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Expressions and equations

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Question 1 of 93

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 93

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 93

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 4 of 93

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 5 of 93

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 6 of 93

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 7 of 93

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 8 of 93

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 9 of 93

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 10 of 93

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 11 of 93

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 12 of 93

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 13 of 93

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 14 of 93

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 15 of 93

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 16 of 93

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 17 of 93

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 18 of 93

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 19 of 93

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 20 of 93

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 21 of 93

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 22 of 93

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 23 of 93

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 24 of 93

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 25 of 93

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 26 of 93

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 27 of 93

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 28 of 93

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 29 of 93

2024 Test A Math, question 7

Question

If 2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x} and x≠0x \neq 0, what expression is represented by □\square?

  • Ayy
  • B2y2y
  • C3y3y
  • D4y4y
Show the answer

Correct answer: C

Use properties of equations to determine the missing expression:

2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

22∙2yx−y2x=□2x\frac{2}{2} \bullet \frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

4y2x−y2x=□2x\frac{4y}{2x} - \frac{y}{2x} = \frac{\square}{2x}

4y−y2x=□2x\frac{4y - y}{2x} = \frac{\square}{2x}

3y2x=□2x\frac{3y}{2x} = \frac{\square}{2x}

The missing expression is 3y3y.

Question analysis

An algebra question with fractions: two fractions with x and 2x in their denominators are subtracted, and students find the numerator that makes the difference equal a fraction over 2x. It tests rewriting a fraction over a common denominator when the denominators contain a variable. Adjusting only part of a fraction is the common mistake.

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 subtract fractions with unlike denominators and combine like terms, but rarely see a variable in a denominator. The steps are the same two as with numbers, so the demand is typical once the form is familiar.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 13, 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

This is subtraction of fractions with unlike denominators, written with letters. The box over 2x shows the target denominator, which helps, but each wrong choice matches a familiar fraction error: subtracting the tops before matching the bottoms, or rewriting the first fraction and stopping. Students who rewrite 2y/x as 4y/2x and then subtract y from the new numerator finish in seconds.

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and there is no number to check unless students substitute values such as x = 1 and y = 2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 2y/x becomes 4y/2x with top and bottom both doubled, while y/2x already has the denominator the box sits over. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite the first fraction over 2x by multiplying its top and bottom by 2, then subtract the numerators, which now share a denominator.

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×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top of the fraction, b is the bottom, and k is any number except zero. Multiplying the top and bottom by the same number does not change the value. It is like multiplying by k over k, which is 1.
Subtracting fractions with the same bottom
ad−bd=a−bd\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}
a and b are the tops, and d is the shared bottom, called the denominator. When two fractions have the same bottom, their pieces are the same size. So subtract the tops and keep the bottom.

  1. Sort out the question

    The box, □\square, stands for a missing expression. An expression is a math phrase made of numbers and letters, like 5y. You need what goes in the box to make the two sides equal. x≠0x \neq 0 means x is not zero, so no fraction divides by zero.

  2. Pick a shared bottom

    You can only subtract fractions when their bottoms match. The bottoms here are x and 2x. The box sits over 2x, so use 2x for both. The second fraction, y2x\frac{y}{2x}, already has it.

  3. Rewrite the first fraction over 2x

    To turn x into 2x, multiply by 2. Multiply the top by 2 as well, so the value stays the same.

    ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}

    2yx=2y×2x×2=4y2x\frac{2y}{x} = \frac{2y \times 2}{x \times 2} = \frac{4y}{2x}

    Watch out: this is only the first fraction. Stopping here and choosing 4y gives choice D.

  4. Subtract the tops

    ad−bd=a−bd\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}

    4y2x−y2x=4y−y2x=3y2x\frac{4y}{2x} - \frac{y}{2x} = \frac{4y - y}{2x} = \frac{3y}{2x}

    4y take away y is like 4 apples take away 1 apple. That leaves 3y. Watch out: do not subtract the tops, 2y − y, before the bottoms match. That gives y, choice A.

  5. Match the box

    The left side equals 3y2x\frac{3y}{2x}. The right side is □2x\frac{\square}{2x}. The bottoms are the same, so the tops must be equal. The box is 3y. So the answer is C, 3y3y.

Check your answer

Try numbers. Let x = 1 and y = 2. The left side is 41−22=4−1=3\frac{4}{1} - \frac{2}{2} = 4 - 1 = 3. With 3y in the box, the right side is 3×22×1=62=3\frac{3 \times 2}{2 \times 1} = \frac{6}{2} = 3. Both sides are 3, so 3y works.

More questions like this

The original DOE question
Question 64 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 77, question 64. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 93

2024 Test A Math, question 10

Question

Simplify:

(39(x−3)3)+3913\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}

  • Axx
  • Bx−2x - 2
  • C13x−3613x - 36
  • Dx+3613x + \frac{36}{13}
Show the answer

Correct answer: A

(39(x−3)3)+3913\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}

Divide the first 39 by 3.

13(x−3)+3913\frac{13(x - 3) + 39}{13}

Distribute 13 through (x−3)(x - 3).

13x−39+3913\frac{13x - 39 + 39}{13}

In the numerator, the terms −39 and 39 cancel out.

13x13\frac{13x}{13}

Finally, divide the numerator and denominator by 13 to get x.

Question analysis

An algebra question that asks students to simplify a stacked fraction: 39 times (x − 3) over 3, plus 39, all divided by 13. It tests dividing, distributing and combining like terms in the right order. The nested fraction bars make it easy to divide only some of the terms or to distribute to only one.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.3 generating equivalent expressions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students expand and simplify linear expressions such as 13(x − 3) + 39. This wraps that work inside two fraction bars, which adds reading load and chances to misstep but no new mathematics.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16+, 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

The expression looks heavier than it is: every number is a multiple of 13, and the constants cancel to leave a single term. What makes it risky is scope, knowing which terms each fraction bar and each multiplier covers. Each wrong choice comes from stretching or shrinking that scope by one term, so working outward from the innermost parentheses, one line per step, is the safe way.

  1. Error-prone arithmetic moderate Working it out

    Two fraction bars are nested: 39 ÷ 3 applies only to the first term, the 13 must reach both x and −3, and only then does the whole numerator divide by 13. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Four moves come in a fixed order, divide 39 by 3, distribute 13, cancel −39 with +39, divide by 13, and each is a chance to apply a step to the wrong terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work from the inside out: divide 39 by 3, distribute through the parentheses, combine the constants in the numerator, and only then divide everything by 13.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c
a is the number outside the parentheses. 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) are a groups of b, take away a groups of c.

  1. Sort out the expression

    You need to simplify the big fraction, which means write it in its shortest form. It has a small fraction inside its top. The top is 39(x−3)3+39\frac{39(x - 3)}{3} + 39, and the bottom is 13. Work from the inside out, one step at a time.

  2. Divide the first 39 by 3

    The small fraction bar is only under 39(x−3)39(x - 3). So divide just that 39 by 3.

    39(x−3)3=13(x−3)\frac{39(x - 3)}{3} = 13(x - 3)

    Now the top is 13(x−3)+3913(x - 3) + 39. Watch out: the small bar does not reach the added 39. Dividing that 39 by 3 as well leads to x−2x - 2, choice B.

  3. Multiply out the parentheses

    a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c

    13(x−3)=13×x−13×3=13x−3913(x - 3) = 13 \times x - 13 \times 3 = 13x - 39

    Watch out: multiply the 3 by 13 too, not just the x. Leaving −3 as it is leads to choice D.

  4. Combine the plain numbers on top

    13x−39+39=13x13x - 39 + 39 = 13x

    −39 and 39 add up to 0, so they cancel. The top is now just 13x, which means 13 times x.

  5. Divide by the bottom

    Now divide the whole top by 13. Dividing by 13 undoes times 13.

    13x13=x\frac{13x}{13} = x

    Watch out: the long bar is under the whole top. Dividing only the last 39 by 13 gives 13x−3613x - 36, choice C. So the answer is A, xx.

Check your answer

Try x = 5. Then x−3=2x - 3 = 2 and 39×2=7839 \times 2 = 78. Next, 78÷3=2678 \div 3 = 26 and 26+39=6526 + 39 = 65. Last, 65÷13=565 \div 13 = 5. You get back 5, the same as x. So x is right.

More questions like this

The original DOE question
Question 67 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 67. Open page 78 of the handbook (PDF)

Question 31 of 93

2024 Test A Math, question 16

Question

A truck rental company charges a one-time fee of $40 plus $1 per mile driven. Dalia rented a truck and used a coupon for 20% off the total rental cost. After the coupon was applied, she spent a total of $60. How many miles did she drive?

  • A8
  • B20
  • C32
  • D35
Show the answer

Correct answer: D

Let m equal the miles driven during Dalia’s rental. The total cost, c, can be expressed as an equation.

c=40+mc = 40 + m

Dalia spent a total of $60, which included a 20% off coupon. Find the cost before the coupon by dividing the final cost ($60) by the percentage paid (80%):

6080%=600.80=75\frac{60}{80\%} = \frac{60}{0.80} = 75

Use $75 as the total cost to find m.

40+m=7540 + m = 75

m=35m = 35

Dalia drove 35 miles.

Question analysis

A multistep word problem about a truck rental that costs a $40 fee plus $1 per mile, with a 20% off coupon applied to the total. Given that Dalia paid $60 after the coupon, students find how many miles she drove. Undoing the discount correctly before removing the fixed fee is the hard part.

NYS standards
  • 7.EE.4a two-step equations
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations and work with discounts, usually separately. This combines them and makes students undo a 20% discount starting from the price paid, a reversal many grade 7 students have not practised.
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 5, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two ordinary skills, undoing a discount and solving a two-step equation, are stacked, and the first runs in reverse. The natural way to undo 20% off is to add 20% back, which gives $72 instead of $75 and a choice three miles from the answer. Students who check by applying the coupon to the full cost of 40 + 35 = 75 dollars see at once that 80% of it is 60.

  1. Turning words into math moderate Understanding the question

    The sentence "After the coupon was applied, she spent a total of $60" makes 60 the price after 20% off, so it is 80% of the full cost, 40 + m. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Combines ideas moderate Working it out

    A percent discount has to be undone before the fee-plus-mileage equation can be solved: 60 ÷ 0.8 = 75, then 75 − 40 = 35 miles. (Embretson, 1998; Daniel & Embretson, 2010)

  3. A very close wrong answer moderate Choosing the answer

    The choices 32 and 35 are three miles apart, too close to tell apart by estimate; only dividing 60 by 0.8, rather than adding 20% of 60, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Since $60 is 80% of the full cost, divide 60 by 0.80 to find the cost before the coupon, then subtract the $40 fee to get the miles.

Traps

  • Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
  • 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

Rental cost
c=40+1×mc = 40 + 1 \times m
c is the full cost in dollars, before any coupon. 40 is the one-time fee, paid once no matter how far you drive. m is the number of miles, and each mile costs 1 dollar.
Price after a discount
price paid=full cost×(1−p100)\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)
The full cost is the price before the discount. p is the percent taken off. You pay what is left: 100% minus p%. For 20% off you pay 80%, so you multiply by 0.80.

  1. Sort out the question

    The truck costs a $40 fee, plus $1 for each mile. Dalia’s coupon took 20% off the total. After the coupon, she paid $60. You need how many miles she drove. The $60 is the price after the discount, not the full cost.

  2. Write what she paid

    20% off means she paid what was left of the full cost.

    100%−20%=80%100\% - 20\% = 80\%

    As a decimal, 80% is 0.80.

    price paid=full cost×(1−p100)\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)

    60=full cost×0.8060 = \text{full cost} \times 0.80

  3. Undo the discount

    To undo times 0.80, divide by 0.80. Move the decimal point one place in both numbers to make it easier.

    full cost=60÷0.80=600÷8=75\text{full cost} = 60 \div 0.80 = 600 \div 8 = 75

    The full cost was $75. Watch out: the 20% was taken off the full cost, not off $60. Adding 20% of $60 gives $72 and leads to 32 miles, choice C. Taking 20% off $60 again leads to 8 miles, choice A.

  4. Find the miles

    c=40+1×mc = 40 + 1 \times m

    75=40+m75 = 40 + m

    m=75−40=35m = 75 - 40 = 35

    She drove 35 miles. Watch out: do not skip the coupon. Using $60 as the full cost gives 20 miles, choice B. So the answer is D, 35.

Check your answer

Go forward. 35 miles cost $35. Add the $40 fee to get $75. The coupon takes off 20% of $75, which is 0.20×75=150.20 \times 75 = 15 dollars. Then 75−15=6075 - 15 = 60. That is exactly what she paid.

More questions like this

The original DOE question
Question 73 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 73. Open page 79 of the handbook (PDF)

Question 32 of 93

2024 Test A Math, question 20

Question

A piece of wood that is 4124\frac{1}{2} feet long is cut into 2 pieces of different lengths. The shorter piece has a length of x feet. Which inequality expresses all possible values of x?

  • A0<x<2140 < x < 2\frac{1}{4}
  • B0≤x≤2140 \leq x \leq 2\frac{1}{4}
  • C0<x<4120 < x < 4\frac{1}{2}
  • D214<x<4122\frac{1}{4} < x < 4\frac{1}{2}
Show the answer

Correct answer: A

The original piece of wood is 4124\frac{1}{2} feet long.

The shorter piece is x feet long, and it must be less than half the length of the original piece of wood.

Since half of 4124\frac{1}{2} is 2142\frac{1}{4}, it follows that x<214x < 2\frac{1}{4}.

The shorter piece of wood cannot be 0 feet long, so it follows that 0<x0 < x.

This gives the compound inequality 0<x<2140 < x < 2\frac{1}{4}.

Question analysis

An inequality question: a piece of wood 4½ feet long is cut into two pieces of different lengths, and the shorter piece is x feet long. Students choose the inequality that gives every possible value of x. They must turn each condition in the words into a boundary and decide whether each end of the range is included.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.8 writing simple inequalities
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph simple inequalities. Writing a compound inequality from a description, where the words decide whether each end is strict, is less familiar and calls for careful reading of every condition.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Halving 4 1/2 feet is the only computation; the rest is deciding whether each end of the range can actually happen. A shorter piece of 2 1/4 feet would make the two pieces equal, and a piece of 0 feet is no piece at all, so both ends are excluded. Students who find the right bounds but never test the ends are drawn to the version with the inclusive symbols.

  1. Turning words into math moderate Understanding the question

    Two phrases set the two ends: "cut into 2 pieces" means x is more than 0, and "of different lengths" with "the shorter piece" means x is less than half of 4 1/2. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    The inequalities 0 < x < 2 1/4 and 0 ≤ x ≤ 2 1/4 share both bounds and differ only in their symbols; only checking whether each end can happen separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The question asks for every possible length of the shorter piece, a range rather than one length, so students must picture the cut moving along the board. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Picture cutting the board in different places, find the largest and smallest values the shorter piece could take, and check whether each boundary value is actually possible.

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".
  • 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

Two pieces make the whole board
x+L=412x + L = 4\frac{1}{2}
x is the length of the shorter piece, and L is the length of the longer piece, both in feet. Together they make the whole board. Since x is smaller than L, x must be less than half of the board.
Inequality signs
a<b and a≤ba < b \text{ and } a \leq b
a < b says a is less than b. a ≤ b says a is less than or equal to b, so a can also equal b. A chain like 0 < x < 5 means x is more than 0 and less than 5.

  1. Sort out the question

    A board is 4124\frac{1}{2} feet long. It is cut into 2 pieces of different lengths. The shorter piece is x feet long. You need an inequality for every possible value of x. An inequality uses signs like <, less than, to show a range of numbers.

  2. Find the top end for x

    If you cut exactly in the middle, each piece is half the board.

    412÷2=92×12=94=2144\frac{1}{2} \div 2 = \frac{9}{2} \times \frac{1}{2} = \frac{9}{4} = 2\frac{1}{4}

    But the pieces have different lengths, so the cut is not in the middle. The shorter piece must be less than half.

    x<214x < 2\frac{1}{4}

  3. Find the bottom end for x

    The shorter piece can be very small, like 0.1 foot. But it can’t be 0 feet. Then nothing would be cut off, and there would not be 2 pieces.

    0<x0 < x

  4. Put the two ends together

    0<x<2140 < x < 2\frac{1}{4}

    This says x is more than 0 and less than 2142\frac{1}{4}. Watch out: choice B uses ≤, which allows x = 0 or x=214x = 2\frac{1}{4}. Neither can happen. Choice C lets x be more than half, so it would not be the shorter piece. Choice D describes the longer piece. So the answer is A, 0<x<2140 < x < 2\frac{1}{4}.

Check your answer

Test values. Cut off 1 foot: the pieces are 1 foot and 3123\frac{1}{2} feet. The shorter one is 1, and 1 is between 0 and 2142\frac{1}{4}. If x were 3, the other piece would be 1121\frac{1}{2} feet, so x would not be the shorter one. Choice A leaves 3 out, as it should.

More questions like this

The original DOE question
Question 77 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 81, question 77. Open page 81 of the handbook (PDF)

Question 33 of 93

2024 Test A Math, question 22

Question

F=95C+32F = \frac{9}{5}C + 32

Yesterday in Centerville, the highest Fahrenheit temperature, F, was 86°, and the lowest was 68°. What was the difference between these temperatures, in degrees Celsius, C?

  • A10.0° C
  • B15.0° C
  • C20.0° C
  • D32.4° C
Show the answer

Correct answer: A

The highest temperature was 86° F.

F=95C+32F = \frac{9}{5}C + 32

86=95C+3286 = \frac{9}{5}C + 32

Subtract 32 from both sides of the equation:

86−32=95C86 - 32 = \frac{9}{5}C

54=95C54 = \frac{9}{5}C

Multiply both sides of the equation by 59\frac{5}{9}:

54(59)=(59)(95C)54\left(\frac{5}{9}\right) = \left(\frac{5}{9}\right)\left(\frac{9}{5}C\right)

30=C30 = C

The lowest temperature was 68° F.

F=95C+32F = \frac{9}{5}C + 32

68=95C+3268 = \frac{9}{5}C + 32

68−32=95C68 - 32 = \frac{9}{5}C

36=95C36 = \frac{9}{5}C

(59)(36)=C\left(\frac{5}{9}\right)(36) = C

20=C20 = C

The difference, in degrees Celsius, is 30 minus 20.

30−20=1030 - 20 = 10

Question analysis

A formula question: given F = 9/5 C + 32 and a day's high and low of 86 and 68 degrees Fahrenheit, students find the difference between the two temperatures in degrees Celsius. It tests solving a two-step equation with a fraction coefficient. Answering with the difference, in the right scale, is what to watch.

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 equations of the form px + q = r with fraction coefficients is grade 7 work. Doing it twice and comparing the results is a typical multistep grade 7 task, with extra care over which scale the answer is in.
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, 29 words.
Corrected from the DOE handbook
  • A stray term in the explanation’s working 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

The formula is written to give Fahrenheit from Celsius, and the question runs it the other way, twice. Plugging 86 and 68 straight into the right-hand side is the tempting misuse, and it produces a listed choice. A shortcut avoids most of the work: the 18-degree Fahrenheit gap times 5/9 is the Celsius gap, because the 32 cancels when two readings are subtracted.

  1. Many steps moderate Working it out

    Each Fahrenheit reading needs its own two-step solve, subtracting 32 and multiplying by 5/9, and then the two Celsius values must be subtracted. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Multiplying 54 and 36 by 5/9 is easy only if the division by 9 comes first; multiplying by 5 first leaves 270/9 and 180/9 to reduce. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Solve the formula for C at 86° and at 68° by subtracting 32 and multiplying by 5/9, then subtract; or convert the 18-degree gap directly with 5/9.

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).
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: After “Subtract 32 from both sides of the equation,” the handbook writes “86 − 32 = (9/5)C = 32”. The extra “= 32” would mean (9/5)C = 32, which is false: 86 − 32 = 54, and the next line is 54 = (9/5)C. The site now reads 86 − 32 = (9/5)C, as the handbook writes the same step for 68°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing Celsius to Fahrenheit
F=95C+32F = \frac{9}{5}C + 32
F is a temperature in degrees Fahrenheit. C is the same temperature in degrees Celsius. Multiply the Celsius number by 9 fifths, then add 32. To go the other way, undo each step: subtract 32, then multiply by 5 ninths.

  1. Sort out the question

    Fahrenheit and Celsius are two ways to measure temperature. The formula changes Celsius, C, into Fahrenheit, F. Yesterday’s high was 86° F and the low was 68° F. You need the difference between them in degrees Celsius. So change each one to Celsius, then subtract.

  2. Undo the + 32 for the high

    Put 86 in for F. Then subtract 32 from both sides. That keeps the equation balanced.

    F=95C+32F = \frac{9}{5}C + 32

    86=95C+3286 = \frac{9}{5}C + 32

    86−32=95C86 - 32 = \frac{9}{5}C

    54=95C54 = \frac{9}{5}C

  3. Solve for the high in Celsius

    To undo times 95\frac{9}{5}, multiply both sides by 59\frac{5}{9}. That is its reciprocal, the fraction flipped upside down.

    C=54×59=6×5=30C = 54 \times \frac{5}{9} = 6 \times 5 = 30

    Dividing 54 by 9 first gives 6, which keeps the numbers small. The high was 30° C.

  4. Change the low to Celsius

    Do the same with 68.

    68=95C+3268 = \frac{9}{5}C + 32

    68−32=95C68 - 32 = \frac{9}{5}C

    36=95C36 = \frac{9}{5}C

    C=36×59=4×5=20C = 36 \times \frac{5}{9} = 4 \times 5 = 20

    The low was 20° C. Watch out: 20° C is the low, not the difference. That is choice C.

  5. Subtract to find the difference

    30−20=1030 - 20 = 10

    The difference is 10 degrees Celsius. Watch out: do not multiply the 18-degree Fahrenheit gap by 95\frac{9}{5}. That goes the wrong way and gives 32.4, choice D. So the answer is A, 10.0° C.

Check your answer

A shortcut: the gap in Fahrenheit is 86−68=1886 - 68 = 18 degrees. The + 32 moves both temperatures the same, so it does not change the gap. Multiply the gap by 59\frac{5}{9}: 18×59=1018 \times \frac{5}{9} = 10. Same answer.

More questions like this

The original DOE question
Question 79 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 81, question 79. Open page 81 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 34 of 93

2024 Test A Math, question 23

Question

Let x be an odd number. In terms of x, what is the sum of the two even numbers closest to x?

  • Axx
  • B2x2x
  • C2x−22x - 2
  • D2x−42x - 4
Show the answer

Correct answer: B

If x is odd, then the two even numbers closest to x are (x−1)(x - 1) and (x+1)(x + 1). Find the sum of these two expressions:

(x−1)+(x+1)=2x(x - 1) + (x + 1) = 2x

Question analysis

An algebra question about odd and even numbers: x is odd, and students write the sum of the two even numbers closest to x in terms of x. It tests representing neighboring integers with expressions and adding them. Deciding exactly which even numbers are closest to an odd number is where errors start.

NYS standards
  • 6.EE.2a writing algebraic expressions
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write and combine linear expressions; here the expressions describe neighboring integers, a familiar idea from consecutive-integer problems. Two short steps and a quick number check put it at typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 3, 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

The algebra is a single addition, so the work is in naming the right two numbers. An odd x sits between two even numbers, one step away on each side, and the minus one and plus one cancel. Wrong choices come from picking both neighbors below x, from treating x as even, or from giving the middle of the pair; a test value catches each of them.

  1. Variables instead of numbers moderate Understanding the question

    Here x stands for any odd number, so the two even neighbors must be written as x − 1 and x + 1; trying x = 7, with neighbors 6 and 8, makes it concrete. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase "the two even numbers closest to x" means one just below and one just above x, not two in a row on the same side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Substitute a real odd number such as 7 for x, find its two closest even numbers and their sum, then see which expression gives that sum.

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

Neighbors of a number
x−1 and x+1x - 1 \text{ and } x + 1
x is any whole number. The number just below it is x − 1, and the number just above it is x + 1. Whole numbers take turns being odd and even. So if x is odd, both of its neighbors are even.
Combining like terms
x+x=2xx + x = 2x
Like terms have the same letter, such as x and 2x. To add them, add the numbers in front. x by itself means 1x, so x + x is 2x, which is two x’s.

  1. Sort out the question

    x is an odd number. You need the sum of the two even numbers closest to x. Sum means the result of adding. In terms of x means write the answer using the letter x.

  2. Try a real number first

    Pick an odd number, like 7. The even numbers closest to 7 are 6 and 8, one on each side.

    6+8=146 + 8 = 14

    Notice that 14 is 2×72 \times 7.

  3. Write the neighbors with x

    Odd and even numbers take turns. So both numbers right next to an odd x are even.

    x−1 and x+1x - 1 \text{ and } x + 1

    For x = 7, these are 6 and 8. Watch out: take one even number below x and one above. Using two even numbers below x, x−1x - 1 and x−3x - 3, gives 2x−42x - 4, choice D.

  4. Add the two neighbors

    (x−1)+(x+1)=x+x−1+1=2x(x - 1) + (x + 1) = x + x - 1 + 1 = 2x

    The −1 and the +1 cancel, and x+x=2xx + x = 2x. Watch out: x is the number in the middle, not the sum, so choice A is wrong. So the answer is B, 2x2x.

Check your answer

Try another odd number, 11. Its even neighbors are 10 and 12, and 10+12=2210 + 12 = 22. The answer 2x gives 2×11=222 \times 11 = 22. They match.

More questions like this

The original DOE question
Question 80 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 80. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 35 of 93

2024 Test A Math, question 25

Question

Points M, N, P and Q, each marked with a dot, lie in that order from left to right on a line. A dimension arrow from M to N is labeled x cm, and a longer dimension arrow from M to P is labeled y cm. No lengths are marked for NP, PQ or MQ.

In the figure above, N is the midpoint of MQ\overline{MQ}. Which segment has length (2x−y)(2x - y) centimeters?

  • APQ\overline{PQ}
  • BNP\overline{NP}
  • CMQ\overline{MQ}
  • DMP\overline{MP}
Show the answer

Correct answer: A

These facts are given:

  • N is the midpoint of MQ\overline{MQ}.
  • MN\overline{MN} has a length of x centimeters.
  • MP\overline{MP} has a length of y centimeters.

Use the first two facts to determine that MQ\overline{MQ} has a length of 2x2x centimeters. So (2x−y)(2x - y) can be represented by MQ−MPMQ - MP, which is PQ\overline{PQ}.

Question analysis

A geometry and algebra question with a labeled figure: points M, N, P and Q lie on a line, MN is x cm, MP is y cm, and N is the midpoint of MQ. Students find which segment has length 2x − y. They must turn the midpoint fact into an expression and read a difference of lengths on the figure.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students evaluate and rewrite expressions but rarely read one as a length on a figure. Combining the midpoint idea from geometry with a difference of expressions, and no numbers, makes this more abstract than typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 9, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question reverses the usual task: instead of finding a length, students match an expression to a segment. The midpoint fact turns into MQ = 2x, and taking away the y of MP leaves the piece from P to Q. The wrong choices reward stopping at 2x, reading only the labeled arrows, or matching the letter y; picking x = 5 and y = 7 lets students check with numbers.

  1. Variables instead of numbers moderate Understanding the question

    Every length is a letter: MN is x, MP is y, and the target is the expression 2x − y, so the matching segment cannot be found by measuring. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The figure marks only MN and MP with arrows; the length of MQ comes from the stem's midpoint fact, not from the drawing, and has to be written as 2x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the midpoint to write the length of the whole segment in terms of x, or pick sample values for x and y, then see which segment's length matches 2x − y.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint
MQ=2×MNMQ = 2 \times MN
N is the midpoint of segment MQ, the point exactly in the middle. MN is the half from M to N, and MQ is the whole segment. The whole is two equal halves, so it is 2 times MN.
Adding segments
MP+PQ=MQMP + PQ = MQ
When points lie in order on a line, the short segments add up to the long one. MP runs from M to P, and PQ runs from P to Q. Put end to end, they make MQ.

  1. Sort out the figure

    Points M, N, P and Q lie in that order on a line. Segment MN is x cm long, and segment MP is y cm long. N is the midpoint of MQ\overline{MQ}, the point exactly halfway from M to Q. You need the segment that is 2x−y2x - y cm long.

  2. Find the length of MQ

    MN is half of MQ, so MQ is twice as long.

    MQ=2×MN=2xMQ = 2 \times MN = 2x

    MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.

  3. Take MP away from MQ

    Subtracting y from 2x means taking MP away from MQ. MP covers the part from M to P. What is left of MQ is the piece from P to Q.

    MP+PQ=MQMP + PQ = MQ

    PQ=MQ−MP=2x−yPQ = MQ - MP = 2x - y

    So PQ is 2x−y2x - y cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is y−xy - x, choice B. So the answer is A, PQ\overline{PQ}.

Check your answer

Try numbers. Say x = 5, so MN = 5 and MQ = 10. Say y = 7, so MP = 7. Then PQ is 10−7=310 - 7 = 3. And 2x−y=10−7=32x - y = 10 - 7 = 3. They match. NP would be 7−5=27 - 5 = 2, which does not.

More questions like this

The original DOE question
Question 82 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 82. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 93

2024 Test A Math, question 27

Question

On Wednesday, a baker produced 100 more loaves of bread than were produced on Tuesday. On Thursday, the baker produced 50 fewer loaves than were produced on Tuesday. If the total number of loaves produced on all three days was 230, how many loaves were produced on Wednesday?

  • A60
  • B80
  • C120
  • D160
Show the answer

Correct answer: D

Let t represent the number of loaves of bread produced on Tuesday.

First, determine the number of loaves of bread produced on Tuesday, t, using the number of loaves produced on Wednesday, t+100t + 100, the number produced on Thursday, t−50t - 50, and the total number produced on these three days, 230:

t+t+100+t−50=230t + t + 100 + t - 50 = 230

t+t+t+100−50=230t + t + t + 100 - 50 = 230

3t+50=2303t + 50 = 230

3t+(50−50)=230−503t + (50 - 50) = 230 - 50

3t=1803t = 180

(3÷3)t=180÷3(3 \div 3)t = 180 \div 3

t=60t = 60

Next, determine the number of loaves of bread produced on Wednesday, t+100t + 100, when t=60t = 60:

t+100=60+100=160t + 100 = 60 + 100 = 160

The number of loaves of bread produced on Wednesday is 160.

Question analysis

A word problem about a baker's bread over three days: Wednesday's output is 100 loaves more than Tuesday's, Thursday's is 50 fewer than Tuesday's, and the total is 230. Students find the number of loaves made on Wednesday. Writing every day in terms of Tuesday, and answering for the right day, are the keys.

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 Writing and solving an equation of the form px + q = r from a word problem is a core grade 7 standard. This is a typical three-step version, with one negative adjustment and an extra step after solving.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 45 words.
Corrected from the DOE handbook
  • A misplaced parenthesis in the explanation’s working 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

Writing all three days in terms of Tuesday turns the story into 3t + 50 = 230, a routine equation. The trap comes after it is solved: the value found is Tuesday's, and it is a listed choice. Students who label t as Tuesday from the start, and reread the question before choosing, add the 100 loaves and are done.

  1. Turning words into math moderate Understanding the question

    Both comparisons are made with Tuesday, 100 more loaves on Wednesday and 50 fewer on Thursday, so the three days become t, t + 100 and t − 50. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    The equation's unknown is Tuesday's count, while the question asks for Wednesday's, so one more step remains after the equation is solved. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let t be Tuesday's loaves, write Wednesday and Thursday in terms of t, set the sum equal to 230, solve for t, and add 100.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes “3t + (50 − 50 = 230 − 50)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 50 − 50 = 230 − 50 is false). The site now reads 3t + (50 − 50) = 230 − 50: subtract 50 from both sides, giving 3t = 180 as before.

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\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. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you divide both sides by the same number that is not zero.

  1. Sort out the question

    The baker made bread on Tuesday, Wednesday and Thursday. Wednesday had 100 more loaves than Tuesday. Thursday had 50 fewer loaves than Tuesday. All three days add up to 230 loaves. You need the number of loaves made on Wednesday.

  2. Write each day using Tuesday

    Both other days are compared with Tuesday. So call Tuesday’s loaves t.

    DayLoaves
    Tuesdaytt
    Wednesdayt+100t + 100
    Thursdayt−50t - 50

    100 more means add 100. 50 fewer means subtract 50.

  3. Write the equation and combine

    t+(t+100)+(t−50)=230t + (t + 100) + (t - 50) = 230

    Group the t’s together and the plain numbers together. Three t’s make 3t, and 100−50=50100 - 50 = 50.

    3t+50=2303t + 50 = 230

    Watch out: do not just subtract 100 and 50 from 230. That gives 80, choice B, and skips writing each day with t.

  4. Solve for Tuesday

    Subtract 50 from both sides. Then divide both sides by 3.

    if a=b, then a−c=b−c\text{if } a = b \text{, then } a - c = b - c

    3t+50−50=230−503t + 50 - 50 = 230 - 50

    3t=1803t = 180

    t=180÷3=60t = 180 \div 3 = 60

    Tuesday had 60 loaves. Watch out: 60 is Tuesday, choice A. The question asks about Wednesday.

  5. Find Wednesday’s loaves

    Wednesday had 100 more loaves than Tuesday, so add 100 to t.

    t+100=60+100=160t + 100 = 60 + 100 = 160

    Wednesday had 160 loaves. So the answer is D, 160.

Check your answer

Find all three days. Tuesday is 60, Wednesday is 160, and Thursday is 60−50=1060 - 50 = 10. Add them: 60+160+10=23060 + 160 + 10 = 230. That is the total in the question, so 160 is right.

More questions like this

The original DOE question
Question 84 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 84. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 37 of 93

2024 Test A Math, question 29

Question

Which graph represents the solution to the inequality x+4≥3x + 4 \geq 3?

  • ANumber line from −4 to 7 with a tick at every integer. A closed dot at −1, with the line shaded from −1 to the right, ending in an arrow.
  • BNumber line from −4 to 7 with a tick at every integer. A closed dot at 1, with the line shaded from 1 to the right, ending in an arrow.
  • CNumber line from −4 to 7 with a tick at every integer. A closed dot at 3, with the line shaded from 3 to the right, ending in an arrow.
  • DNumber line from −4 to 7 with a tick at every integer. A closed dot at 4, with the line shaded from 4 to the right, ending in an arrow.
Show the answer

Correct answer: A

Use the properties of equations to solve the inequality for x.

x+4≥3x + 4 \geq 3

Subtract 4 from both sides of the inequality.

x+4−4≥3−4x + 4 - 4 \geq 3 - 4

x≥−1x \geq -1

The graph of x≥−1x \geq -1 will have a point at −1 and will include all values that are greater than, or to the right of, −1. The greater than or equal sign is used, which indicates that the solution includes −1. So the point at −1 will be solid. Option A is the correct answer.

Question analysis

This inequality question asks which number line graph shows the solution set of x + 4 ≥ 3. Students solve a one-step inequality by undoing the addition, then match the result to a graph. All four graphs use a solid dot and an arrow pointing right, so the whole question rests on where the shaded ray begins.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 inequality work is usually two-step, such as 3x − 5 ≥ 10, and asks students to choose an open or closed dot and a direction. This one needs a single subtraction, and all four graphs share a closed dot and a rightward arrow.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 13, 9 words.
Students, last 30 days
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 a single subtraction, and the graphs differ only in where the solid dot sits, so the question comes down to 3 − 4. Students who lose the negative sign land on the graph that starts at 1, while those who never isolate x pick a graph that starts at a number printed in the inequality. Testing 0, which satisfies 0 + 4 ≥ 3, shows that the shading must include 0.

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

    All four graphs share a closed dot and a rightward arrow, so the only thing to read is where the dot sits, against tick labels printed with raised minus signs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic minor Working it out

    Undoing the + 4 means computing 3 − 4, the one step that crosses zero; a dropped minus sign moves the dot from −1 to 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 4 from both sides to get x alone, then find the graph whose solid dot sits at that number; test one shaded value in the original inequality to confirm.

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.)
  • 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

Keeping an inequality balanced
if a≥b, then a−c≥b−c\text{if } a \geq b \text{, then } a - c \geq b - c
a and b are the two sides of an inequality, and c is any number. The sign ≥ stands for greater than or equal to. Taking the same number away from both sides keeps the comparison true, like taking the same weight off both pans of a scale. Use it to get x by itself.

  1. Sort out the question

    An inequality is a math sentence that compares two amounts. The sign ≥\geq means is greater than or equal to. So x+4≥3x + 4 \geq 3 says that x plus 4 is 3 or more.

    You need the graph that shows every number x that makes this true. All four graphs have a solid dot and an arrow to the right. Only the place where the dot sits is different.

  2. Get x by itself

    Right now 4 is added to x. To undo adding 4, subtract 4 from both sides.

    if a≥b, then a−c≥b−c\text{if } a \geq b \text{, then } a - c \geq b - c

    x+4−4≥3−4x + 4 - 4 \geq 3 - 4

    x≥−1x \geq -1

    So x is −1 or any number greater than −1. Watch out: 3 − 4 is −1, not 1. Dropping the minus sign puts the dot at 1, choice B. Putting the dot at 3 or 4 means you read a number off the inequality without solving. That leads to C or D.

  3. Match the solution to a graph

    On a number line, greater numbers are to the right. So the shading starts at −1 and goes right. The arrow shows that it keeps going.

    The dot is solid, or filled in, because of the or equal to part. A solid dot means −1 itself is a solution, a number that makes the inequality true.

    Graph A has a solid dot at −1 and is shaded to the right. So the answer is A, the graph with a solid dot at −1.

Check your answer

Test a number from the shaded part, such as 0. 0+4=40 + 4 = 4, and 4 is more than 3. Now test the dot: −1+4=3-1 + 4 = 3, and 3 is equal to 3, so −1 works. A number left of the dot, −2, gives −2+4=2-2 + 4 = 2. That is less than 3, so −2 should not be shaded, and it isn’t.

More questions like this

The original DOE question
Question 86 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 86. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 38 of 93

2024 Test A Math, question 33

Question

If (35−12)x=14+23\left(\frac{3}{5} - \frac{1}{2}\right)x = \frac{1}{4} + \frac{2}{3}, what is the value of x?

  • A11120\frac{11}{120}
  • B27\frac{2}{7}
  • C56\frac{5}{6}
  • D556\frac{55}{6}
Show the answer

Correct answer: D

(35−12)x=14+23\left(\frac{3}{5} - \frac{1}{2}\right)x = \frac{1}{4} + \frac{2}{3}

Rewrite with equivalent fractions that have like denominators.

(3×25×2−1×52×5)x=1×34×3+2×43×4\left(\frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5}\right)x = \frac{1 \times 3}{4 \times 3} + \frac{2 \times 4}{3 \times 4}

(610−510)x=312+812\left(\frac{6}{10} - \frac{5}{10}\right)x = \frac{3}{12} + \frac{8}{12}

(110)x=1112\left(\frac{1}{10}\right)x = \frac{11}{12}

Multiply both sides of the equation by 10.

10(110)x=10(1112)10\left(\frac{1}{10}\right)x = 10\left(\frac{11}{12}\right)

x=11012=556x = \frac{110}{12} = \frac{55}{6}

Question analysis

This equation has fractions on both sides: the coefficient of x is 3/5 minus 1/2, and the right side is 1/4 plus 2/3. Students simplify each side using common denominators, then solve for x. Each step is routine, but four fraction operations in a row leave many places to slip, especially when undoing the coefficient.

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 Grade 7 students solve equations such as (2/3)x = 4/5 with fractional coefficients. Here both sides must first be simplified over unlike denominators, so four fraction operations come before x is found: on level, but long.
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 8, 12 words.
Corrected from the DOE handbook
  • A minus sign printed in place of an equals sign 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

Nothing here is new, but the question strings together four fraction operations, and the last one runs against habit. Having found that one tenth of x is 11/12, students must multiply by 10; multiplying by 1/10 instead gives a tiny listed value. The size of the answer is a useful check: if a tenth of x is almost 1, then x must be almost 10.

  1. Error-prone arithmetic moderate Working it out

    Both sides need common denominators, tenths on the left and twelfths on the right, and the coefficient comes out as 1/10, a fraction to divide by. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps moderate Working it out

    Four operations come before x appears: 3/5 − 1/2, then 1/4 + 2/3, then 11/12 divided by 1/10, then reducing 110/12 to 55/6. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Simplify the parentheses on the left and the sum on the right over common denominators, then undo the coefficient by multiplying both sides by its reciprocal.

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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes “(1/10)x − 11/12”, an expression, where the working needs the equation (1/10)x = 11/12: the line before is (6/10 − 5/10)x = 3/12 + 8/12, and the next step multiplies both sides of the equation by 10. The site now reads (1/10)x = 11/12, which gives x = 110/12 = 55/6.

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 is the top of a fraction and b is the bottom. n is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, cut into smaller pieces. Use it to give two fractions the same bottom before you add or subtract.
Keeping an equation balanced
if a=b, then n×a=n×b\text{if } a = b \text{, then } n \times a = n \times b
a and b are the two sides of an equation, and n is any number. An equation is like a balanced scale. If you multiply both sides by the same number, it stays balanced. Use it to undo a number that multiplies x.

  1. Sort out the equation

    An equation says two things are equal. On the left, the subtraction in the parentheses multiplies x. On the right is a sum. You need the value of x. First make each side as simple as you can. Then get x by itself.

  2. Simplify the left side

    To subtract, the fractions need the same bottom. Both 5 and 2 go into 10, so use tenths.

    ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}

    35−12=3×25×2−1×52×5=610−510=110\frac{3}{5} - \frac{1}{2} = \frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5} = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

    Watch out: this is a minus sign. Adding instead gives 1110\frac{11}{10}, and that leads to 56\frac{5}{6}, choice C.

  3. Simplify the right side

    Both 4 and 3 go into 12, so use twelfths.

    14+23=1×34×3+2×43×4=312+812=1112\frac{1}{4} + \frac{2}{3} = \frac{1 \times 3}{4 \times 3} + \frac{2 \times 4}{3 \times 4} = \frac{3}{12} + \frac{8}{12} = \frac{11}{12}

    Now the equation is much simpler:

    110x=1112\frac{1}{10}x = \frac{11}{12}

  4. Get x by itself

    x is multiplied by 110\frac{1}{10}. To undo that, multiply both sides by 10. That is the reciprocal of 110\frac{1}{10}, the fraction flipped upside down.

    if a=b, then n×a=n×b\text{if } a = b \text{, then } n \times a = n \times b

    10×110x=10×111210 \times \frac{1}{10}x = 10 \times \frac{11}{12}

    x=11012x = \frac{110}{12}

    Watch out: multiplying 1112\frac{11}{12} by 110\frac{1}{10} does not undo anything. It gives 11120\frac{11}{120}, choice A.

  5. Simplify the fraction

    Both 110 and 12 are even, so divide the top and the bottom by 2.

    x=110÷212÷2=556x = \frac{110 \div 2}{12 \div 2} = \frac{55}{6}

    So the answer is D, 556\frac{55}{6}.

Check your answer

Put 556\frac{55}{6} back in for x. 110×556=5560=1112\frac{1}{10} \times \frac{55}{6} = \frac{55}{60} = \frac{11}{12}. That matches the right side. Also estimate. A tenth of x is almost 1, so x should be almost 10. 556\frac{55}{6} is about 9.2.

More questions like this

The original DOE question
Question 90 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 90. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 39 of 93

2024 Test A Math, question 36

Question

CELL PHONE SALES BY COLOR

Color Percentage of Cell Phones Sold
White 8k8k
Black 30
Blue 30−2k30 - 2k
Red k+5k + 5
Total 100

The table above shows cell phone sales by color. What percentage of the cell phones sold were blue?

  • A18%
  • B20%
  • C22%
  • D28%
Show the answer

Correct answer: B

First, use the information given in the table and properties of equations to determine the value of k:

8k+30+30−2k+k+5=1008k + 30 + 30 - 2k + k + 5 = 100

8k−2k+k+30+30+5=1008k - 2k + k + 30 + 30 + 5 = 100

7k+65=1007k + 65 = 100

7k+(65−65)=100−657k + (65 - 65) = 100 - 65

7k=357k = 35

k=5k = 5

Then, determine the percentage of cell phones sold that were blue by finding the value of the expression 30−2k30 - 2k (given in the table), when k=5k = 5:

30−2k=30−2(5)=30 - 2k = 30 - 2(5) =

30−10=2030 - 10 = 20

The percentage of cell phones sold that were blue is 20%.

Question analysis

A table of cell phone sales by color gives each color's percentage as an expression in k, such as 8k for white and 30 − 2k for blue, with a total of 100. Students write and solve an equation for k, then find the percentage for blue. Combining terms with a negative coefficient, and remembering that k itself is not the answer, both need care.

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 Grade 7 students solve two-step equations that are written out for them. Here the equation must be built from a table whose percentages total 100, with like terms to combine, and the value found must then be substituted back.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 4, 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 hides an equation in a table, and the key insight is that the four percentages must add to 100. After that the algebra is routine, and no listed choice rewards stopping early, since neither k nor any other color's percentage appears. The wrong choices sit near the answer or use the blue expression with k ignored, so students who cannot set up the equation are left guessing.

  1. Many steps moderate Working it out

    The answer needs a chain: add the four expressions, set the sum to 100, collect 8k − 2k + k into 7k, solve for k, then substitute into 30 − 2k. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    No equation is printed; the table's Total row of 100 is what turns the four percentages written in k into one equation to solve. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Error-prone arithmetic minor Working it out

    Collecting like terms includes the negative coefficient in the blue row, −2k, and the constants 30, 30 and 5 have to be gathered too. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four color expressions and set the sum equal to 100, combine like terms to solve for k, then substitute k into the blue row's expression.

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

Combining like terms
ak+bk=(a+b)kak + bk = (a + b)k
k is the unknown number. a and b are the numbers in front of k, called coefficients. Terms with the same letter are called like terms. You can add their coefficients, because a groups of k plus b groups of k make a + b groups of k.
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 \div c = b \div 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. Doing the same thing to both sides keeps it balanced. Use it to undo each step and get k alone.

  1. Sort out the table

    The table gives the percent of phones sold in each color. Some entries use k, a letter that stands for an unknown number. 8k means 8 times k. You need the percent that were blue, 30−2k30 - 2k. To find it, you first need k.

  2. Write an equation for the total

    Every phone sold is one of the four colors. So the four percents add up to the total, 100, as the last row of the table shows.

    8k+30+(30−2k)+(k+5)=1008k + 30 + (30 - 2k) + (k + 5) = 100

  3. Combine like terms

    Terms with k in them are called like terms, so they can be combined. Group the k terms together and the plain numbers together. A k on its own means 1k.

    ak+bk=(a+b)kak + bk = (a + b)k

    8k−2k+k=(8−2+1)k=7k8k - 2k + k = (8 - 2 + 1)k = 7k

    30+30+5=6530 + 30 + 5 = 65

    So the equation becomes 7k+65=1007k + 65 = 100. Watch out: the blue row takes away 2k. Keep its minus sign, or the k terms will be wrong.

  4. Solve for k

    Undo the adding first, then the multiplying.

    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 \div c = b \div c

    7k+65−65=100−657k + 65 - 65 = 100 - 65

    7k=357k = 35

    k=35÷7=5k = 35 \div 7 = 5

    So k is 5. But k is not the answer. You still need the blue row.

  5. Find the blue percent

    Put 5 in for k in the blue row.

    30−2k=30−2×5=30−10=2030 - 2k = 30 - 2 \times 5 = 30 - 10 = 20

    So 20% of the phones sold were blue. Watch out: 2k means 2 times k, not 2. Reading 30−2k30 - 2k as 30−230 - 2 gives 28%, choice D.

    So the answer is B, 20%.

Check your answer

Put k = 5 into every row. White: 8×5=408 \times 5 = 40. Black: 30. Blue: 20. Red: 5+5=105 + 5 = 10. Now add them: 40+30+20+10=10040 + 30 + 20 + 10 = 100. The total matches the table, so k = 5 is right.

More questions like this

The original DOE question
Question 93 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 93. Open page 85 of the handbook (PDF)

Question 40 of 93

2024 Test B Math, question 1

Question

35(2x+5)−2x\frac{3}{5}(2x+5)-2x

After the expression above is simplified, what is the coefficient of x, expressed as a decimal?

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

Show the answer

Correct answer: -0.8

First, distribute 35\frac{3}{5} over (2x+5)(2x+5) to get the expression 65x+3−2x\frac{6}{5}x+3-2x.

Express −2x-2x as −105x-\frac{10}{5}x and then add like terms:

65x+3+(−105x)=−45x+3\frac{6}{5}x+3+\left(-\frac{10}{5}x\right)=-\frac{4}{5}x+3.

The coefficient of x is −45-\frac{4}{5}.

When expressed as a decimal,

−45=−0.8-\frac{4}{5}=-0.8.

Question analysis

An algebra grid-in on simplifying an expression: three fifths times the quantity 2x + 5, minus 2x. Students distribute a fraction, combine the x terms and report the coefficient of x as a decimal. Multiplying by a fraction, the subtraction that follows, and gridding the decimal correctly all leave room for slips.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.2b parts of an expression
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Expanding and simplifying linear expressions with fraction coefficients is a core grade 7 skill, and this is a typical short example. Writing the coefficient as a decimal in the grid adds care but no new idea.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 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 distribution itself is routine; points are lost after it. Students who stop at 6/5x forget the −2x outside the parentheses, and those who combine the x terms correctly can still drop the sign of a negative result. Rewriting 2x as 10/5x keeps the subtraction in fifths, and substituting x = 1 into both forms confirms the coefficient before gridding.

  1. Error-prone arithmetic moderate Working it out

    Multiplying 2x by 3/5 gives 6/5x, and taking away 2x means rewriting it as 10/5x; the result is a negative fraction that must then become a decimal with its sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    There are no choices to check against, and the coefficient must be gridded as a decimal with its minus sign, so −4/5 has to become −0.8. (Enright, Morley & Sheehan, 2002)

Strategy: Multiply both terms inside the parentheses by 3/5, rewrite 2x in fifths so the x terms share a denominator, combine them, and convert that fraction to a decimal.

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.)
  • 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

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Multiplying a fraction by a whole number
ab×c=a×cb\frac{a}{b} \times c = \frac{a \times c}{b}
a over b is the fraction, and c is the whole number. Multiply only the top number, the numerator, by c. The bottom number, the denominator, stays the same. The size of each piece does not change. You just have more pieces.

  1. Sort out the question

    You start with this expression. An expression is a math phrase with numbers, letters and signs, but no equals sign.

    35(2x+5)−2x\frac{3}{5}(2x + 5) - 2x

    To simplify it means to write it in its shortest form. The coefficient of x is the number multiplied by x. In 7x, the coefficient is 7. You need the coefficient of x, written as a decimal.

  2. Multiply out the parentheses

    The 3/5 multiplies everything inside the parentheses. Use the distributive property.

    a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c

    35(2x+5)=35×2x+35×5\frac{3}{5}(2x + 5) = \frac{3}{5} \times 2x + \frac{3}{5} \times 5

    Multiply only the top of 3/5 by 2: 35×2=65\frac{3}{5} \times 2 = \frac{6}{5}. Also, 35×5=3\frac{3}{5} \times 5 = 3. Now the whole expression is:

    65x+3−2x\frac{6}{5}x + 3 - 2x

    Watch out: do not double the bottom number too. Writing 6/10 leads to −1.4.

  3. Write 2x in fifths

    There are now two x terms: 65x\frac{6}{5}x and −2x-2x. To combine them, they need the same denominator (bottom number). One whole is 5/5, so 2 wholes are 10/5.

    2x=105x2x = \frac{10}{5}x

    Watch out: the −2x is outside the parentheses, but it still counts. Stopping at 6/5 and gridding 1.2 is a common mistake.

  4. Combine the x terms

    Six fifths take away ten fifths leaves negative four fifths.

    65x−105x=−45x\frac{6}{5}x - \frac{10}{5}x = -\frac{4}{5}x

    The 3 has no x, so it stays as it is. The simplified expression is −45x+3-\frac{4}{5}x + 3. So the coefficient of x is −45-\frac{4}{5}. Watch out: 6 − 10 is negative. Dropping the minus sign gives 0.8.

  5. Change it to a decimal

    A fraction means the top divided by the bottom: 4÷5=0.84 \div 5 = 0.8. Keep the minus sign.

    −45=−0.8-\frac{4}{5} = -0.8

    So you would grid -0.8, with the minus sign in its own column.

Check your answer

Pick an easy value, x = 1, and put it into both forms. The first form: 35×7−2=4.2−2=2.2\frac{3}{5} \times 7 - 2 = 4.2 - 2 = 2.2. The simplified form: −0.8+3=2.2-0.8 + 3 = 2.2. Both give 2.2, so the simplifying worked.

More questions like this

The original DOE question
Question 58 as printed on page 198 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 198, question 58. Open page 198 of the handbook (PDF)

Question 41 of 93

2024 Test B Math, question 8

Question

If 92x=3y8\frac{9}{2x}=\frac{3y}{8}, where x≠0x\neq 0 and y≠0y\neq 0, what is the product of x and y?

  • A4
  • B6
  • C12
  • D16
Show the answer

Correct answer: C

92x=3y8\frac{9}{2x}=\frac{3y}{8}

9∙8=729\bullet 8=72

2x∙3y=6xy2x\bullet 3y=6xy

72=6xy72=6xy

12=xy12=xy

Question analysis

An algebra question gives a proportion with x in one denominator and y in one numerator, 9 over 2x equal to 3y over 8, and asks for the product of x and y. Neither variable can be found on its own, so students must clear the fractions and treat the product as a single quantity.

NYS standards
  • 8.EE.7 linear equations in one variable
  • 6.EE.2b parts of an expression
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students cross-multiply numeric proportions to find one unknown. Here two variables sit in the proportion and the question wants their product, a literal-equation idea from grade 8 that many grade 7 students have not met.
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, 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 looks harder than it is, because it asks for a product rather than for x or y. Cross-multiplying turns the proportion into 72 = 6xy, and dividing by 6 finishes it, so the only real hurdle is trusting that xy can be solved for as one quantity. The wrong choices come from grabbing numbers along the way, not from a nearly complete solution.

  1. No obvious method moderate Working it out

    Neither x nor y can be found alone, so students must see that cross-multiplying gives 72 = 6xy, where the product xy appears as one block to solve for. (Schoenfeld, 1985)

  2. Variables instead of numbers minor Understanding the question

    Both unknowns stay letters and only their product is asked for, so there is no value of x or y to check a result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Cross-multiply to clear both fractions, collect the numbers on one side and x and y together on the other, then divide so that only the product is left.

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.
  • 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

Cross-multiplying
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a, b, c and d are numbers or letters, and b and d are not zero. When two fractions are equal, multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.

  1. Sort out the question

    The equation says that two fractions are equal:

    92x=3y8\frac{9}{2x} = \frac{3y}{8}

    Here 2x means 2 times x, and 3y means 3 times y. You need the product of x and y. A product is the answer to a multiplication. So you need x×yx \times y, written xy. You do not need x or y alone.

  2. Cross-multiply the fractions

    An equation without fractions is easier to work with. So cross-multiply.

    ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c

    9×8=2x×3y9 \times 8 = 2x \times 3y

    Watch out: do not just set the bottoms equal. Writing 2x = 8 gives x = 4, choice A. That is x, not the product.

  3. Multiply out both sides

    On the left, 9×8=729 \times 8 = 72. On the right, multiply the numbers and keep the letters together.

    2x×3y=2×3×x×y=6xy2x \times 3y = 2 \times 3 \times x \times y = 6xy

    So the equation is now:

    72=6xy72 = 6xy

    The xy is the product you want, all in one piece.

  4. Divide both sides by 6

    xy is being multiplied by 6. To undo that, divide both sides by 6.

    xy=72÷6=12xy = 72 \div 6 = 12

    Watch out: 6 is just the number in front of xy. Choosing it gives B.

    So the answer is C, 12.

Check your answer

Pick x = 4. Then y must be 3, so that xy = 12. Left side: 92×4=98\frac{9}{2 \times 4} = \frac{9}{8}. Right side: 3×38=98\frac{3 \times 3}{8} = \frac{9}{8}. The two sides match, so a product of 12 works.

More questions like this

The original DOE question
Question 65 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 65. Open page 200 of the handbook (PDF)

Question 42 of 93

2024 Test B Math, question 13

Question

p+2r=r(p+1)+1p+2r=r(p+1)+1

In the equation above, if r=2r=2, what is the value of p?

  • A0
  • B1
  • C2
  • D3
Show the answer

Correct answer: B

First substitute 2 for r in the equation.

p+2r=r(p+1)+1p+2r=r(p+1)+1

p+2(2)=2(p+1)+1p+2(2)=2(p+1)+1

Then solve for p.

p+2(2)=2(p+1)+1p+2(2)=2(p+1)+1

p+4=2p+2+1p+4=2p+2+1

p+4=2p+3p+4=2p+3

p+(4−3)=2p+(3−3)p+(4-3)=2p+(3-3)

p+1=2pp+1=2p

(p−p)+1=2p−p(p-p)+1=2p-p

1=p1=p

The value of p is 1.

Question analysis

An equation question: p + 2r equals r times (p + 1), plus 1, and r is given as 2. Students substitute for r, expand the parentheses and solve the resulting linear equation for p. With p on both sides and a product to expand, distributing is where a slip is most likely.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 Equations with the variable on both sides are formally grade 8, but after substituting, this is a short solve with small whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 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

Substituting r = 2 is immediate; the points are lost in expanding 2(p + 1) + 1. Each wrong choice solves a slightly wrong expansion: forgetting to multiply the 1 inside, folding the outer 1 into the parentheses, or losing a term. Putting the answer back into the original equation, with both r and p, is a quick and complete check.

  1. Error-prone arithmetic moderate Working it out

    After substituting, 2(p + 1) + 1 must become 2p + 3, multiplying both terms in the parentheses before adding the 1 outside, with p left on both sides. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Replace every r with 2, multiply 2 by both terms in the parentheses, gather the p terms on one side, and check by putting the value back into the original equation.

Traps

  • 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

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Keeping an equation balanced
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. If you subtract the same thing from both sides, it stays balanced. Adding the same thing, or dividing both sides by the same number that is not zero, works too.

  1. Sort out the question

    The equation has two letters, p and r. You are told that r = 2. You need the value of p. Here 2r means 2 times r, and r(p + 1) means r times the group p + 1.

  2. Put 2 in for r

    Replace every r with 2. Keep the parentheses so you can see what gets multiplied.

    p+2(2)=2(p+1)+1p + 2(2) = 2(p + 1) + 1

    On the left, 2(2)=42(2) = 4, so the left side is p+4p + 4.

  3. Multiply out the parentheses

    The 2 multiplies both the p and the 1 inside the parentheses. Then add the 1 that is outside.

    a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c

    2(p+1)+1=2p+2+1=2p+32(p + 1) + 1 = 2p + 2 + 1 = 2p + 3

    Watch out: multiply the 1 inside by 2 as well. Writing 2p + 1 + 1 gives p = 2, choice C. Adding the outside 1 inside first gives p = 0, choice A.

  4. Get p by itself

    Now the equation is p+4=2p+3p + 4 = 2p + 3. Subtract p from both sides. Then subtract 3 from both sides.

    if a=b, then a−c=b−c\text{if } a = b \text{, then } a - c = b - c

    p+4−p=2p+3−pp + 4 - p = 2p + 3 - p

    4=p+34 = p + 3

    4−3=p+3−34 - 3 = p + 3 - 3

    1=p1 = p

    So p = 1, and the answer is B, 1.

Check your answer

Put p = 1 and r = 2 back into the first equation. Left side: 1+2×2=51 + 2 \times 2 = 5. Right side: 2×(1+1)+1=52 \times (1 + 1) + 1 = 5. Both sides are 5, so p = 1 works.

More questions like this

The original DOE question
Question 70 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 70. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 43 of 93

2024 Test B Math, question 25

Question

Which number line represents the values of x that satisfy the inequality −5≤1−3x≤4-5\leq 1-3x\leq 4?

  • ANumber line from −3 to 3 with a tick at every integer. Open circles at −1 and 2, with the segment between them shaded.
  • BNumber line from −3 to 3 with a tick at every integer. Closed circles at −1 and 2, with the segment between them shaded.
  • CNumber line from −3 to 3 with a tick at every integer. Open circles at −1 and 2, with the line shaded from −1 to the left and from 2 to the right.
  • DNumber line from −3 to 3 with a tick at every integer. A closed circle at −1 and an open circle at 2, with the line shaded from −1 to the left and from 2 to the right.
Show the answer

Correct answer: B

Split the compound inequality into two inequalities.

−5≤1−3x-5\leq 1-3x and 1−3x≤41-3x\leq 4

For each inequality, subtract 1 from each side.

−6≤−3x-6\leq -3x and −3x≤3-3x\leq 3

For each inequality, divide both sides by −3. Note that dividing each side of an inequality by a negative number changes the direction of the inequality.

2≥x2\geq x and x≥−1x\geq -1

Combining the pieces gives the compound inequality −1≤x≤2-1\leq x\leq 2.

The solution is that x is greater than or equal to −1 and less than or equal to 2, so choose the number line with filled circles at −1 and 2, with the filled segment between the circles.

Question analysis

An inequality question: choose the number line that shows the solutions of the compound inequality −5 ≤ 1 − 3x ≤ 4. Students solve an inequality whose x term has a negative coefficient, then match the endpoints, open or closed circles, and shading to one of four graphs.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph two-step inequalities one at a time. This compound inequality needs two solved together and a division by a negative number, which reverses the signs, a step many grade 7 classes only touch on.
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 12, 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

The question stacks two classic traps: reversing the inequality signs when dividing by −3, and matching filled or open circles to ≤. All four graphs use the endpoints −1 and 2, so finding those numbers proves little; the choice turns on the circles and the direction of the shading. Testing x = 0, x = 3 and the endpoint x = 2 in the original inequality checks both decisions in under a minute.

  1. Error-prone arithmetic moderate Working it out

    Dividing all three parts of −6 ≤ −3x ≤ 3 by −3 reverses both signs and the order, giving 2 ≥ x ≥ −1, which must be rewritten as −1 ≤ x ≤ 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    Two graphs share the endpoints −1 and 2 and the shading between them; only the circles, filled for ≤ and open for <, separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Each graph must be read for three things at once: the endpoints, the fill of each circle, and whether the shading runs between them or outward. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Subtract 1 from all three parts, divide all three by −3 while reversing both inequality signs, then rewrite the result from smallest to largest before matching circles and shading.

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).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting from every part
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. Subtracting the same number from every part keeps the inequality true. The sign does not change.
Dividing by a negative number
if a≤b, then a÷(−c)≥b÷(−c)\text{if } a \leq b \text{, then } a \div (-c) \geq b \div (-c)
a and b are numbers, and c is a positive number, so −c is negative. When you divide both sides by a negative number, the inequality sign flips. For example, 2 is less than 3, but −2 is greater than −3.

  1. Sort out the question

    The inequality −5≤1−3x≤4-5 \leq 1 - 3x \leq 4 says that 1−3x1 - 3x is between −5 and 4. The sign ≤ means less than or equal to. So −5 and 4 themselves are allowed. You need the number line that shows every x that works. To find them, get x alone in the middle.

  2. Subtract 1 from every part

    First take away the 1 in the middle. Subtract 1 from all three parts.

    if a≤b, then a−c≤b−c\text{if } a \leq b \text{, then } a - c \leq b - c

    −5−1≤1−3x−1≤4−1-5 - 1 \leq 1 - 3x - 1 \leq 4 - 1

    −6≤−3x≤3-6 \leq -3x \leq 3

  3. Divide by −3 and flip

    Now x is multiplied by −3. Divide all three parts by −3. Dividing by a negative number flips each sign.

    if a≤b, then a÷(−c)≥b÷(−c)\text{if } a \leq b \text{, then } a \div (-c) \geq b \div (-c)

    −6−3≥−3x−3≥3−3\frac{-6}{-3} \geq \frac{-3x}{-3} \geq \frac{3}{-3}

    2≥x≥−12 \geq x \geq -1

    Watch out: if you forget to flip, x seems to lie outside −1 and 2. That leads to choices C and D, which are shaded outward.

  4. Read the result

    Turn it around so the smaller number is on the left. It means the same thing.

    −1≤x≤2-1 \leq x \leq 2

    So x can be any number from −1 to 2. The ends, −1 and 2, are included, because the signs are ≤.

  5. Pick the number line

    On a number line, a closed (filled-in) circle means the end number is included. An open circle means it is not. So you need closed circles at −1 and 2, with the part between them shaded. Watch out: choice A has the right shading but open circles, which leave out −1 and 2.

    So the answer is B.

Check your answer

Test a few numbers in 1−3x1 - 3x. For x = 0 you get 1, which is between −5 and 4, and 0 is shaded. For x = 2 you get −5, which is allowed, so the circle at 2 is filled in. For x = 3 you get −8, which is too small, so 3 is not shaded.

More questions like this

The original DOE question
Question 82 as printed on page 206 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 82. Open page 206 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 93

2024 Test B Math, question 27

Question

Simplify:

8x−(7+2.5x)+28x-(7+2.5x)+2

  • A5.5x−95.5x-9
  • B5.5x−55.5x-5
  • C10.5x−910.5x-9
  • D10.5x−510.5x-5
Show the answer

Correct answer: B

8x−(7+2.5x)+28x-(7+2.5x)+2

First, distribute −1 over 7+2.5x7+2.5x:

8x−7−2.5x+28x-7-2.5x+2

Then combine the like terms:

5.5x−55.5x-5

The simplified form of the expression is 5.5x−55.5x-5.

Question analysis

An algebra question asks students to simplify 8x minus the quantity 7 + 2.5x, plus 2. They remove the parentheses and combine the x terms and the constants. Decimal coefficients and a subtraction in front of a group make sign errors likely.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simplifying a linear expression with a subtracted group and decimal coefficients is standard grade 7 work. This is a typical two-step textbook exercise, with the usual sign trap and nothing extra.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 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

The question checks one skill, removing parentheses that follow a minus sign, and its choices are built from the two ways to get it wrong. A student who applies the minus to too few terms gets 10.5x, and one who extends it past the parentheses gets −9. Handling the x terms and the constants as two separate small checks makes each slip easy to catch.

  1. Error-prone arithmetic moderate Working it out

    The minus sign before the parentheses must turn + 2.5x into − 2.5x and 7 into − 7, while the + 2 written after the parentheses keeps its sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The choices pair two x terms, 5.5x and 10.5x, with two constants, −9 and −5, so each wrong one differs from the right one in a single part. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Change the minus in front of the parentheses into a minus on each term inside, then combine like terms, keeping the + 2 outside the group as it is.

Traps

  • 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

Subtracting a group in parentheses
−(a+b)=−a−b-(a + b) = -a - b
a and b are the terms inside the parentheses. A minus sign in front means take away the whole group. So each term inside is taken away, and its sign flips. Terms outside the parentheses keep their own signs.
Combining like terms
ax+bx=(a+b)xax + bx = (a + b)x
a and b are numbers, and x is the letter. Like terms have the same letter part. You add or subtract their numbers and keep the letter. Plain numbers combine only with plain numbers.

  1. Sort out the problem

    You need to simplify 8x−(7+2.5x)+28x - (7 + 2.5x) + 2. To simplify means to write it in its shortest form. The tricky part is the minus sign in front of the parentheses. It takes away everything inside.

  2. Remove the parentheses

    −(a+b)=−a−b-(a + b) = -a - b

    8x−(7+2.5x)+2=8x−7−2.5x+28x - (7 + 2.5x) + 2 = 8x - 7 - 2.5x + 2

    Both 7 and 2.5x get a minus sign. The + 2 is outside the parentheses, so it stays + 2. Watch out: 2.5x must become −2.5x. Leaving it as + 2.5x gives 10.5x, as in choices C and D.

  3. Combine the x terms

    Like terms have the same letter part. Put the two x terms together.

    ax+bx=(a+b)xax + bx = (a + b)x

    8x−2.5x=(8−2.5)x=5.5x8x - 2.5x = (8 - 2.5)x = 5.5x

  4. Combine the plain numbers

    −7+2=−5-7 + 2 = -5

    Watch out: the + 2 is not inside the parentheses. Making it −2 gives −9, as in choice A.

    Put the parts together: 5.5x−55.5x - 5. So the answer is B, 5.5x−55.5x - 5.

Check your answer

Pick x = 2 and try both forms. The first form: 16−(7+5)+2=616 - (7 + 5) + 2 = 6. Choice B: 5.5×2−5=11−5=65.5 \times 2 - 5 = 11 - 5 = 6. They match. Choice D gives 10.5×2−5=1610.5 \times 2 - 5 = 16, so D is wrong.

More questions like this

The original DOE question
Question 84 as printed on page 206 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 84. Open page 206 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 45 of 93

2024 Test B Math, question 31

Question

If y=4xy=4x, what is the value of 3y+2(3y+5)−x3y+2(3y+5)-x in terms of x?

  • A8x+108x+10
  • B32x+1032x+10
  • C35x+1035x+10
  • D36x+1036x+10
Show the answer

Correct answer: C

First, substitute 4x4x for each y in the expression:

3y+2(3y+5)−x=3y+2(3y+5)-x=

3(4x)+2[3(4x)+5]−x3(4x)+2[3(4x)+5]-x

Then simplify the expression:

3(4x)+2[3(4x)+5]−x3(4x)+2[3(4x)+5]-x

12x+2(12x+5)−x12x+2(12x+5)-x

12x+24x+10−x12x+24x+10-x

12x+24x−x+1012x+24x-x+10

35x+1035x+10

The simplified expression is 35x+1035x+10.

Question analysis

Given that y equals 4x, students rewrite the expression 3y + 2(3y + 5) − x using only x. It tests substitution, the distributive property and combining like terms. Errors creep in when a term is dropped or when x terms and y terms are merged before y has been replaced.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting an expression for a variable and simplifying with the distributive property is standard grade 7 work. Putting the substitution inside parentheses calls for care, but the steps are routine for students who combine like terms fluently.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 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

Substitution and distribution are routine here, so points are lost to bookkeeping rather than method. The expression has three x pieces, 12x, 24x and a lone − x, and the last is easy to overlook after the larger terms. The other two wrong choices come from collecting 3y, 6y and x as if they were like terms, a reason to replace y with 4x before simplifying anything.

  1. A very close wrong answer moderate Choosing the answer

    Every choice ends in + 10, and 36x + 10 differs from 35x + 10 by a single x, so only remembering the last − x in the expression separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    The substitution sits inside nested grouping, 2[3(4x) + 5], so the 2 must reach both the 12x and the 5 before the x terms are collected. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Replace every y with 4x in parentheses before doing anything else, then distribute the 2 and collect the x terms and the constant separately.

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
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. Terms are the parts joined by plus or minus signs. Multiply a by each term inside. It works because 2 groups of (12x + 5) hold 2 of the 12x and 2 of the 5.

  1. Sort out the question

    y=4xy = 4x means y is always 4 times x. You need to rewrite the expression with no y in it. That is what “in terms of x” means. The answer will use only x and numbers.

  2. Replace each y with 4x

    Put 4x in parentheses everywhere you see y. The parentheses keep the multiplying clear.

    3y+2(3y+5)−x=3(4x)+2(3(4x)+5)−x3y + 2(3y + 5) - x = 3(4x) + 2(3(4x) + 5) - x

    3(4x) means 3 times 4x, which is 12x. It shows up twice.

    =12x+2(12x+5)−x= 12x + 2(12x + 5) - x

  3. Multiply out the parentheses

    Use the distributive property. The 2 must multiply both the 12x and the 5.

    2(12x+5)=2×12x+2×5=24x+102(12x + 5) = 2 \times 12x + 2 \times 5 = 24x + 10

    The whole expression is now:

    12x+24x+10−x12x + 24x + 10 - x

  4. Combine the like terms

    Like terms have the same letter part. Here the x terms are 12x, 24x and −x. The −x means take away one x.

    12x+24x−x=35x12x + 24x - x = 35x

    The 10 has no x, so it stays as it is. The result is 35x+1035x + 10.

    Watch out: Don’t forget the −x at the end. Stopping at 12x + 24x gives 36x + 10, choice D. Also, swap in 4x before you combine. Adding 3y, 6y and −x as if they were all y’s mixes unlike terms. It leads to choice B.

    So the answer is C, 35x+1035x + 10.

Check your answer

Test a number. Let x = 1, so y = 4. The original expression gives 3(4)+2(3×4+5)−1=12+34−1=453(4) + 2(3 \times 4 + 5) - 1 = 12 + 34 - 1 = 45. Your answer gives 35(1)+10=4535(1) + 10 = 45. They match. Choice D would give 46.

More questions like this

The original DOE question
Question 88 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 88. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 46 of 93

2024 Test B Math, question 40

Question

The value of a particular car decreases at a constant rate. If the car is worth $15,000 three years after its original purchase date and $10,000 five years after its original purchase date, what was the value of the car on its original purchase date?

  • A$18,000
  • B$20,000
  • C$22,500
  • D$25,000
Show the answer

Correct answer: C

The value of the car decreased $5,000 in two years: $15,000 to $10,000.

5,000÷2=2,5005{,}000\div 2=2{,}500

Therefore, the value of the car decreases $2,500 each year. After the first 3 years, the car’s value had decreased $7,500.

So 7,500+15,000=22,5007{,}500+15{,}000=22{,}500.

Question analysis

A car loses value at a constant rate, and its worth three years and five years after purchase is given. Students find what the car was worth on the day it was bought. It tests rate of change and initial value in a linear relationship, and the tricky part is counting the right number of years when working back.

NYS standards
  • 8.F.4 modeling with linear functions
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Finding a starting value from two later values of a steady change is grade 8 linear-function work. A grade 7 student can reason with a yearly rate, but extending it back to the purchase date goes beyond typical grade 7 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, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is a straight-line pattern read backward, and its numbers are friendly; the difficulty is counting years correctly. The $5,000 drop happens over 2 years, not 5, and the purchase lies 3 years before the $15,000 value, not 2. Each miscount gives a listed answer, so sketching a timeline from year 0 to year 5 before computing pays off.

  1. Turning words into math moderate Understanding the question

    The phrases "three years after its original purchase date" and "five years after" must become points at year 3 and year 5 on a timeline whose start, year 0, holds the unknown value. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three moves are chained: the $5,000 drop over 2 years, the $2,500 yearly rate, and three years of that rate added back to $15,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the yearly drop from the change between the two given years, then add that drop back once for each year between the purchase and year 3.

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".
  • 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

Rate of change
rate=change in valuechange in time\text{rate} = \frac{\text{change in value}}{\text{change in time}}
change in value is how many dollars the value went up or down. change in time is how many years passed. rate is the change for one year. A constant rate means the change is the same every year.

  1. Sort out the question

    The car’s value “decreases at a constant rate.” That means the car loses the same number of dollars each year. Year 0 is the day it was bought.

    YearValue
    0?
    3$15,000
    5$10,000

    You need the value at year 0.

  2. Find the drop for one year

    From year 3 to year 5 is 2 years. In that time, the value fell by 15,000−10,000=5,00015{,}000 - 10{,}000 = 5{,}000 dollars.

    rate=5,0002=2,500\text{rate} = \frac{5{,}000}{2} = 2{,}500

    The car loses $2,500 each year. Watch out: Divide by the 2 years between the values, not by 5. Dividing by 5 leads to $18,000, choice A.

  3. Go back to year 0

    From year 0 to year 3 is 3 years. The car lost $2,500 in each of them. So add that amount back 3 times.

    15,000+3×2,500=15,000+7,500=22,50015{,}000 + 3 \times 2{,}500 = 15{,}000 + 7{,}500 = 22{,}500

    Watch out: Adding back only one $5,000 gives $20,000, choice B. That is the value at year 1, not year 0.

    So the answer is C, $22,500.

Check your answer

Count forward from $22,500, taking off $2,500 each year. Year 1 is $20,000. Year 2 is $17,500. Year 3 is $15,000. Year 4 is $12,500. Year 5 is $10,000. Both given values match.

More questions like this

The original DOE question
Question 97 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 97. Open page 210 of the handbook (PDF)

Question 47 of 93

2024 Test B Math, question 44

Question

Which inequality is equivalent to 6x−20<5x−2(5−3x)6x-20<5x-2(5-3x)?

  • Ax<16x<16
  • Bx<52x<\frac{5}{2}
  • Cx<−2x<-2
  • Dx>−2x>-2
Show the answer

Correct answer: D

6x−20<5x−2(5−3x)6x-20<5x-2(5-3x)

Distribute the coefficient −2 on the right-hand side:

6x−20<5x+(−2)(5)+(−2)(−3x)6x-20<5x+(-2)(5)+(-2)(-3x)

6x−20<5x−10+6x6x-20<5x-10+6x

Subtract 6x6x from both sides:

−20<5x−10-20<5x-10

Add 10 to both sides:

−10<5x-10<5x

Divide both sides by 5:

−2<x-2<x

This is equivalent to

x>−2x>-2

Question analysis

Students choose the simplest equivalent form of a linear inequality that has x on both sides and a negative number multiplied into parentheses. It tests distributing, combining like terms and solving inequalities. Sign errors in the distribution, and reading the direction of the final inequality, are where most mistakes happen.

NYS standards
  • 7.EE.4b two-step inequalities
  • 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 inequalities have the form px + q < r. This one has x on both sides and a negative factor to distribute across a subtraction, steps that belong to grade 8 equation work, so it is harder than typical grade 7 practice.
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 16+, 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

Every move is standard algebra, but the inequality is built so that small slips change the answer completely. The −2 has to reach both terms in the parentheses, and once the boundary −2 is found its direction still has to be read correctly, since the variable ends up on the right as −2 < x. Rewriting that as x > −2, or substituting x = 0 into the original, catches the trap.

  1. Error-prone arithmetic moderate Working it out

    Distributing −2 over (5 − 3x) gives −10 + 6x, two sign decisions in one move, and the x terms then sit on both sides of the inequality. (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, which depends on reading −2 < x correctly or flipping the sign when dividing by a negative. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    Four moves follow one another, distributing, collecting 5x + 6x, moving the x terms and dividing by 5, and a slip at any of them changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Distribute the −2 to both terms in the parentheses, combine the x terms on the right, then gather the x terms on the side that keeps their coefficient positive.

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

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. Multiply a by each one. A minus sign inside counts as a negative term, so 5 − 3x is 5 + (−3x). A negative times a negative is positive.
Keeping an inequality true
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. Adding or subtracting the same number on both sides keeps the inequality true. Dividing both sides by a positive number also keeps it true. Only multiplying or dividing by a negative number flips the sign.

  1. Sort out the question

    An inequality compares two sides with a sign like <, which means is less than. To solve it, you get x alone on one side.

    You need the short inequality that has the same answers as the long one.

  2. Distribute the −2

    Multiply −2 by both terms inside the parentheses, 5 and −3x.

    −2(5−3x)=(−2)×5+(−2)×(−3x)=−10+6x-2(5 - 3x) = (-2) \times 5 + (-2) \times (-3x) = -10 + 6x

    A negative times a negative is positive, so the second part is +6x. Watch out: If the −2 multiplies only the 5, you end up with x<52x < \frac{5}{2}, choice B.

  3. Combine the x terms

    The right side is now 5x − 10 + 6x. Put the x terms together: 5x + 6x = 11x.

    6x−20<11x−106x - 20 < 11x - 10

  4. Gather the x terms on one side

    Subtract 6x from both sides. That leaves the x terms on the right, where they stay positive.

    6x−20−6x<11x−10−6x6x - 20 - 6x < 11x - 10 - 6x

    −20<5x−10-20 < 5x - 10

  5. Get x by itself

    Add 10 to both sides. Then divide both sides by 5.

    −10<5x-10 < 5x

    −2<x-2 < x

    You divided by a positive 5, so the sign does not flip.

  6. Read it with x first

    −2<x-2 < x says −2 is less than x. That is the same as saying x is greater than −2, just as 3 < 5 means 5 > 3.

    x>−2x > -2

    Watch out: x<−2x < -2, choice C, has the same number but points the wrong way.

    So the answer is D, x>−2x > -2.

Check your answer

Test x = 0, which is greater than −2. The left side is 6(0)−20=−206(0) - 20 = -20. The right side is 5(0)−2(5−0)=−105(0) - 2(5 - 0) = -10. And −20 < −10 is true. Now test x = −3. The left side is −38 and the right side is −43. That is false, as it should be.

More questions like this

The original DOE question
Question 101 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 101. Open page 211 of the handbook (PDF)

Question 48 of 93

2024 Test B Math, question 53

Question

Which graph represents the solution to |x+1|≥2|x+1|\geq 2?

  • ANumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1. The line is shaded from −3 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • BNumber line from −3 to 3 with a tick at every integer. Closed dots at −1 and 1. The line is shaded from −1 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • CNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1, with the segment between them shaded.
  • DNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 3. The line is shaded from −3 to the left, ending in an arrow, and from 3 to the right, ending in an arrow.
Show the answer

Correct answer: A

Since |x+1|≥2|x+1|\geq 2, either x+1≤−2x+1\leq -2 or x+1≥2x+1\geq 2.

Solving x+1≤−2x+1\leq -2 for x gives x≤−3x\leq -3.

Solving x+1≥2x+1\geq 2 for x gives x≥1x\geq 1.

A filled circle on a number line represents “or equal to.” So the correct number line representation has filled circles at −3 and 1, with the number line filled to the left of −3 for “less than” and to the right of 1 for “greater than.”

Question analysis

Students choose the number line graph that shows every solution of an inequality in which x + 1 sits inside absolute value bars. It tests absolute value as distance and graphing inequalities. The tricky part is turning one absolute value statement into two pieces and placing each boundary point and shaded ray correctly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above typical grade 7 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 9, 7 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Reading |x + 1| as the distance from x to −1 makes this quick: the points 2 units away are −3 and 1, and greater than or equal means outside them. Students who solve only one case and mirror it across 0 land on graphs centered on 0, and students who find both endpoints may still shade between them. The graphs differ in just one dot or one shading choice, so each must be read carefully.

  1. Something you must already know moderate Working it out

    Solving |x + 1| ≥ 2 needs two cases, x + 1 ≤ −2 or x + 1 ≥ 2, joined by or; absolute value inequalities go beyond grade 7, and the stem gives no hint. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The right graph and the segment from −3 to 1 share both filled endpoints, so only deciding between outside and between the boundaries separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Four number lines must be read for filled dots and arrow directions, and two of them differ from the right one only in where a single dot sits. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the absolute value as distance from −1 on the number line, find the points exactly 2 units away, and decide whether the solutions lie between or outside them.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value as distance
|x−a|=distance from x to a|x - a| = \text{distance from } x \text{ to } a
x and a are points on the number line. The bars mean absolute value. |x − a| is how far apart the two points are, which is never negative. Since x + 1 = x − (−1), |x + 1| is the distance from x to −1.

  1. Sort out the question

    You need the graph of every x that makes |x+1|≥2|x + 1| \geq 2 true. The sign ≥ means greater than or equal to. The bars mean absolute value.

    On a number line, a filled dot means that point counts. A thick line with an arrow means all the numbers that way count too.

  2. Read it as a distance

    Adding 1 is the same as subtracting −1, so x+1=x−(−1)x + 1 = x - (-1). By the formula, |x+1||x + 1| is the distance from x to −1.

    So the inequality says that x is at least 2 units away from −1.

  3. Find the two boundary points

    Start at −1 and go 2 units each way.

    −1−2=−3-1 - 2 = -3

    −1+2=1-1 + 2 = 1

    At −3 and at 1, the distance is exactly 2. The words or equal to let them count, so they get filled dots.

  4. Decide which side to shade

    Being at least 2 away means being farther out. Numbers between −3 and 1 are too close to −1. For example, 0 is only 1 unit away. So shade to the left of −3 and to the right of 1.

    In algebra, that is x+1≤−2x + 1 \leq -2 or x+1≥2x + 1 \geq 2. These give x≤−3x \leq -3 or x≥1x \geq 1.

  5. Match the graph

    Choice A has filled dots at −3 and 1, with arrows going outward. That matches.

    Watch out: Choice C shades between −3 and 1. Those are the points within 2 of −1, the answer to |x+1|≤2|x + 1| \leq 2. Choices B and D are centered on 0, but the center here is −1.

    So the answer is A, the graph shaded outward from −3 and 1.

Check your answer

Test a point from each part. For x = 2, |2+1|=3|2 + 1| = 3, and 3 is at least 2. For x = 0, |0+1|=1|0 + 1| = 1, which is less than 2. For x = −3, |−3+1|=2|-3 + 1| = 2, which counts. Graph A agrees with all three.

More questions like this

The original DOE question
Question 110 as printed on page 214 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 110. Open page 214 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 49 of 93

2024 Test B Math, question 55

Question

Typically, an amusement park sells 3 times as many tickets on Sunday as on any weekday, and 4 times as many tickets on Saturday as on Sunday. If the number of tickets sold on Saturday and Sunday totaled 30,000, how many tickets are expected to be sold on the following Monday?

  • A2,000
  • B2,500
  • C3,000
  • D4,285
Show the answer

Correct answer: A

First, determine the number of tickets sold on Sunday, x, using the total number of tickets sold on Saturday and Sunday, 30,000, and the number of tickets sold on Saturday, 4x4x.

30,000=4x+x30{,}000=4x+x

30,000=5x30{,}000=5x

6,000=x6{,}000=x

Then determine the number of tickets expected to be sold on Monday, y, using the number of tickets sold on Sunday, 6,000.

6,000=3y6{,}000=3y

2,000=y2{,}000=y

The number of tickets expected to be sold on Monday is 2,000.

Question analysis

An amusement park sells 3 times as many tickets on Sunday as on a weekday and 4 times as many on Saturday as on Sunday. Given the weekend total, students find the expected ticket sales for the following Monday. It tests writing an equation from multiplicative comparisons, and the trap is comparing Saturday with the wrong day.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving a one-variable equation from a word problem is grade 7 work. Two linked multiplicative comparisons, with Saturday measured against Sunday rather than a weekday, add a layer of translation, but the equation itself stays simple.
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 12, 48 words.
Students, last 30 days
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 two comparisons chained through Sunday, and most wrong answers come from linking them incorrectly. Students who read Saturday's 4 times as 4 weekdays, or who leave Sunday out of the 30,000, find listed answers, one of them from a division that does not come out even. Letting a weekday be one share, so Sunday is 3 and Saturday 12, makes the weekend 15 shares and the check easy.

  1. Turning words into math moderate Understanding the question

    "4 times as many tickets on Saturday as on Sunday" compares Saturday with Sunday, not with a weekday, so Saturday is 4 × 3 = 12 weekdays' worth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Three days have to be held in relation at once, a weekday, Sunday and Saturday, with the 30,000 covering only the last two. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let Sunday's sales be the unknown, write Saturday as 4 times it, solve the weekend equation for Sunday, then divide Sunday's figure by 3 for a weekday.

Traps

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

Undoing multiplication
a×x=b means x=b÷aa \times x = b \text{ means } x = b \div a
x is the unknown number. a and b are known numbers. Division undoes multiplication. So if a times x is b, you find x by dividing b by a. This is the same as dividing both sides by a.

  1. Sort out the question

    Sunday sells 3 times as many tickets as a weekday. Saturday sells 4 times as many as Sunday. Saturday and Sunday together sell 30,000.

    Monday is a weekday. You need Monday’s tickets. Notice that Saturday is compared with Sunday, not with a weekday.

  2. Write an equation for the weekend

    Let x be Sunday’s tickets. Then Saturday’s tickets are 4x. Together they make 30,000.

    4x+x=30,0004x + x = 30{,}000

    5x=30,0005x = 30{,}000

  3. Solve for Sunday

    Use the rule for undoing multiplication. Divide both sides by 5.

    x=30,000÷5=6,000x = 30{,}000 \div 5 = 6{,}000

    Sunday sells 6,000 tickets. Saturday sells 4×6,000=24,0004 \times 6{,}000 = 24{,}000.

  4. Find Monday’s tickets

    Sunday sells 3 times as many as a weekday. Let y be Monday’s tickets.

    3×y=6,000 means y=6,000÷3=2,0003 \times y = 6{,}000 \text{ means } y = 6{,}000 \div 3 = 2{,}000

    Watch out: If you read Saturday as 4 times a weekday, you split 30,000 into 7 shares. That gives about 4,285, choice D. Dividing 30,000 by 12 leaves out Sunday and gives 2,500, choice B.

    So the answer is A, 2,000.

Check your answer

Think in shares. A weekday is 1 share. Sunday is 3 shares, and Saturday is 4×3=124 \times 3 = 12 shares. The weekend is 15 shares, and 30,000÷15=2,00030{,}000 \div 15 = 2{,}000. It matches.

More questions like this

The original DOE question
Question 112 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 112. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 93

2023 Test A Math, question 3

Question

Solve for x:

7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13

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

Show the answer

Correct answer: 4

7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13 Apply the distributive property; multiply the −2 by 2x2x and 1.
7x+3−4x−2=137x + 3 - 4x - 2 = 13 Combine like terms.
3x+1=133x + 1 = 13 Apply the additive inverse property; subtract 1 from both sides of the equation.
3x=123x = 12 Apply the multiplicative inverse property; divide both sides of the equation by 3.
x=4x = 4

Question analysis

This grid-in gives the equation 7x + 3 − 2(2x + 1) = 13 and asks students to solve for x. It tests the distributive property with a negative factor, combining like terms and solving a two-step equation. The minus sign in front of the parentheses has to reach both terms inside, a step many students rush.

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 Grade 7 students expand expressions with negative factors and solve equations such as p(x + q) = r. This equation needs both skills plus combining like terms, which is routine work with one negative sign that is easy to lose.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
4 steps
Reading level
Question and choices: about grade 3, 4 words.
Students, last 30 days
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 standard multi-step linear equation, and the work is short once the parentheses are cleared. Everything rests on distributing −2 correctly; the later steps, combining 7x with −4x and 3 with −2, then undoing the + 1 and the × 3, are routine. Substituting 4 back into the left side, 28 + 3 − 18 = 13, is a fast check that catches a sign slip.

  1. Error-prone arithmetic moderate Working it out

    The −2 in front of (2x + 1) must reach both terms, giving −4x − 2; the second product, −2 × 1, is the one most often written as +2 or left as +1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    As a grid-in with no choices, a value such as 3 or 8/3 from a distributing slip looks just as finished as the right one; only substituting back exposes it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute −2 to both terms inside the parentheses, combine the x terms and the constants, then undo the addition and the multiplication in turn.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. Multiply each term inside by a. If a is negative, every product gets its sign too.
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 \div c = b \div c
a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

  1. Sort out the equation

    You need the value of x that makes 7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13 true. x is the unknown number. The tricky part is the −2 in front of the parentheses. It multiplies everything inside them.

  2. Multiply out the parentheses

    Use the distributive property with a = −2, b = 2x and c = 1.

    −2(2x+1)=(−2)×2x+(−2)×1=−4x−2-2(2x + 1) = (-2) \times 2x + (-2) \times 1 = -4x - 2

    So the equation becomes this:

    7x+3−4x−2=137x + 3 - 4x - 2 = 13

    Watch out: the −2 must multiply the 1 too. Leaving + 1 gives x = 3. Writing + 2 instead of − 2 gives x=83x = \frac{8}{3}.

  3. Combine like terms

    Like terms are terms of the same kind. Put the x terms together. Put the plain numbers together.

    7x−4x=3x7x - 4x = 3x

    3−2=13 - 2 = 1

    3x+1=133x + 1 = 13

  4. Get x by itself

    Undo the + 1 first. Subtract 1 from both sides, so the equation stays balanced.

    3x+1−1=13−13x + 1 - 1 = 13 - 1

    3x=123x = 12

    3x means 3 times x. So divide both sides by 3.

    x=12÷3=4x = 12 \div 3 = 4

    Watch out: don’t stop at 3x = 12. 12 is three times the answer. So you would grid 4.

Check your answer

Put x = 4 back into the first equation: 7×4+3−2(2×4+1)=28+3−2×9=31−18=137 \times 4 + 3 - 2(2 \times 4 + 1) = 28 + 3 - 2 \times 9 = 31 - 18 = 13. That matches the 13 on the right side, so x = 4 is right.

More questions like this

The original DOE question
Question 60 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 2024 Admissions, page 75, question 60. Open page 75 of the handbook (PDF)

Question 51 of 93

2023 Test A Math, question 18

Question

Sheila is saving money for her summer vacation. She starts the summer with a balance of $90.00 and plans to save 15% of her earnings each week. She earns the same amount each week. After 12 weeks, Sheila has saved a total of $472.59. How much money does Sheila earn each week?

  • A$212.55
  • B$262.55
  • C$302.55
  • D$312.55
Show the answer

Correct answer: A

Let x be the amount that Sheila needs to earn each week.

$90.00+0.15×12x=$472.59\$90.00 + 0.15 \times 12x = \$472.59

First, subtract 90 from both sides:

0.15×12x=$382.590.15 \times 12x = \$382.59

1.8x=$382.591.8x = \$382.59

Last, divide both sides by 1.8:

x=$212.55x = \$212.55

Question analysis

Sheila starts the summer with $90 and saves 15% of what she earns each week, earning the same amount every week, and after 12 weeks she has $472.59. Students find her weekly earnings. They must write an equation that separates the starting balance from the savings and then work with a percent and long decimals.

NYS standards
  • 7.EE.4a two-step equations
  • 7.RP.3 multistep ratio and percent problems
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 core grade 7 work. Here the coefficient combines a percent with 12 weeks, and the decimals are long, which adds steps but no new idea.
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 4, 48 words.
Students, last 30 days
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 story has three layers, a starting balance, a weekly saving rate and a 12-week total, and the unknown sits under the percent. Students lose the point by forgetting the $90 was already there, or by moving it the wrong way. Testing a choice is a dependable alternative: 15% of $212.55 is about $31.88, and 12 of those plus $90 gives $472.59.

  1. Turning words into math moderate Understanding the question

    The total of $472.59 includes the $90.00 she started with, and only 15% of each week's pay is saved, so the equation is 90 + 12 × 0.15x = 472.59. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Dividing $382.59 by 1.8 by hand means shifting the decimal point to get 3,825.9 ÷ 18, which comes out to exactly $212.55 only if every digit is placed correctly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Many steps minor Working it out

    Subtract the $90, then divide by 12 weeks and by 0.15, or by their product 1.8; doing the division before the subtraction gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the starting $90 from the final total, then divide by 12 weeks and by 0.15, or test each choice by taking 15% of it, multiplying by 12 and adding $90.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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 of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Savings over time
total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}
The start is the money she had at the beginning. Weekly savings is what she adds each week. Weeks is how many weeks pass. Each week adds the same amount, so multiply. Then add what she started with.

  1. Sort out the question

    Sheila starts with $90.00. Each week she earns the same amount and saves 15% of it. After 12 weeks she has $472.59 in all. That total includes the $90 she started with. You need how much she earns each week, not how much she saves.

  2. Write an equation

    Call her weekly earnings x. She saves 15% of x each week. That is 0.15×x0.15 \times x.

    total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}

    472.59=90+0.15x×12472.59 = 90 + 0.15x \times 12

  3. Take away the starting $90

    Only the money added during the 12 weeks came from her pay. Subtract 90 from both sides.

    0.15x×12=472.59−90=382.590.15x \times 12 = 472.59 - 90 = 382.59

    She saved $382.59 from her pay. Watch out: subtract the $90, don’t add it. Adding leads to $312.55, choice D. Forgetting the $90 leads to $262.55, choice B.

  4. Combine the numbers next to x

    0.15×12=1.80.15 \times 12 = 1.8

    1.8x=382.591.8x = 382.59

    Each week she saves 0.15 of one week’s pay. Over 12 weeks, that adds up to 1.8 times one week’s pay.

  5. Divide by 1.8

    Divide both sides by 1.8. First move the decimal point one place right in both numbers. Then you divide by the whole number 18.

    x=382.59÷1.8=3,825.9÷18=212.55x = 382.59 \div 1.8 = 3{,}825.9 \div 18 = 212.55

    Sheila earns $212.55 each week. So the answer is A, $212.55.

Check your answer

Work forward. 12 weeks of saving 15% is 1.8 weeks of pay: 1.8×212.55=382.591.8 \times 212.55 = 382.59. Add the $90 she started with: 382.59+90=472.59382.59 + 90 = 472.59. That is the total in the question.

More questions like this

The original DOE question
Question 75 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 75. Open page 80 of the handbook (PDF)

Question 52 of 93

2023 Test A Math, question 22

Question

The numbers m, n, p, and q are different, and each is equal to one of the numbers 1, 2, 3, 6, or 12. If 2m=6q=12n=p2m = 6q = \frac{1}{2}n = p, what is the value of p?

  • A2
  • B3
  • C6
  • D12
Show the answer

Correct answer: C

Substitute each answer option for p to see which value will work:

Option A:

If p=2p = 2, then 12n=2\frac{1}{2}n = 2, so n=4n = 4. This cannot be the answer, because n cannot equal 4.

Option B:

If p=3p = 3, then 6q=36q = 3, so q=0.5q = 0.5. This cannot be the answer, because q cannot equal 0.5.

Option C:

If p=6p = 6, then 2m=62m = 6, so m=3m = 3; 6q=66q = 6, so q=1q = 1; and 12n=6\frac{1}{2}n = 6, so n=12n = 12. This is the correct answer.

Option D:

If p=12p = 12, then 12n=12\frac{1}{2}n = 12, so n=24n = 24. This cannot be the answer, because n cannot equal 24.

Question analysis

Four letters, m, n, p and q, stand for different numbers from the list 1, 2, 3, 6 and 12, and the expressions 2m, 6q, half of n, and p are all equal. Students find the value of p. The puzzle-like setup has no single equation to solve, so testing values against every condition is the reliable route.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 The equations involve only doubling, halving and multiplying by 6, all grade 6 skills. What goes beyond grade 7 practice is the chain of four equal expressions, with every letter limited to a short list of different numbers.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 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 chain looks forbidding, but it gives each letter as a simple multiple of p: m is half of p, q a sixth of it, and n twice it. Seen that way, q = p/6 forces p to be 6 or 12, and n = 2p rules out 12, so only one value fits. Students who check one or two of the letters, but not all three, can accept 2 or 12.

  1. Variables instead of numbers moderate Understanding the question

    Four letters are tied by one chain of equalities, 2m = 6q = ½n = p, and the only numbers are the five allowed values, so each letter has to be written in terms of p. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Each choice for p must be tested against three letters, m = p/2, q = p/6 and n = 2p, and all of them must land on different numbers from 1, 2, 3, 6 and 12. (Embretson & Wetzel, 1987)

Strategy: Test each choice as p: find m, q and n from it, and keep the value for which all three are different numbers from the list 1, 2, 3, 6 and 12.

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

Undoing a multiplication
if a×m=p, then m=p÷a\text{if } a \times m = p \text{, then } m = p \div a
a is a number that multiplies m, and p is the result. Dividing by a undoes multiplying by a. Use it to write each letter using p.

  1. Sort out the question

    m, n, p and q are four different numbers. Each one is 1, 2, 3, 6 or 12. The chain 2m=6q=12n=p2m = 6q = \frac{1}{2}n = p says that four amounts are all equal to p. You need the value of p.

  2. Write each letter using p

    Each part of the chain equals p. Undo each multiplication.

    2m=p, so m=p÷22m = p \text{, so } m = p \div 2

    6q=p, so q=p÷66q = p \text{, so } q = p \div 6

    12n=p, so n=2×p\frac{1}{2}n = p \text{, so } n = 2 \times p

    Half of n is p, so n is twice p.

  3. Test each choice for p

    pmqnAll on the list?
    2113\frac{1}{3}4No
    31.50.56No
    63112Yes
    126224No

    Only p = 6 puts every letter on the list.

  4. Check that the letters are different

    With p = 6, you get m = 3, q = 1 and n = 12. All four numbers are on the list, and no two are the same. Watch out: p = 12 makes m = 6 and q = 2, which look fine. But n = 24 is not on the list, so choice D fails. Also, the question asks for p, not m. Picking 3 is choice B. So the answer is C, 6.

Check your answer

Put the values into the chain: 2×3=62 \times 3 = 6, 6×1=66 \times 1 = 6, 12×12=6\frac{1}{2} \times 12 = 6, and p = 6. All four parts equal 6. The numbers 3, 1, 12 and 6 are all different and all on the list.

More questions like this

The original DOE question
Question 79 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 79. Open page 81 of the handbook (PDF)

Question 53 of 93

2023 Test A Math, question 33

Question

If m=5m = 5 and t=−1t = -1, what is the value of 6−8(2−t)2m+4(3−m)\frac{6 - 8(2 - t)}{2m + 4(3 - m)}?

  • A−9
  • B−1
  • C1
  • D9
Show the answer

Correct answer: A

6−8(2−t)2m+4(3−m)=\frac{6 - 8(2 - t)}{2m + 4(3 - m)} =

6−8[2−(−1)]2(5)+4(3−5)=\frac{6 - 8[2 - (-1)]}{2(5) + 4(3 - 5)} =

6−8(3)10+4(−2)=\frac{6 - 8(3)}{10 + 4(-2)} =

6−2410+(−8)=\frac{6 - 24}{10 + (-8)} =

−182=−9\frac{-18}{2} = -9

Question analysis

This algebra question gives m = 5 and t = −1 and asks for the value of a fraction whose numerator and denominator both contain parentheses. Students substitute the values and follow the order of operations. Subtracting a negative number and multiplying by a negative difference make sign mistakes likely, and the four choices differ only in sign and size.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting values into expressions with negative numbers is everyday grade 7 work. This fraction has a numerator and a denominator to evaluate separately, with a minus sign before a negative and a negative difference, so it is a longer version of a standard exercise.
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 8, 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 plain evaluation with no story, and every point is lost on signs. Subtracting the negative t, then handling the two negative intermediate results, each invite a dropped minus, and the four choices are built so that each such slip lands on a listed value. Writing −1 in parentheses when substituting, and finishing the top and the bottom before dividing, keeps the signs in view.

  1. Error-prone arithmetic moderate Working it out

    The value t = −1 sits inside 2 − t, so the numerator needs 2 − (−1) = 3, and the work then passes through two negatives, 6 − 24 = −18 and 3 − 5 = −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The numerator and the denominator each take a substitution, a parenthesis, a product and a sum, worked separately before one final division, so a slip in either half changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The choices pair each size with each sign, −9 with 9 and −1 with 1, so getting 18 and 2 right still leaves the sign to decide between two choices. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Substitute with parentheses around −1, simplify the numerator and the denominator separately, innermost parentheses first, and divide only at the end, checking 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.)

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 and b are numbers. Taking away a negative number is the same as adding the positive number. Think of a debt of 1 dollar being taken away: you are 1 dollar better off.
Dividing a negative by a positive
−ab=−ab\frac{-a}{b} = -\frac{a}{b}
a and b are positive numbers. A negative number divided by a positive number gives a negative answer. The answer is positive only when both signs are the same.

  1. Sort out the question

    You know m = 5 and t = −1. You need the value of the big fraction. Work out the top, called the numerator, and the bottom, called the denominator, one at a time. Then divide the top by the bottom.

  2. Put in the numbers

    6−8(2−t)2m+4(3−m)=6−8(2−(−1))2(5)+4(3−5)\frac{6 - 8(2 - t)}{2m + 4(3 - m)} = \frac{6 - 8(2 - (-1))}{2(5) + 4(3 - 5)}

    Put each number in parentheses, so the minus signs stay clear. Work inside the parentheses first, then multiply, then add or subtract.

  3. Work out the top

    a−(−b)=a+ba - (-b) = a + b

    2−(−1)=2+1=32 - (-1) = 2 + 1 = 3

    6−8(3)=6−24=−186 - 8(3) = 6 - 24 = -18

    The top is −18. It is negative because 24 is more than 6. Watch out: 2 − (−1) is 3, not 1. Using 1 makes the top −2 and the answer −1, choice B.

  4. Work out the bottom

    2(5)+4(3−5)=10+4(−2)=10−8=22(5) + 4(3 - 5) = 10 + 4(-2) = 10 - 8 = 2

    Here 3 − 5 = −2, because 5 is more than 3. Then 4 times −2 is −8. The bottom is 2.

  5. Divide the top by the bottom

    −ab=−ab\frac{-a}{b} = -\frac{a}{b}

    −182=−9\frac{-18}{2} = -9

    Watch out: if you lost the minus sign in 6 − 24, you would get 9, choice D. So the answer is A, −9.

Check your answer

Check the sign first. The top, 6 − 24, is clearly negative. The bottom, 10 − 8, is positive. Negative over positive is negative, so only −1 or −9 can work. Without the signs, the top is 18 and the bottom is 2, and 18 ÷ 2 = 9. So −9 fits.

More questions like this

The original DOE question
Question 90 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 90. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 54 of 93

2023 Test A Math, question 35

Question

A number line with tick marks every 2 units, labeled −6, −4, −2, 0, 2, 4 and 6. Point J is the dot at −4 and point K is the dot at 6.

On the number line above, point L (not shown) is located on line segment JK so that JL=23LK\mathrm{JL} = \frac{2}{3}\mathrm{LK}. What is the position of point L?

  • A−2
  • B0
  • C2
  • D4
Show the answer

Correct answer: B

JK=6−(−4)=10\mathrm{JK} = 6 - (-4) = 10 units. Let x=JLx = \mathrm{JL}. Then LK=10−x\mathrm{LK} = 10 - x. Substitute these expressions in the given equation:

x=23(10−x)x = \frac{2}{3}(10 - x)

Solve for x:

x=23(10−x)x = \frac{2}{3}(10 - x) Apply the multiplicative inverse property; multiply both sides of the equation by 3.
3x=2(10−x)3x = 2(10 - x) Apply the distributive property; multiply the 2 by 10 and x.
3x=20−2x3x = 20 - 2x Apply the additive inverse property; add 2x2x to both sides of the equation.
5x=205x = 20 Apply the multiplicative inverse property; divide both sides of the equation by 5.
x=4x = 4

So, point L is 4 units to the right of point J: −4+4=0-4 + 4 = 0. Point L is located at 0 on the number line.

Question analysis

This number line question puts J at −4 and K at 6 and asks for the position of a point L on segment JK where JL is two thirds of LK. Students find the segment's length, divide it in a 2-to-3 ratio, and turn a distance into a coordinate. Reading the ratio backwards, or giving a distance instead of a position, leads to wrong answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 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

The ratio is stated between two unknown pieces, which is less familiar than a fraction of a whole, and the final step turns a length into a position. Students who stop at JL's length, read 2/3 the wrong way round, or use its 2 as a distance all reach listed points. Sketching five equal shares of 2 units from −4 to 6, two for JL and three for LK, makes the position plain.

  1. Turning words into math moderate Understanding the question

    JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length of JK, split it into five equal shares with two for JL and three for LK, then count JL's length to the right from J.

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).
  • 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 on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
Subtract the position on the left from the position on the right. The answer is how many units apart the two points are. A distance is never negative.
Splitting a length in a ratio
one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}
A ratio a : b compares two pieces. If a length is split in the ratio a : b, think of it as a + b equal parts. The first piece gets a of them and the second gets b. Divide the whole length by a + b to find one part.

  1. Sort out the question

    Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is 23\frac{2}{3} of LK. You need the position of L.

  2. Find the length of JK

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    K is on the right, so subtract J’s position from K’s.

    JK=6−(−4)=6+4=10\mathrm{JK} = 6 - (-4) = 6 + 4 = 10

    Taking away −4 is the same as adding 4. J and K are 10 units apart.

  3. Turn the rule into equal parts

    JL is 23\frac{2}{3} of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.

    one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}

    one part=102+3=2\text{one part} = \frac{10}{2 + 3} = 2

    Each part is 2 units long.

  4. Find the two lengths

    JL=2×2=4\mathrm{JL} = 2 \times 2 = 4

    LK=3×2=6\mathrm{LK} = 3 \times 2 = 6

    Watch out: JL is 23\frac{2}{3} of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.

  5. Find the position of L

    L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.

    −4+4=0-4 + 4 = 0

    Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

Put L at 0 and test the rule. JL = 0 − (−4) = 4, and LK = 6 − 0 = 6. Then 23×6=4\frac{2}{3} \times 6 = 4, which is JL. Also 4 + 6 = 10, the whole length JK.

More questions like this

The original DOE question
Question 92 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 92. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 55 of 93

2023 Test A Math, question 36

Question

Josef and Mai divided some stamps between themselves. Josef got 60% of the stamps. If Josef received 500 more stamps than Mai, how many stamps did Josef receive?

  • A300
  • B800
  • C1,500
  • D2,500
Show the answer

Correct answer: C

Let j be the number of stamps Josef has. Then Mai has j−500j - 500 stamps. The total number of stamps is j+(j−500)=2j−500j + (j - 500) = 2j - 500. Since Josef has 60% of the stamps, set up an equation to solve for j:

j2j−500=60100\frac{j}{2j - 500} = \frac{60}{100} Cross multiply.
100j=60(2j−500)100j = 60(2j - 500) Apply the distributive property; multiply the 60 by 2j2j and 500.
100j=120j−30,000100j = 120j - 30{,}000 Apply the additive inverse property; subtract 120j120j from both sides of the equation.
−20j=−30,000-20j = -30{,}000 Apply the multiplicative inverse property; divide both sides of the equation by −20.
j=1,500j = 1{,}500

Question analysis

This percent word problem says Josef and Mai split some stamps, Josef got 60% of them, and he received 500 more than Mai; it asks how many stamps Josef received. Students must connect a difference in counts to a difference in percents. Stopping at the total, or taking a percent of 500, gives tempting wrong answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
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 4, 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

The arithmetic is easy once the 500 is seen as the gap between two percents of one total; seeing that is the hard part. The tempting move is to apply the 60% to the only number given, the 500, which treats a difference as a whole. Students who do find the total must still take Josef's share of it, since the total waits among the choices. Checking that 1,500 and 1,000 differ by 500 confirms the answer.

  1. Turning words into math moderate Understanding the question

    The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method moderate Working it out

    No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)

  3. Many steps minor Working it out

    Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: If Josef has 60%, Mai has the other 40%, so the 500-stamp difference is a known percent of the total; use it to find the total, then take Josef's share.

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=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the total amount, and the part is the piece you want. Percent means “out of 100,” so 60% is 60 out of every 100. Use it to find a part, or work backward to find the whole.

  1. Sort out the question

    Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.

  2. Find Mai’s percent

    All the stamps make 100%. Josef has 60%, so Mai has the rest.

    100%−60%=40%100\% - 60\% = 40\%

    Mai has 40% of the stamps.

  3. Find what the 500 stamps are

    Josef’s extra 500 stamps are the gap between his share and Mai’s share.

    60%−40%=20%60\% - 40\% = 20\%

    So 500 stamps are 20% of all the stamps.

  4. Find the total number of stamps

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    500=20100×whole500 = \frac{20}{100} \times \text{whole}

    whole=500÷20100=500×5=2,500\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500

    There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.

  5. Find Josef’s share

    part=60100×2,500=1,500\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500

    Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

Mai gets 40% of 2,500, which is 1,000 stamps. Josef’s 1,500 minus Mai’s 1,000 is 500, just as the question says. And 1,500 + 1,000 = 2,500, the whole amount.

More questions like this

The original DOE question
Question 93 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 93. Open page 84 of the handbook (PDF)

Question 56 of 93

2023 Test A Math, question 38

Question

Which graph represents the solution to −3x−7>−4-3x - 7 > -4?

  • ANumber line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the left, ending in an arrow.
  • BNumber line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the left, ending in an arrow.
  • CNumber line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the right, ending in an arrow.
  • DNumber line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the right, ending in an arrow.
Show the answer

Correct answer: A

−3x−7>−4-3x - 7 > -4 Apply the additive inverse property; add 7 to both sides of the inequality.
−3x>3-3x > 3 Apply the multiplicative inverse property; divide both sides of the inequality by −3. Note: when dividing an inequality by a negative number, the inequality sign is reversed.
x<−1x < -1

The solution is that x is less than −1, so choose the graph that starts at −1 and continues to the left.

Question analysis

This inequality question asks which number line graph shows the solution of −3x − 7 > −4. Students solve a two-step inequality whose x term has a negative coefficient, then match the boundary point and the direction of the shading. Handling the inequality symbol at the division step is where most mistakes happen, and each slip has a matching graph.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving and graphing a two-step inequality is a core grade 7 standard. This one has a negative coefficient on x, the classic grade 7 inequality trap, and four similar graphs test whether the sign and the direction were both handled.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 7 words.
Corrected from the DOE handbook
  • A wrong term 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 solving takes two moves, but the second carries the rule the question tests: dividing an inequality by a negative number flips its direction. The four graphs cross two features, boundary at −1 or 1 and shading left or right, so forgetting the flip, dropping a sign, or both each land on a graph. Testing a shaded value such as x = −2 in the original inequality, where 6 − 7 = −1 is greater than −4, confirms the choice.

  1. Something you must already know moderate Working it out

    The step that decides the question, reversing the symbol when dividing both sides by −3, is a rule students must bring themselves; −3x > 3 becomes x < −1, not x > −1. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The graph shaded to the right from an open circle at −1 matches the right graph's boundary exactly and differs only in direction, the result of skipping the reversal. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Each graph must be read for two things, where its open circle sits, at −1 or 1, and which way the arrow runs, so a boundary alone does not identify the answer. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add 7 to both sides, divide by −3 and reverse the inequality symbol, then match both the boundary point and the shading direction; test one shaded value in the original.

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 handbook’s steps say “add 7 to both sides of the equation” and “divide both sides of the equation by −3” (p. 142, question 95). But −3x − 7 > −4 is an inequality, not an equation, and the step’s own note is the point of the question: dividing an inequality by a negative number reverses the sign. The site now says “both sides of the inequality” in both steps, as the DOE’s later guides word this step (“Subtract 4 from both sides of the inequality” in the 2024–2025 guide). The working and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding 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. Adding the same amount to both sides keeps the comparison true, like adding the same weight to both pans of a scale.
Dividing an inequality by a negative number
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, and c is a negative number. When you divide both sides by a negative number, the greater-than sign flips to less-than. That is because negatives reverse order: 2 is more than 1, but −2 is less than −1.

  1. Sort out the question

    An inequality compares two amounts. The sign > means is greater than. You need the graph of every x that makes −3x−7>−4-3x - 7 > -4 true. The graphs differ in two ways: where the open circle sits, at −1 or 1, and which way the shading goes.

  2. Add 7 to both sides

    if a>b, then a+c>b+c\text{if } a > b\text{, then } a + c > b + c

    −3x−7+7>−4+7-3x - 7 + 7 > -4 + 7

    −3x>3-3x > 3

    Adding 7 undoes the −7, so the x term is alone on the left.

  3. Divide by −3 and flip the sign

    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}

    −3x−3<3−3\frac{-3x}{-3} < \frac{3}{-3}

    x<−1x < -1

    So x is any number less than −1. Watch out: if you forget to flip the sign, you get x > −1, and that graph is choice C. Also, 3 ÷ (−3) is −1, not 1. Dropping that minus sign leads to B or D.

  4. Match the solution to a graph

    Less than −1 means to the left of −1 on the number line. The circle at −1 is open, or empty, because −1 itself does not work. The sign means greater than, not greater than or equal to.

    Graph A has an open circle at −1 and is shaded to the left. So the answer is A, the graph shaded left from an open circle at −1.

Check your answer

Test a number on each side of −1. For x = −2: −3(−2)−7=6−7=−1-3(-2) - 7 = 6 - 7 = -1, and −1 is greater than −4, so −2 works. For x = 0: −3(0)−7=−7-3(0) - 7 = -7, and −7 is less than −4, so 0 does not work. The shading must go left.

More questions like this

The original DOE question
Question 95 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 95. Open page 85 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 57 of 93

2023 Test A Math, question 40

Question

Ken has k video games, and Jeff has j video games. If Ken gives 6 video games to Jeff, Ken will have twice as many video games as Jeff. Which equation shows the relationship between k and j?

  • Ak−6=2(j+6)k - 6 = 2(j + 6)
  • Bk−6=2j+6k - 6 = 2j + 6
  • C2(k−6)=j2(k - 6) = j
  • D2(k−6)=j+62(k - 6) = j + 6
Show the answer

Correct answer: A

If Ken gives 6 video games to Jeff, the number Ken has is k−6k - 6, and the number Jeff has is j+6j + 6. After the exchange, Ken has twice as many video games as Jeff. The equation is:

k−6=2(j+6)k - 6 = 2(j + 6)

Question analysis

This algebra word problem says Ken has k video games and Jeff has j, and if Ken gives Jeff 6 games, Ken will have twice as many as Jeff. Students choose the equation that models the situation. Getting the comparison the right way round, and grouping each new total correctly, is where most errors happen.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.2a writing algebraic expressions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write equations from word problems. This one tracks two people's counts after a transfer and states a times-as-many comparison, so both counts change and the factor of 2 must go on the right side: typical but error-prone.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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 is all setup: no solving is needed, just an equation that says what the story says. The trade moves games out of one count and into the other, and the comparison has to be written with the 2 on the smaller side. The wrong equations reverse the comparison, forget Jeff's gain, or drop the parentheses, and all four look alike. Plugging in sample counts is the surest check.

  1. Turning words into math moderate Understanding the question

    After the gift both counts change, Ken's to k − 6 and Jeff's to j + 6, and “twice as many video games as Jeff” must double Jeff's whole new count, putting the 2 on his side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    All four equations share the pieces k − 6, j + 6 and a 2; the right one differs from 2(k − 6) = j + 6 only in where the 2 sits, and from k − 6 = 2j + 6 by parentheses. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    There are no numbers to check against, only k and j, so testing an equation means inventing counts that fit the story, such as Jeff with 4 games and Ken with 26. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Write each boy's number of games after the exchange as an expression, then translate "Ken will have twice as many as Jeff" into an equation and test it with sample numbers.

Traps

  • 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.
  • 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

Twice as many
A=2×BA = 2 \times B
A and B are two amounts. “A has twice as many as B” means A equals 2 times B. The 2 goes with the smaller amount, because that is the one you double to match the bigger one.
The distributive property
a(b+c)=ab+aca(b + c) = ab + ac
a, b and c are numbers. Multiplying a by a sum means multiplying a by each part, then adding. For example, 2(4 + 6) = 8 + 12 = 20.

  1. Sort out the question

    Ken has k video games, and Jeff has j. Then Ken gives 6 of his games to Jeff. After that, Ken has twice as many games as Jeff. You need the equation that links k and j.

  2. Write each new amount

    Ken gives away 6 games, so he has k − 6 left. Jeff gets those 6 games, so he has j + 6. Both amounts change, so use both new amounts.

  3. Write “twice as many”

    After the gift, Ken has more games. So Jeff’s new amount is the one that gets doubled.

    A=2×BA = 2 \times B

    k−6=2(j+6)k - 6 = 2(j + 6)

    Watch out: putting the 2 on Ken’s side, as in 2(k−6)=j+62(k - 6) = j + 6, would mean Jeff has twice as many. That is choice D.

  4. Keep the parentheses

    a(b+c)=ab+aca(b + c) = ab + ac

    2(j+6)=2j+122(j + 6) = 2j + 12

    Twice Jeff’s new amount is 2j + 12, not 2j + 6. Watch out: dropping the parentheses gives choice B. So the answer is A, k−6=2(j+6)k - 6 = 2(j + 6).

Check your answer

Make up numbers that fit the story. Say Jeff has 4 games and Ken has 26. After the gift, Ken has 20 and Jeff has 10, and 20 is twice 10. Choice A gives 26−6=2(4+6)26 - 6 = 2(4 + 6), or 20 = 20. Choice B gives 20 = 14, which is false.

More questions like this

The original DOE question
Question 97 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 97. Open page 85 of the handbook (PDF)

Question 58 of 93

2023 Test A Math, question 44

Question

For what value of x is the equation x5−4=3(4−2x)−1\frac{x}{5} - 4 = 3(4 - 2x) - 1 true?

  • A7511\frac{75}{11}
  • B7531\frac{75}{31}
  • C157\frac{15}{7}
  • D6531\frac{65}{31}
Show the answer

Correct answer: B

x5−4=3(4−2x)−1\frac{x}{5} - 4 = 3(4 - 2x) - 1 Apply the distributive property; multiply the 3 by 4 and −2x-2x.
x5−4=12−6x−1\frac{x}{5} - 4 = 12 - 6x - 1 Combine like terms.
x5−4=11−6x\frac{x}{5} - 4 = 11 - 6x Apply the additive inverse property; add 6x6x and 4 to both sides of the equation.
x5+6x=15\frac{x}{5} + 6x = 15 Get a common denominator.
x+30x5=15\frac{x + 30x}{5} = 15 Apply the multiplicative inverse property; multiply both sides of the equation by 5.
31x=7531x = 75 Apply the multiplicative inverse property; divide both sides of the equation by 31.
x=7531x = \frac{75}{31}

Question analysis

This equation question asks for the value of x that makes x/5 − 4 = 3(4 − 2x) − 1 true. Students use the distributive property, collect variable terms on one side, and clear a fraction before solving. Each wrong choice matches a common slip with the parentheses or with the fraction, so careful bookkeeping matters more than speed.

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 equations have the form px + q = r. This one has x on both sides, a fractional coefficient and a product to expand, which are grade 8 equation skills, and every wrong choice matches a plausible slip along the way.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
5 steps
Reading level
Question and choices: about grade 8, 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

Every step is a routine move in solving a linear equation, but there are five of them and the fraction raises the cost of each. The listed wrong answers come from distributing the 3 to only one term, clearing the fraction from one term only, and stretching the parentheses over the −1. Substituting the result back is slow with a denominator of 31, so careful step-by-step work matters more than checking.

  1. Error-prone arithmetic moderate Working it out

    The term x/5 sits beside whole-number terms and a negative −6x, so clearing the fraction means multiplying every term by 5, and the answer is the unfriendly 75/31. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps moderate Working it out

    Five moves in a row, distributing the 3, combining 12 and −1, collecting the x terms, clearing the fifth and dividing by 31, each give a slip a place to enter. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The fraction 75/11 shares its numerator with 75/31 and differs by one digit of the denominator, so only the coefficient of x, 6 1/5 rather than 2 1/5, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Distribute the 3 to both terms in the parentheses, combine constants, collect the x terms on one side, then multiply every term by 5 to clear the fraction; check by substituting.

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.

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)=ab−aca(b - c) = ab - ac
a, b and c are numbers or terms. Multiplying a by a difference means multiplying a by each part. The number outside must reach every term inside the parentheses.
Doing the same to both sides
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. Adding the same amount to both sides, or multiplying both sides by the same number, keeps it balanced.

  1. Sort out the question

    You need the value of x that makes both sides of the equation equal. The plan: clear the parentheses, gather the x terms on one side, and get rid of the fraction. Then divide.

  2. Clear the parentheses

    a(b−c)=ab−aca(b - c) = ab - ac

    3(4−2x)=12−6x3(4 - 2x) = 12 - 6x

    x5−4=12−6x−1=11−6x\frac{x}{5} - 4 = 12 - 6x - 1 = 11 - 6x

    Watch out: the 3 multiplies both 4 and 2x. Writing 12 − 2x leads to 7511\frac{75}{11}, choice A. The −1 is outside the parentheses, so the 3 does not touch it.

  3. Gather the x terms

    Add 6x to both sides to move the x terms to the left. Add 4 to both sides to move the plain numbers to the right.

    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

    x5+6x=11+4=15\frac{x}{5} + 6x = 11 + 4 = 15

  4. Clear the fraction

    Multiply both sides by 5. Every term gets multiplied, not only the fraction.

    5×x5+5×6x=5×155 \times \frac{x}{5} + 5 \times 6x = 5 \times 15

    x+30x=75x + 30x = 75

    31x=7531x = 75

    Watch out: multiplying only the fraction by 5 gives x + 6x = 15. That leads to 157\frac{15}{7}, choice C.

  5. Divide by 31

    x=7531x = \frac{75}{31}

    75 and 31 share no factor except 1, so the fraction is already simplified. So the answer is B, 7531\frac{75}{31}.

Check your answer

Put x=7531x = \frac{75}{31} back in. Then x5=1531\frac{x}{5} = \frac{15}{31} and 2x=150312x = \frac{150}{31}. Write 4 as 12431\frac{124}{31}. The left side is 1531−12431=−10931\frac{15}{31} - \frac{124}{31} = -\frac{109}{31}. On the right, 4−15031=−26314 - \frac{150}{31} = -\frac{26}{31}, and 3×(−2631)−1=−7831−3131=−109313 \times \left(-\frac{26}{31}\right) - 1 = -\frac{78}{31} - \frac{31}{31} = -\frac{109}{31}. Both sides match.

More questions like this

The original DOE question
Question 101 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 101. Open page 86 of the handbook (PDF)

Question 59 of 93

2023 Test B Math, question 5

Question

Solve the equation for x:

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

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

Show the answer

Correct answer: 7

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

(2149)(1448)x=0.875\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875

(37)(724)x=0.875\left(\frac{3}{7}\right)\left(\frac{7}{24}\right)x = 0.875

(18)x=0.875\left(\frac{1}{8}\right)x = 0.875

x=(0.875)(8)=7x = (0.875)(8) = 7

Question analysis

This equation puts x in a fraction with (21)(14) on top and (49)(48) below, set equal to the decimal 0.875. The grid-in asks for the value of x. Students do best to cancel common factors before multiplying anything out, and it helps to know the fraction that 0.875 equals.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.2c properties for multiplying and dividing
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 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

This is a one-unknown equation dressed in large numbers, and the work is in reducing them without slipping. Students who cancel early find that the whole coefficient is one eighth and that 0.875 is seven eighths, which makes the last step mental. Those who multiply out face a four-digit product and a decimal division, and with nothing to check against, an early slip goes unnoticed.

  1. Error-prone arithmetic moderate Working it out

    Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. No obvious method minor Working it out

    The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)

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

    The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

Strategy: Cancel before multiplying: reduce 21 over 49 and 14 over 48, multiply the simplified fractions, then undo that coefficient, writing 0.875 as a fraction if it helps.

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).
  • 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

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, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

  1. Sort out the question

    You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.

    Multiplying 49 by 48 makes a big number. It is easier to simplify first.

  2. Split the fraction into two

    Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.

    21×1449×48=2149×1448\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}

    Now the equation says these two fractions times x equal 0.875.

  3. Simplify each fraction

    To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.

    2149=371448=724\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}

    Watch out: 48 ÷ 2 is 24, not 12. Writing 712\frac{7}{12} leads to 3.5.

  4. Multiply the simple fractions

    37×724=3×77×24=324=18\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}

    The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get 18\frac{1}{8}. So the equation becomes:

    18×x=0.875\frac{1}{8} \times x = 0.875

  5. Solve for x

    One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.

    x=0.875×8=7x = 0.875 \times 8 = 7

    Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as 78\frac{7}{8}. One eighth of 7 is 78\frac{7}{8}, so x = 7 works. You can also multiply it all out: 21×14×7=2,05821 \times 14 \times 7 = 2{,}058 and 49×48=2,35249 \times 48 = 2{,}352. Then 2,058÷2,352=0.8752{,}058 \div 2{,}352 = 0.875.

More questions like this

The original DOE question
Question 62 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 62. Open page 192 of the handbook (PDF)

Question 60 of 93

2023 Test B Math, question 8

Question

Solve for x in terms of y if 2x+4=y2x + 4 = y.

  • Ax=y−42x = \frac{y - 4}{2}
  • Bx=2(y+4)x = 2(y + 4)
  • Cx=y+42x = \frac{y + 4}{2}
  • Dx=2(4−y)x = 2(4 - y)
Show the answer

Correct answer: A

2x+4=y2x + 4 = y Apply the additive inverse property; subtract 4 from both sides of the equation.
2x=y−42x = y - 4 Apply the multiplicative inverse property; divide both sides of the equation by 2.
x=y−42x = \frac{y - 4}{2}

Question analysis

Given the equation 2x + 4 = y, students solve for x in terms of y. This literal-equation question offers four choices that mix multiplying and dividing by 2 with adding and subtracting 4. Undoing the two operations on x in the right order, with the right inverse each time, is what it tests.

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
At grade 7 Grade 7 students solve two-step equations such as 2x + 4 = 10. Here the right side is the letter y, so the same two inverse steps leave an expression, a first taste of the grade 8 skill of rearranging.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 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

The two inverse steps are routine, subtract 4 and then divide by 2, but the letters hide whether they were done correctly. The main hazard is moving the 4 across the equals sign while keeping its plus sign, which produces an expression nearly identical to the right one. Substituting x = 1 and y = 6 tests every choice in seconds.

  1. A very close wrong answer moderate Choosing the answer

    (y − 4)/2 and (y + 4)/2 differ only in the sign before 4, and that sign depends on undoing the +4 by subtracting it from both sides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Variables instead of numbers minor Understanding the question

    With y on the right instead of a number, the answer is an expression, so there is no value to check unless the student substitutes a pair such as x = 1 and y = 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Treat y as if it were a number and undo the operations on x in reverse order, using the inverse of each, then check by substituting a value.

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.
  • 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 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. Doing the same thing to both sides keeps it balanced. Use this to undo, one at a time, each step that was done to x.

  1. Sort out the question

    The equation is 2x+4=y2x + 4 = y. “Solve for x in terms of y” means get x alone on one side. The other side will have y in it. Treat y as if it were a number.

  2. Undo the adding of 4

    Right now, x is doubled and then 4 is added. Undo those steps in reverse order. First, subtract 4 from both sides.

    2x+4−4=y−42x + 4 - 4 = y - 4

    2x=y−42x = y - 4

    Watch out: the 4 must be subtracted from y. Writing 2x=y+42x = y + 4 leads to choice C.

  3. Undo the doubling

    Next, divide both sides by 2. Divide the whole right side, not only the y.

    2x2=y−42\frac{2x}{2} = \frac{y - 4}{2}

    x=y−42x = \frac{y - 4}{2}

  4. Give the answer

    Watch out: choices B and D multiply by 2. But x was multiplied by 2, so you undo that by dividing. So the answer is A, x=y−42x = \frac{y - 4}{2}.

Check your answer

Pick a value, such as x = 1. Then y=2×1+4=6y = 2 \times 1 + 4 = 6. Put y = 6 into choice A: 6−42=1\frac{6 - 4}{2} = 1. That gives back x = 1. Choice C would give 6+42=5\frac{6 + 4}{2} = 5, which is wrong.

More questions like this

The original DOE question
Question 65 as printed on page 193 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 65. Open page 193 of the handbook (PDF)

Question 61 of 93

2023 Test B Math, question 17

Question

m∙t=72m \bullet t = 72

For the equation above, what is the sum of the values of m when t=1t = 1, when t=2t = 2, and when t=3t = 3?

  • A12
  • B72
  • C132
  • D135
Show the answer

Correct answer: C

First, find the value of m for each value of t:

For t=1t = 1:

1m=721m = 72

m=72m = 72

For t=2t = 2:

2m=722m = 72

m=36m = 36

For t=3t = 3:

3m=723m = 72

m=24m = 24

Then, add those three values together:

72+36+24=13272 + 36 + 24 = 132

Question analysis

The equation m • t = 72 links two variables, and the raised dot means multiplication. Students find m when t is 1, 2 and 3, and then add those three values of m. Each value is a one-step division, and the trap is combining the t values before dividing.

NYS standards
  • 6.EE.7 one-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Solving equations like 3m = 72 is grade 6 work. This item repeats that step three times and adds the results, more bookkeeping than difficulty, so it is below typical 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, 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 value of m comes from one division, so the work is bookkeeping rather than algebra. The trap is structural: adding 1, 2 and 3 first and dividing once feels efficient but answers a different question, since m does not change in step with t. Writing the three values of m in a row before adding them keeps the steps separate.

  1. Turning words into math minor Understanding the question

    “The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Divide 72 by each value of t in turn to get three separate values of m, then add the three quotients.

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

Division undoes multiplication
if a×b=c, then a=c÷b\text{if } a \times b = c \text{, then } a = c \div b
a, b and c are numbers, and b is not 0. If you know a product and one of the numbers that made it, divide to find the other one. For example, 4 × 5 = 20, so 4 = 20 ÷ 5.

  1. Sort out the question

    The equation is m∙t=72m \bullet t = 72. The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.

  2. Get m by itself

    Since m times t is 72, m is 72 divided by t.

    m=72÷tm = 72 \div t

    Now put in each value of t, one at a time.

  3. Find m for each value of t

    Each time, divide 72 by the value of t. A table keeps the three answers in order.

    Value of tValue of m
    1m=72÷1=72m = 72 \div 1 = 72
    2m=72÷2=36m = 72 \div 2 = 36
    3m=72÷3=24m = 72 \div 3 = 24

  4. Add the three values

    72+36+24=13272 + 36 + 24 = 132

    Watch out: don’t add the t values first. 72÷(1+2+3)=1272 \div (1 + 2 + 3) = 12 is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.

Check your answer

Check each pair by multiplying: 72×1=7272 \times 1 = 72, 36×2=7236 \times 2 = 72 and 24×3=7224 \times 3 = 72. All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.

More questions like this

The original DOE question
Question 74 as printed on page 196 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 74. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 62 of 93

2023 Test B Math, question 19

Question

Orange juice costs $2 per gallon, and grapefruit juice costs $3 per gallon. A recent shipment of equal amounts of the two types of juice costs a total of $250. How many gallons of orange juice are in the shipment?

  • A50
  • B62.5
  • C100
  • D125
Show the answer

Correct answer: A

Let x represent the number of gallons of orange juice. Since the shipment contains equal amounts of the two types of juice, x also represents the number of gallons of grapefruit juice. Write an equation to solve for x:

2x+3x=2502x + 3x = 250

5x=2505x = 250

x=50x = 50

Question analysis

Orange juice costs $2 a gallon and grapefruit juice costs $3 a gallon, and a shipment with equal amounts of each costs $250. Students find how many gallons of orange juice are in the shipment. The word problem leads to a simple equation; the key phrase is equal amounts of juice, not equal amounts of money.

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 Writing and solving an equation from a short word problem is standard grade 7 work. This one has a single unknown, since both juices come in the same number of gallons, and only two short steps.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 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 equation is short once the key phrase is read as equal gallons, since one gallon of each juice costs $5 together and $250 buys 50 such pairs. Read as equal spending, the same phrase leads to a listed answer that is not a whole number of gallons. The other trap is giving the gallons of both juices together when only orange juice is asked for.

  1. Turning words into math moderate Understanding the question

    “Equal amounts of the two types of juice” means the same number of gallons of each, x, so the total cost is 2x + 3x, not two equal halves of $250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Price one gallon of each juice together, then see how many such pairs $250 buys; that is the number of gallons of each kind.

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

Total cost
cost=price per gallon×number of gallons\text{cost} = \text{price per gallon} \times \text{number of gallons}
The price per gallon is what one gallon costs. Multiply it by the number of gallons to get what you pay in all. For example, 3 gallons at $2 a gallon cost $6.

  1. Sort out the question

    Orange juice costs $2 a gallon. Grapefruit juice costs $3 a gallon. The shipment has equal amounts of each, so it has the same number of gallons of each juice. It costs $250 in all. You need the number of gallons of orange juice.

  2. Write the cost of each juice

    Let x be the number of gallons of each juice. Use the cost formula for each one.

    cost=price per gallon×number of gallons\text{cost} = \text{price per gallon} \times \text{number of gallons}

    orange juice cost=2×x=2x\text{orange juice cost} = 2 \times x = 2x

    grapefruit juice cost=3×x=3x\text{grapefruit juice cost} = 3 \times x = 3x

  3. Add the costs and solve

    The two costs add up to $250. Two x’s plus three x’s make five x’s.

    2x+3x=2502x + 3x = 250

    5x=2505x = 250

    x=250÷5=50x = 250 \div 5 = 50

    There are 50 gallons of each juice.

  4. Give the answer

    Watch out: equal amounts means equal gallons, not equal dollars. Spending $125 on each juice gives 62.5 gallons, choice B. Also, 100 gallons is both juices together, choice C. So the answer is A, 50.

Check your answer

Find each cost. 50 gallons of orange juice cost 50×2=10050 \times 2 = 100 dollars. 50 gallons of grapefruit juice cost 50×3=15050 \times 3 = 150 dollars. Together, 100+150=250100 + 150 = 250 dollars, the total in the question.

More questions like this

The original DOE question
Question 76 as printed on page 196 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 76. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 63 of 93

2023 Test B Math, question 31

Question

If xy=30xy=30, what is the value of 10÷x10\div x in terms of y?

  • A110y\frac{1}{10}y
  • B13y\frac{1}{3}y
  • C3y3y
  • D10y10y
Show the answer

Correct answer: B

Solve xy=30xy=30, the given equation, for x:

xy=30xy=30

x=30yx=\frac{30}{y}

Then:

10÷x=10÷30y=10×y30=13y10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y

Question analysis

An algebra question: given that x times y equals 30, students express 10 divided by x using only y. It tests solving for one variable and dividing by an algebraic fraction. Working with letters instead of numbers, and dividing in the right order, are what trip students up.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.1 dividing fractions by fractions
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is three short moves, but each is a place to turn a fraction upside down: solving xy = 30 for x, dividing 10 by that fraction, and reducing 10/30. Students who reach for the bigger-over-smaller division they know from whole numbers land on 3y. Choosing two numbers with a product of 30, such as 3 and 10, and testing every choice settles the question in seconds.

  1. Variables instead of numbers moderate Understanding the question

    x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer minor Choosing the answer

    The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Solve xy = 30 for x, substitute that fraction into 10 ÷ x, and divide by multiplying by its reciprocal; or choose values of x and y with a product of 30 and test each choice.

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

Dividing by a fraction
a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}
a is the number you divide. b over c is the fraction you divide by: b is its top and c is its bottom. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. For example, 6 ÷ 1/2 = 6 × 2 = 12, because 12 halves fit in 6.

  1. Sort out the question

    xyxy means x times y. So x times y is 30. You need 10÷x10 \div x “in terms of y.” That means your answer can use y and numbers, but no x.

  2. Solve for x

    To swap x out, first get x by itself. Divide both sides of xy=30xy = 30 by y.

    x=30yx = \frac{30}{y}

    This says x is 30 divided by y.

  3. Put the fraction in place of x

    Replace x in 10÷x10 \div x with this fraction.

    10÷x=10÷30y10 \div x = 10 \div \frac{30}{y}

    Now use the rule for dividing by a fraction. Flip the fraction and multiply.

    10÷30y=10×y30=10y3010 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}

  4. Simplify the fraction

    Both 10 and 30 can be divided by 10.

    10y30=y3=13y\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y

    Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.

    So the answer is B, 13y\frac{1}{3}y.

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then 10÷x=10÷3=10310 \div x = 10 \div 3 = \frac{10}{3}. Choice B gives 13×10=103\frac{1}{3} \times 10 = \frac{10}{3}. They match. Choice C would give 30, and choice D would give 100.

More questions like this

The original DOE question
Question 88 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 88. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 64 of 93

2023 Test B Math, question 37

Question

Which expression is equivalent to 12x+7−(x−92)\frac{1}{2}x+7-\left(x-\frac{9}{2}\right)?

  • A12(x−5)\frac{1}{2}(x-5)
  • B12(x+5)\frac{1}{2}(x+5)
  • C−12(x+23)-\frac{1}{2}(x+23)
  • D−12(x−23)-\frac{1}{2}(x-23)
Show the answer

Correct answer: D

First, distribute the negative sign, which is the same as multiplying both terms in the parentheses by −1.

Next, combine like terms:

12x+7−(x−92)\frac{1}{2}x+7-\left(x-\frac{9}{2}\right)

12x+7−x+92\frac{1}{2}x+7-x+\frac{9}{2}

−12x+232-\frac{1}{2}x+\frac{23}{2}

Then, factor the common factor of −12-\frac{1}{2} from both terms, paying careful attention to the signs:

−12x+232=−12(x−23)-\frac{1}{2}x+\frac{23}{2}=-\frac{1}{2}(x-23)

Question analysis

An algebra question asking which factored expression equals one half x plus 7 minus the quantity x minus nine halves. Students distribute the subtraction, combine like terms with fractions, and match a factored form. Signs are the danger at every step, and the choices differ only in the signs and numbers inside the parentheses.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students expand and simplify linear expressions with fractions. Here a subtracted group must be distributed and the result factored by a negative fraction, two places where a sign is easily lost, a longer chain than typical grade 7 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 15, 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

Two separate sign decisions decide this question: distributing the subtraction over x − 9/2, and factoring a negative half out of +23/2. Students who handle the first correctly often stumble on the second and choose the look-alike with +23. Expanding each choice, rather than factoring by hand, is quicker and safer, since it turns the comparison into matching −1/2x + 23/2 term by term.

  1. Error-prone arithmetic moderate Working it out

    The minus before (x − 9/2) must change two signs, and the result mixes halves with whole numbers: 1/2x − x and 7 + 9/2 both need care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The simplified result, −1/2x + 23/2, appears nowhere; every choice is factored by ±1/2, so students must factor out −1/2 themselves or expand each choice to compare. (Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    The answer, −1/2(x − 23), and −1/2(x + 23) differ only in the sign inside the parentheses, and the simplified form’s +23/2 makes the plus sign look right. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Change the sign of both terms inside the parentheses, combine the x terms and the constants, then expand each choice to see which one matches rather than factoring yourself.

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.)
  • 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

Subtracting a group
−(b−c)=−b+c-(b - c) = -b + c
b and c are the two terms inside the parentheses. Terms are the parts joined by plus or minus signs. A minus sign in front of parentheses takes away the whole group. So it changes the sign of every term inside, not only the first one.
Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. Multiply a by each term inside. It works because a groups of (b + c) hold a of each. Use it to multiply out the choices and compare them.

  1. Sort out the question

    Equivalent means equal for every value of x. The choices are written in a different form from the problem. So simplify the problem first. Then multiply out the choices and compare.

  2. Remove the parentheses

    The minus sign in front of the parentheses takes away both terms inside. So x becomes −x, and −92-\frac{9}{2} becomes +92+\frac{9}{2}.

    12x+7−(x−92)=12x+7−x+92\frac{1}{2}x + 7 - \left(x - \frac{9}{2}\right) = \frac{1}{2}x + 7 - x + \frac{9}{2}

    Watch out: If you change only the sign of x, the numbers become 7−92=527 - \frac{9}{2} = \frac{5}{2}. That mistake leads to choice A or B.

  3. Combine the x terms

    Like terms have the same letter part. The x terms are 12x\frac{1}{2}x and −x. Write x as 22x\frac{2}{2}x so the fractions match.

    12x−22x=−12x\frac{1}{2}x - \frac{2}{2}x = -\frac{1}{2}x

    The result is negative, because you take away more x than you have.

  4. Combine the numbers

    Now combine the plain numbers, 7 and 92\frac{9}{2}. Write 7 as fourteen halves so the bottoms match. Then add.

    7+92=142+92=2327 + \frac{9}{2} = \frac{14}{2} + \frac{9}{2} = \frac{23}{2}

    The expression simplifies to −12x+232-\frac{1}{2}x + \frac{23}{2}.

  5. Multiply out choices C and D

    Only C and D start with −12-\frac{1}{2}, so test those. Use the distributive property. A negative times a negative is positive.

    −12(x+23)=−12x−232-\frac{1}{2}(x + 23) = -\frac{1}{2}x - \frac{23}{2}

    −12(x−23)=−12x+232-\frac{1}{2}(x - 23) = -\frac{1}{2}x + \frac{23}{2}

    The first line is choice C, and the second is choice D. Only D matches.

    Watch out: Choice C looks almost the same. But it gives −232-\frac{23}{2}, not +232+\frac{23}{2}.

    So the answer is D, −12(x−23)-\frac{1}{2}(x - 23).

Check your answer

Try x = 1 in the problem: 12+7−(1−92)=712+312=11\frac{1}{2} + 7 - \left(1 - \frac{9}{2}\right) = 7\frac{1}{2} + 3\frac{1}{2} = 11. Now put x = 1 into choice D: −12(1−23)=−12×(−22)=11-\frac{1}{2}(1 - 23) = -\frac{1}{2} \times (-22) = 11. Both give 11, so they match.

More questions like this

The original DOE question
Question 94 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 94. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 65 of 93

2023 Test B Math, question 38

Question

How is 1x−12x\frac{1}{x}-\frac{1}{2x} written as a single fraction? (Assume x≠0x\neq 0.)

  • A12x\frac{1}{2x}
  • B13x\frac{1}{3x}
  • C−1x-\frac{1}{x}
  • D−12x-\frac{1}{2x}
Show the answer

Correct answer: A

1x−12x=22x−12x=12x\frac{1}{x}-\frac{1}{2x}=\frac{2}{2x}-\frac{1}{2x}=\frac{1}{2x}

Question analysis

This algebra question asks students to write 1/x minus 1/(2x) as a single fraction, assuming x is not zero. It tests finding a common denominator when the denominators contain a variable. Getting both fractions over the same denominator before subtracting, and keeping track of the sign, are what matter.

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 subtract fractions with unlike denominators and work with linear expressions, but seldom see a variable in a denominator. The steps match those for numerical fractions, so it is a short task once the common denominator is found.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 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

This is subtraction of unlike fractions with a letter in the bottom, and every wrong choice is a familiar fraction error: combining the denominators straight across by adding or subtracting, or subtracting the numerators the wrong way round. Substituting x = 1 makes it 1 − 1/2 = 1/2, which rules out the negative choices and 1/3 at once.

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and no number checks the result unless students substitute a value such as x = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 1/x becomes 2/2x, and then the subtraction 2 − 1 must be done in the order written, first numerator minus second. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite 1/x with the denominator 2x by multiplying its top and bottom by 2, then subtract the numerators and keep the common denominator.

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.
  • 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

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top, or numerator. b is the bottom, or denominator. k is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, because k over k is 1.
Subtracting fractions with the same bottom
ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}
a and b are the tops. c is the shared bottom. When the bottoms match, the pieces are the same size. So subtract the tops and keep the bottom.

  1. Sort out the question

    You need to write 1x−12x\frac{1}{x} - \frac{1}{2x} as one fraction. The problem says x is not 0. That matters because you can’t divide by 0.

  2. Find a common bottom

    To subtract fractions, their bottoms must match. 2x is 2 times x. So multiply the top and bottom of 1x\frac{1}{x} by 2.

    1x=1×2x×2=22x\frac{1}{x} = \frac{1 \times 2}{x \times 2} = \frac{2}{2x}

    Now both fractions have the bottom 2x.

  3. Subtract the tops

    Subtract the tops in order. Keep the bottom.

    22x−12x=2−12x=12x\frac{2}{2x} - \frac{1}{2x} = \frac{2 - 1}{2x} = \frac{1}{2x}

    Watch out: Never add or subtract the bottoms. Adding them gives 13x\frac{1}{3x}, choice B. Subtracting them gives −1x-\frac{1}{x}, choice C. And keep the order: 2 − 1, not 1 − 2. The wrong order gives choice D.

    So the answer is A, 12x\frac{1}{2x}.

Check your answer

Try x = 1. The problem becomes 1−12=121 - \frac{1}{2} = \frac{1}{2}. Choice A gives 12×1=12\frac{1}{2 \times 1} = \frac{1}{2}. It matches. Choices B, C and D give 13\frac{1}{3}, −1 and −12-\frac{1}{2}.

More questions like this

The original DOE question
Question 95 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 95. Open page 202 of the handbook (PDF)

Question 66 of 93

2023 Test B Math, question 45

Question

The Barnes family and the Ramirez family each have a pizza of equal size. The first pizza is cut into 18 equal slices, and the second pizza is cut into 15 equal slices. If the Barnes family eats 11 slices from the first pizza, what is the greatest number of whole slices from the second pizza that the Ramirez family can eat without eating a greater percentage of a pizza than the Barnes family ate?

  • A6
  • B7
  • C8
  • D9
Show the answer

Correct answer: D

The fraction of the pizza that the Barnes family eats is 1118\frac{11}{18}. Let x represent the number of slices of pizza that the Ramirez family eats. Then the fraction of the pizza that the Ramirez family eats is x15\frac{x}{15}. Write an inequality and solve for x:

x15≤1118\frac{x}{15}\leq\frac{11}{18}

x≤11×1518x\leq\frac{11\times 15}{18}

x≤16518x\leq\frac{165}{18}

x≤916x\leq 9\frac{1}{6}

So, the greatest number of whole slices of pizza the Ramirez family can eat is 9.

Question analysis

Two pizzas of the same size are cut into 18 and 15 equal slices, and one family eats 11 slices of the first. Students find the most whole slices of the second pizza the other family can eat without eating a larger share. It tests comparing fractions with different denominators and rounding to a whole number in the right direction.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 inequality problems usually ask for the greatest whole number that fits a budget. This one wraps that idea in a long stem about shares of two pizzas sliced differently, with several conditions to keep straight at once.
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 11, 72 words.
Students, last 30 days
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 math is one inequality between fractions, but the long final sentence hides it, so students who skim fall back on comparing slice counts. Reasoning that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, is the strongest pull. Comparing 11/18 with x/15, or turning 11/18 into about 61% and taking 61% of 15, keeps the comparison about shares of a pizza.

  1. Complex sentences moderate Understanding the question

    The question is one 42-word sentence ending in a nested condition, “without eating a greater percentage of a pizza than the Barnes family ate”, a reading hurdle rather than more math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

  2. Turning words into math moderate Understanding the question

    The limit of no “greater percentage of a pizza” must become the inequality x/15 ≤ 11/18, comparing fractions of two pizzas cut into different numbers of slices, not slice counts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Solving gives x ≤ 11 × 15 ÷ 18 = 165/18, a division by 18 that leaves 9 1/6, which must then be cut back to whole slices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Barnes family's share as a fraction of their pizza, then find the largest whole number of fifteenths that does not exceed it, by cross-multiplying or by converting to decimals.

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.
  • 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

Fraction of a pizza
fraction eaten=slices eatenslices in the pizza\text{fraction eaten} = \frac{\text{slices eaten}}{\text{slices in the pizza}}
The top counts the slices eaten. The bottom counts all the equal slices the pizza was cut into. The fraction tells how much of the whole pizza was eaten. Use fractions to compare pizzas cut into different numbers of slices.

  1. Sort out the question

    The two pizzas are the same size. The Barnes pizza has 18 slices, and they eat 11. The Ramirez pizza has 15 slices, so its slices are bigger. You need the most whole slices the Ramirez family can eat without eating a bigger share of their pizza.

  2. Find the Barnes family’s share

    Their share is the slices they ate out of all the slices in their pizza. Use the fraction formula.

    fraction eaten=1118\text{fraction eaten} = \frac{11}{18}

    The Barnes family ate 1118\frac{11}{18} of their pizza.

  3. Write an inequality

    Let x be the number of Ramirez slices. Their share is x15\frac{x}{15}. It must be no more than the Barnes share. The sign ≤ means less than or equal to.

    x15≤1118\frac{x}{15} \leq \frac{11}{18}

  4. Solve for x

    Multiply both sides by 15 to get x by itself.

    x≤11×1518=16518x \leq \frac{11 \times 15}{18} = \frac{165}{18}

    Now divide. 18 goes into 165 nine times, since 18 × 9 = 162. There are 3 left over.

    x≤9318=916x \leq 9\frac{3}{18} = 9\frac{1}{6}

  5. Pick the whole number

    x must be a whole number no bigger than 9169\frac{1}{6}. So the most is 9 slices. 10 slices would be too many.

    Watch out: The Ramirez pizza has 3 fewer slices, but that does not mean they eat 3 fewer. Their slices are bigger. Eating 11 − 3 = 8 slices is choice C, which is not the most they can eat.

    So the answer is D, 9.

Check your answer

Use decimals. The Barnes share is 11÷18≈0.61111 \div 18 \approx 0.611. Nine Ramirez slices are 9÷15=0.69 \div 15 = 0.6, just under it. Ten slices are 10÷15≈0.66710 \div 15 \approx 0.667, too much. So 9 is the most.

More questions like this

The original DOE question
Question 102 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 102. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 67 of 93

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is 35\frac{3}{5}?

  • A8
  • B10
  • C14
  • D20
Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to 35\frac{3}{5} and solve the equation for w:

16+w14+10+16+w=35\frac{16+w}{14+10+16+w}=\frac{3}{5}

16+w40+w=35\frac{16+w}{40+w}=\frac{3}{5}

5(16+w)=3(40+w)5(16+w)=3(40+w)

80+5w=120+3w80+5w=120+3w

2w=402w=40

w=20w=20

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards
  • 6.SP.8a uniform probability models
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
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 8, 34 words.
Students, last 30 days
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 tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

  1. Turning words into math moderate Understanding the question

    “How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P(white)=white marblesall marblesP(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}
P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

  1. Sort out the box

    The box has 14 red, 10 black and 16 white marbles.

    14+10+16=4014 + 10 + 16 = 40

    That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is 35\frac{3}{5}.

  2. Write the new probability

    Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.

    16+w40+w=35\frac{16 + w}{40 + w} = \frac{3}{5}

    Watch out: Taking 35\frac{3}{5} of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.

  3. Cross multiply

    Fractions are hard to solve with. Clear them with cross multiplication.

    5×(16+w)=3×(40+w)5 \times (16 + w) = 3 \times (40 + w)

    Then multiply each outside number by both terms inside the parentheses.

    80+5w=120+3w80 + 5w = 120 + 3w

  4. Solve for w

    Subtract 3w from both sides. Then subtract 80 from both sides.

    2w=402w = 40

    Divide both sides by 2.

    w=20w = 20

    You need to add 20 white marbles.

    So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: 3660=35\frac{36}{60} = \frac{3}{5}. It works.

More questions like this

The original DOE question
Question 114 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 68 of 93

2018 Test A Math, question 3

Question

For what value of ww is 4w=2w−84w=2w-8?

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

Show the answer

Correct answer: -4

4w=2w−84w=2w-8

2w=−82w=-8

w=−4w=-4

Question analysis

A one-line algebra grid-in: find the value of w that makes 4w equal to 2w minus 8. Students solve a linear equation with the variable on both sides by collecting the w terms and dividing. Errors come from handling the subtraction sign or stopping before w stands alone.

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 Equations with the variable on both sides are formally a grade 8 topic, but this one takes two short steps with whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 7, 7 words.
Students, last 30 days
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 of the easiest questions on the test: collect the w terms on one side and divide. Errors come from carelessness rather than method, such as dropping the negative sign or gridding −8 after the first step because it already looks like a finished answer. Substituting the result back into both sides of the equation takes seconds and catches both.

  1. Error-prone arithmetic minor Working it out

    The only constant is −8, so the answer is negative; dividing −8 by the coefficient 2 is the step where the sign is most often lost. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The grid-in needs the negative sign filled in its own column, and there are no choices to show that the answer lies below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract 2w from both sides so the variable appears only once, divide by the remaining coefficient, and check by substituting into both sides.

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.)
  • 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

Keeping an equation balanced
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. If you subtract the same thing from both sides, it stays balanced. The same is true if you add the same thing, or divide both sides by the same number that is not zero.

  1. Sort out the equation

    An equation says two things are equal. You need the value of w that makes 4w4w equal to 2w−82w - 8. Here 4w means 4 times w, and 2w means 2 times w. The answer may be a negative number, so keep track of the signs.

  2. Get w on one side

    w is on both sides. Subtract 2w from both sides. Then w is only on the left, and the equation stays balanced.

    4w−2w=2w−8−2w4w - 2w = 2w - 8 - 2w

    2w=−82w = -8

    Now 2 times w is −8.

  3. Divide both sides by 2

    To undo times 2, divide both sides by 2. A negative number divided by a positive number is negative.

    w=−8÷2=−4w = -8 \div 2 = -4

    Watch out: keep the minus sign. Dropping it gives 4. Also, don’t stop at 2w=−82w = -8 and grid −8. So you would grid -4, with the minus sign in its own column.

Check your answer

Put −4 back in for w. Left side: 4×(−4)=−164 \times (-4) = -16. Right side: 2×(−4)−8=−8−8=−162 \times (-4) - 8 = -8 - 8 = -16. Both sides are −16, so −4 works.

More questions like this

The original DOE question
Question 60 as printed on page 81 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 60. Open page 81 of the handbook (PDF)

Question 69 of 93

2018 Test A Math, question 6

Question

The set of possible values of mm is {5,7,9}\left \{ 5,7,9 \right \}. What is the set of possible values of kk if 2k=m+32k=m+3?

  • A{3,4,5}\left \{ 3,4,5 \right \}
  • B{4,5,6}\left \{ 4,5,6 \right \}
  • C{8,10,12}\left \{ 8,10,12 \right \}
  • D{10,14,18}\left \{ 10,14,18 \right \}
Show the answer

Correct answer: B

2k=m+32k=m+3

so

k=m+32k=\cfrac{m+3}{2}

Substitute each value of m to find the values of k:

k=5+32=82=4k=\cfrac{5+3}{2}=\cfrac{8}{2}=4

k=7+32=102=5k=\cfrac{7+3}{2}=\cfrac{10}{2}=5

k=9+32=122=6k=\cfrac{9+3}{2}=\cfrac{12}{2}=6

The set k is { 4, 5, 6}.

Question analysis

An algebra question about substitution: m can be 5, 7 or 9, and an equation links k to m. Students find every possible value of k by substituting each value of m and solving. The curly-brace set notation can look unfamiliar, and each value has to be carried all the way through the solve.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Each substitution leads to a one-step grade 6 equation, easier than a typical grade 7 two-step problem. What adds demand is doing it three times and reading the answer as a set of values, which grade 7 rarely uses.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 9, 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 equation is simple, and the main hazard is stopping one step short: m + 3 gives 8, 10 and 12, a listed choice even though those are values of 2k. Checking a single value of m is the other risk, since k = 4 and k = 5 both appear in the first two sets. Students who substitute all three values and halve each result finish quickly.

  1. Variables instead of numbers minor Understanding the question

    There is no single unknown: m stands for the set {5, 7, 9}, and the answer is the matching set for k, so the equation has to be applied to three values. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the equation so k equals m plus 3, all divided by 2, then evaluate it for each of the three values of m.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Getting k by itself
k=m+32k = \frac{m + 3}{2}
k and m are the letters from the equation 2k = m + 3. The fraction bar means divide. If 2 times k equals m + 3, then k is half of m + 3. You get this by dividing both sides of the equation by 2.

  1. Sort out the question

    The letter m can be 5, 7 or 9. The curly braces { } show a set, which is a list of numbers. The equation 2k=m+32k = m + 3 links k to m. You need every value k can have, one for each value of m.

  2. Get k by itself

    2k means 2 times k. To undo times 2, divide both sides by 2. Doing the same thing to both sides keeps the equation true.

    k=m+32k = \frac{m + 3}{2}

  3. Try each value of m

    Put each value of m into the formula. Add 3 first, then divide by 2.

    k=5+32=82=4k = \frac{5 + 3}{2} = \frac{8}{2} = 4

    k=7+32=102=5k = \frac{7 + 3}{2} = \frac{10}{2} = 5

    k=9+32=122=6k = \frac{9 + 3}{2} = \frac{12}{2} = 6

    Watch out: 8, 10 and 12 are the values of m + 3, which equal 2k, not k. Stopping before you divide by 2 gives choice C.

  4. Pick the matching set

    The values of k are 4, 5 and 6. Watch out: try all three values of m. If you only try m = 5, you get k = 4, and choice A has a 4 too. So the answer is B, {4, 5, 6}.

Check your answer

Put each k back into the equation. 2×4=8=5+32 \times 4 = 8 = 5 + 3. 2×5=10=7+32 \times 5 = 10 = 7 + 3. 2×6=12=9+32 \times 6 = 12 = 9 + 3. All three work, so the set is right.

More questions like this

The original DOE question
Question 63 as printed on page 82 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 63. Open page 82 of the handbook (PDF)

Question 70 of 93

2018 Test A Math, question 7

Question

7+(3n+6)−(4n+8)=7+\left ( 3n+6 \right )-\left ( 4n+8 \right )=

  • A5−n5-n
  • B5+n5+n
  • C21−n21-n
  • D21+n21+n
Show the answer

Correct answer: A

7+3n+6−4n−8=(7+6−8)+(3n−4n)=5−n7+3n+6-4n-8=(7+6-8)+(3n-4n)=5-n

Question analysis

An expression question: simplify a sum and difference of two linear expressions in n. Students remove the parentheses, subtract every term inside the second group, and combine like terms. The minus sign in front of the second parentheses is the trap, since it has to change the sign of both terms inside.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simplifying linear expressions with a subtracted group is standard grade 7 work. This is a typical textbook exercise in length, with the usual sign trap and no extra steps or real-world context.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 5 words.
Corrected from the DOE handbook
  • The handbook’s answer key for this question 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

There are only five terms, but two signs decide the answer: the minus in front of the second parentheses and the sign left after 3n − 4n. Each wrong choice comes from one or both of those sign decisions, so all of them look plausible, and checking needs care rather than time. Even the DOE handbook keyed the version with the wrong sign on n.

  1. Error-prone arithmetic moderate Working it out

    The minus sign before (4n + 8) has to reach both terms, making −4n and −8, and 3n − 4n leaves a negative n term. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    5 − n and 5 + n are written alike and share the constant 5; only the sign of the n term, from 3n − 4n, separates them, and the DOE handbook's own key picked the wrong one. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite the subtraction as adding the opposite of each term in the second parentheses, then group the constants and the n terms separately.

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.)

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Answer key: The handbook (its question 64) keys this question as F, which is 5 + n (choice B here), in its answer key and in the heading of its explanation. Its own working gives 7 + 3n + 6 − 4n − 8 = 5 − n, which is the handbook’s choice E and choice A here, and that is correct. The key here is A.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a group in parentheses
−(a+b)=−a−b-(a + b) = -a - b
a and b are the terms inside the parentheses. A term is a number, a letter, or a number times a letter. The minus sign in front means you subtract the whole group. So you subtract every term inside, not just the first one.
Combining like terms
3n−4n=(3−4)n3n - 4n = (3 - 4)n
n stands for an unknown number, and 3n means 3 times n. Like terms have the same letter part, like 3n and 4n. To combine them, add or subtract the numbers in front and keep the letter. Plain numbers combine only with plain numbers.

  1. Sort out the question

    You need to simplify the expression, which means write it in a shorter form. An expression is a math phrase made of numbers, letters and signs. Here n stands for an unknown number. Find the choice that matches the short form.

  2. Remove the parentheses

    The first group has a plus sign in front. Adding a group changes nothing, so +(3n+6)+(3n + 6) is just +3n+6+3n + 6. The second group has a minus sign in front. You must subtract both terms inside it.

    −(4n+8)=−4n−8-(4n + 8) = -4n - 8

    7+3n+6−4n−87 + 3n + 6 - 4n - 8

    Watch out: the minus reaches the 8 too. Adding the 8 instead gives 21 for the plain numbers. That leads to choice C or D.

  3. Combine the plain numbers

    Group the numbers without a letter. Work them out from left to right.

    7+6−8=57 + 6 - 8 = 5

    The plain numbers make 5.

  4. Combine the n terms

    You have 3 n’s, and you take away 4 n’s. The numbers in front give 3 − 4 = −1. So you are left with −1n, which is written −n.

    3n−4n=(3−4)n=−1n=−n3n - 4n = (3 - 4)n = -1n = -n

    Watch out: it is easy to write +n here. That gives 5 + n, choice B.

  5. Put the two parts together

    Join the number part and the n part. Adding a negative is the same as subtracting.

    5+(−n)=5−n5 + (-n) = 5 - n

    So the answer is A, 5 − n.

Check your answer

Pick any number for n and test it. Try n = 1. The original: 7+(3+6)−(4+8)=7+9−12=47 + (3 + 6) - (4 + 8) = 7 + 9 - 12 = 4. Choice A: 5−1=45 - 1 = 4. They match. Choice B gives 5+1=65 + 1 = 6, which does not.

More questions like this

The original DOE question
Question 64 as printed on page 82 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 64. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 71 of 93

2018 Test A Math, question 9

Question

Jenny starts a game with twice as many marbles as Keiko. Jenny gives Keiko 5 marbles, but she still has 10 more than Keiko. How many marbles did Jenny have to start with?

  • A25
  • B30
  • C35
  • D40
Show the answer

Correct answer: D

Set up some equations. Jenny (J) has twice as many marbles as Keiko (K):

J=2KJ=2K

Jenny gives Keiko 5 marbles, so now they each have: J − 5 and K + 5 marbles. Jenny still has 10 more than Keiko:

J−5=(K+5)+10J-5=(K+5)+10

To find how many marbles Jenny had to start with, solve J = 2K for K and substitute that into the second equation:

In equation J = 2K, solve for K:

K=J2K=\frac{J}{2}

Substitute J2\frac{J}{2} in for K.

J−5=(K+5)+10J-5=(K+5)+10

J−5=(J2+5)+10J-5=(\frac{J}{2}+5)+10

J−5=J2+15J-5=\frac{J}{2}+15

J=J2+20J=\frac{J}{2}+20

J2=20\frac{J}{2}=20

J=40J=40 marbles

Question analysis

A two-variable word problem about marbles: Jenny starts with twice as many as Keiko, gives Keiko 5, and still has 10 more. Students set up equations for before and after the trade and solve for Jenny's starting number. The trade changes both players' counts, and the question asks about the start, not the end.

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. This needs two related unknowns and a before-and-after change, which leads to a system of equations or a variable on both sides, both grade 8 skills.
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 4, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Most of the difficulty is in setting up the equations, not solving them. The gift of 5 marbles changes both counts, so the gap between the girls shrinks by 10, not 5, and students who take the marbles from Jenny without giving them to Keiko reach a consistent but wrong answer. Two other choices are true counts from after the trade, so finishing the algebra is not the same as answering the question. Testing the choices backward is a fast, safe alternative.

  1. Turning words into math major Understanding the question

    Three relations hide in two sentences: "twice as many", the gift of 5 that lowers Jenny's count and raises Keiko's, and "still has 10 more", which compares the counts after the gift. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of moderate Working it out

    Each girl has a count before and after the trade, four quantities in all, and the question asks for only one of them, Jenny's at the start. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Work backward from the choices: halve each one for Keiko's start, move 5 marbles from Jenny to Keiko, and check whether Jenny ends with exactly 10 more.

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

Twice as many
J=2KJ = 2K
J is the number of marbles Jenny starts with, and K is the number Keiko starts with. Twice as many means 2 times as many, so J equals 2 times K. Because J and 2K are equal, you can write 2K anywhere you see J.
Keeping an equation balanced
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. If you subtract the same thing from both sides, it stays balanced. The same is true if you add the same thing, or divide both sides by the same number that is not zero.

  1. Sort out the question

    Let K be Keiko’s marbles at the start. Jenny starts with twice as many, so she has 2K. Then Jenny gives Keiko 5 marbles. After that, “she still has 10 more than Keiko”. You need Jenny’s count at the start, before the gift.

  2. Write the counts after the gift

    The gift changes both counts. Jenny loses 5, and Keiko gains 5.

    • Jenny now has 2K−52K - 5.
    • Keiko now has K+5K + 5.

    Jenny has 10 more than Keiko. So Jenny’s count equals Keiko’s count plus 10.

    2K−5=(K+5)+102K - 5 = (K + 5) + 10

    Watch out: Keiko gains the 5 marbles too. Taking 5 from Jenny but not adding 5 to Keiko leads to 30, choice B.

  3. Solve for K

    Add the numbers on the right side first.

    2K−5=K+152K - 5 = K + 15

    Subtract K from both sides. Then add 5 to both sides. Each move keeps the equation balanced.

    K−5=15K - 5 = 15

    K=20K = 20

    Keiko started with 20 marbles.

  4. Find Jenny’s starting count

    Jenny started with twice as many as Keiko.

    J=2K=2×20=40J = 2K = 2 \times 20 = 40

    Watch out: 25 is Keiko’s count after the gift, choice A. 35 is Jenny’s count after the gift, choice C. The question asks about the start. So the answer is D, 40.

Check your answer

Start: Jenny has 40 and Keiko has 20, and 40 is twice 20. After the gift, Jenny has 35 and Keiko has 25. 35−25=1035 - 25 = 10, so Jenny still has 10 more. Every fact fits.

More questions like this

The original DOE question
Question 66 as printed on page 82 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 66. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 72 of 93

2018 Test A Math, question 14

Question

Which number line below shows the solution to the inequality −4<x2<2-4< \frac{x}{2}< 2 ?

  • ANumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −2 and 1, with the segment between them shaded.
  • BNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −4 and 2, with the segment between them shaded.
  • CNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the line shaded to the left of −8 and to the right of 4.
  • DNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the segment between them shaded.
Show the answer

Correct answer: D

Multiply each term by 2 to eliminate the fraction, and isolate xx:

−4(2)<(x2)(2)<2(2)-4(2)< \left(\frac{x}{2}\right)(2)< 2(2)

−8<x<4-8< x< 4

Therefore, xx must be between −8 and 4.

Question analysis

An inequality question: choose the number line that shows the solution of a compound inequality in which x divided by 2 lies between −4 and 2. Students undo the division in every part and match the graph's endpoints and shading. The four number lines differ in their endpoints and in which part of the line is shaded.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve two-step inequalities and graph them on a number line. This compound inequality needs only one operation applied to all three parts at once, so it matches typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 11 words.
Students, last 30 days
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 one step, multiplying through by 2, so most of the work is matching the result to a picture. The choices pair each common slip with a graph: the bounds left as written, the bounds halved, and the right bounds shaded on the wrong side. Reading the open circles against the labeled ticks at −5 and 5 settles it.

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

    The four number lines are labeled only at −10, −5, 0, 5 and 10, so open circles at −8 and 4, or at −4 and 2, have to be located by counting ticks. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic minor Working it out

    All three parts of −4 < x/2 < 2 must be multiplied by 2, and the negative bound −4 becomes −8; halving instead gives −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply all three parts of the inequality by 2 to isolate x, then pick the graph whose open circles and shading match the new bounds.

Traps

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

Multiplying an inequality by a positive number
if a<b, then a×2<b×2\text{if } a < b \text{, then } a \times 2 < b \times 2
a and b are numbers, and the less-than sign means the left side is smaller. If you multiply both sides by the same positive number, like 2, the smaller side stays smaller. So the sign keeps pointing the same way. A negative number would flip it, but 2 is positive.

  1. Sort out the inequality

    An inequality compares two amounts that may not be equal. The sign << means less than. Here x2\frac{x}{2} means x divided by 2. The inequality says x divided by 2 is between −4 and 2. You need the values of x itself. Then you pick the number line that shows them.

  2. Multiply all three parts by 2

    To undo dividing by 2, multiply by 2. Multiply all three parts, so the whole statement stays true.

    −4×2<x2×2<2×2-4 \times 2 < \frac{x}{2} \times 2 < 2 \times 2

    −8<x<4-8 < x < 4

    Watch out: multiply, don’t divide. Dividing gives −2 and 1, choice A. Leaving the ends at −4 and 2 is choice B.

  3. Read what the answer means

    x is greater than −8 and less than 4. So x is any number between −8 and 4. x cannot equal −8 or 4 exactly. On a number line, that means open circles at −8 and 4, with the part between them shaded.

  4. Find the matching number line

    The number lines are labeled only at −10, −5, 0, 5 and 10. Count the tick marks. −8 is 3 ticks left of −5, and 4 is 1 tick left of 5. Watch out: choice C has circles at −8 and 4 but shades the parts outside them. So the answer is D.

Check your answer

Test a number from the shaded part, like 0. 0÷2=00 \div 2 = 0, and 0 is between −4 and 2. Now test 6, which is outside. 6÷2=36 \div 2 = 3, and 3 is not less than 2. So only the part between −8 and 4 works.

More questions like this

The original DOE question
Question 71 as printed on page 84 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 71. Open page 84 of the handbook (PDF)

Question 73 of 93

2018 Test A Math, question 20

Question

Mrs. Cranston bought five bottles of water for $0.90 each and 8 pounds of meat. She paid a total of $26.90 for these items, not including tax. What was the price per pound of the meat?

  • A$2.80
  • B$3.25
  • C$14.40
  • D$22.40
Show the answer

Correct answer: A

Let xx be the price per pound for the meat. Set up an equation to show what Mrs. Cranston spent:

5(0.90)+8x=26.905(0.90)+8x=26.90

4.50+8x=26.904.50+8x=26.90

8x=22.408x=22.40

x=2.80x=2.80

The price per pound for the meat is $2.80.

Question analysis

This money word problem has a shopper buying five bottles of water at a set price plus 8 pounds of meat, for a known total. Students write and solve a two-step equation to find the price per pound of the meat. The decimals call for care, and the question wants a price per pound rather than a total.

NYS standards
  • 7.EE.4a two-step equations
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Word problems that lead to an equation like px + q = r are the core of grade 7 algebra. This is a standard example with money decimals and no extra twist.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 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

This is a two-step money problem: take away the water, then divide by the pounds. The listed wrong answers come from doing one of those steps wrongly or not at all, especially stopping at the total cost of the meat, which looks like a finished dollar amount. Checking that 8 pounds at the chosen price plus $4.50 makes $26.90 confirms the answer.

  1. Turning words into math minor Understanding the question

    "five bottles of water for $0.90 each" has to become 5 × 0.90 before the total is split, and the unknown is a price per pound, not the meat's total cost. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Every amount is a decimal of dollars, so 26.90 − 4.50 and 22.40 ÷ 8 must be lined up and worked carefully without a calculator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the cost of all five bottles from the total, then divide what is left by the 8 pounds of meat.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
C=p×qC = p \times q
C is the total cost. p is the price of one item, or of one pound. q is the quantity, how many items or pounds you buy. Each one costs p, so q of them cost p times q.

  1. Sort out the question

    Mrs. Cranston bought five bottles of water “for $0.90 each”. She also bought 8 pounds of meat. Everything cost $26.90 in all. You need the price per pound of the meat, the cost of just 1 pound.

  2. Find the cost of the water

    Each bottle costs $0.90, and there are 5 bottles. Use the cost formula.

    C=p×qC = p \times q

    C=0.90×5=4.50C = 0.90 \times 5 = 4.50

    The water cost $4.50. Watch out: each means every bottle costs $0.90. Taking away just one $0.90 leads to $3.25, choice B.

  3. Find the cost of the meat

    Take the water away from the total. What is left paid for the meat.

    26.90−4.50=22.4026.90 - 4.50 = 22.40

    All 8 pounds of meat cost $22.40. Watch out: this is the cost of all the meat, not of one pound. Stopping here gives choice D.

  4. Find the price per pound

    Use the cost formula again. Now C is 22.40 and q is 8 pounds. Divide to find p.

    22.40=p×822.40 = p \times 8

    p=22.40÷8=2.80p = 22.40 \div 8 = 2.80

    Each pound costs $2.80. So the answer is A, $2.80.

Check your answer

Put it all back together. Meat: 8×2.80=22.408 \times 2.80 = 22.40. Water: 4.50. Total: 22.40+4.50=26.9022.40 + 4.50 = 26.90. That is exactly what she paid.

More questions like this

The original DOE question
Question 77 as printed on page 85 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 77. Open page 85 of the handbook (PDF)

Question 74 of 93

2018 Test A Math, question 30

Question

Today, Tien’s age is 14\frac{1}{4} of Jordan’s age. In 2 years, Tien’s age will be 13\frac{1}{3} of Jordan’s age. How old is Jordan today? 

  • A4 years old
  • B6 years old
  • C12 years old
  • D16 years old
Show the answer

Correct answer: D

First, set up an equation to express Tien’s age (TT) and Jordan’s age (JJ) today:

T=14JT=\frac{1}{4}J

Two years from now, Tien’s age will be T+2T+2, and Jordan’s age will be J+2J+2.

Set up an equation about the relationship between Tien’s age and Jordan’s age in two years:

T+2=13(J+2)T+2=\frac{1}{3}(J+2)

Solve the above equation for TT:

T=13(J+2)−2T=\frac{1}{3}(J+2)-2

Now set the two equations equal to each other and solve for JJ:

14J=13(J+2)−2\frac{1}{4}J=\frac{1}{3}(J+2)-2

14J=13J−43\frac{1}{4}J=\frac{1}{3}J-\frac{4}{3}

−112J=−43-\frac{1}{12}J=-\frac{4}{3}

J=−43(−121)J=-\frac{4}{3}\left(-\frac{12}{1}\right)

J=16J=16

Question analysis

This age word problem says Tien is one fourth of Jordan's age today and will be one third of Jordan's age in two years, then asks for Jordan's current age. Students must turn two fractional relationships into equations and solve them together. Both people age by the same two years, and it is easy to solve for the wrong person.

NYS standards
  • 8.EE.8c systems in real-world problems
  • 8.EE.7b linear equations with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 equations have one unknown on one side, like 3x + 5 = 20. This one links two unknown ages with fractional coefficients and leads to a variable on both sides, which is grade 8 algebra.
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 1, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The difficulty is in the setup: two fractional relationships at two different times, with both people aging. Once the equations are right, the algebra is short, but it ends with two ages and only one is asked for. Testing the choices against both conditions is a reliable alternative that avoids the algebra entirely.

  1. Turning words into math major Understanding the question

    The second sentence must become T + 2 = (J + 2)/3; writing J/3, or forgetting that Jordan also ages two years, breaks the equation. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Substituting J/4 for T leaves J/4 + 2 = (J + 2)/3, an equation with fractions on both sides that must be cleared, usually by multiplying by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Let Jordan's age be J, write Tien's age as J/4, and set J/4 + 2 equal to one third of (J + 2); or test each choice against both conditions.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a×(b+c)=a×b+a×ca \times (b + c) = a \times b + a \times c
a, b and c are any numbers. To multiply a sum by a number, multiply each part of the sum, then add. It works because a groups of (b + c) is the same as a groups of b plus a groups of c.

  1. Sort out what you know

    Today, Tien’s age is 14\frac{1}{4} of Jordan’s age. That means Jordan is 4 times as old as Tien.

    In 2 years, both of them will be 2 years older. Then Tien’s age will be 13\frac{1}{3} of Jordan’s. So Jordan will be 3 times as old as Tien. You need Jordan’s age today.

  2. Name the ages with a letter

    Let t stand for Tien’s age today. Jordan is 4 times as old, so his age is 4t, which means 4 times t. Add 2 years to each age:

    PersonTodayIn 2 years
    Tienttt+2t + 2
    Jordan4t4t4t+24t + 2

    Watch out: Jordan gets 2 years older too. Do not forget to add 2 to his age.

  3. Write an equation for later

    In 2 years, Jordan’s age will be 3 times Tien’s age. Write that as an equation:

    4t+2=3×(t+2)4t + 2 = 3 \times (t + 2)

    Use the distributive property on the right side. Multiply both t and 2 by 3:

    4t+2=3t+64t + 2 = 3t + 6

  4. Solve for Tien’s age

    Take 3t away from both sides. Doing the same thing to both sides keeps them equal:

    t+2=6t + 2 = 6

    Now take 2 away from both sides:

    t=4t = 4

    So Tien is 4 years old today. Watch out: 4 is Tien’s age, not Jordan’s. Choice A, 4 years old, is a trap.

  5. Find Jordan’s age

    Jordan is 4 times as old as Tien:

    4×4=164 \times 4 = 16

    So the answer is D, 16 years old.

Check your answer

Put 16 back into the story. Today, Tien is 16 ÷ 4 = 4. In 2 years, Tien is 6 and Jordan is 18. Is 6 one third of 18? Yes, because 18 ÷ 3 = 6. Choice C fails: Tien would be 3, then 5, and 5 is not a third of 14.

More questions like this

The original DOE question
Question 87 as printed on page 87 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 87. Open page 87 of the handbook (PDF)

Question 75 of 93

2018 Test A Math, question 32

Question

The least of 5 consecutive integers is ll, and the greatest is gg. What is the value of l+g2\frac{l+g}{2} in terms of ll?

  • A2l2l
  • B3l3l
  • Cl+2l+2
  • Dl+5l+5
Show the answer

Correct answer: C

The first integer is ll, so the second is l+1l+1, the third is l+2l+2, then l+3l+3, and finally l+4l+4.

Since gg is the fifth and greatest of the integers,

g=l+4g=l+4

Substitute l+4l+4 for gg and simplify:

l+g2=l+l+42=2l+42=l+2\frac{l+g}{2}=\frac{l+l+4}{2}=\frac{2l+4}{2}=l+2

Question analysis

This algebra question about five consecutive integers calls the least one l and the greatest g, then asks for (l + g)/2 written in terms of l. Students must express g using l and simplify a fraction that contains a variable. Counting how far the fifth integer is from the first is where many students slip.

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 combine like terms in expressions such as 3(x + 4) − x. This item first asks them to build the expression from a description of consecutive integers, then simplify it, with only letters to work with.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 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

The algebra is short once g is written as l + 4, so most errors come from the fence-post count or from testing numbers carelessly. The wrong choices in front of the answer are exactly the ones a single test value cannot rule out, which punishes students who stop at the first match. Using two test values, or simplifying (2l + 4)/2 directly, avoids both traps.

  1. Variables instead of numbers moderate Understanding the question

    Every expression is in l, so the simplification of (l + l + 4)/2 has to be done with letters and cannot be checked against a single number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The description "the least of 5 consecutive integers is l, and the greatest is g" must become g = l + 4, four steps up, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Every choice must be checked minor Choosing the answer

    A single test value fits two choices: l = 1 makes 3l and l + 2 both 3, and l = 2 makes 2l and l + 2 both 4. (Embretson & Wetzel, 1987)

Strategy: Write the integers as l, l + 1, and so on up to the fifth, substitute for g and simplify; then test two different starting values, since one value can match a wrong choice by coincidence.

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).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest of 5 consecutive integers
g=l+4g = l + 4
l is the least (smallest) integer and g is the greatest. Consecutive integers go up by 1 each time. Five integers have only 4 jumps from the first to the last, so you add 4, not 5.
Dividing a sum
a+bc=ac+bc\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. You must divide every part of the top, not just one.

  1. Sort out the words

    Integers are whole numbers, like −2, 0 and 7. Consecutive integers come one right after another, like 8, 9, 10. There are 5 of them here.

    The smallest is called l and the biggest is called g. You must write l+g2\frac{l + g}{2} using only the letter l.

  2. Write all five integers

    Each integer is 1 more than the one before:

    l, l+1, l+2, l+3, l+4l,\ l+1,\ l+2,\ l+3,\ l+4

    The fifth one is the greatest, so g = l + 4. Watch out: five integers have only four jumps between them. Writing g = l + 5 leads to choice D.

  3. Put l + 4 in place of g

    Swap g for l + 4 in the fraction:

    l+g2=l+l+42=2l+42\frac{l + g}{2} = \frac{l + l + 4}{2} = \frac{2l + 4}{2}

    Here l + l is 2l, which means 2 times l.

  4. Split the fraction

    Divide each part of the top by 2:

    2l2+42=l+2\frac{2l}{2} + \frac{4}{2} = l + 2

    So the answer is C, l+2l + 2.

Check your answer

Test two starting numbers. If l = 1, the integers are 1 to 5, and (1 + 5) ÷ 2 = 3, which is 1 + 2. If l = 10, they are 10 to 14, and (10 + 14) ÷ 2 = 12, which is 10 + 2. Use two tests, because l = 1 also makes 3l equal 3, choice B.

More questions like this

The original DOE question
Question 89 as printed on page 87 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 89. Open page 87 of the handbook (PDF)

Question 76 of 93

2018 Test A Math, question 39

Question

Suppose M=wxM=\frac{w}{x} , N=yzN=\frac{y}{z} , and w,x,yw,x,y, and zz do not equal 0. What is MN\frac{M}{N} in terms of w,x,yw,x,y, and zz ?

  • Awxyz\frac{wx}{yz}
  • Bwyxz\frac{wy}{xz}
  • Cwzxy\frac{wz}{xy}
  • Dxywz\frac{xy}{wz}
Show the answer

Correct answer: C

It is easier to rewrite

MN\frac{M}{N}

as

M÷NM\div N

since  they are both fractions.

M÷N=wx÷yz=wx×zy=wzxyM\div N=\frac{w}{x}\div \frac{y}{z}=\frac{w}{x}\times \frac{z}{y}=\frac{wz}{xy}

Question analysis

This algebra question defines M and N as fractions built from the letters w, x, y and z, and asks for M divided by N in terms of those letters. Students must divide one fraction by another using only variables. Without numbers to check against, it is easy to multiply instead of divide or flip the wrong fraction.

NYS standards
  • 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 divide numeric fractions such as 2/3 ÷ 4/5. This item applies the same rule with four letters instead of numbers, which adds abstraction but no new steps.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 10, 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

This is fraction division in disguise, and students who recognize M/N as M ÷ N need one rule. The difficulty is that every choice is built from the same four letters, so a wrong rule still produces something that looks like a reasonable answer. Substituting small numbers, such as w = 1, x = 2, y = 3 and z = 4, checks the result in seconds.

  1. Variables instead of numbers moderate Understanding the question

    All four letters stand for unknown nonzero numbers, so dividing w/x by y/z must be done symbolically, with no value to check the result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    All four choices arrange w, x, y and z as one fraction, and xy/wz is wz/xy turned over, so only carrying out the division in the right order tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite M/N as M ÷ N, then multiply by the reciprocal of the second fraction; substituting small numbers for the letters 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.
  • 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 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 0. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing by a number is the same as multiplying by 1 over that number.
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 and b and d are the bottoms. Multiply the tops together and the bottoms together. Use it after you flip the second fraction.

  1. Sort out what you know

    M is the fraction w over x. N is the fraction y over z. The letters stand for numbers that are not 0.

    The fraction bar in MN\frac{M}{N} means M divided by N. So you must divide one fraction by another.

  2. Write it as a division

    Rewrite the big fraction as a division. Then put in the fractions for M and N. This lets you use the rule for dividing fractions:

    MN=M÷N=wx÷yz\frac{M}{N} = M \div N = \frac{w}{x} \div \frac{y}{z}

  3. Flip the second fraction

    Use the rule for dividing fractions. Keep the first fraction. Flip only the second one, and change divide to multiply:

    wx×zy\frac{w}{x} \times \frac{z}{y}

    Watch out: flipping the first fraction instead gives choice D, which is N ÷ M. Multiplying without flipping at all gives choice B.

  4. Multiply across

    Multiply the tops and the bottoms:

    w×zx×y=wzxy\frac{w \times z}{x \times y} = \frac{wz}{xy}

    Here wz means w times z. So the answer is C, wzxy\frac{wz}{xy}.

Check your answer

Try small numbers. Let w = 1, x = 2, y = 3 and z = 4. Then M = 1/2 and N = 3/4. So M ÷ N = 1/2 × 4/3 = 4/6. Choice C gives (1 × 4) ÷ (2 × 3) = 4/6 too. They match.

More questions like this

The original DOE question
Question 96 as printed on page 88 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 96. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 77 of 93

2018 Test A Math, question 43

Question

wx=yz\frac{w}{x}=\frac{y}{z}

In the equation above, w,x,yw,x,y , and zz are positive numbers. Which of these is equal to zz?

  • Axx
  • Bxyxy
  • Cwxy\frac{w}{xy}
  • Dxyw\frac{xy}{w}
Show the answer

Correct answer: D

Solve the equation for zz:

wx=yz\frac{w}{x}=\frac{y}{z}

wz=xywz=xy

z=xywz=\frac{xy}{w}

Question analysis

This algebra question gives the proportion w/x = y/z, with all four letters positive, and asks which expression equals z. Students must solve the equation for one letter in terms of the others. With only letters on the page, it is easy to multiply when you should divide or to stop one step early.

NYS standards
  • 8.EE.7 linear equations in one variable
  • 6.EE.7 one-step equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students cross-multiply numeric proportions such as 3/4 = 9/x. Solving a proportion made only of letters for one of them is a literal equation, a grade 8 algebra skill, though it takes just two steps.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 10, 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

The method, cross-multiply and divide, is the same one students use on numeric proportions, but letters hide whether each step went the right way. The two most attractive wrong choices are both one step from the answer: one never divides, and the other divides upside down. A quick check with 2/3 = 4/6, where z = 6, settles it.

  1. Variables instead of numbers moderate Understanding the question

    The proportion has only letters, so cross-multiplying to wz = xy and dividing by w must be done without numbers to show whether w goes on top or bottom. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    The choices w/(xy) and xy/w use the same three letters with the fraction turned over, so only dividing in the right direction after wz = xy separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cross-multiply to clear both fractions, then divide both sides by whatever multiplies z; check with a true proportion such as 2/3 = 4/6.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a, b, c and d are numbers that are not 0. When two fractions are equal, each top times the other bottom gives the same answer. It works because you multiply both sides by both bottoms, which clears the fractions.

  1. Sort out what you need

    The equation says w over x equals y over z. All four letters are positive numbers. You need to get z alone on one side. Then the other side shows what z equals.

  2. Cross-multiply the fractions

    Multiply each top by the other fraction’s bottom:

    w×z=x×yw \times z = x \times y

    So wz = xy, and z is no longer on the bottom. Watch out: stopping here and picking xy gives choice B. z is not alone yet.

  3. Get z alone

    z is being multiplied by w. To undo that, divide both sides by w. Doing the same to both sides keeps them equal:

    w×zw=x×yw\frac{w \times z}{w} = \frac{x \times y}{w}

    z=xywz = \frac{xy}{w}

    Watch out: dividing the wrong way round gives w over xy, choice C. So the answer is D, xyw\frac{xy}{w}.

Check your answer

Use a true pair of equal fractions: 2/3 = 4/6. So w = 2, x = 3, y = 4 and z = 6. Choice D gives 3 × 4 ÷ 2 = 6, which is z. Choice C gives 2 ÷ 12, which is not 6.

More questions like this

The original DOE question
Question 100 as printed on page 89 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 100. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 78 of 93

2018 Test A Math, question 55

Question

If 4x−3y=124x-3y=12 , what is xx in terms of yy?

  • Ax=34y+12x=\frac{3}{4}y+12
  • Bx=−34y+12x=-\frac{3}{4}y+12
  • Cx=34y+3x=\frac{3}{4}y+3
  • Dx=−34y+3x=-\frac{3}{4}y+3
Show the answer

Correct answer: C

4x−3y=124x-3y=12

4x=12+3y4x=12+3y

x=34y+124x=\frac{3}{4}y+\frac{12}{4}

x=34y+3x=\frac{3}{4}y+3

Question analysis

This algebra question gives the linear equation 4x − 3y = 12 and asks for x in terms of y. Students must isolate the x term and then divide to solve for x. With two variables and a negative coefficient, small sign and division slips each lead to a different wrong choice.

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 one-variable equations like 4x − 3 = 12. Rearranging an equation with two variables to express one in terms of the other is grade 8 algebra, with sign and division slips to avoid.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 10, 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

The solution is two moves, add 3y to both sides and divide everything by 4, yet each move has a classic slip attached. The four choices are the four combinations of getting the sign right or wrong and dividing the 12 or not, so every partial mistake has a home. Checking one point that satisfies the original equation catches any of them.

  1. Variables instead of numbers moderate Understanding the question

    With two variables there is no number to check against unless students invent one; the point x = 6, y = 4 satisfies the equation and separates the choices. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    The choices x = 3/4 y + 3 and x = −3/4 y + 3 are identical except for the sign of the y term, which only correctly moving −3y across the equals sign settles. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Error-prone arithmetic minor Working it out

    Dividing 12 + 3y by 4 must reach both terms, and the negative coefficient −3 has to change sign when it moves, two slips in a short line. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Move the y term to the right side, watching its sign, so only the x term remains on the left, then divide the entire right side, term by term, by 4.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced
a=b means a+c=b+ca = b \text{ means } a + c = b + c
a and b are the two sides of an equation, and c is any number. An equation is like a balance. If you do the same thing to both sides, such as adding 3y or dividing by 4, they stay equal.
Dividing a sum
a+bc=ac+bc\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. It works because sharing a + b into c groups shares a and b each into c groups.

  1. Sort out what you need

    The question asks for x in terms of y. That means get x alone on one side. The other side may have y and numbers, but no x. Start with 4x−3y=124x - 3y = 12.

  2. Move the y term

    The 3y is being subtracted on the left. Add 3y to both sides to clear it:

    4x−3y+3y=12+3y4x - 3y + 3y = 12 + 3y

    4x=12+3y4x = 12 + 3y

    Watch out: the term was minus 3y. Adding 3y to both sides makes it plus 3y on the right. Keeping the minus sign leads to choice D.

  3. Divide both sides by 4

    x is being multiplied by 4. Divide both sides by 4 to undo it:

    x=12+3y4x = \frac{12 + 3y}{4}

  4. Split the fraction

    Divide each part of the top by 4:

    x=124+3y4=3+34yx = \frac{12}{4} + \frac{3y}{4} = 3 + \frac{3}{4}y

    Watch out: the 12 must be divided by 4 too. Leaving it as 12 gives choice A. The choices put the y term first. So the answer is C, x=34y+3x = \frac{3}{4}y + 3.

Check your answer

Pick y = 4. Choice C gives x = 3 + 3 = 6. Put x = 6 and y = 4 into the start: 4 × 6 − 3 × 4 = 24 − 12 = 12. It works. Choice D gives x = 0, and 4 × 0 − 12 = −12, not 12.

More questions like this

The original DOE question
Question 112 as printed on page 91 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 112. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 79 of 93

2018 Test B Math, question 2

Question

147−x12=12\frac{147-x}{12}=12

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

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

Show the answer

Correct answer: 3

147−x12=12\frac{147-x}{12}=12

147−x=144147-x=144

x=3x=3

Question analysis

This grid-in asks students to solve a two-step linear equation in which an expression containing x is divided by 12 and set equal to 12. The algebra is short, but x is being subtracted from 147, so undoing the operations in the right order and keeping the minus sign straight is where students slip.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 This is exactly the two-step equation practice of grade 7: undo a division, then undo a subtraction. The only twist is that the unknown is being subtracted, so the last step needs care with its sign.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 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

This is a two-move equation that most students finish quickly: multiply both sides by 12, then find what must be taken from 147 to leave 144. Points are lost only at the last move, where the minus sign in front of x invites 144 − 147 or 147 + 144. Substituting the result back into the fraction catches either slip in a few seconds.

  1. Error-prone arithmetic minor Working it out

    After multiplying by 12, the unknown sits behind a minus sign in 147 − x = 144, so students must ask what is taken from 147 rather than subtract in the order the numbers appear. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply both sides by 12 to clear the fraction, then ask what number must be taken away from 147 to leave that product.

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

Undoing division
ab=c means a=b×c\frac{a}{b} = c \text{ means } a = b \times c
a is the top of the fraction, b is the bottom, and c is the result. Division and multiplication undo each other. So if a divided by b gives c, then b times c gives back a. This is the same as multiplying both sides by b.

  1. Read the equation in words

    The equation says: start with 147, take away a mystery number x, then divide by 12. The result is 12.

    You need to find x. To get x alone, undo the steps in reverse order. Dividing by 12 was done last, so undo it first.

  2. Undo the division by 12

    Multiply both sides by 12. Whatever you do to one side, you must do to the other. That keeps the equation balanced.

    147−x12=12\frac{147 - x}{12} = 12

    147−x=12×12=144147 - x = 12 \times 12 = 144

    Watch out: Do not divide 12 by 12. That gives 147 − x = 1 and the wrong answer 146.

  3. Find what was taken away

    Now ask: 147 minus what number leaves 144? The gap between the two numbers is x.

    x=147−144=3x = 147 - 144 = 3

    Watch out: Do not work out 144 − 147. That gives −3. Since 147 is bigger than 144, x must be positive.

    So you would grid 3.

Check your answer

Put 3 back in for x. First, 147−3=144147 - 3 = 144. Then 144÷12=12144 \div 12 = 12. That matches the right side of the equation, so x = 3 is correct.

More questions like this

The original DOE question
Question 59 as printed on page 153 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 59. Open page 153 of the handbook (PDF)

Question 80 of 93

2018 Test B Math, question 6

Question

If x=9x=9 and y=−7y=-7, what is the value of x(x−2y)x(x-2y)?

  • A18
  • B45
  • C144
  • D207
Show the answer

Correct answer: D

x(x−2y)=9[9−2(−7)]=9(9+14)=9(23)=207x(x-2y)=9[9-2(-7)]=9(9+14)=9(23)=207

Question analysis

This algebra question asks students to evaluate the expression x(x − 2y) when x is 9 and y is negative 7. It practices substitution and the order of operations with integers. The risk is the sign, since subtracting twice a negative number turns into addition, and dropping that sign or the factor of 2 is easy.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 7.NS.2a rational number multiplication
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting values, including a negative one, into an expression is everyday grade 7 work with integers. This has three short steps and one sign decision, matching a typical classroom exercise.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 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

Substitution itself is routine, so the question tests one thing: subtracting a negative inside the parentheses. A student who turns 9 − 2(−7) into 9 − 14 gets −45 and may then choose 45, since it is the only choice that looks close. Working the parentheses first and writing −(−14) as +14 before multiplying by 9 avoids every wrong choice.

  1. Error-prone arithmetic moderate Working it out

    The middle step is 9 − 2(−7): the product 2(−7) is −14, and subtracting it means adding 14, so two negatives must be handled in one short move. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work inside the parentheses first: find 2y, subtract it from x by rewriting minus a negative as plus, and multiply that result by x last.

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
a−(−b)=a+ba - (-b) = a + b
a and b are numbers. Taking away a negative number is the same as adding the positive number. For example, 9 − (−14) = 9 + 14. Think of taking away a debt: you end up richer.
Multiplying by a negative
positive×negative=negative\text{positive} \times \text{negative} = \text{negative}
When one number is positive and the other is negative, their product is negative. The product is the answer to a multiplication. So 2 × (−7) = −14.

  1. Write down what you know

    You know x = 9 and y = −7. You need the value of x(x−2y)x(x - 2y).

    A number written right next to parentheses means multiply. So this is x times the amount inside the parentheses.

  2. Put in the numbers

    Replace each letter with its value. Put −7 in parentheses so you do not lose its minus sign.

    x(x−2y)=9×(9−2×(−7))x(x - 2y) = 9 \times (9 - 2 \times (-7))

  3. Work inside the parentheses

    Always work inside parentheses first. Multiply before you subtract. A positive times a negative is negative, so 2×(−7)=−142 \times (-7) = -14.

    9−(−14)=9+14=239 - (-14) = 9 + 14 = 23

    Taking away a negative is the same as adding.

    Watch out: If you write 9 − 14 = −5, you end with −45. Then choice B, 45, looks close, but it is wrong.

  4. Multiply by x last

    Now multiply 9 by the 23 you found inside the parentheses.

    9×23=2079 \times 23 = 207

    Watch out: Do not drop the 2. Using 9 − (−7) = 16 gives 144, choice C.

    So the answer is D, 207.

Check your answer

Check the inside another way. 9−2×(−7)9 - 2 \times (-7) means start at 9 and move 14 to the right on the number line. You land on 23. Then 9×23=180+27=2079 \times 23 = 180 + 27 = 207.

More questions like this

The original DOE question
Question 63 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 63. Open page 154 of the handbook (PDF)

Question 81 of 93

2018 Test B Math, question 8

Question

M=3N=P4=Q+5=R7>0M=3N=\frac{P}{4}=Q+5=\frac{R}{7}> 0

Based on the statement above, which variable has the greatest value?

  • AMM
  • BNN
  • CPP
  • DRR
Show the answer

Correct answer: D

Break the equations apart to each equal MM:

M=3NM=3N

M=P4M=\frac{P}{4}

M=Q+5M=Q+5

M=R7M=\frac{R}{7}

Pick a number to substitute into the equations, and solve the equations to find the values of MM, NN, PP, QQ, and RR.

Let M=2M=2. Since all the equations are equal to 2, substitute 2 to find each variable.

M=3NM=3N

2=3N2=3N

23=N\frac{2}{3}=N

M=P4M=\frac{P}{4}

2=P42=\frac{P}{4}

8=P8=P

M=Q+5M=Q+5

2=Q+52=Q+5

−3=Q-3=Q

M=R7M=\frac{R}{7}

2=R72=\frac{R}{7}

14=R14=R

Variable RR has the greatest value.

Question analysis

This algebra question sets five expressions equal to one another and to a positive number, then asks which variable has the greatest value. It tests whether students can rewrite each variable in terms of one shared value and compare them. What makes it tricky is that there are no numbers to compute, so each variable's size has to be reasoned out.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students rewrite expressions to see how quantities relate, usually one equation at a time. This chain of five linked expressions, with no numbers to compute and a positivity condition to use, is a reasoning task beyond routine grade 7 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 10, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards a strategy, not a calculation: once the shared value is named, each variable follows in one line and R is clearly largest. Without that step, students judge by how the expressions look, and the look is backwards, since the 3 in 3N makes N small and the 7 under R makes R large. The positivity condition at the end is easy to skip but is what fixes the order.

  1. Variables instead of numbers major Understanding the question

    There are no numbers to compute: five variables are tied together in the one chain M = 3N = P/4 = Q + 5 = R/7, and each must be compared through the shared value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. No obvious method moderate Working it out

    Nothing says how to compare; the efficient move is to name the common value, or pick one such as 12, and rewrite each variable, which students must think of themselves. (Schoenfeld, 1985)

  3. Much to keep track of minor Working it out

    Five relations and the condition > 0 must be held at once; the condition matters, since with a negative common value the order of M, N, P and R would reverse. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Call the shared value k, a positive number, and rewrite each variable as an expression in k; or substitute a convenient positive value such as 12 and compare the results.

Traps

  • 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

Undoing division
ab=c means a=b×c\frac{a}{b} = c \text{ means } a = b \times c
a is the top of the fraction, b is the bottom, and c is the result. Multiplying undoes dividing. So if a divided by b is c, then a is b times c.
Undoing multiplication
b×a=c means a=c÷bb \times a = c \text{ means } a = c \div b
b is the number multiplied by a, and c is the result. Dividing undoes multiplying. So to get a alone, divide c by b.

  1. Understand the chain

    The statement is a chain of equal signs. It says five expressions all have the same value. The last part says that value is greater than 0, so it is positive.

    You need to find which letter has the largest value.

  2. Pick a value to test

    All five expressions are equal, so give them one shared value. Pick a positive number that 3 divides evenly. That way N will be a whole number.

    Let the shared value be 12. Then M=12M = 12, and each of the other four expressions also equals 12.

  3. Solve for each letter

    Undo each operation to find each letter. Divide to undo times. Multiply to undo divide. Subtract to undo plus.

    3N=12, so N=12÷3=43N = 12 \text{, so } N = 12 \div 3 = 4

    P4=12, so P=12×4=48\frac{P}{4} = 12 \text{, so } P = 12 \times 4 = 48

    Q+5=12, so Q=12−5=7Q + 5 = 12 \text{, so } Q = 12 - 5 = 7

    R7=12, so R=12×7=84\frac{R}{7} = 12 \text{, so } R = 12 \times 7 = 84

  4. Compare the values

    The values are M = 12, N = 4, P = 48, Q = 7 and R = 84. R is the largest.

    Watch out: 3N does not make N big. It means 3 times N equals M, so N is only one third of M. That trap points to choice B. Also, P is big, but R is bigger. Check every letter before you pick choice C.

    So the answer is D, R.

Check your answer

Think without numbers. R divided by 7 is M, so R is 7 times M. P is only 4 times M. N is a third of M, and Q is 5 less than M. Since M is positive, 7 times M is the biggest.

More questions like this

The original DOE question
Question 65 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 65. Open page 154 of the handbook (PDF)

Question 82 of 93

2018 Test B Math, question 10

Question

To make party invitations, Macie could buy a package of paper for $10.50, or she could buy x individual sheets of the same paper for $0.15 each. What is the largest value of x that would make buying the individual sheets less expensive than buying the package?

  • A60
  • B65
  • C69
  • D70
Show the answer

Correct answer: C

Set up an inequality to compare the costs:

0.15x < 10.50
x < 70

Therefore, 70 individual sheets of paper would cost $10.50, so 69 is the greatest number of individual sheets of paper that Macie can buy that would be less expensive than the package.

Question analysis

This inequality word problem compares buying a package of paper with buying single sheets at 15 cents each, and asks for the largest number of sheets for which the single sheets cost less. Students write and solve an inequality with decimals. The tricky part is interpreting the solution as a whole number of sheets.

NYS standards
  • 7.EE.4b two-step inequalities
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an inequality from a real-world situation is a grade 7 standard. This one is short, but it asks students to interpret a strict inequality as a count of sheets, the kind of interpretation grade 7 tests expect.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 10, 45 words.
Corrected from the DOE handbook
  • An inequality sign 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 computation leads straight to 70, and the whole question is about what to do with that number. Students who treat the boundary as the answer forget that equal cost is not less cost, while the other choices only catch those who misdivide or ignore the word largest. The question rewards checking the boundary case in the story: 70 sheets cost exactly the package price.

  1. A very close wrong answer moderate Choosing the answer

    70 and 69 sheets differ by one, and 70 is exactly where the two costs meet at $10.50; only the strict “less expensive” separates them, ruling out 70. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math minor Understanding the question

    The words “less expensive than buying the package” must become a strict inequality, 0.15x < 10.50, and “largest value” must then pick the biggest whole number below the boundary. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Dividing 10.50 by 0.15 means shifting the decimal point in both numbers, a step where a misplaced digit gives a wrong boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide the package price by the price of one sheet to see where the two costs meet, then find the largest whole number of sheets that still costs less than $10.50.

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).

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook sets up 0.15x ≤ 10.50 and x ≤ 70, which would allow 70 sheets, and then says that 70 sheets cost exactly $10.50. Buying the sheets has to be less expensive than the package, so the site uses strict inequalities, 0.15x < 10.50 and x < 70, which make 69 the largest number of sheets.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
cost=price of one×number bought\text{cost} = \text{price of one} \times \text{number bought}
Price of one is what a single sheet costs, here $0.15. Number bought is how many sheets, here x. Multiplying gives the total you pay, because you pay the same price for every sheet.

  1. Sort out the question

    The package costs $10.50. Single sheets cost $0.15 each. Macie buys x sheets.

    You need the largest x that makes the sheets cost less than the package. Costing the same is not enough.

  2. Write the inequality

    An inequality is a math sentence that compares two amounts. The sign < means is less than.

    The cost of x sheets is 0.15 times x. It must be less than 10.50.

    cost=price of one×number bought\text{cost} = \text{price of one} \times \text{number bought}

    0.15×x<10.500.15 \times x < 10.50

  3. Find where the costs are equal

    First find how many sheets cost exactly $10.50. Divide the package price by the price of one sheet. Multiply both numbers by 100 to clear the decimals.

    10.50÷0.15=1050÷15=7010.50 \div 0.15 = 1050 \div 15 = 70

    So 70 sheets cost exactly $10.50.

  4. Pick the largest number below 70

    x must be less than 70. You can only buy whole sheets. The largest whole number below 70 is 69.

    Watch out: 70 sheets cost the same as the package, not less. That is why choice D is wrong. Choices A and B are cheaper too, but they are not the largest.

    So the answer is C, 69.

Check your answer

Test both numbers. 69×0.15=10.3569 \times 0.15 = 10.35, and $10.35 is less than $10.50. But 70×0.15=10.5070 \times 0.15 = 10.50, which is not less. So 69 is the largest number that works.

More questions like this

The original DOE question
Question 67 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 67. Open page 154 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 83 of 93

2018 Test B Math, question 18

Question

PRICES FOR NEWSPAPER ADVERTISING

Page Space Price
14\frac{1}{4} page $200
12\frac{1}{2} page $350
full page $600

The table above shows prices for newspaper advertising. A store purchased 14\frac{1}{4} pages, 12\frac{1}{2} pages, and full pages of page space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?

  • A1341\frac{3}{4} pages
  • B10 pages
  • C161216\frac{1}{2} pages
  • D171217\frac{1}{2} pages
Show the answer

Correct answer: D

The question says that an equal number (xx) of each type of page space was purchased. To find the number of each type of page space that was purchased, multiply the price per type by xx and set it equal to the total amount spent, then solve for xx:

200x+350x+600x=11,500200x+350x+600x=11,500

1,150x=11,5001,150x=11,500

x=10x=10

Thus, the store purchased 10 units of each type of page space. To find the total amount of page space purchased, multiply each type of page space by 10, and add:

(10×14 page)+(10×12 page)+(10×1 page)\left(10\times\frac{1}{4}\text{ page}\right)+\left(10\times\frac{1}{2}\text{ page}\right)+(10\times 1\text{ page})

=1712 pages=17\frac{1}{2}\text{ pages}

Question analysis

This word problem gives a table of newspaper advertising prices for quarter-page, half-page and full-page ads, and says a store bought equal numbers of each for a set total. Students write and solve an equation for how many ads of each size were bought. The question wants total page space, which takes a second calculation with fractions.

NYS standards
  • 7.EE.4a two-step equations
  • 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 write and solve equations like px = r from word problems. This adds a table, combines three prices into one coefficient and needs a second step with fractions after solving, a longer chain than typical grade 7 work.
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 5, 59 words.
Corrected from the DOE handbook
  • An arithmetic slip in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Solving for the number of ads is the obvious goal, and finding it feels like the finish, but the question asks for page space, which needs one more pass through the table. The awkward wording about buying pages in equal numbers slows students before any arithmetic starts. The wrong choices mark each place to stop: one set of ads, the number of each ad, or a slip in adding the fractional pages.

  1. Many steps major Working it out

    The work has two phases: turn the money into a count, 1,150x = 11,500 so x = 10, then turn the count into space by multiplying 10 by 1/4, 1/2 and 1 and adding. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math moderate Understanding the question

    The phrase “purchased 1/4 pages, 1/2 pages, and full pages of page space in equal numbers” has to be read as the same number of ads of each size, so the three prices add to one coefficient. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    Both columns of the table are needed: the price column builds the equation, and the page-space column turns the count of ads into the final total. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Let x be the number of each ad size, add the three prices to write one equation for the total, solve it, then multiply x by the page space in one set of ads.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook adds 200x + 350x + 600x and writes 1,500x = 11,500. The sum is 1,150x, and the handbook’s next line, x = 10, only follows from 1,150x = 11,500. The site shows 1,150x = 11,500.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal sets
number of sets=total spentcost of one set\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}}
A set here means one ad of each size. The cost of one set is what one of each costs together. Dividing the total money by it tells how many sets were bought, because every set costs the same.

  1. Sort out the question

    The store bought the same number of each ad size: quarter page, half page and full page. It spent $11,500 in all.

    You need the total page space. That is not the number of ads, and not the money.

  2. Find the cost of one set

    Think of one quarter page, one half page and one full page as a set. The store bought equal numbers of each, so it bought whole sets.

    200+350+600=1,150200 + 350 + 600 = 1{,}150

    One set costs $1,150.

  3. Find how many sets

    number of sets=total spentcost of one set=11,5001,150=10\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}} = \frac{11{,}500}{1{,}150} = 10

    The store bought 10 sets. That means 10 ads of each size.

    Watch out: 10 is the number of each ad, not the page space. Stopping here gives choice B.

  4. Find the page space in one set

    Add the sizes in one set. Write them in quarters so they have the same bottom number.

    14+12+1=14+24+44=74\frac{1}{4} + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}

    One set is 74\frac{7}{4}, or 1341\frac{3}{4}, pages.

    Watch out: 1341\frac{3}{4} pages is only one set. That is choice A.

  5. Multiply by the number of sets

    There are 10 sets, so multiply the space in one set by 10.

    10×74=704=171210 \times \frac{7}{4} = \frac{70}{4} = 17\frac{1}{2}

    So the answer is D, 171217\frac{1}{2} pages.

Check your answer

Add the sizes another way. 10 quarter pages make 2.5 pages, not 1.5. 10 half pages make 5 pages. 10 full pages make 10 pages. Then 2.5+5+10=17.52.5 + 5 + 10 = 17.5, the same answer.

More questions like this

The original DOE question
Question 75 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 75. Open page 156 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 84 of 93

2018 Test B Math, question 19

Question

How many positive odd numbers satisfy the inequality 3x+8≤923x+8\leq 92 ?

  • A13
  • B14
  • C17
  • D28
Show the answer

Correct answer: B

Solve the inequality for x.

3x + 8 ≤ 92

3x ≤ 84

x ≤ 28

The positive odd numbers less than or equal to 28 are 1, 3, 5, …, 25, and 27. There are 14 of them.

Question analysis

This question asks how many positive odd numbers satisfy the two-step inequality 3x + 8 ≤ 92. Students solve the inequality and then count the odd whole numbers in the solution. The counting after the algebra is where errors creep in, especially at the end of the list.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality is grade 7 work. Asking for a count of the odd solutions adds a number-property step and an endpoint check, but the total demand stays close to a typical grade 7 problem.
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 16, 9 words.
Corrected from the DOE handbook
  • The explanation’s description of the solutions 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

Solving the inequality is routine, and that is the trap: x ≤ 28 looks like an answer and 28 is among the choices. The real task is counting odd numbers from 1 to 27, where the usual shortcut, (27 − 1) ÷ 2, misses the first term. Pairing each odd number with the even number after it, up to 28, gives the count directly and avoids the fencepost slip.

  1. A very close wrong answer moderate Choosing the answer

    13 and 14 differ by one, and a rough count of the odd numbers up to 27 cannot separate them; only counting both ends, 1 and 27, gives the exact 14. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Combines ideas minor Working it out

    Solving 3x + 8 ≤ 92 is only half the work; counting the positive odd whole numbers up to 28 is a separate number-sense task. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Solve the inequality as you would an equation, keeping the direction of the sign, then count odd numbers from 1 up to the boundary by pairing or halving.

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.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook solves the inequality to x ≤ 28 and then lists “the positive odd numbers less than 28.” The solutions include 28 itself, so the site says “less than or equal to 28.” The count does not change, because 28 is even: the odd numbers are still 1, 3, 5, …, 27, which is 14 numbers.

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\text{if } a \leq b \text{, then } a - c \leq b - c
a, b and c are numbers. The sign ≤ means less than or equal to. You may subtract the same number from both sides. You may also divide both sides by the same positive number. The sign stays the same either way.
Counting odd numbers
odd numbers from 1 to n=n2\text{odd numbers from 1 to } n = \frac{n}{2}
n is an even number. Split the numbers from 1 to n into pairs: 1 and 2, 3 and 4, and so on. There are n ÷ 2 pairs, and each pair holds exactly one odd number.

  1. Sort out the question

    You need to count the positive odd numbers that make 3x+8≤923x + 8 \leq 92 true.

    The sign ≤ means is less than or equal to. Positive means greater than 0. Odd numbers are 1, 3, 5, 7, and so on.

  2. Solve the inequality

    Solve it like an equation. Subtract 8 from both sides. Then divide both sides by 3. Dividing by a positive number keeps the sign the same.

    3x+8≤923x + 8 \leq 92

    3x≤843x \leq 84

    x≤28x \leq 28

    Watch out: 28 is the largest x allowed, not the count. That trap is choice D.

  3. List the odd numbers

    The positive odd numbers up to 28 are 1, 3, 5, and so on, up to 25 and 27. The number 28 is even, so it does not count.

  4. Count them by pairing

    Split 1 to 28 into pairs: 1 and 2, 3 and 4, and so on, up to 27 and 28. Each pair has one odd number.

    odd numbers from 1 to 28=282=14\text{odd numbers from 1 to 28} = \frac{28}{2} = 14

    Watch out: Working out (27−1)÷2=13(27 - 1) \div 2 = 13 counts the jumps of 2 from 1 to 27, not the numbers. You must add 1 for the first number. That mistake gives choice A.

    So the answer is B, 14.

Check your answer

Test the edges. For x = 27: 3×27+8=893 \times 27 + 8 = 89, which is at most 92, so it works. For x = 29: 3×29+8=953 \times 29 + 8 = 95, which is too big. So 27 is the last odd number that works.

More questions like this

The original DOE question
Question 76 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 76. Open page 156 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 85 of 93

2018 Test B Math, question 20

Question

If 36y=4x\frac{36}{y}=4x, what is the value of xx when y=3y=3?

  • A3
  • B4
  • C9
  • D12
Show the answer

Correct answer: A

Substitute 3 for yy and solve for xx:

36y=4x\frac{36}{y}=4x

363=4x\frac{36}{3}=4x

12=4x12=4x

3=x3=x

Question analysis

This equation question gives 36 divided by y equal to 4x and asks for the value of x at one given value of y. Students substitute, simplify the fraction and solve a one-step equation. It is short, so the risk lies in stopping one step early or dividing by the wrong number.

NYS standards
  • 6.EE.7 one-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Substituting a value and solving a one-step equation are grade 6 skills. With two quick steps and whole numbers, this is lighter than the multistep equations typical of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 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 substitution followed by a one-step equation, with small whole numbers throughout, so most students answer correctly. The only real risk is stopping at 12, which is the value of 4x rather than of x. Writing 12 = 4x as its own line before reading the choices makes the last division hard to skip.

  1. Variables instead of numbers minor Understanding the question

    The variable y sits in a denominator, 36/y, which looks less familiar than an equation in one unknown, though substituting y = 3 removes it at once. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Replace y with its value, simplify the left side to a single number, and then divide by the coefficient of x.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing multiplication
a×x=b means x=b÷aa \times x = b \text{ means } x = b \div a
a is the number multiplied by x, and b is the result. Division undoes multiplication. So to get x alone, divide both sides by a.

  1. Write down what you know

    The equation is 36y=4x\frac{36}{y} = 4x. You are told y = 3. You need the value of x.

    The letter y is on the bottom of a fraction. Putting in its value turns the fraction into a plain number.

  2. Put in the value of y

    Replace y with 3.

    363=4x\frac{36}{3} = 4x

    The fraction means 36 divided by 3, which is 12.

    12=4x12 = 4x

    Watch out: 12 is the value of 4x, not x. Stopping here gives choice D.

  3. Divide both sides by 4

    4x means 4 times x. To undo multiplying by 4, divide both sides by 4.

    x=12÷4=3x = 12 \div 4 = 3

    So the answer is A, 3.

Check your answer

Put x = 3 and y = 3 back in. The left side is 36÷3=1236 \div 3 = 12. The right side is 4×3=124 \times 3 = 12. Both sides match, so x = 3 is right.

More questions like this

The original DOE question
Question 77 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 77. Open page 157 of the handbook (PDF)

Question 86 of 93

2018 Test B Math, question 27

Question

Nicole’s age now is three times Carmen’s age. If Carmen will be 17 in two years, how old was Nicole 5 years ago?

  • A38 yr
  • B40 yr
  • C45 yr
  • D50 yr
Show the answer

Correct answer: B

First, find out how old Nicole and Carmen are now.

Let NN = Nicole’s age now. Let CC = Carmen’s age now.

C+2=17C+2=17

C=15C=15 (Carmen’s age now)

N=3CN=3C

N=3(15)=45N=3(15)=45 (Nicole’s age now)

N−5=45−5=40N-5=45-5=40 (Nicole’s age 5 years ago)

Question analysis

This age word problem says Nicole is three times as old as Carmen now, and that Carmen will be 17 in two years. Students work out both present ages and then go back five years for Nicole. Keeping the times straight is the tricky part: two years ahead for Carmen, five years back for Nicole.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write simple equations from word problems. Age problems like this one take three short steps, and the shifts in time are the only real hurdle, so the demand is typical of grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 1, 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

The arithmetic is short, but the question moves through three points in time, and only one of them carries the three-times relationship. Students who find Nicole's present age often stop there, since it is a choice. Labeling each age with its time, now or then, keeps the two-year and five-year shifts with the right person and in the right direction.

  1. Turning words into math moderate Understanding the question

    Three points in time appear, now, in two years and 5 years ago, and “three times Carmen's age” holds only for now, so each shift must be applied to the right person. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Students must hold Carmen's future age 17, her present age 15, Nicole's present age 45 and the 5-year shift while deciding which to report. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find Carmen's age now by going back two years from 17, triple it for Nicole's current age, then subtract five.

Traps

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

Age at another time
age in n years=age now+n\text{age in } n \text{ years} = \text{age now} + n
n is a number of years. To find an age in the future, add n to the age now. To find an age in the past, subtract n instead. Everyone gets older at the same speed, one year per year.

  1. Sort out the times

    The question talks about three times: now, two years from now, and 5 years ago. Nicole is 3 times as old as Carmen now.

    Carmen will be 17 in two years. You need Nicole’s age 5 years ago.

  2. Find Carmen's age now

    In two years Carmen will be 17. So right now she is 2 years younger.

    17−2=1517 - 2 = 15

    Carmen is 15 now.

  3. Find Nicole's age now

    Nicole’s age now is three times Carmen’s age now.

    3×15=453 \times 15 = 45

    Nicole is 45 now.

    Watch out: 45 is her age now, choice C. The question asks about 5 years ago.

  4. Go back 5 years

    Going back in time means subtracting.

    45−5=4045 - 5 = 40

    Nicole was 40 five years ago.

    Watch out: Adding 5 gives 50, choice D. Also, the 2 years only apply to Carmen. Taking 2 more years off Nicole gives 38, choice A.

    So the answer is B, 40 yr.

Check your answer

Work forward from your answer. Nicole was 40 five years ago, so she is 45 now. Carmen is a third of that, 15. In two years Carmen will be 17. That matches the question.

More questions like this

The original DOE question
Question 84 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 84. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 87 of 93

2018 Test B Math, question 31

Question

Simplify this expression:

4(7−3x)−(5−x)4\left ( 7-3x \right )-\left ( 5-x \right )

  • A23−4x23-4x
  • B23−11x23-11x
  • C28−4x28-4x
  • D28−12x28-12x
Show the answer

Correct answer: B

4( 7 − 3x ) −( 5 − x) = 28 − 12x − 5 + x = 23 − 11x

Question analysis

This question asks students to simplify an algebraic expression by distributing a number into one set of parentheses, subtracting another set, and combining like terms. It checks the distributive property with a minus sign in front of parentheses. The usual slip is subtracting only the first term inside the second parentheses, which changes the sign of an x term.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Expanding and combining linear expressions is core grade 7 work. This is a standard example, two distributions and one subtraction, with nothing beyond what a grade 7 lesson on the distributive property would cover.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16+, 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 subtracting a whole group is where students go wrong: the minus sign has to reach the −x as well as the 5. Each choice pairs a constant with an x coefficient, so getting the constant right only narrows the field to two. The question rewards writing out both distributions in full rather than doing them mentally.

  1. Error-prone arithmetic moderate Working it out

    The minus sign in front of (5 − x) must change both terms, so −x becomes +x and then combines with −12x from the first group. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Two distributions, then collecting the constants 28 and −5 and the x terms separately: four small operations, each a place where a sign can slip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    23 − 4x shares the right constant with 23 − 11x, so getting 23 does not settle it; only the x coefficient, −12x + x rather than −3x − x, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite the subtraction as adding −1 times the second group, distribute both multipliers, then collect the constant terms and the x terms separately.

Traps

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. b and c are the terms inside. You multiply a by each term inside, not only the first one. This works because a groups of (b − c) is a groups of b take away a groups of c.
Subtracting a group
−(b−c)=−b+c-(b - c) = -b + c
b and c are the terms inside the parentheses. A minus sign in front means take away the whole group. So you change the sign of each term inside. Taking away a minus c is the same as adding c.

  1. Sort out the question

    You need to simplify 4(7−3x)−(5−x)4(7-3x)-(5-x). Simplify means write it in the shortest way. A term is one piece of the expression, like 7 or −3x-3x. Like terms have the same letter part, so −12x-12x and xx are like terms. Plain numbers are like terms too.

  2. Multiply the 4 into the first group

    Use the distributive property. The 4 multiplies both terms inside.

    4(7−3x)=4×7−4×3x=28−12x4(7-3x) = 4 \times 7 - 4 \times 3x = 28 - 12x

    Watch out: the 4 must multiply the 3x3x too. Writing only 28−3x28-3x is the first slip that leads to choice A.

  3. Take away the second group

    The minus sign in front of (5−x)(5-x) means subtract the whole group. So change the sign of each term inside.

    −(5−x)=−5+x-(5-x) = -5 + x

    Taking away a minus xx gives plus xx.

    Watch out: if you stop before this step, you get 28−12x28-12x. That is choice D.

  4. Combine the like terms

    Now you have 28−12x−5+x28-12x-5+x. Put the plain numbers together. Then put the xx terms together.

    28−5=2328 - 5 = 23

    −12x+x=−11x-12x + x = -11x

    So the expression simplifies to 23−11x23-11x.

  5. Give the answer

    The plain number is 23, and the xx term is −11x-11x. So the answer is B, 23−11x23-11x.

Check your answer

Put in an easy number, like x=1x = 1. The original gives 4(7−3)−(5−1)=16−4=124(7-3)-(5-1) = 16-4 = 12. Choice B gives 23−11=1223-11 = 12. They match. Choice A would give 23−4=1923-4 = 19, which does not match.

More questions like this

The original DOE question
Question 88 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 88. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 88 of 93

2018 Test B Math, question 33

Question

A large container is partially filled with nn liters of water. Ito adds 10 liters of water to the container, making it 60% full. If Ignacio adds 6 more liters of water, the container will be 75% full. What is the value of nn ?

  • A14
  • B15
  • C26
  • D30
Show the answer

Correct answer: A

Let xx be the total number of liters the container can hold.

n+10x=60% and n+16x=75%\frac{n+10}{x}=60\%\text{ and }\frac{n+16}{x}=75\%

First, solve each equation for xx:

Equation 1:

n+10x=60100\frac{n+10}{x}=\frac{60}{100}

n+10x=35\frac{n+10}{x}=\frac{3}{5}

3x=5(n+10)3x=5(n+10)

x=5n+503x=\frac{5n+50}{3}

Equation 2:

n+16x=75100\frac{n+16}{x}=\frac{75}{100}

n+16x=34\frac{n+16}{x}=\frac{3}{4}

3x=4(n+16)3x=4(n+16)

x=4n+643x=\frac{4n+64}{3}

Now, set the two equations equal to each other and solve for nn.

5n+503=4n+643\frac{5n+50}{3}=\frac{4n+64}{3}

5n+50=4n+645n+50=4n+64

n+50=64n+50=64

n=14 litersn=14\text{ liters}

Question analysis

Water is poured into a large container in two stages: 10 liters make it 60% full, and 6 more liters bring it to 75% full. Students must work out how much water was in it to begin with. This percent word problem hides two unknowns, the starting amount and the capacity, and several in-between numbers look like answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here the whole is unknown and must be found from the change between two percents before a second step reaches the answer.
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 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

The container's size is the hidden key, and nothing in the stem names it. Students who write an equation for each fill level, or who notice that 6 liters equal 15 percent, get there; others grab a number from the story or stop at the 75% level. Because every wrong choice is a number that appears in the problem or arises while solving it, a careless finish is rewarded with a match.

  1. Turning words into math major Understanding the question

    Two percent-full statements about a container of unknown size have to become equations, n + 10 = 60% of the capacity and n + 16 = 75% of it, and the capacity is never given. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method moderate Working it out

    The quick route is to see that Ignacio's 6 liters raise the level by 15 percentage points, which fixes the capacity at 40 liters; the wording does not point to it. (Schoenfeld, 1985)

  3. Many steps moderate Working it out

    After the capacity, the solution still needs the 60% amount, 24 liters, and a subtraction of Ito's 10 liters, so two unknowns are chased through four steps. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  4. A very close wrong answer minor Choosing the answer

    15 sits one unit from the right 14, so a rough estimate of the starting amount cannot choose between them; only solving exactly, 24 − 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compare the two fill levels: the extra 6 liters account for the change in percent, which gives the capacity. Then use the 60% level to work back to the starting amount.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
  • 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 of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the full amount, here how much the container holds. The part is p percent of it. Percent means out of 100, so you divide p by 100 and then multiply.

  1. Sort out what you know

    The container starts with nn liters of water. Ito adds 10 liters, and it is 60% full. Then 6 more liters make it 75% full. You do not know how much the container holds when full. Call that its capacity. You need nn, the starting amount.

  2. See what the 6 liters did

    The extra 6 liters raised the level from 60% to 75%.

    75−60=1575 - 60 = 15

    So 6 liters is 15% of the capacity.

    Watch out: 15 is a percent, not a number of liters. Picking 15 gives choice B.

  3. Find the full capacity

    If 15% is 6 liters, then 1% is 6÷15=0.46 \div 15 = 0.4 liters. A full container is 100%.

    100×0.4=40100 \times 0.4 = 40

    The container holds 40 liters when full.

  4. Find the water at 60 percent

    Use the percent formula with 60 for p and 40 liters for the whole.

    part=60100×40=24\text{part} = \frac{60}{100} \times 40 = 24

    After Ito adds his water, there are 24 liters.

  5. Undo Ito’s 10 liters

    Ito added 10 liters to the starting amount. To go back, subtract them.

    n=24−10=14n = 24 - 10 = 14

    The container started with 14 liters.

    Watch out: 75% of 40 is 30 liters. That is the amount after both people add water, not at the start. It is choice D.

  6. Give the answer

    The question asks for nn, the amount before anyone added water. So the answer is A, 14.

Check your answer

Run the story forward. Start with 14 liters. Add 10 to get 24 liters, and 24 is 60% of 40. Add 6 more to get 30 liters, and 30 is 75% of 40. Both facts match.

More questions like this

The original DOE question
Question 90 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 90. Open page 159 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 89 of 93

2018 Test B Math, question 34

Question

5x3+3x+9+1x25x^{3}+3x+9+\frac{1}{x^{2}}

If x=10x=10, what is the value of the expression above?

  • A2,539.01
  • B5,039.01
  • C5,039.1
  • D5,139
Show the answer

Correct answer: B

5x3+3x+9+1x25x^3+3x+9+\frac{1}{x^2}

=5(103)+3(10)+9+1102=5(10^3)+3(10)+9+\frac{1}{10^2}

=5,000+30+9+1100=5,039.01=5{,}000+30+9+\frac{1}{100}=5{,}039.01

Question analysis

Students substitute 10 for x in an expression with a cube, a linear term, a constant and a fraction with x squared in its denominator, then add the results. It tests evaluating expressions with exponents and decimals. The fraction term is where most slips happen, because its place value is easy to get wrong.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Evaluating expressions with whole-number exponents is grade 6 work. The only extra care here is a reciprocal term that turns into a decimal, a single routine substitution that typical grade 7 students handle comfortably.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 11 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Substituting 10 is easy; the risk lies in the two unlike terms at the ends. The cube must be applied to x before multiplying by 5, and the reciprocal of 100 must become a hundredth, not a tenth. With choices that differ by a single decimal place, the question rewards writing each term as a number before adding.

  1. Error-prone arithmetic moderate Working it out

    The last term, 1/x² at x = 10, is 1/100, which must be written as 0.01 and lined up beside 5,039; one place too far gives 0.1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    5,039.1 sits beside the right 5,039.01 and differs only in the hundredths place; only writing 1/100 as 0.01, not 0.1, tells them apart, and estimating cannot. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work term by term: compute 10 cubed, then the reciprocal of 10 squared as a decimal, and line up the place values carefully before adding.

Traps

  • 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

Exponents
x3=x×x×xx^3 = x \times x \times x
x is the base number. The small raised 3 is the exponent. It tells how many times to use x in a multiplication. In the same way, x squared, written with a raised 2, means x times x.

  1. Sort out the question

    You need the value of 5x3+3x+9+1x25x^3+3x+9+\frac{1}{x^2} when x=10x = 10. Put 10 in place of every xx. Then work out each term, one at a time. A term is a piece joined to the others by a plus sign.

  2. Work out the first term

    Do the exponent first. Then multiply by 5.

    103=10×10×10=1,00010^3 = 10 \times 10 \times 10 = 1{,}000

    5×1,000=5,0005 \times 1{,}000 = 5{,}000

    Watch out: the exponent belongs only to the xx, not to the 5. Squaring 5×105 \times 10 gives 2,500, which leads to choice A.

  3. Work out the middle terms

    The next term means 3 times xx.

    3×10=303 \times 10 = 30

    The 9 has no xx in it, so it stays 9.

  4. Work out the last term

    First find the bottom. Then divide.

    102=10×10=10010^2 = 10 \times 10 = 100

    1100=0.01\frac{1}{100} = 0.01

    One hundredth is 0.01, with the 1 two places after the decimal point.

    Watch out: 0.1 is one tenth, not one hundredth. That slip gives choice C. Adding 100 instead of 1100\frac{1}{100} gives choice D.

  5. Add the four terms

    Line up the decimal points and add. The 0.01 goes in the hundredths place.

    5,000+30+9+0.01=5,039.015{,}000 + 30 + 9 + 0.01 = 5{,}039.01

    So the answer is B, 5,039.01.

Check your answer

The first three terms are whole numbers. They add up to 5,039. The last term is a small part of 1. So the answer is just over 5,039. Only B and C fit. The last term is hundredths, so B is right.

More questions like this

The original DOE question
Question 91 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 91. Open page 159 of the handbook (PDF)

Question 90 of 93

2018 Test B Math, question 45

Question

Phan chose an Internet service that charges $18.00 per month plus $0.024 per minute. Deion chose an Internet service that charges $30.00 per month for unlimited usage. At the end of the month, Phan’s and Deion’s charges were identical. For how many minutes did Phan use the Internet service that month?

  • A50
  • B60
  • C100
  • D500
Show the answer

Correct answer: D

Let x be the number of minutes Phan used his internet service in the month. Phan’s monthly charges were

18 + 0.024x

Since Deion’s charges were the same as Phan’s, set the expression equal to 30 and solve for x:

18 + 0.024x = 30

0.024x = 12

x = 500

Phan used his service for 500 minutes.

Question analysis

Phan pays $18 a month plus 2.4 cents per minute for Internet service, Deion pays a flat $30, and their bills come out the same. Students find how many minutes Phan used. It tests writing and solving a two-step linear equation, and the small decimal rate makes the division error-prone.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving equations of the form px + q = r from a word problem is central grade 7 work. This is a typical example; only the three-decimal-place rate adds some care to the division.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 48 words.
Students, last 30 days
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 up the equation is routine, so the question is mostly a test of decimal division. Dividing 12 by 0.024 invites a misplaced decimal point, and one such slip gives a choice exactly ten times too small. A quick check, 500 × 0.024 = 12, confirms the answer in seconds.

  1. Error-prone arithmetic moderate Working it out

    The last step divides 12 by 0.024, a three-place decimal, and moving the decimal point one place too few gives 50 instead of 500. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    "Phan's and Deion's charges were identical" has to become 18 + 0.024m = 30, with the flat $30 on one side and Phan's fee plus per-minute cost on the other. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Set Phan's cost, the monthly fee plus the per-minute charge times the minutes, equal to Deion's flat fee, then subtract and divide, checking the decimal places.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost with a monthly fee
C=f+r×mC = f + r \times m
C is the total cost for the month. f is the fixed fee you pay no matter what. r is the price for each minute, and m is the number of minutes. Use it when a bill has a set fee plus a charge for each unit.

  1. Sort out what you know

    Phan pays $18.00 a month plus $0.024 for each minute. Deion pays $30.00 a month, however much he uses. Their bills were the same. You need the number of minutes Phan used. Call it mm.

  2. Write an equation

    Use the cost formula for Phan. His bill equals Deion’s $30.

    18+0.024×m=3018 + 0.024 \times m = 30

    An equation says two amounts are equal. Here it says the two bills are equal.

  3. Take away the monthly fee

    Subtract 18 from both sides. That leaves the minutes part alone.

    0.024×m=30−18=120.024 \times m = 30 - 18 = 12

    So Phan paid $12 for his minutes.

  4. Divide to find the minutes

    Divide 12 by 0.024. To get rid of the decimal, move the point 3 places right in both numbers.

    m=12÷0.024=12,000÷24=500m = 12 \div 0.024 = 12{,}000 \div 24 = 500

    Phan used 500 minutes.

    Watch out: 0.024 is not 0.24. Using 0.24 gives 12÷0.24=5012 \div 0.24 = 50, choice A. Move the point the same number of places in both numbers.

  5. Give the answer

    Phan used 500 minutes that month. That is the only choice that makes the two bills equal. So the answer is D, 500.

Check your answer

Put 500 back in. The minutes cost 500×0.024=12500 \times 0.024 = 12 dollars. Add the fee: 18+12=3018 + 12 = 30 dollars. That is Deion’s bill, so the bills match.

More questions like this

The original DOE question
Question 102 as printed on page 161 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 102. Open page 161 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 91 of 93

2018 Test B Math, question 50

Question

Which number line below shows the solution set for 2x−2≤y≤4x+102x-2\leq y\leq 4x+10 when y=1y=1?

  • AA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to −1 1/2; both ends are closed dots.
  • BA number line from −3 to 3 marked in quarter units, with a solid segment from −1 1/2 to 2 1/4; both ends are closed dots.
  • CA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to 1 1/2; both ends are closed dots.
  • DA number line from −3 to 3 marked in quarter units, with a closed dot at 1 1/2 and a solid ray from it running left past −2 1/2 to an arrowhead.
Show the answer

Correct answer: C

Separate the compound inequality into two pieces:

2x−2≤y2x-2\leq y and y≤4x+10y\leq 4x+10

Substitute y=1y=1 into each inequality and solve for xx:

2x−2≤12x-2\leq1

2x≤32x\leq 3

x≤32x\leq\frac{3}{2}

1≤4x+101\leq 4x+10

−9≤4x-9\leq4x

−94≤x-\frac{9}{4}\leq x

The solution is the number line that shows that xx is greater than or equal to −214-2\frac{1}{4} and less than or equal to 1121\frac{1}{2} .

Question analysis

Students substitute y = 1 into the compound inequality 2x − 2 ≤ y ≤ 4x + 10 and pick the number line that shows the values of x that work. It tests solving two-step inequalities and graphing the solution. The four graphs differ in their endpoints and in whether the shading is closed off on both sides, so each endpoint has to be found exactly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph one two-step inequality. This compound inequality in two variables requires substituting, solving two inequalities and combining them into one interval, a step toward algebra 1.
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 9, 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

Substituting y = 1 is easy; the work is solving two inequalities and matching a fractional interval to a graph. Students who solve only the left inequality find a matching ray, and sign slips on either endpoint produce intervals that shift or flip across zero. The graphs differ by quarter-unit ticks, so reading them carefully matters as much as the algebra.

  1. Many steps moderate Working it out

    The compound inequality must be split into 2x − 2 ≤ 1 and 1 ≤ 4x + 10, each solved for x, and the two results joined into one interval. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Both endpoints are fractions, 3/2 and −9/4, and the second comes from dividing −9 by 4, where a lost sign moves the endpoint across zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The graphs are marked in quarter units, so −2 1/4 and 1 1/2 have to be found by counting small ticks between the labeled integers. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. A very close wrong answer minor Choosing the answer

    The ray ending at 1 1/2 looks almost like the right segment, sharing its right end; only the left end, an arrow instead of a closed dot at −2 1/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Split the compound inequality into two separate inequalities, solve each for x after putting 1 in for y, then combine the two results into a single interval and match it to a graph.

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.)
  • 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

Solving an inequality
if a≤b, then a+c≤b+c\text{if } a \leq b, \text{ then } a + c \leq b + c
a, b and c are numbers. The sign ≤ means is less than or equal to. Adding or subtracting the same number c on both sides keeps it true. Dividing both sides by the same positive number also keeps it true.

  1. Sort out the question

    The sign ≤\leq means is less than or equal to. An inequality compares two amounts that may not be equal. The statement 2x−2≤y≤4x+102x-2 \leq y \leq 4x+10 is two inequalities joined together. You put 1 in place of yy. Then you need every xx that makes both parts true.

  2. Split it into two parts

    With y=1y = 1, the statement becomes two inequalities.

    2x−2≤12x - 2 \leq 1

    1≤4x+101 \leq 4x + 10

    The answer must make both of them true at the same time.

  3. Solve the first part

    Add 2 to both sides. Then divide both sides by 2.

    2x≤32x \leq 3

    x≤32x \leq \frac{3}{2}

    So xx is at most 1121\frac{1}{2}. This is the right end of the answer.

    Watch out: the right end is a positive number. Choice A ends its bar at −112-1\frac{1}{2} instead.

  4. Solve the second part

    Subtract 10 from both sides. Then divide both sides by 4.

    −9≤4x-9 \leq 4x

    −94≤x-\frac{9}{4} \leq x

    So xx is at least −214-2\frac{1}{4}. This is the left end.

    Watch out: keep the minus sign. Choice B uses the opposite signs and ends at 2142\frac{1}{4}.

  5. Put the two parts together

    The answer is every number from −214-2\frac{1}{4} to 1121\frac{1}{2}, with both ends included. On the graph, that is a solid bar with a dot at each end. The small ticks are quarters. So count 1 tick left of −2 and 2 ticks right of 1.

    Watch out: choice D uses only the first part. Its arrow goes left forever.

  6. Give the answer

    Choice C shows a bar from −214-2\frac{1}{4} to 1121\frac{1}{2}, with dots at both ends. So the answer is C.

Check your answer

Test a number on the bar, like x=0x = 0. Then 2(0)−2=−22(0)-2 = -2 and 4(0)+10=104(0)+10 = 10, and 1 is between them. Now test x=−3x = -3, which is off the bar. Then 4(−3)+10=−24(-3)+10 = -2, and 1 is not less than −2. So the bar must stop on the left, which rules out D.

More questions like this

The original DOE question
Question 107 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 107. Open page 163 of the handbook (PDF)

Question 92 of 93

2018 Test B Math, question 54

Question

Raoul is at least 3 years older than Vahn. Which of the following inequalities gives the relationship between Raoul’s age (rr ) and Vahn’s age ( vv )?

  • Ar−v≥3r-v\geq 3
  • Br−v≤3r-v\leq 3
  • C3−v≤r3-v\leq r
  • D3−r≤v3-r\leq v
Show the answer

Correct answer: A

Raoul is at least 3 years older than Vahn, which can be written as

r ≥ v + 3

Rewrite this inequality to match the answer options:

r − v ≥ 3

Question analysis

Raoul is at least 3 years older than Vahn, and students choose the inequality that links their ages, r and v. It tests translating a phrase like "at least" into an inequality. The correct form may not match the one students write first, so they need to rearrange carefully and keep the direction of the inequality.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.8 writing simple inequalities
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing inequalities from words is grade 6 and 7 work. The extra demand is that "at least" must become the right symbol and the choices are rearranged forms, so students must manipulate inequalities rather than just recognize one.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 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

Two skills are tested at once: translating "at least … older than" into an inequality, and recognizing that inequality after a rearrangement. The two choices that begin with 3 − v and 3 − r look like legitimate moves but say only that the two ages add up to at least 3. Testing one pair of ages that fits the sentence and one that does not, such as 20 and 12 against 13 and 12, sorts the choices without algebra.

  1. Turning words into math major Understanding the question

    "Raoul is at least 3 years older than Vahn" has to become r ≥ v + 3; "at least" calls for ≥, though the word "least" seems to point the other way. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    No choice is written as r ≥ v + 3, so the natural translation has to be rearranged to r − v ≥ 3 before it can be matched. (Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    r − v ≤ 3 is the right r − v ≥ 3 with the symbol reversed; only deciding that "at least" means greater than or equal to separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the sentence directly as Raoul's age compared with Vahn's age plus 3, then rearrange each choice into that form to see which one matches.

Traps

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

Greater than or equal to
a≥ba \geq b
a and b are two amounts. The sign ≥ means a is greater than or equal to b, so a is b or more. The words at least mean the same thing. The sign ≤ means less than or equal to, or at most.

  1. Sort out the question

    The letter rr is Raoul’s age, and vv is Vahn’s age. Raoul is “at least 3 years older than Vahn”. That means the gap between their ages is 3 years or more. An inequality is a math sentence that uses a sign like ≥\geq instead of =. You need the one that says this.

  2. Write the age gap

    Raoul is older, so the gap is his age minus Vahn’s age. The gap is r−vr - v. For example, ages 15 and 10 have a gap of 5.

  3. Turn at least into a sign

    At least 3 means 3 or more. So use ≥\geq, which means greater than or equal to.

    r−v≥3r - v \geq 3

    This is choice A.

    Watch out: the word least does not mean less. Choice B, r−v≤3r - v \leq 3, says the gap is 3 or less. Then Raoul could even be younger.

  4. Check choices C and D

    Add vv to both sides of choice C. You get 3≤r+v3 \leq r + v. That is about the sum of their ages, not the gap. Choice D works the same way. Neither one says Raoul is older.

  5. Give the answer

    Only one choice says the gap is 3 years or more. So the answer is A, r−v≥3r-v \geq 3.

Check your answer

Test ages that do not fit: Raoul 11 and Vahn 10. He is only 1 year older. Choice A gives 11−10=111 - 10 = 1, and 1 is not 3 or more. So A rightly says no. Choice B says yes, so B is wrong.

More questions like this

The original DOE question
Question 111 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 111. Open page 164 of the handbook (PDF)

Question 93 of 93

2018 Test B Math, question 56

Question

There are now xx cans stacked on a shelf that holds 36 cans when full. If 4 of these cans were removed, the shelf would be half full. What is the value of xx ?

  • A14
  • B16
  • C18
  • D22
Show the answer

Correct answer: D

The shelf, when full, holds 36 cans. When the shelf is half full, it holds 18 cans.

x − 4 = 18

x = 22

Question analysis

A shelf holds 36 cans when full, and removing 4 of the cans now on it would leave it half full. Students find how many cans are on the shelf now. It tests writing and solving a one-step equation from words, and the risk is undoing the removal in the wrong direction.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 6 students solve equations like x − p = q, and this one needs only that and halving 36. It is a quick, routine word problem, easier than the two-step equations typical of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 3, 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

Short and routine, this question still punishes a quick reading. Students must keep three amounts apart, the 36-can capacity, the 18 cans on a half-full shelf and the x cans there now, and then undo the removal by adding. Writing x − 4 = 18 before computing prevents every wrong choice.

  1. Turning words into math minor Understanding the question

    The 36 cans are the shelf's capacity, not its contents, and "if 4 of these cans were removed" describes a hypothetical, so the equation is x − 4 = 18. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    18 sits just below the right 22, and both are near half of 36; only remembering that x counts the cans before the 4 were removed separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out how many cans make a half-full shelf, then undo the removal of the 4 cans; write it as an equation in x if that helps.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing subtraction
x−a=b means x=b+ax - a = b \text{ means } x = b + a
x is the number you want. a is the amount taken away, and b is what is left. To undo taking away a, add a back. Adding the same number to both sides keeps the equation balanced.

  1. Sort out what you know

    The shelf holds 36 cans when it is full. That is its size, not how many cans are on it now. Right now it has xx cans. If 4 cans were taken off, the shelf would be half full. You need xx.

  2. Find half of a full shelf

    A full shelf holds 36 cans, so half full is half of that.

    36÷2=1836 \div 2 = 18

    A half-full shelf has 18 cans.

  3. Write an equation

    Taking 4 cans away from xx leaves 18 cans.

    x−4=18x - 4 = 18

    Watch out: 18 is the number after the cans are taken off. Picking 18 gives choice C.

  4. Add the 4 cans back

    Undo the subtraction by adding 4 to both sides.

    x=18+4=22x = 18 + 4 = 22

    Watch out: subtracting the 4 again gives 18−4=1418 - 4 = 14, choice A. The cans were taken from xx, so xx is more than 18.

  5. Give the answer

    There are 22 cans on the shelf now, before any are taken off. So the answer is D, 22.

Check your answer

Start with 22 cans and take off 4. That leaves 22−4=1822 - 4 = 18 cans. Half of 36 is 18, so the shelf is half full. The answer works.

More questions like this

The original DOE question
Question 113 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 113. Open page 164 of the handbook (PDF)