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Practice Test

8.EE expressions, equations, and inequalities

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Question 1 of 22

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 22

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 3 of 22

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 4 of 22

2026 Practice Test C Math, question 18

Question

I. s−ts - t
II. st2st^{2}
III. sts^{t}

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

  • AII only
  • BI and II only
  • CII and III only
  • DI, II, and III
Show the answer

Correct answer: D

Each of the three expressions must be considered:

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

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

Expression II: st2st^{2}, where s is a positive integer and t is a negative integer:

Squaring a negative integer results in a positive integer, so s(−t)2s(-t)^{2} is equivalent to finding the value of st2st^{2}, which must result in a positive number.

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know moderate Working it out

    Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.

  2. Try one pair of numbers

    Try s = 2 and t = −3 in each expression:

    • I: 2−(−3)=2+3=52 - (-3) = 2 + 3 = 5
    • II: 2×(−3)2=2×9=182 \times (-3)^{2} = 2 \times 9 = 18
    • III: 2−3=123=182^{-3} = \frac{1}{2^{3}} = \frac{1}{8}

    All three are positive this time. Now see why that always happens.

  3. Explain expression I

    Use the rule for subtracting:

    s−t=s+(−t)s - t = s + (-t)

    t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.

    Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.

  4. Explain expression II

    t2t^{2} means t × t. A negative times a negative is positive, so t2t^{2} is always positive. Then s×t2s \times t^{2} is a positive times a positive. That is always positive too.

  5. Explain expression III

    Use the rule for a negative exponent:

    st=1s−ts^{t} = \frac{1}{s^{-t}}

    −t is a positive integer, so s−ts^{-t} is a positive integer. So sts^{t} is 1 over a positive number. That is a positive fraction.

    Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.

  6. Choose the answer

    Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.

    So the answer is D, I, II, and III.

Check your answer

Try the values closest to 0: s = 1 and t = −1. I gives 1−(−1)=21 - (-1) = 2. II gives 1×(−1)2=11 \times (-1)^{2} = 1. III gives 1−1=11=11^{-1} = \frac{1}{1} = 1. All three are still positive.

More questions like this

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

Question 5 of 22

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 6 of 22

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 7 of 22

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 8 of 22

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 9 of 22

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 10 of 22

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 11 of 22

2024 Test B Math, question 34

Question

The volume of a cube is 512 cubic centimeters. What is the length, in centimeters, of one edge of this cube?

  • A8
  • B422342\frac{2}{3}
  • C851385\frac{1}{3}
  • D128
Show the answer

Correct answer: A

The volume of a cube is V=s3V=s^{3}, where s is the length of one edge.

So 512=s3512=s^{3}, and s=5123s=\sqrt[3]{512}.

If you are not sure how to find the cube root, you can estimate.

We know 10×10×10=103=1,00010\times 10\times 10=10^{3}=1{,}000, so 1,0003=10\sqrt[3]{1{,}000}=10.

Since 512 is less than 1,000, the answer must be less than 10. The only given option less than 10 is 8. Double-check to make sure this is correct: 83=8×8×8=5128^{3}=8\times 8\times 8=512.

Question analysis

A cube has a volume of 512 cubic centimeters, and students work backward to find the length of one edge. It tests the volume formula for a cube and the idea of a cube root. The tricky part is going from a volume to a length, the reverse of the usual volume problem.

NYS standards
  • 8.EE.2 square and cube roots
  • 6.G.5 perfect squares and perfect cubes
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 volume work goes from edge lengths to a volume. Going backward needs a cube root, a grade 8 idea, and 512 is beyond the perfect cubes students are expected to know, though checking a guess by multiplying keeps it short.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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

This is a one-fact question: a cube's volume is its edge used three times as a factor. Students who remember that need only the number whose cube is 512, and the choices help, because only one is under 10 while 10 × 10 × 10 = 1,000. The wrong choices divide 512 by numbers tied to other measures of a cube or a square, so they catch students who do not recall the formula.

  1. Something you must already know moderate Working it out

    Neither V = s³ nor the cube root of 512 is given, so students must recall the formula and know, or find by trial, that 8 × 8 × 8 = 512. (Anderson & Krathwohl, 2001)

Strategy: Write the volume as edge × edge × edge, then find the number that, multiplied by itself three times, gives 512, and confirm it by multiplying.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube
V=s3V = s^3
V is the volume, the space inside, in cubic units. s is the length of one edge. s³ means s × s × s. A cube’s length, width and height are all s, so length × width × height becomes s × s × s.

  1. Sort out the question

    The cube’s volume is 512 cubic centimeters. Volume is the space inside a solid. You need the length of one edge.

    A cube has 12 edges, and they are all the same length. So one edge tells you all of them.

  2. Write the volume formula

    V=s3V = s^3

    512=s×s×s512 = s \times s \times s

    You need a number that gives 512 when you multiply it by itself three times. That number is called the cube root of 512.

  3. Try numbers to find the edge

    Start with an easy one: 10×10×10=1,00010 \times 10 \times 10 = 1{,}000. That is too big, so the edge is less than 10. Try 8.

    8×8×8=64×8=5128 \times 8 \times 8 = 64 \times 8 = 512

    It works. Each edge is 8 centimeters.

  4. Rule out the other choices

    Watch out: The volume is not shared out among the faces or the edges. Dividing 512 by the 6 faces gives 851385\frac{1}{3}, choice C. Dividing by the 12 edges gives 422342\frac{2}{3}, choice B. Dividing by 4 gives 128, choice D.

    So the answer is A, 8.

Check your answer

Test the biggest choice. 128×128×128128 \times 128 \times 128 is over 2 million, far more than 512. Only an edge less than 10 can work, and 8 is the only choice less than 10. And 83=5128^3 = 512.

More questions like this

The original DOE question
Question 91 as printed on page 209 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 91. Open page 209 of the handbook (PDF)

Question 12 of 22

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 13 of 22

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 14 of 22

2023 Test A Math, question 57

Question

Chanelle selects a marble from a container and then returns it to the container. When she does this 3 times, the probability of choosing a red marble all 3 times is 1216\frac{1}{216}. Based on this information, what is the probability of Chanelle choosing a red marble the first time she selects a marble?

  • A172\frac{1}{72}
  • B136\frac{1}{36}
  • C18\frac{1}{8}
  • D16\frac{1}{6}
Show the answer

Correct answer: D

Let 1x\frac{1}{x} be the probability of Chanelle selecting a red marble the first time. The probability of selecting a red marble three times is 1x×1x×1x\frac{1}{x} \times \frac{1}{x} \times \frac{1}{x}, which is given as 1216\frac{1}{216}. Since 216=6×6×6216 = 6 \times 6 \times 6, the value of x is 6, and the probability of Chanelle selecting a red marble the first time is 16\frac{1}{6}.

Question analysis

This probability question says Chanelle draws a marble and puts it back, three times, and the chance of red all three times is 1/216. It asks for the chance of red on her first draw. Students reverse the multiplication rule for repeated independent draws, which means finding a cube root, and dividing by 3 is the tempting shortcut.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 8.EE.2 square and cube roots
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Multiplying probabilities for repeated draws with replacement is grade 7 compound probability. Running it backward, from three draws to one, needs a cube root, a grade 8 idea, though the numbers are friendly and the work is short once set up.
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 7, 55 words.
Students, last 30 days
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 three draws are seen as one chance cubed, the question is a cube root, and 216 is a perfect cube many students know. The risk lies in undoing the cube the wrong way: dividing by 3 because there are three draws, or stopping at 36 after one division by 6. A quick check, 1/6 × 1/6 × 1/6 = 1/216, confirms the answer.

  1. Something you must already know moderate Working it out

    Students must know that three draws with the marble returned each time multiply, p × p × p, and must recognize 216 as 6 × 6 × 6; the stem states neither. (Anderson & Krathwohl, 2001)

  2. No obvious method minor Working it out

    The usual direction runs from one draw to three; here the three-draw chance is given and the bold first asks for one draw, so students work backward and undo a cube. (Schoenfeld, 1985)

Strategy: Call the chance of red on one draw p, so three draws with replacement give p × p × p; then find the number whose cube equals the given probability.

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

Probability of repeated picks
P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p
p is the chance of red on one pick. Chanelle puts the marble back each time, so every pick has the same chance p, and one pick doesn’t change the next. The chance of red on all three picks is the three chances multiplied.

  1. Sort out the question

    Chanelle picks a marble and puts it back, 3 times. The chance of red all 3 times is 1216\frac{1}{216}. You need the chance of red on just the first pick. Call that chance p.

  2. Write the three picks as a product

    She puts each marble back, so the container is the same every time. Each pick has the same chance, p.

    P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p

    p×p×p=1216p \times p \times p = \frac{1}{216}

  3. Find the number that fits

    You need a fraction that, multiplied by itself three times, makes 1216\frac{1}{216}. Its top is 1, since 1 × 1 × 1 = 1. For the bottom, find a number that times itself three times makes 216.

    6×6×6=36×6=2166 \times 6 \times 6 = 36 \times 6 = 216

    So the bottom is 6, and p is 16\frac{1}{6}.

  4. Avoid the tempting shortcuts

    Watch out: dividing 216 by 3 gives 72, choice A. Three picks means multiplying three times, not dividing by 3. Also, 136\frac{1}{36} is the chance of red twice in a row, choice B.

    So the answer is D, 16\frac{1}{6}.

Check your answer

Multiply it out: 16×16×16=1216\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}. That matches the question. Choice A fails fast: 172×172\frac{1}{72} \times \frac{1}{72} is already far smaller than 1216\frac{1}{216}.

More questions like this

The original DOE question
Question 114 as printed on page 91 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 114. 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 15 of 22

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 16 of 22

2023 Test B Math, question 43

Question

A certain square puzzle has an area of 64 square inches, and a cubic block has a volume of 64 cubic inches. What is the ratio of the length of a side of the puzzle to the length of a side of the block?

  • A1:31:3
  • B1:21:2
  • C2:32:3
  • D2:12:1
Show the answer

Correct answer: D

The length of one side of the puzzle is 64=8\sqrt{64}=8 inches. The length of one side of the block is 643=4\sqrt[3]{64}=4 inches.

The ratio of the side of the puzzle to the side of the block is 8:48:4, which reduces to 2:12:1.

Question analysis

A square puzzle has an area of 64 square inches, and a cube-shaped block has a volume of 64 cubic inches. Students find the ratio of a side of the puzzle to a side of the block. Both figures are described by the number 64, but turning an area and a volume back into lengths takes different steps, and the order of the ratio matters.

NYS standards
  • 8.EE.2 square and cube roots
  • 6.G.5 perfect squares and perfect cubes
  • 6.RP.1 ratio concepts and language
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 area and volume problems go from side lengths to an area or a volume. Going backward needs a square root and a cube root, and cube roots are grade 8 content, although 64 is both a perfect square and a perfect cube.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 46 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Because 64 is both a perfect square and a perfect cube, the computation is easy once students know to undo squaring and cubing. The risk is at the end: with sides of 8 and 4 inches in hand, writing the block first turns the ratio upside down. Students who never find the side lengths are left with ideas such as two dimensions against three.

  1. Something you must already know moderate Working it out

    Neither formula is given: the puzzle’s side is the number whose square is 64 and the block’s edge the number whose cube is 64, and cube roots are grade 8 work. (Anderson & Krathwohl, 2001)

  2. Combines ideas minor Working it out

    A square and a cube share the number 64 in different units, so two different inverse operations feed one ratio, which must list the puzzle’s side first. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the side of the square from its area and the edge of the cube from its volume by testing small whole numbers, then write the puzzle's side first and simplify.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square
A=s×sA = s \times s
A is the area, in square units. s is the length of one side. All sides of a square are equal, so the area is the side times itself. To find the side from the area, ask which number times itself gives the area.
Volume of a cube
V=s×s×sV = s \times s \times s
V is the volume, the space inside, in cubic units. s is the length of one edge. A cube’s length, width and height are all equal, so the volume is the edge used three times. To go backward, ask which number used three times gives the volume.

  1. Sort out the shapes

    The puzzle is a square with an area of 64 square inches. The block is a cube with a volume of 64 cubic inches. A cube is a box with every edge the same length. You need the ratio of the puzzle’s side to the block’s side, with the puzzle first.

  2. Find the puzzle’s side

    Use the area formula backward. Which number times itself makes 64?

    8×8=648 \times 8 = 64

    The puzzle’s side is 8 inches. This number is the square root of 64.

  3. Find the block’s side

    Use the volume formula backward. Which number used three times makes 64? Try small numbers. 3 × 3 × 3 = 27 is too small.

    4×4×4=644 \times 4 \times 4 = 64

    The block’s side is 4 inches. This number is the cube root of 64.

  4. Write and simplify the ratio

    Put the puzzle first, as the question asks. Then divide both parts by 4.

    8:4=(8÷4):(4÷4)=2:18 : 4 = (8 \div 4) : (4 \div 4) = 2 : 1

    Watch out: Writing the block’s side first gives 1 : 2, choice B. The order in a ratio matters.

    So the answer is D, 2 : 1.

Check your answer

Check both sides. 8 × 8 = 64 square inches, and 4 × 4 × 4 = 64 cubic inches. The puzzle’s side is longer, so its number in the ratio must be bigger. Only choice D puts the bigger number first.

More questions like this

The original DOE question
Question 100 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 100. Open page 203 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 17 of 22

2023 Test B Math, question 46

Question

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

What is the value of the expression above, in decimal form?

  • A0
  • B0.5
  • C1.5
  • D2.5
Show the answer

Correct answer: D

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

=0+1+12+1=0+1+\frac{1}{2}+1

=212=2.5=2\frac{1}{2}=2.5

Question analysis

An exponents question: students find the decimal value of a sum of four powers, zero cubed, four to the zero power, two to the negative first power and negative one squared. Each of the four terms follows a different rule, and a slip on any one of them changes the total.

NYS standards
  • 8.EE.1 integer exponent properties
  • 7.NS.2a rational number multiplication
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 work uses whole-number exponents and signed numbers. Zero and negative exponents are grade 8 content, and this expression tests both alongside a negative base, each with its own common misconception.
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 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

Every term tests a different exponent rule, and each wrong choice comes from breaking exactly one of them. The zero exponent is the most common casualty, since 4⁰ looks as if it should be 0 like 0³; a negative exponent read as a negative number is close behind. Writing each term’s value above it before adding, 0, 1, 1/2 and 1, keeps the four rules separate.

  1. Something you must already know major Working it out

    Three rules the test does not give decide the sum: any nonzero number to the 0 power is 1, 2 to the −1 power is 1/2, and (−1)² is positive 1. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    The four terms carry a zero base, a zero exponent, a negative exponent and a negative base, so a sign or a stray 0 can slip in at each one. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Evaluate each power on its own, using the rules for a zero exponent and a negative exponent and the sign of a squared negative number, then add the four values.

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

Zero exponent
a0=1a^0 = 1
a is any number except 0. An exponent of 0 always gives 1. It works because each step down in the exponent divides by a. For example, 4 to the first power is 4, and 4 ÷ 4 = 1.
Negative exponent
a−n=1ana^{-n} = \frac{1}{a^n}
a is the base, and n is a positive whole number. A negative exponent does not make the number negative. It means 1 divided by the base to the positive power. For example, 2 to the −1 power is one half.

  1. Sort out the question

    The expression adds four powers. A power is a base with an exponent, a small raised number. The exponent tells how many times to use the base as a factor, a number you multiply. Find each power, then add.

  2. Find the first two powers

    Zero times itself is still zero.

    03=0×0×0=00^3 = 0 \times 0 \times 0 = 0

    Next, use the zero exponent rule.

    40=14^0 = 1

    Watch out: 404^0 is 1, not 0. Taking it as 0 gives choice C.

  3. Find the negative power

    Use the negative exponent rule.

    2−1=121=12=0.52^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5

    Watch out: 2−12^{-1} is not −2. That mistake gives choice A.

  4. Find the last power

    The base is −1, inside the parentheses. Multiply it by itself. A negative times a negative is positive.

    (−1)2=(−1)×(−1)=1(-1)^2 = (-1) \times (-1) = 1

    Watch out: The answer is not −1. That mistake gives choice B.

  5. Add the four values

    Now add them. The question wants a decimal, so write one half as 0.5.

    0+1+0.5+1=2.50 + 1 + 0.5 + 1 = 2.5

    So the answer is D, 2.5.

Check your answer

See why 40=14^0 = 1 with a pattern. 42=164^2 = 16 and 41=44^1 = 4. Each step down divides by 4, so 40=4÷4=14^0 = 4 \div 4 = 1. In the same way, 20=12^0 = 1 and 2−1=1÷2=122^{-1} = 1 \div 2 = \frac{1}{2}.

More questions like this

The original DOE question
Question 103 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 103. Open page 204 of the handbook (PDF)

Question 18 of 22

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 19 of 22

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 20 of 22

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 21 of 22

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 22 of 22

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).