every row where x is not zero, and check that the quotient never changes and that y is 0 when x is 0.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Checks only one number: Tries one value in each choice and takes the first that works, when that value happens to work for a wrong choice too; a second value would rule it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship
y=kxy = kx
y and x are the two amounts in each row. k is the constant of proportionality, the same number in every row. Two tests follow from it. When x is 0, y must be 0. And y ÷ x must equal k in every row where x is not 0.

  1. Sort out the question

    Each table gives y for x = 0, 4 and 8. You need the one that shows a proportional relationship. That means y is always the same number times x.

    Use two tests. First, y must be 0 when x is 0. Second, y ÷ x must be the same in the other rows.

  2. Test the row where x is 0

    Look at the top row of each table:

    • Table A: y = 0. It passes.
    • Table B: y = 1. It fails.
    • Table C: y = −1. It fails.
    • Table D: y = 0. It passes.

    Watch out: tables B and C go up by equal steps. Steady steps are not enough, because they do not start at 0. Table C does have a 0, but it is next to x = 8.

  3. Test the ratio in table A

    Divide y by x. Dividing by 4 is the same as multiplying by 14\frac{1}{4}:

    32÷4=32×14=38\frac{3}{2} \div 4 = \frac{3}{2} \times \frac{1}{4} = \frac{3}{8}

    3÷8=383 \div 8 = \frac{3}{8}

    Both rows give 38\frac{3}{8}. Table A passes both tests.

  4. Test the ratio in table D

    Divide y by x in the same way:

    12÷4=316÷8=212 \div 4 = 3 \qquad 16 \div 8 = 2

    The ratios are different, so table D is not proportional.

    Watch out: table D starts at (0, 0) like a proportional table. If you check only that row, or only the row with 4, you will pick choice D.

  5. Give the answer

    Only table A starts at (0, 0) and keeps y ÷ x the same.

    So the answer is A, the table with y = 0, 32\frac{3}{2} and 3.

Check your answer

Use the rule for table A, y=38xy = \frac{3}{8}x. When x = 4, y=38×4=128=32y = \frac{3}{8} \times 4 = \frac{12}{8} = \frac{3}{2}. When x = 8, y=38×8=3y = \frac{3}{8} \times 8 = 3. Both match. Doubling x from 4 to 8 also doubles y.

More questions like this

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

Question 67 of 120

2026 Practice Test C Math, question 49

Question

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

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

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

The range of screen time hours is hours.

The IQR is hours.

The difference between the range and the IQR is hours.

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

Show the answer

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

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

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

20 hr−5 hr=15 hr20\text{ hr} - 5\text{ hr} = 15\text{ hr}

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

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

15 hr−8 hr=7 hr15\text{ hr} - 8\text{ hr} = 7\text{ hr}

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

15 hr−7 hr=8 hr15\text{ hr} - 7\text{ hr} = 8\text{ hr}

Question analysis

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

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Reading a box plot and finding its range and interquartile range are grade 7 statistics skills. The values sit on labeled whole-number ticks and the three sentences lead through the steps, a typical grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 69 words.
Corrected from the DOE handbook
  • A missing comma in the explanation was restored.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

    The sentences name the range and the IQR but never define them; students must know the range spans the whiskers and the IQR spans only the box. (Anderson & Krathwohl, 2001)

  2. Many steps minor Working it out

    Three boxes must all be right, the third depends on the first two, and each tile can be used once, so one mix-up spreads to a second box. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The box plot shows the hours of screen time for eighth graders in a week. You need three numbers: the range, the IQR, and the difference between them.

    There are four tiles: 5, 7, 8 and 15. Each tile is used at most once, and one tile is left over.

  2. Read the five values

    The number line has a tick at every hour, and every value sits on a tick:

    • The left whisker starts at 5. This is the minimum.
    • The box starts at 8. This is the lower quartile, the middle of the lower half of the data.
    • The line inside the box is at 12. This is the median, the middle value.
    • The box ends at 15. This is the upper quartile, the middle of the upper half.
    • The right whisker ends at 20. This is the maximum.

  3. Find the range

    Use the range formula with the two ends of the whiskers:

    range=20−5=15\text{range} = 20 - 5 = 15

    The screen times spread over 15 hours, from the least to the most.

  4. Find the IQR

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

    IQR=15−8=7\text{IQR} = 15 - 8 = 7

    The middle half of the students spend between 8 and 15 hours, a spread of 7 hours.

    Watch out: the IQR is a width, not a point on the plot. The tile 8 is the lower quartile. Putting it in the IQR box also makes the last box 7, which is wrong.

  5. Find the difference

    Subtract the IQR from the range:

    15−7=815 - 7 = 8

    Watch out: do not swap the two measures. The range uses the whiskers’ ends, and the IQR uses only the box.

  6. Fill in the three boxes

    On the test, you drag a tile into each box. On this site, you choose each tile from its box’s menu. The range box gets 15, the IQR box gets 7, and the difference box gets 8. The tile 5 is not used.

    So the boxes get 15, 7 and 8.

Check your answer

The difference is just the two whiskers added together. The left whisker is 8−5=38 - 5 = 3 hours long. The right whisker is 20−15=520 - 15 = 5 hours long. 3+5=83 + 5 = 8, which matches the third box.

More questions like this

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

Question 68 of 120

2026 Practice Test C Math, question 50

Question

Terry wants to pour cement around the edge of the circular patio in her backyard. The patio has a radius of 5 feet. What is the distance, in feet, around the edge of the patio? Use 3.14 for π\pi.

  • A15.7
  • B31.4
  • C49.3
  • D78.5
Show the answer

Correct answer: B

To find the distance around a circle if the radius is known, use the formula for circumference, C=2πrC = 2\pi r.

C=2(3.14)(5)=31.4C = 2(3.14)(5) = 31.4

Question analysis

Terry wants to pour cement around the edge of a circular patio with a radius of 5 feet. Students find the distance around the edge, using 3.14 for π. It tests the circumference of a circle, and because the formulas for a circle's area and circumference both use the radius, they are easy to mix up.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying a circle formula is a grade 7 standard, and this is a one-formula problem with a given value for π. The only demand beyond routine is recalling the right formula, which the test does not print.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One formula and one multiplication finish the problem once the right formula is chosen. The cement in the story pulls toward area, and the area, 78.5, is listed, as is half the circumference, from multiplying 3.14 by the radius alone. Remembering that the circumference is π times the diameter, here 10 feet, and that the answer is in feet, not square feet, settles it.

  1. Something you must already know moderate Working it out

    The circumference formula is not printed, and the area formula uses the same radius and π, so recalling the wrong one gives a number that looks just as finished. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    Pouring cement suggests covering a surface, but the question asks for “the distance, in feet, around the edge”, which is the circumference, not the area. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Decide which circle measurement the distance around the edge is, recall its formula, and substitute the radius and 3.14, doubling the radius wherever the formula calls for the diameter.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle
C=2πrC = 2 \pi r
C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, about 3.14. 2r is the diameter, the distance all the way across. The distance around is always about 3.14 times that.
Area of a circle
A=πr2A = \pi r^2
A is the area, the space inside the circle, in square units. r² means r times r. It is here so you can tell the two formulas apart. This question does not use it.

  1. Sort out the question

    The patio is a circle with a radius of 5 feet. The radius is the distance from the center to the edge. Terry wants cement around the edge.

    You need the distance around the edge, in feet. The distance around a circle is called its circumference. Use 3.14 for π.

  2. Choose the right formula

    The distance around is a length, so use the circumference formula, C=2πrC = 2\pi r.

    Watch out: pouring cement may make you think of area. The area, 3.14×5×5=78.53.14 \times 5 \times 5 = 78.5 square feet, is choice D. That is the space inside the patio, not the distance around it.

  3. Put in the numbers

    Put 3.14 in for π and 5 in for r. Multiply 2 by 5 first, because that is easy:

    C=2×3.14×5=10×3.14=31.4C = 2 \times 3.14 \times 5 = 10 \times 3.14 = 31.4

    The distance around the patio is 31.4 feet.

    Watch out: 3.14×5=15.73.14 \times 5 = 15.7, choice A, forgets the 2. It goes only halfway around.

  4. Give the answer

    The distance around the edge of the patio is 31.4 feet.

    So the answer is B, 31.4.

Check your answer

Estimate. The diameter is 10 feet. The distance around a circle is a little more than 3 times its diameter. 3×10=303 \times 10 = 30, so about 31 feet makes sense. Choice D, 78.5, is far too big, and it is in square feet.

More questions like this

The original DOE question
Question 50 as shown on the DOE’s digital test, page 123 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 123, question 50.

Question 69 of 120

2026 Practice Test D Math, question 1

Question

What is the prime factorization of 3,575?

  • A5⋅11⋅135 \cdot 11 \cdot 13
  • B52⋅11⋅135^{2} \cdot 11 \cdot 13
  • C5⋅7155 \cdot 715
  • D52⋅1435^{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÷25=1433{,}575 \div 25 = 143

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

143÷11=13143 \div 11 = 13; both 11 and 13 are prime numbers.

The first factor, 25, is not prime but is the product of primes, 5×5=525 \times 5 = 5^{2}.

The prime factorization of 3,575 is 52×11×135^{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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisibility by 25
last two digits 00,25,50 or 75→divisible 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
a2=a×aa^{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.

  1. 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.

  2. Divide by 25 first

    3,575 ends in 75. So it is divisible by 25:

    3,575÷25=1433{,}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.

  3. 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÷11=13143 \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.

  4. 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=5×5×11×13=52⋅11⋅133{,}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, 52⋅11⋅135^{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 70 of 120

2026 Practice Test D Math, question 2

Question

What is the value of z in x10+y−x5=z10\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}?

  • A1
  • B5
  • Cyy
  • D2y−x2y - 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.

x10+y−x5=z10\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}

10(x10+y−x5)=10(z10)10\left(\frac{x}{10} + \frac{y - x}{5}\right) = 10\left(\frac{z}{10}\right)

10x10+10(y−x)5=10z10\frac{10x}{10} + \frac{10(y - x)}{5} = \frac{10z}{10}

x+2(y−x)=zx + 2(y - x) = z

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

x+2(y−x)=zx + 2(y - x) = z

x+2y−2x=zx + 2y - 2x = z

2y−x=z2y - 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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a=b, then 10×a=10×b\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×b−a×ca(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.

  1. 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.

  2. 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×x10+10×y−x5=10×z1010 \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)=zx + 2(y - x) = z

  3. Distribute the 2

    Use the distributive property. Multiply the 2 by both y and x:

    2(y−x)=2y−2x2(y - x) = 2y - 2x

    So the equation becomes:

    x+2y−2x=zx + 2y - 2x = z

  4. 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=−xx - 2x = -x

    That leaves:

    2y−x=z2y - 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−x2y - 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 71 of 120

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.

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

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

What is the value of x−yx - 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+85 + x + 3 + 8, which is x+16x + 16.

The sum for set S is 6+y+4+16 + y + 4 + 1, which is y+11y + 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+11x + 16 = y + 11

x=y−5x = y - 5

x−y=−5x - y = -5

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

For example, if x=1x = 1, then y=6y = 6 because 1+16=6+111 + 16 = 6 + 11. If x=2x = 2, then y=7y = 7 because 2+16=7+112 + 16 = 7 + 11. Then find the value of x−yx - y. No matter which values are chosen, x−yx - y will always be −5-5.

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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a=b, then a−c=b−c\text{if } a = 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.

  1. 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−yx - y, the gap between them.

  2. Add up each set

    Add the known numbers in each set, and keep the letter:

    5+x+3+8=x+165 + x + 3 + 8 = x + 16

    6+y+4+1=y+116 + y + 4 + 1 = y + 11

    The two sums are equal, so:

    x+16=y+11x + 16 = y + 11

  3. Rearrange to get x − y

    You want x−yx - y on one side. Subtract y from both sides:

    x−y+16=11x - y + 16 = 11

    Then subtract 16 from both sides:

    x−y=11−16=−5x - y = 11 - 16 = -5

  4. 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−yx - y negative.

    Watch out: 16 − 11 = 5 is y−xy - x, not x−yx - 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=−5x - 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 72 of 120

2026 Practice Test D Math, question 4

Question

If x=14x = \frac{1}{4}, what is the value of 22−x\frac{2}{2 - x}?

  • A34\frac{3}{4}
  • B89\frac{8}{9}
  • C1171\frac{1}{7}
  • D3123\frac{1}{2}
Show the answer

Correct answer: C

The first step is to substitute the value for x into the expression to get 22−14\frac{2}{2 - \frac{1}{4}}.

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

2(74)=2÷(74)=2(47)=87=117\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction
a÷bc=a×cba \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
ab=qrb\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.

  1. Sort out the question

    You know x=14x = \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.

  2. Work out the bottom

    Put in 1/4 for x. To subtract, write 2 as fourths, since 2 = 8/4:

    2−14=84−14=742 - \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.

  3. 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÷74=2×47=872 \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.

  4. Write it as a mixed number

    The choices use mixed numbers. 7 fits into 8 once, with 1 left over:

    87=117\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, 1171\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 73 of 120

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 125\frac{1}{25}.

The probability that a randomly selected student will receive Card C17 is 1100\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, 125⋅1100\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome
P=outcomes you wantall possible outcomesP = \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 and B)=P(A)×P(B)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.

  1. 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.

  2. 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(win)=125P(\text{win}) = \frac{1}{25}

  3. 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(C17)=1100P(\text{C17}) = \frac{1}{100}

  4. 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(win and C17)=125⋅1100P(\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.

  5. 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 74 of 120

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≥3p \geq 3.

  • A12+2(p+3)12 + 2(p + 3)
  • B12+2(p−3)12 + 2(p - 3)
  • C14p+314p + 3
  • D14p−314p - 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)=1412 + (4 - 3)(2) = 14

  • 5 pounds is

12+(5−3)(2)=1612 + (5 - 3)(2) = 16

  • p pounds is

12+(p−3)(2)=12+2(p−3)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.
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.

  1. 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)

  2. 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)

  3. 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

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
total=base fee+rate×extra pounds\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.

  1. 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.

  2. Count the extra pounds

    Only the pounds above 3 cost extra. To find them, take 3 away from the weight:

    extra pounds=p−3\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.

  3. Build the expression

    Each extra pound costs $2, so the extra pounds cost 2(p − 3) dollars. Add the $12 base fee:

    total=12+2(p−3)\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.

  4. 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:

    ChoiceValue when p = 5
    A12 + 2(8) = 28
    B12 + 2(2) = 16
    C14(5) + 3 = 73
    D14(5) − 3 = 67

    Only B gives $16. So the answer is B, 12+2(p−3)12 + 2(p - 3).

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 75 of 120

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 140=0.025\frac{1}{40} = 0.025.

Convert the decimal to a percentage: 0.025×100=2.5%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
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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome
P=outcomes you wantall possible outcomesP = \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
percent=ab×100\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%.

  1. 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.

  2. Write the probability as a fraction

    Only 1 book is number 223, and there are 40 books in all:

    P=140P = \frac{1}{40}

    The numbers on the labels do not matter beyond this. They just give each book its own name.

  3. 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÷40=0.0251 \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.

  4. Change the decimal to a percent

    Multiply by 100 to get a percent. That moves the decimal point two places to the right:

    0.025×100=2.50.025 \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.

  5. 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 76 of 120

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=s2A = s^{2}.

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

To determine the area of the circle, use A=πr2A = \pi r^{2}.

The radius of the circle (the hole) is one-half the diameter or 2 in.÷2=1 in.2\text{ in.} \div 2 = 1\text{ in.}, so the area of the entire circle is π(1 in.)2≈3.14 sq in.\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 sq in.−π sq in.=9\text{ sq in.} - \pi\text{ sq in.} =

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

5.86 sq in.5.86\text{ sq in.}

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

  1. 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)

  2. 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)

  3. 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

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=s2A = 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=πr2A = \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.

  1. 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.

  2. Find the area of the square

    Use the formula for the area of a square, with s = 3 inches:

    A=s2=32=3×3=9A = s^{2} = 3^{2} = 3 \times 3 = 9

    The whole plate, before the hole is cut, is 9 square inches.

  3. 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÷2=1r = 2 \div 2 = 1

    The radius is 1 inch.

  4. Find the area of the hole

    Use the formula for the area of a circle, with r = 1 inch:

    A=πr2=π×12=π≈3.14A = \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.

  5. Subtract the hole from the square

    The shaded metal is the square with the circle taken away:

    9−π≈9−3.14=5.869 - \pi \approx 9 - 3.14 = 5.86

    The shaded region is about 5.86 square inches.

  6. 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 77 of 120

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-m + k
  • B−m⋅k-m \cdot k
  • Cm−km - k
  • D−k⋅m-k \cdot m
  • Emk\frac{m}{k}
Show the answer

Correct answer: A, B, and D

Each of the expressions must be considered:

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

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

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

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

m−km - 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⋅m-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.

mk\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 + k, −m⋅k-m \cdot k, and −k⋅m-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.
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.

  1. 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)

  2. 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)

  3. 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

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) = 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)×(−b)=a×b(-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.

  1. 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.

  2. 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.

  3. Check choices A and B

    A: −m+k-m + k is a positive plus a positive, so it is always positive. Test: 2 + 3 = 5.

    B: −m⋅k-m \cdot k is a positive times a positive, so it is always positive. The dot · means times. Test: 2 × 3 = 6.

  4. Check choices C and E

    C: m−km - k starts below 0 and moves further down, so it is always negative. Test: −2 − 3 = −5.

    E: mk\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.

  5. Check choice D

    D: −k⋅m-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.

  6. 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 78 of 120

2026 Practice Test D Math, question 10

Question

2x+11>3x+92x + 11 > 3x + 9

For what values of x is the above inequality true?

  • Ax<2x < 2
  • Bx>2x > 2
  • Cx<20x < 20
  • Dx>20x > 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+92x + 11 > 3x + 9

2x−3x+11>3x−3x+92x - 3x + 11 > 3x - 3x + 9

−x+11>9-x + 11 > 9

−x+11−11>9−11-x + 11 - 11 > 9 - 11

−x>−2-x > -2

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

−x>−2-x > -2

x<2x < 2

Another way to solve is to keep the coefficient of x positive and then recognize that 2>x2 > x is the same as x<2x < 2.

2x+11>3x+92x + 11 > 3x + 9

11>x+911 > x + 9

2>x2 > x

x<2x < 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.
What makes it hard
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.

  1. 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)

  2. 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

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
if a>b, then a−c>b−c\text{if } a > b \text{, then } a - c > b - c
a and b are the two sides, and c is any number 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
if a>b, then −a<−b\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.

  1. 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.

  2. 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+9−2x2x + 11 - 2x > 3x + 9 - 2x

    11>x+911 > x + 9

  3. Get x by itself

    Subtract 9 from both sides:

    11−9>x+9−911 - 9 > x + 9 - 9

    2>x2 > 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.

  4. Read it from the x side

    2>x2 > x says 2 is greater than x. That is the same as saying x is less than 2:

    x<2x < 2

    Watch out: when you swap the sides, the sign must point the other way. Writing x>2x > 2 gives choice B. If you subtract 3x instead, you get −x>−2-x > -2. Dividing by −1 then flips the sign, and you get x<2x < 2 again.

    So the answer is A, x<2x < 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 79 of 120

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=205 + 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 inches(5 miles4 inches)=6.25 miles5\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
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a triangle
P=a+b+cP = a + b + c
P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. You add them because going once around the triangle covers each side once.
Using a scale
actual length=drawing length×miles per inch\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.

  1. 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.

  2. Find the drawing’s perimeter

    Add the three sides of the drawing:

    P=5+7+8=20P = 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.

  3. Find the scale

    20 inches stand for 25 miles. Divide to find how many miles 1 inch stands for:

    25÷20=1.2525 \div 20 = 1.25

    So each inch on the drawing is 1.25 miles in the park.

  4. Scale the shortest side

    The shortest side on the drawing is 5 inches. Multiply it by the scale:

    5×1.25=6.255 \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 80 of 120

2026 Practice Test D Math, question 12

Question

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

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

  • A12
  • B15
  • C17
  • D19
Show the answer

Correct answer: D

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

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

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

x+5=24x + 5 = 24

x=19x = 19

Question analysis

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

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

Why it is hard

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

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

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

  2. No obvious method minor Working it out

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. See that SP is a radius

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

  3. Find the diameter RT

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

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

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

  4. Solve for x

    RT is also split into x and 5, so:

    x+5=24x + 5 = 24

    Subtract 5 from both sides:

    x=24−5=19x = 24 - 5 = 19

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

  5. Work from the numbers, not the picture

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

    So the answer is D, 19.

Check your answer

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

More questions like this

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

Question 81 of 120

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

5x−5=905x - 5 = 90

5x=955x = 95

x=19x = 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=472x + 9 = 2(19) + 9 = 47

3x−14=3(19)−14=433x - 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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles in a triangle
a+b+c=180a + b + c = 180
a, b and c are the three angles 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=90a + 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.

  1. 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.

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

  3. Solve for x

    Combine the x terms, 2x + 3x = 5x. Combine the numbers, 9 − 14 = −5:

    5x−5=905x - 5 = 90

    Add 5 to both sides, then divide both sides by 5:

    5x=955x = 95

    x=95÷5=19x = 95 \div 5 = 19

  4. Find both angles

    Put x = 19 into each expression:

    2x+9=2×19+9=38+9=472x + 9 = 2 \times 19 + 9 = 38 + 9 = 47

    3x−14=3×19−14=57−14=433x - 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.

  5. 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 82 of 120

2026 Practice Test D Math, question 14

Question

James must read 84 pages of a book for a class assignment. If he reads 14\frac{1}{4} of the pages in the morning and 13\frac{1}{3} of the remaining pages in the afternoon, how many pages will he still need to read?

  • A35
  • B42
  • C45
  • D56
Show the answer

Correct answer: B

Pages read in the morning: (14)(84)=21\left(\frac{1}{4}\right)(84) = 21

Pages remaining after the morning: 84−21=6384 - 21 = 63

Pages read in the afternoon: (13)(63)=21\left(\frac{1}{3}\right)(63) = 21

Pages he still needs to read: 63−21=4263 - 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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of a number
ab of n=ab×n\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.

  1. 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.

  2. Find the morning pages

    Take 1/4 of 84:

    14×84=84÷4=21\frac{1}{4} \times 84 = 84 \div 4 = 21

    He reads 21 pages in the morning. That leaves:

    84−21=6384 - 21 = 63

    63 pages are left after the morning.

  3. Find the afternoon pages

    The afternoon fraction is 1/3 of the 63 pages left, not of all 84:

    13×63=63÷3=21\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.

  4. Find the pages still to read

    Take the afternoon pages away from the 63 pages left after the morning:

    63−21=4263 - 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 83 of 120

2026 Practice Test D Math, question 15

Question

111+122+133\frac{1}{11} + \frac{1}{22} + \frac{1}{33}

What is the value of the expression shown above?

  • A122\frac{1}{22}
  • B111\frac{1}{11}
  • C16\frac{1}{6}
  • D322\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.

111=1×611×6=666\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}

122=1×322×3=366\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}

133=1×233×2=266\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}

666+366+266=1166=16\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×nb×n\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
ad+bd+cd=a+b+cd\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.

  1. 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.

  2. 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:

    2×3×11=662 \times 3 \times 11 = 66

  3. Rename each fraction

    Multiply the top and bottom of each fraction by the number that makes the bottom 66:

    111=1×611×6=666\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}

    122=1×322×3=366\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}

    133=1×233×2=266\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}

  4. Add the tops

    Now every piece is a sixty-sixth. Add the tops and keep the bottom:

    666+366+266=1166\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.

  5. 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:

    1166=11÷1166÷11=16\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, 16\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 84 of 120

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=pxy = px.

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

1=2p1 = 2p

12=2p2\frac{1}{2} = \frac{2p}{2}

0.5=p0.5 = 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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality
y=kxsok=yxy = 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.

  1. 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.

  2. 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=yx=1.02=0.5k = \frac{y}{x} = \frac{1.0}{2} = 0.5

    So there is 0.5 ounce of raisins for each cup of oatmeal.

  3. Check the other rows

    A constant must be the same in every row:

    1.53=0.53.06=0.5\frac{1.5}{3} = 0.5 \qquad \frac{3.0}{6} = 0.5

    Every row gives 0.5, so the rule is y=0.5xy = 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.

  4. 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 85 of 120

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?

  • A1621\frac{16}{21}
  • B1019\frac{10}{19}
  • C1023\frac{10}{23}
  • D516\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=8438 + 46 = 84

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

84−64=2084 - 64 = 20

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

2064=516\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.
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.

  1. 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)

  2. 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)

  3. 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

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
total=O+C−both\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
fraction=partwhole\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.

  1. 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.

  2. Add the two groups

    Add the number in the orchestra and the number in the choir:

    38+46=8438 + 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.

  3. Find the students in both

    The extra amount over 64 is the students who were counted twice:

    84−64=2084 - 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.

  4. Write the fraction and simplify

    The part is 20 students, and the whole is all 64 students. Both numbers divide by 4:

    2064=20÷464÷4=516\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, 516\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 86 of 120

2026 Practice Test D Math, question 18

Question

Which of the following decimals is equivalent to 123+2491\frac{2}{3} + 2\frac{4}{9}?

  • A3.13.\overline{1}
  • B3.63.\overline{6}
  • C4.14.\overline{1}
  • D4.94.\overline{9}
Show the answer

Correct answer: C

First, add the mixed numbers. Then, convert the fraction 19\frac{1}{9} to a decimal.

123+2491\frac{2}{3} + 2\frac{4}{9}

169+2491\frac{6}{9} + 2\frac{4}{9}

3+1093 + \frac{10}{9}

3+1193 + 1\frac{1}{9}

419=4.111…=4.14\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×nb×n\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
19=0.111…=0.1\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…

  1. 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.

  2. 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:

    23=2×33×3=69\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.

  3. Add the mixed numbers

    Add the whole numbers, then add the fractions:

    1+2=369+49=1091 + 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+119=4193 + 1\frac{1}{9} = 4\frac{1}{9}

    Watch out: forgetting to carry the 1 gives 3 1/9, choice A.

  4. Change it to a decimal

    One ninth is 0.111…, with the 1 repeating forever. So:

    419=4.111…=4.14\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, 4.14.\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 87 of 120

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:

456×1.25=570456 \times 1.25 = 570

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

456+570=1,026456 + 570 = 1{,}026

The sales tax on the selling price is:

1,026(0.08)=82.081{,}026(0.08) = 82.08

The customer pays the selling price plus the sales tax:

1,026+82.08=1,108.081{,}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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
p% of n=p100×np\% \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
new price=old price+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.

  1. 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.

  2. 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:

    456×1.25=456+114=570456 \times 1.25 = 456 + 114 = 570

    The increase is $570.

  3. Find the selling price

    Increases by means the increase is added to the cost:

    456+570=1,026456 + 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.

  4. Find the sales tax

    As a decimal, 8% is 0.08. Find 8% of the selling price:

    1,026×0.08=82.081{,}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.

  5. Add the tax to the price

    The customer pays the selling price plus the tax:

    1,026+82.08=1,108.081{,}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 88 of 120

2026 Practice Test D Math, question 20

Question

What is the value of −9(−4)2+36(−4)+304−4\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.

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

−9(16)+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.

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

−144−144+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.

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

−288+304−4=16−4\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.

16−4=−4\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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations
grouping→exponents→×,÷→+,−\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)×(−a)=a2(-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
positivenegative=negative\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.

  1. 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.

  2. Do the exponent first

    The only exponent is (−4)². Squaring means multiplying a number by itself:

    (−4)2=(−4)×(−4)=16(-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.

  3. Do the multiplication

    Now multiply. A negative times a positive is negative:

    −9×16=−144-9 \times 16 = -144

    36×(−4)=−14436 \times (-4) = -144

    So the top becomes:

    −144+(−144)+304-144 + (-144) + 304

  4. Add from left to right

    Add the two negatives first:

    −144+(−144)=−288-144 + (-144) = -288

    Then add 304. Since 304 is more than 288, the result is positive:

    −288+304=16-288 + 304 = 16

    The whole top is 16.

  5. Divide by −4

    Now divide the top by the bottom. A positive divided by a negative is negative:

    16−4=−4\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 89 of 120

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=23.257.75 + 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=1.7525.00 - 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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Money left over
left=money she has−money she owes\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.

  1. 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.

  2. Add up what she borrowed

    Line up the decimal points and add:

    7.75+11.00+4.50=23.257.75 + 11.00 + 4.50 = 23.25

    She owes $23.25 in all. That is the amount she must pay back.

  3. 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.

  4. Subtract to find what is left

    Take the debt away from the gift:

    25.00−23.25=1.7525.00 - 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 90 of 120

2026 Practice Test D Math, question 22

Question

Seth grows strawberries in his garden.

  • He picked 2122\frac{1}{2} cups of strawberries on Monday.
  • He picked 4 cups of strawberries on Tuesday.
  • He ate 34\frac{3}{4} cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.

How many cups of strawberries did Seth have left?

  • A3123\frac{1}{2}
  • B4144\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.

212+4+(3)(−34)2\frac{1}{2} + 4 + (3)\left(-\frac{3}{4}\right)

212+4−2142\frac{1}{2} + 4 - 2\frac{1}{4}

612−2146\frac{1}{2} - 2\frac{1}{4}

624−2146\frac{2}{4} - 2\frac{1}{4}

4144\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.

212+4+(−34)+(−34)+(−34)2\frac{1}{2} + 4 + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right)

4144\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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total of a repeated amount
total=number of times×amount each time\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×ab=n×abn \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.

  1. 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.

  2. Add what he picked

    Add the two days of picking:

    212+4=6122\frac{1}{2} + 4 = 6\frac{1}{2}

    He picked 6 1/2 cups in all, before he ate any.

  3. Find how much he ate

    He ate 3/4 cup on each of 3 days. Multiply:

    3×34=94=2143 \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.

  4. 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:

    612−214=624−214=4146\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, 4144\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 91 of 120

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+5, and giving a friend $5 is represented by −5-5. These values combine to equal zero because +5−5=0+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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Opposites add to zero
a+(−a)=0a + (-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.

  1. 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.

  2. Write each choice with signs

    Turn each situation into signed numbers and add them up:

    ChoiceAmountsTotal
    A+5, then −50
    B−10, then −10−20
    C+3, then +3+6
    Dstart at −20, then +20, then +20+20

  3. 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.

  4. 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)=05 + (-5) = 0

    So the answer is A: receiving $5 as a gift and then giving $5 to a friend.

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 92 of 120

2026 Practice Test D Math, question 24

Question

An athlete runs 18\frac{1}{8} kilometer in 34\frac{3}{4} minute. At this rate, how many kilometers would the athlete run in 1 minute?

  • A332\frac{3}{32}
  • B16\frac{1}{6}
  • C1121\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: 18 kilometer34 minute\frac{\frac{1}{8}\text{ kilometer}}{\frac{3}{4}\text{ minute}}.

To evaluate the fraction, use the reciprocal when dividing to get 16\frac{1}{6} kilometer per 1 minute.

18÷34=18×43=424=16\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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
rate=distance÷time\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
ab÷cd=ab×dc\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.

  1. 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.

  2. Set up the division

    Kilometers per minute means kilometers divided by minutes. So the distance comes first and the time second:

    rate=18÷34\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.

  3. Divide the fractions

    Flip 3/4 to get 4/3, then multiply. Multiply the tops and multiply the bottoms:

    18×43=424\frac{1}{8} \times \frac{4}{3} = \frac{4}{24}

    Simplify by dividing the top and bottom by 4:

    424=16\frac{4}{24} = \frac{1}{6}

    Watch out: multiplying 1/8 × 3/4 without flipping gives 3/32, choice A.

  4. 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, 16\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 93 of 120

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?

  • A8
  • B10
  • C12
  • D14
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÷45=10 hours450 \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
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, speed and time
d=r×td = 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÷rt = 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.

  1. 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.

  2. 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=50×9=450d = 50 \times 9 = 450

    The trip is 450 miles long.

  3. 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÷45=10t = 450 \div 45 = 10

    The trip takes 10 hours.

  4. 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 94 of 120

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⋅x\$18 \cdot x, written as 18x18x.

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+18x30 + 18x.

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

This is represented as 30+18x≤12030 + 18x \leq 120 or 120≥30+18x120 \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
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.
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.

  1. 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)

  2. 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

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
total cost=fee+rate×x\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
total cost≤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.

  1. 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.

  2. Work out a few totals

    Find the total cost for a few hours. Each hour adds one more $18:

    HoursHourly charges ($)Total cost ($)
    118×1=1818 \times 1 = 1830+18=4830 + 18 = 48
    218×2=3618 \times 2 = 3630+36=6630 + 36 = 66
    x18x18x30+18x30 + 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.

  3. 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≤12030 + 18x \leq 120

    Watch out: the sign < means less than, but not equal. 30+18x<12030 + 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+1830x + 18 charges $30 an hour.

  4. Stop before you solve

    You could solve the inequality. Subtract 30 from both sides to get 18x≤9018x \leq 90. Then divide both sides by 18 to get x≤5x \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.

  5. 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.

Check your answer

Test two numbers of hours. For 5 hours, 30+18×5=12030 + 18 \times 5 = 120. Is 120 ≤ 120? Yes, so 5 hours fit the budget. For 6 hours, 30+18×6=13830 + 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 95 of 120

2026 Practice Test D Math, question 27

Question

A birdseed mixture is 25\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.

25=1025\frac{2}{5} = \frac{10}{25}

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

($0.101 ounce)(16 ounces1 pound)=$1.60\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.

$1.601 pound×10 pounds1=$16.00\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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of an amount
part=fraction×whole\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
ounces=pounds×16\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
cost=number of units×price of one unit\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.

  1. Sort out the question

    You know:

    • The birdseed is 25\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.

  2. Find the weight of sunflower seeds

    Only 2/5 of the birdseed is sunflower seeds. Find 2/5 of 25 pounds:

    25×25=505=10\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.

  3. Change pounds to ounces

    The price is per ounce, so turn the 10 pounds into ounces. Each pound is 16 ounces:

    10×16=16010 \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.

  4. Find the cost

    Each ounce costs $0.10. Multiply:

    160×0.10=16.00160 \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 400×0.10=40.00400 \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 16×0.10=1.6016 \times 0.10 = 1.60 dollars. Ten pounds cost 10×1.60=16.0010 \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 96 of 120

2026 Practice Test D Math, question 28

Question

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

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

Correct answer: C

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. A very close wrong answer minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The graph links two amounts:

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

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

  2. Match each number to its letter

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

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

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

  3. Read the pair as a sentence

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

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

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

Check your answer

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

More questions like this

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

Question 97 of 120

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?

  • A116\frac{1}{16}
  • B316\frac{3}{16}
  • C47\frac{4}{7}
  • D716\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:

11121314212223243132333441424344\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 316\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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event
P=number of outcomes you wanttotal number of outcomesP = \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
total outcomes=a×b\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.

  1. 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.

  2. 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:

    4×4=164 \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.

  3. Make a table of products

    Write each product in a table. Red goes down the side and blue goes across the top.

    Red × Blue1234
    11234
    22468
    336912
    4481216

    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.

  4. Find the probability

    Use the probability rule. Three outcomes give 4, out of 16 in all:

    P=316P = \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, 316\frac{3}{16}.

Check your answer

List the ways to multiply two numbers from 1 to 4 and get 4: 1×41 \times 4, 2×22 \times 2 and 4×14 \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 98 of 120

2026 Practice Test D Math, question 30

Question

A recipe uses 23\frac{2}{3} cup of sugar for every 1141\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?

  • A815\frac{8}{15}
  • B56\frac{5}{6}
  • C1781\frac{7}{8}
  • D3233\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:

23 cup of sugar114 cups of flour\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.

2354=23÷54=23×45=815\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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=cups of sugarcups of flour\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
wnd=w×d+ndw\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
ab÷cd=ab×dc\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.

  1. Sort out the question

    The recipe uses 23\frac{2}{3} cup of sugar for every 1141\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.

  2. Set up the unit rate

    You want sugar for each cup of flour. So sugar goes on top and flour goes on the bottom:

    unit rate=23÷114\text{unit rate} = \frac{2}{3} \div 1\frac{1}{4}

    Watch out: dividing the other way, flour by sugar, gives 158=178\frac{15}{8} = 1\frac{7}{8}. That is cups of flour for each cup of sugar, choice C.

  3. Change the mixed number

    You cannot divide by a mixed number directly. Change it to a fraction first. Each whole has 4 fourths:

    114=1×4+14=541\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}

  4. Divide the fractions

    Multiply by the reciprocal of 5/4, which is 4/5. Multiply the tops and multiply the bottoms:

    23÷54=23×45=815\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 23×54=1012=56\frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}, choice B.

    So the answer is A, 815\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 23=1015\frac{2}{3} = \frac{10}{15}, the answer 8/15 is a bit less. Also, 815×54=4060=23\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 99 of 120

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.

28=100x\frac{2}{8} = \frac{100}{x}

2x=8002x = 800

x=400x = 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
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios (a proportion)
red in samplegreen in sample=red in boxgreen in box\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.

  1. 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.

  2. 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:

    28=100x\frac{2}{8} = \frac{100}{x}

  3. Scale up the sample

    How do you get from 2 red to 100 red? Multiply by 50, since 2×50=1002 \times 50 = 100. The green marbles grow by the same factor:

    8×50=4008 \times 50 = 400

    So the box has about 400 green marbles.

  4. 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=400500 - 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 100500=15\frac{100}{500} = \frac{1}{5}. The sample had 210=15\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 100 of 120

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.

  • A2142\frac{1}{4}
  • B4124\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 36×2=7236 \times 2 = 72 cups.

Convert 72 cups to gallons:

72 cups×1 quart4 cups×1 gallon4 quarts=72 \text{ cups} \times \frac{1 \text{ quart}}{4 \text{ cups}} \times \frac{1 \text{ gallon}}{4 \text{ quarts}} =

7216 gallons=\frac{72}{16} \text{ gallons} =

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

412 gallons4\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
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.
What makes it hard
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.

  1. 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)

  2. 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

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
total=number of people×amount each\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
cups in 1 gallon=4×4=16\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
gallons=cups÷16\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.

  1. 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.

  2. Find the total in cups

    Each of the 36 people drank 2 cups:

    36×2=7236 \times 2 = 72

    They drank 72 cups in all.

    Watch out: skipping the word each uses only 36 cups. Then 36÷16=21436 \div 16 = 2\frac{1}{4} gallons, choice A.

  3. Find the cups in a gallon

    A gallon is 4 quarts, and each quart is 4 cups:

    4×4=164 \times 4 = 16

    So 1 gallon holds 16 cups.

    Watch out: adding the two 4s gives 8 cups in a gallon. Then 72÷8=972 \div 8 = 9, choice C.

  4. Change cups to gallons

    Divide the cups by 16. Simplify the fraction by dividing the top and bottom by 8:

    72÷16=7216=92=41272 \div 16 = \frac{72}{16} = \frac{9}{2} = 4\frac{1}{2}

    They drank 4124\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, 4124\frac{1}{2}.

Check your answer

Work backward. 4124\frac{1}{2} gallons is 412×4=184\frac{1}{2} \times 4 = 18 quarts. 18 quarts is 18×4=7218 \times 4 = 72 cups. Shared among 36 people, that is 72÷36=272 \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 101 of 120

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

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

$25,920−$10,000=$15,920\$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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\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
amount left=price+tax−down payment\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.

  1. 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.

  2. Find the sales tax

    Change 8% to the decimal 0.08. Multiply it by the price:

    0.08×24,000=1,9200.08 \times 24{,}000 = 1{,}920

    A quick way: 1% of 24,000 is 240, so 8% is 8×240=1,9208 \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.

  3. Find the full cost

    Add the tax to the price:

    24,000+1,920=25,92024{,}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.

  4. Subtract the down payment

    Bindu already paid $10,000. Take it away:

    25,920−10,000=15,92025{,}920 - 10{,}000 = 15{,}920

    She has $15,920 left to pay.

    Watch out: leaving out the tax gives 24,000−10,000=14,00024{,}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: 24,000×1.08=25,92024{,}000 \times 1.08 = 25{,}920. Then 25,920−10,000=15,92025{,}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 102 of 120

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?

  • A15\frac{1}{5}
  • B725\frac{7}{25}
  • C1225\frac{12}{25}
  • D710\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 35\frac{3}{5}. The probability that Emily will ride the bus home on any given afternoon is 45\frac{4}{5}. Therefore, the probability that Emily will ride the bus in the morning and the afternoon is 35×45=1225\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.
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.

  1. 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)

  2. 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

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=times it happenedtimes it could have happenedP = \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 and B)=P(A)×P(B)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.

  1. 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.

  2. Write each chance

    Use last week to find each chance:

    • Bus to school on a morning: 35\frac{3}{5}
    • Bus home on an afternoon: 45\frac{4}{5}

    Keep them apart. The morning rides go to school, and the afternoon rides go home.

  3. Multiply the two chances

    Both rides must happen, so multiply:

    35×45=1225\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 3×4=123 \times 4 = 12 of them, she rides both times. So the chance is 12 out of 25.

  4. 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, 1225\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 103 of 120

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?

  • A4
  • B6
  • C12
  • D24
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.
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting with two choices
number of groups=ways to pick the man×ways to pick the women\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.

  1. 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.

  2. 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.

  3. 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 4×3=124 \times 3 = 12 pairs, each one twice. Then doubling for the men gives 24, choice D.

  4. Add a man to each pair

    Each pair of women can go with man 1 or man 2. Use the counting rule:

    6×2=126 \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=126 + 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 104 of 120

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(90 - 45) + 1 = 46 pages

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

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

Leah read a total of 128 pages, since 46+36+46=12846 + 36 + 46 = 128.

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

128500=128÷4500÷4=32125\frac{128}{500} = \frac{128 \div 4}{500 \div 4} = \frac{32}{125}

32÷125=0.256=25.6%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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range
pages=last page−first page+1\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
percent=partwhole×100\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.

  1. 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.

  2. Count the pages each day

    Use the rule for an inclusive range: last page minus first page, plus 1.

    90−45+1=4690 - 45 + 1 = 46

    135−100+1=36135 - 100 + 1 = 36

    190−145+1=46190 - 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.

  3. Add the three days

    Add the pages from all three days to get her total:

    46+36+46=12846 + 36 + 46 = 128

    Leah read 128 pages in all. This is the part you compare with the goal.

  4. Change to a percent

    Divide the pages read by the goal, then multiply by 100:

    128500×100=0.256×100=25.6\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.

  5. 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: 0.256×500=1280.256 \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 105 of 120

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÷8=80 and 640÷5=128640 \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÷8=79R3 and 635÷5=127635 \div 8 = 79R3 \text{ and } 635 \div 5 = 127

    645÷8=80R5 and 645÷5=129645 \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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisible by 5
last digit=0 or 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
LCM(5,8)=5×8=40\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.

  1. 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.

  2. Use the rule for 5

    A number divisible by 5 ends in 0 or 5. Between 630 and 650, only three numbers do:

    635,640,645635, \; 640, \; 645

    Only these need checking for 8.

  3. 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 8×798 \times 79, passes for 8 but ends in 2, so it fails for 5.

  4. 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. 40×16=64040 \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 106 of 120

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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sample space
sample space=the list of every possible outcome\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.

  1. 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.

  2. 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.

  3. 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

  4. 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 107 of 120

2026 Practice Test D Math, question 39

Question

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

HOCKEY GOAL PROBABILITIES

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

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

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

Correct answer: C

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

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

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

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

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

Question analysis

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

NYS standards
  • 6.SP.8 probability models for simple events
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 8, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Reading a figure, table or graph minor Understanding the question

    The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding probabilities
P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)
A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
The chance it does not happen
P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.

  1. Sort out the question

    The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.

  2. Pick the right rows

    Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:

    • 3 goals: 0.21
    • 4 goals: 0.09
    • 5 goals: 0.04

    Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.

  3. Add the chances

    A game can end with only one number of goals. So add the three chances:

    0.21+0.09+0.04=0.340.21 + 0.09 + 0.04 = 0.34

    Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.

    So the answer is C, 0.34.

Check your answer

Find the chance of fewer than 3 goals, the other rows: 0.16+0.21+0.29=0.660.16 + 0.21 + 0.29 = 0.66. Then the chance of 3 or more is 1−0.66=0.341 - 0.66 = 0.34. Both ways give 0.34.

More questions like this

The original DOE question
Question 39 as shown on the DOE’s digital test, page 112 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 112, question 39.

Question 108 of 120

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?

  • A5+53x≤1005 + 53x \leq 100
  • B5+53x≥1005 + 53x \geq 100
  • C53+5x≤10053 + 5x \leq 100
  • D53+5x≥10053 + 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+5x53 + 5x. The team wants to earn at least (≥) 100 points, so they need 53+5x≥10053 + 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
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total with a start and a rate
total=start+rate×x\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
total≥goal\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.

  1. 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.

  2. Write the total points

    Try 2 cans first. The team gets 5 points for each, so 53+5×2=6353 + 5 \times 2 = 63 points. For x cans, the total is:

    53+5x53 + 5x

    Watch out: the 53 points do not repeat for each can. 5+53x5 + 53x gives 53 points for each can, as in choices A and B.

  3. Choose the sign

    At least 100 means 100 or more, so use ≥:

    53+5x≥10053 + 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≥10053 + 5x \geq 100.

Check your answer

Test two numbers of cans. With 10 cans, 53+5×10=10353 + 5 \times 10 = 103, and 103 ≥ 100 is true. The team wins. With 9 cans, 53+5×9=9853 + 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 109 of 120

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?

  • A3
  • B30
  • C60
  • D90
Show the answer

Correct answer: B

The constant of proportionality is the ratio of y (90) to x (3). 90÷3=3090 \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.
What makes it hard
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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship
y=kx, so k=yxy = 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.

  1. 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.

  2. Divide the distance by the time

    Use the formula. The distance is y and the time is x:

    k=903=30k = \frac{90}{3} = 30

    The constant is 30. It means the eagle flies 30 miles in each hour.

  3. 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 30×3=9030 \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 110 of 120

2026 Practice Test D Math, question 42

Question

A student reads 112\frac{1}{12} of a book in 13\frac{1}{3} hour. What fraction of the book can the student read in 1 hour?

  • A136\frac{1}{36}
  • B14\frac{1}{4}
  • C512\frac{5}{12}
  • D123\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.

112 book13 hour\frac{\frac{1}{12} \text{ book}}{\frac{1}{3} \text{ hour}}

Then use division to find the unit rate.

11213=112÷13=112×31=312=14\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 13\frac{1}{3} hour times 3 equals 1 hour.

3×112 book3×13 hour=14 book1 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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=part of the bookhours\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
ab÷cd=ab×dc\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.

  1. Sort out the question

    The student reads 112\frac{1}{12} of a book in 13\frac{1}{3} of an hour. You need the part of the book read in 1 whole hour. That is a unit rate.

  2. Scale up to one hour

    One hour holds three thirds of an hour. So in 1 hour the student reads 3 times as much:

    3×112=312=143 \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.

  3. 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:

    112÷13=112×31=312=14\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.

  4. Avoid the wrong operations

    Watch out: multiplying gives 112×13=136\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, 14\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 14×13=112\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 111 of 120

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:

$12.0024=$0.50 per bar\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:

$6.5010=$0.65 per bar\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 per bar−$0.50 per bar=$0.15 per bar\$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
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.
What makes it hard
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.

  1. 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)

  2. 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

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
cost per bar=price of the pack÷number of bars\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.

  1. 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.

  2. Find the Art Club’s cost

    Share $12.00 among 24 bars:

    12.00÷24=0.5012.00 \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.

  3. 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÷10=0.656.50 \div 10 = 0.65

    Each bar costs $0.65. Pick $0.65 in the second menu.

  4. Find the difference

    Subtract the smaller cost from the larger one:

    0.65−0.50=0.150.65 - 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.

  5. 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. 24×0.50=12.0024 \times 0.50 = 12.00 and 10×0.65=6.5010 \times 0.65 = 6.50, so both costs per bar match the pack prices. And 0.50+0.15=0.650.50 + 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 112 of 120

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+1.25cp = 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?

  • A4
  • B9
  • C16
  • D18
Show the answer

Correct answer: C

Using the specified equation p=2.50+1.25cp = 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+1.25c22.50 = 2.50 + 1.25c

22.50−2.50=1.25c22.50 - 2.50 = 1.25c

20.00÷1.25=c20.00 \div 1.25 = c

16=c16 = 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.
What makes it hard
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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a=b, then a−n=b−n and an=bn\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.

  1. Sort out the question

    The equation p=2.50+1.25cp = 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.

  2. 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+1.25c22.50 = 2.50 + 1.25c

  3. Subtract the tray

    Take 2.50 away from both sides. That removes the cost of the tray:

    22.50−2.50=1.25c22.50 - 2.50 = 1.25c

    20.00=1.25c20.00 = 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.

  4. Divide by the price of one cupcake

    Divide both sides by 1.25:

    c=20.00÷1.25=16c = 20.00 \div 1.25 = 16

    An easy way: 4 cupcakes cost 4×1.25=5.004 \times 1.25 = 5.00 dollars. $20 is 4 groups of $5. So Samantha bought 4×4=164 \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. 1.25×16=20.001.25 \times 16 = 20.00, and 2.50+20.00=22.502.50 + 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 113 of 120

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⋅w3 \cdot w or 3w3w.

The total amount saved can be represented by the expression 8+3w8 + 3w.

After 3 weeks, Marcus and Isla have $26. This can be represented by the equation 8+3w=268 + 3w = 26.

Solve the equation:

8+3w=268 + 3w = 26

3w=183w = 18 (subtracting 8 from both sides)

w=6w = 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
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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total savings
total=start+weeks×amount each week\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
if a=b, then a−c=b−c and ac=bc\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.

  1. 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.

  2. Write an equation

    Let w be the dollars saved each week. Three weeks add 3w dollars to the $8 they started with:

    8+3w=268 + 3w = 26

  3. Subtract the starting money

    Take 8 away from both sides:

    3w=26−8=183w = 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.

  4. Divide by the number of weeks

    Divide both sides by 3:

    w=18÷3=6w = 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.

  5. 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=148 + 6 = 14. After week 2: 14+6=2014 + 6 = 20. After week 3: 20+6=2620 + 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 114 of 120

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)(x - 1) represents the number of pounds that cost $0.50 each.

The cost equation can be written as Cost=$1.50+$0.50(x−1)\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\$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\$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)\$2.50 = \$1.50 + \$0.50(x - 1)

    $1.00=$0.50(x−1)\$1.00 = \$0.50(x - 1)

    2=x−12 = x - 1

    x=3 poundsx = 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)\$5.50 = \$1.50 + \$0.50(x - 1)

    $4.00=$0.50(x−1)\$4.00 = \$0.50(x - 1)

    8=x−18 = x - 1

    x=9 poundsx = 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)\$3.00 = \$1.50 + \$0.50(x - 1)

$1.50=$0.50(x−1)\$1.50 = \$0.50(x - 1)

3=x−13 = x - 1

x=4x = 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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Shipping cost
C=1.50+0.50×(w−1)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.

  1. 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.

  2. 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×(4−1)=1.50+1.50=3.00C = 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.

  3. Find the two costs

    For 2 pounds, 1 pound is over the first:

    1.50+0.50×1=2.001.50 + 0.50 \times 1 = 2.00

    For 6 pounds, 5 pounds are over the first:

    1.50+0.50×5=1.50+2.50=4.001.50 + 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.

  4. Work back to the two weights

    For $2.50, take off the base fee: 2.50−1.50=1.002.50 - 1.50 = 1.00. At $0.50 a pound, $1.00 pays for 2 extra pounds. Add the first pound: 2+1=32 + 1 = 3 pounds.

    For $5.50: 5.50−1.50=4.005.50 - 1.50 = 4.00. $4.00 pays for 8 extra pounds. Add the first pound: 8+1=98 + 1 = 9 pounds.

    Watch out: forgetting to add back the first pound gives 2 and 8.

  5. 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 115 of 120

2026 Practice Test D Math, question 47

Question

A tall right rectangular prism standing on end. The bottom front edge is labeled 2 in., the bottom edge running back on the right is labeled 2 in., and the height is labeled 6 in. Right-angle marks are drawn at the lower left corner of the front face and at the front left corner of the top face.

What shape will result from a horizontal slice of the figure above?

  • Aa square that has 2-inch sides
  • Ba triangle that has 2-inch sides
  • Ca rectangle that has a 2-inch side and a 6-inch side
  • Da triangle that has one 2-inch side and two 6-inch sides
Show the answer

Correct answer: A

Based on the diagram, the figure shown is a right rectangular prism with a square base. A horizontal slice of the figure will be a plane parallel to the square base, as in this example.

The same prism, 2 in. by 2 in. by 6 in., drawn with its hidden edges dashed and a horizontal slice partway up: a square parallel to the base and the same size as the base.

Since the horizontal slice is parallel to the base, it will be congruent to the base, which is a square with 2-inch sides.

Question analysis

A figure shows a right rectangular prism measuring 2 inches by 2 inches by 6 inches, standing on end. This geometry question asks what two-dimensional shape a horizontal slice of the prism makes. It tests visualizing cross-sections of solids, using the labeled lengths to name both the shape and its size.

NYS standards
  • 7.G.3 slicing three-dimensional solids
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Describing the shape made by slicing a prism parallel to its base is exactly the grade 7 standard. This is its simplest case, a single slice of a labeled right prism, typical of a classroom example.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 2, 46 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A horizontal cut runs parallel to the ground, and this prism stands on a square base, so the slice copies the base. The main danger is picturing the cut along the tall front face, which gives a 2-by-6 rectangle, the shape a vertical slice makes. The two triangles need a slanted cut through a corner and catch only students guessing.

  1. Reading a figure, table or graph minor Understanding the question

    The prism is drawn standing on end, 6 inches tall on a 2-by-2 base, so the cut has to be pictured in a flat drawing where the tall front face is the most visible shape. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Students must know that a slice parallel to the base of a prism has the same shape and size as the base, which here is the square at the bottom. (Anderson & Krathwohl, 2001)

Strategy: Picture a flat cut across the prism parallel to the ground, ask which face of the prism that cut runs parallel to, and read that face's edge lengths from the labels.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A slice parallel to the base
slice parallel to the base≅base\text{slice parallel to the base} \cong \text{base}
The sign ≅ means congruent: the same shape and the same size. A prism has the same shape all the way up, like a stack of matching cards. So a cut parallel to its base, at any height, is a copy of the base.

  1. Sort out the question

    The figure is a right rectangular prism, a box shape whose faces are all rectangles. It stands on end. The two edges along the bottom are labeled 2 in. The tall edge is labeled 6 in.

    You need the shape you get from a horizontal slice. Horizontal means flat, like the floor or a tabletop.

  2. Find the base

    The base is the face the prism stands on. The two bottom edges you can see are both labeled 2 in. The right-angle mark on the top face shows its corners are square. The top and the bottom match. So the base is a square with 2-inch sides.

  3. Picture the slice

    A horizontal slice cuts straight across the prism, parallel to the base. The walls of the prism go straight up, so the slice is a copy of the base. It is a square with 2-inch sides, at whatever height you cut.

    The 6-inch height does not matter. It only tells how tall the stack of squares is.

  4. Rule out the other shapes

    Watch out: a 2-inch by 6-inch rectangle, choice C, comes from a vertical slice. That cut goes straight down, through the 6-inch height. A triangle, choice B or D, needs a slanted cut through a corner.

    So the answer is A, a square that has 2-inch sides.

Check your answer

Picture the slice sliding up and down. At the bottom, it matches the base, 2 in. by 2 in. At the top, it matches the top face, the same square. In between, the walls are straight, so its size never changes.

More questions like this

The original DOE question
Question 47 as shown on the DOE’s digital test, page 120 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 120, question 47.

Question 116 of 120

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=212 : 1 = \frac{2}{1}

Sides:

2413=24÷13=\frac{\frac{2}{4}}{\frac{1}{3}} = \frac{2}{4} \div \frac{1}{3} =

24×31=64=\frac{2}{4} \times \frac{3}{1} = \frac{6}{4} =

32=3:2\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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a square
P=4sP = 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=3tP = 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:ba : 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.

  1. 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.

  2. 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=3×2=6P = 3 \times 2 = 6

  3. Find the square’s side

    The square’s perimeter is twice the triangle’s:

    2×6=122 \times 6 = 12

    The square has 4 equal sides, so divide by 4:

    s=12÷4=3s = 12 \div 4 = 3

    Each side of the square is 3 units.

  4. 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÷4=624 \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 117 of 120

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?

  • A3
  • B5
  • C15
  • D45
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=3×3×545 = 3 \times 3 \times 5

75=3×5×575 = 3 \times 5 \times 5

The common factors can be seen as 3×53 \times 5.

The greatest possible side length, in inches, will be 3×5=153 \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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the prime factors both numbers share\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.

  1. 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.

  2. Break each number into primes

    Write each length as a product of prime numbers:

    45=3×3×545 = 3 \times 3 \times 5

    75=3×5×575 = 3 \times 5 \times 5

    The two lists share one 3 and one 5.

  3. Multiply the shared primes

    Multiply the primes the two numbers share:

    GCF=3×5=15\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.

  4. 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÷15=345 \div 15 = 3 and 75÷15=575 \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 118 of 120

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=630 - 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.

630=0.20\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent decrease
percent decrease=old−newold×100\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.

  1. 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.

  2. Find the drop

    Subtract Tuesday’s count from Monday’s:

    30−24=630 - 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.

  3. Compare the drop with Monday

    Use the formula. Divide the drop by Monday’s count, then multiply by 100:

    630×100=0.20×100=20\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 624=0.25\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: 0.20×30=60.20 \times 30 = 6. Then 30−6=2430 - 6 = 24, which is Tuesday’s count. The answer checks out.

More questions like this

The original DOE question
Question 50 as shown on the DOE’s digital test, page 123 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 123, question 50.

Question 119 of 120

2024 Test A Math, question 1

Question

Simplify:

−3.8+2.3−(−1.1)-3.8 + 2.3 - (-1.1)

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: -0.4

Subtracting a negative number is the same as adding a positive number. Rewrite the expression to simplify:

−3.8+2.3−(−1.1)-3.8 + 2.3 - (-1.1)

−3.8+2.3+1.1-3.8 + 2.3 + 1.1

−3.8+3.4-3.8 + 3.4

−0.4-0.4

Question analysis

This grid-in asks students to simplify a short expression that mixes negative and positive decimals and ends by subtracting a negative number. It tests adding and subtracting rational numbers in order. The double negative, and keeping track of the sign of the final result, are where errors creep in.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding and subtracting negative decimals is core grade 7 content, and this is a routine three-term version. It matches a typical classroom exercise; the only extra demand is gridding a signed decimal correctly.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 16+, 2 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a three-term expression with small decimals, so the method is never in doubt; the points are lost on signs. The double negative at the end is the classic place to slip; the other is the result itself, which sits just below zero and is easy to grid without its minus sign. Grouping the positive terms first, 2.3 + 1.1, and comparing that sum with 3.8 settles both the size and the sign.

  1. Error-prone arithmetic moderate Working it out

    The last term, − (−1.1), has to turn into + 1.1, and the total then crosses zero, since 2.3 + 1.1 = 3.4 falls just short of the 3.8 below zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    As a grid-in it shows no choices, so a double negative read as a single one, or a result gridded without its minus sign, gets no warning from the page. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −1.1 as adding 1.1, combine the two positive decimals first, then compare their total with 3.8 to decide the sign and size.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a is any number, and b is a positive number, so −b is negative. Taking away a negative number is the same as adding the positive number. Think of money: if someone takes away a debt of b dollars, you are b dollars better off.
Adding a negative and a positive
−a+b=−(a−b)-a + b = -(a - b)
a and b are positive numbers, and a is the larger one. A negative and a positive partly cancel out. So find the difference of their sizes, a − b. The result keeps the sign of the number farther from zero, which here is the negative one.

  1. Sort out the expression

    You need to simplify −3.8+2.3−(−1.1)-3.8 + 2.3 - (-1.1). Simplify means work it out to one number. Two of the numbers are negative: −3.8 and −1.1. One is positive: 2.3. The tricky part is at the end. It says minus negative 1.1. The answer can be negative, so keep track of every sign.

  2. Change minus a negative to plus

    Taking away a negative is the same as adding a positive. Use the rule with b = 1.1.

    a−(−b)=a+ba - (-b) = a + b

    −3.8+2.3−(−1.1)=−3.8+2.3+1.1-3.8 + 2.3 - (-1.1) = -3.8 + 2.3 + 1.1

    Now only one negative number is left. Watch out: do not just subtract 1.1. That gives −3.8+2.3−1.1=−2.6-3.8 + 2.3 - 1.1 = -2.6, which is wrong.

  3. Add the two positive numbers

    Group the positive numbers first. This makes the rest of the work short.

    2.3+1.1=3.42.3 + 1.1 = 3.4

    Now the problem is −3.8+3.4-3.8 + 3.4. One number is negative and one is positive.

  4. Combine the negative and the positive

    They partly cancel. Use the rule with a = 3.8 and b = 3.4.

    −a+b=−(a−b)-a + b = -(a - b)

    −3.8+3.4=−(3.8−3.4)=−0.4-3.8 + 3.4 = -(3.8 - 3.4) = -0.4

    The negative number, 3.8, is farther from zero than 3.4 is. So the result is negative. Watch out: keep the minus sign. Dropping it gives 0.4, which is wrong. So you would grid -0.4, with the minus sign in its own column.

Check your answer

Use a number line. Start at −3.8. Adding 2.3 moves you right, to −1.5. Subtracting −1.1 also moves you right, by 1.1, to −0.4. You end just to the left of zero. So a small negative answer makes sense.

More questions like this

The original DOE question
Question 58 as printed on page 75 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 120 of 120

2024 Test A Math, question 2

Question

Angle M and angle R are supplementary. The measure of angle R is 5 times the measure of angle M. What is the measure of angle R, in degrees?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 150

Angle M and angle R are supplementary, so the sum of the angles is 180°.

Let m equal the measure of angle M, and let r equal the measure of angle R.

r+m=180r + m = 180

The measure of angle R is 5 times the measure of angle M.

r=5mr = 5m

5m+m=1805m + m = 180

6m=1806m = 180

m=1806=30m = \frac{180}{6} = 30

r=5m=5(30)=150r = 5m = 5(30) = 150

Therefore, the measure of angle R is 150°.

Question analysis

A geometry grid-in about two supplementary angles, where one angle is five times the size of the other. Students must recall what supplementary means, write an equation linking the two angles, and solve it. The question asks for the larger angle, so stopping at the smaller one is the usual slip.

NYS standards
  • 7.G.5 angle relationships and equations
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students regularly write and solve simple equations for unknown angles using supplementary and complementary facts. This is a two-step version of exactly that task, with whole numbers and no figure to read.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once supplementary is read as a 180° total, the problem is a one-variable equation with whole numbers. Students lose the point in two ways: by stopping at the smaller angle, which the equation produces first, or by using 90° for the total. Checking that the gridded angle is the larger of the pair, and that the two angles add to 180°, catches both.

  1. Something you must already know moderate Working it out

    The stem never says what supplementary means; the 180° total has to be recalled, and a student who confuses it with complementary's 90° halves every angle. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The sentence "The measure of angle R is 5 times the measure of angle M" has to become R = 5M, so the 180° splits into six equal parts, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. An answer in an unexpected form minor Choosing the answer

    As a grid-in there is nothing to compare against, so gridding angle M's 30°, the first value the equation gives, looks just as finished as angle R. (Enright, Morley & Sheehan, 2002)

Strategy: Call angle M one part and angle R five parts; together the six parts make the supplementary total, so find one part and then multiply for angle R.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Supplementary angles
M+R=180M + R = 180
M and R are the sizes of two angles, in degrees. Supplementary means the two angles add up to 180 degrees. Put side by side, they make a straight line. Do not mix this up with complementary, which means a total of 90 degrees.

  1. Sort out the question

    Two angles, M and R, are supplementary. Angle R is 5 times as big as angle M. You need the measure of angle R, in degrees. Measure means its size. Notice that the question asks for R, not M.

  2. Write the facts as equations

    Supplementary angles add up to 180°.

    M+R=180M + R = 180

    The sentence “The measure of angle R is 5 times the measure of angle M” becomes this:

    R=5×MR = 5 \times M

    Watch out: supplementary means 180°, not 90°. Using 90° gives R = 75, which is wrong.

  3. Find angle M

    Put 5×M5 \times M in place of R. Then the equation has only one unknown, M.

    M+5×M=180M + 5 \times M = 180

    6×M=1806 \times M = 180

    M=180÷6=30M = 180 \div 6 = 30

    Think of 6 equal parts. M is 1 part and R is 5 parts. So angle M is 30°.

  4. Find angle R

    Angle R is 5 times angle M.

    R=5×30=150R = 5 \times 30 = 150

    Angle R is 150°. Watch out: do not stop at M. Gridding 30 gives the wrong angle. So you would grid 150.

Check your answer

Add the two angles: 30+150=18030 + 150 = 180. So they are supplementary. Also, 150÷30=5150 \div 30 = 5, so R really is 5 times M. Both facts in the question check out.

More questions like this

The original DOE question
Question 59 as printed on page 75 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)