Practice Test

Variables instead of numbers

Question 1 of 51

2026 Practice Test B Math, question 24

Question

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

A40%
B60%
C64%
D80%

Show the answer

Correct answer: C

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

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

End of second week: $0.80(0.80x) = 0.64x$ or 64%

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 A typical grade 7 percent task applies one increase or decrease to one amount. This one applies a weekly decrease twice in a row, an early form of exponential decay that goes beyond typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer major
Turning words into math moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

A very close wrong answer major Choosing the answer
The choices 60% and 64% sit four percentage points apart, so a rough estimate cannot choose between them; only seeing that the second 20% is taken from 80, not 100, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math moderate Understanding the question
The phrase “the amount at the beginning of that same week” sets each week’s base, so the second week’s 20% is measured from what was left after the first week. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
No starting amount is given, so students must pick one, such as 100 grams, or work with a percent of a percent, 0.8 × 0.8. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Adds the weekly losses, 20% + 20% = 40%, and gives that figure, which describes the amount lost rather than the percentage still left.
B Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Subtracts 20% of the original amount twice, 100 − 20 − 20 = 60, though the second week’s 20% comes from the 80% left, not the starting amount.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that 80% remains after the first week and stops, never taking the second week’s 20% from that remainder.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Each week the chemical loses 20% of what it has. The key words are “the amount at the beginning of that same week.” So each week’s 20% comes from that week’s starting amount. No amount is given. You need the percent of the original amount that is left after two weeks.
Pick an easy starting amount
Pretend the chemical starts at 100 grams. With 100, the number of grams left is the same as the percent left. Any starting amount gives the same percent, so choose the easiest one.
Work out the first week
Use the percent formula. The first week starts with 100 grams:
$\frac{20}{100} \times 100 = 20$
The chemical loses 20 grams, so $100 - 20 = 80$ grams are left. That is 80% after one week.
Watch out: 80% is only the first week. Stopping here gives choice D.
Work out the second week
The second week starts with 80 grams, not 100. So take 20% of 80:
$\frac{20}{100} \times 80 = 16$
This time the chemical loses only 16 grams, so $80 - 16 = 64$ grams are left.
Watch out: taking away 20 grams again uses the original amount. That gives 60%, choice B. Choice A, 40%, adds the two losses. But the question asks what is left, and only 36% was lost.
Read the answer as a percent
Week Start (grams) Loss (grams) Left (grams)
1 100 20 80
2 80 16 64
Another way: each week keeps 80%, or 0.8, of its start. Two weeks gives $0.8 \times 0.8 = 0.64$ , which is 64%.
64 grams out of the original 100 grams is 64%. So the answer is C, 64%.

Week	Start (grams)	Loss (grams)	Left (grams)
1	100	20	80
2	80	16	64

Check your answer

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

More questions like this

The original DOE question

Question 24 as shown on the DOE’s digital test, page 96 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 96, question 24.

Question 2 of 51

2026 Practice Test C Math, question 5

Question

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

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

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

Show the answer

Correct answer: 4n – 12

Let n represent the number of pages that Maya read.

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

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

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

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

Ali’s pages $= 4n - 12$

Question analysis

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

NYS standards: 6.EE.6 variables representing unknown numbers
6.EE.2a writing algebraic expressions
6.EE.3 generating equivalent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Writing an expression from words is grade 6 content, but this chains three relationships, and the last compares the pair with Ali rather than describing Ali directly. Typing the answer with no choices to compare adds risk.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 52 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 3 of 51

2026 Practice Test C Math, question 18

Question

I. $s - t$
II. $st^{2}$
III. $s^{t}$

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

AII only
BI and II only
CII and III only
DI, II, and III

Show the answer

Correct answer: D

Each of the three expressions must be considered:

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

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

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

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know moderate Working it out
Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)
Every choice must be checked minor Choosing the answer
Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Trusts only the squared expression, believing that subtracting t must make s smaller and that a negative exponent makes a negative number.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Handles the subtraction and the square correctly but believes a negative exponent makes the result negative, though s to the power t is a positive fraction such as 1/8.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads s − t as taking a number away from s, so thinks it could be negative, missing that subtracting the negative t adds to s.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.
Try one pair of numbers
Try s = 2 and t = −3 in each expression:
- I: $2 - (-3) = 2 + 3 = 5$
- II: $2 \times (-3)^{2} = 2 \times 9 = 18$
- III: $2^{-3} = \frac{1}{2^{3}} = \frac{1}{8}$
All three are positive this time. Now see why that always happens.
Explain expression I
Use the rule for subtracting:
$s - t = s + (-t)$
t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.
Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.
Explain expression II
$t^{2}$ means t × t. A negative times a negative is positive, so $t^{2}$ is always positive. Then $s \times t^{2}$ is a positive times a positive. That is always positive too.
Explain expression III
Use the rule for a negative exponent:
$s^{t} = \frac{1}{s^{-t}}$
−t is a positive integer, so $s^{-t}$ is a positive integer. So $s^{t}$ is 1 over a positive number. That is a positive fraction.
Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.
Choose the answer
Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.
So the answer is D, I, II, and III.

Check your answer

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

More questions like this

The original DOE question

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

Question 4 of 51

2026 Practice Test C Math, question 28

Question

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

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

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

z =

Show the answer

Correct answer: z = 17w

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

First, multiply both sides by 4:

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

$z + 3w = 20w$

Then subtract $3w$ from both sides:

$z = 17w$

Question analysis

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

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve two-step equations with numbers, such as (x + 3)/4 = 5. Solving for one letter in terms of another is grade 8 work, and the typed answer gives no choices to check against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 5 of 51

2026 Practice Test C Math, question 35

Question

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

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

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

a =

Show the answer

Correct answer: a = 10b

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

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

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

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

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

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

$2a = 20b$

$a = 10b$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 6 of 51

2026 Practice Test C Math, question 38

Question

This formula can be used to determine p, the number of pages that will be in a book that contains w words, with an average of r words per page, and f extra pages in the front for the title and publishing information.

$p = \frac{1}{r}w + f$

If the formula is solved for w, it can be used to find the number of words in a book with p pages. What will the formula be if it is solved for w?

A $w = r(p - f)$
B $w = \frac{1}{r}p + f$
C $w = f + rp$
D $w = r + p + f$

Show the answer

Correct answer: A

Apply the properties of equality to solve for w.

$p = \frac{1}{r}w + f$

Subtract f from both sides of the equation.

$p - f = \frac{1}{r}w$

Multiply both sides by r.

$r(p - f) = r\left(\frac{1}{r}w\right)$

$r(p - f) = w$

$w = r(p - f)$

Question analysis

A formula gives the number of pages p in a book from its number of words w, the average words per page r, and the extra front pages f. Students choose the formula solved for w. It tests rearranging a formula with inverse operations, and with only letters there are no numbers to check, so the order of the steps matters.

NYS standards: 8.EE.7 linear equations in one variable
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve two-step equations such as 5 = x/4 + 2 with numbers. Rearranging a formula with four letters to solve for a different one is grade 8 work, with no numbers to check against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 82 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The formula is short, but solving it for w means undoing two operations in reverse order, with letters only. The tempting shortcut multiplies p by r and leaves the f untouched, which yields a formula that looks as finished as the right one. Swapping p and w without undoing anything is the other quick trap. Substituting a small book, 5,000 words at 250 a page with 4 front pages, shows which formula gives back 5,000.

Variables instead of numbers moderate Understanding the question
Four letters and no numbers mean each choice has to be judged by its structure, unless students pick values such as r = 250, w = 5,000 and f = 4 to test it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Two inverse operations must come in order, subtracting f before multiplying by r; multiplying first means the f would have to be multiplied by r as well. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Treat it like a two-step equation with numbers: undo what was done to w last, then the step before, applying each operation to the whole of the other side.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats solving for w as trading the places of p and w, keeping the 1/r and the + f exactly where they were.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies p by r to undo the 1/r but leaves the + f alone, neither subtracting it first nor multiplying it by r too.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the other three letters, r + p + f, keeping the formula’s letters but none of its multiplication or division.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying by the reciprocal: $r \times \frac{1}{r} = 1$; r is any number except 0. The reciprocal of r is 1/r, the number flipped upside down. A number times its reciprocal is always 1. So multiplying by r undoes the 1/r in front of w and leaves w by itself.

Sort out the question
The formula is $p = \frac{1}{r}w + f$ . Here p is the number of pages, w the number of words, r the words on each page, and f the extra front pages.
You need the formula solved for w. That means w alone on one side of the equal sign.
See what happens to w
Read the formula like a recipe. Start with w. First it is multiplied by $\frac{1}{r}$ , which is the same as dividing by r. Then f is added.
To get w back, undo the steps in reverse order. Take away f first. Then undo the $\frac{1}{r}$ .
Subtract f from both sides
Taking f from both sides keeps them equal:
$p - f = \frac{1}{r}w$
Now the only thing left to undo is the $\frac{1}{r}$ .
Multiply both sides by r
Use the reciprocal rule. Multiply both whole sides by r:
$r(p - f) = r \times \frac{1}{r}w = w$
So $w = r(p - f)$ .
Watch out: $w = f + rp$ , choice C, multiplies only p by r. The f must be taken away before you multiply. Or, if you multiply first, every term gets multiplied by r.
Give the answer
Choice B only trades the places of p and w, and it undoes nothing. Choice D just adds the letters.
So the answer is A, $w = r(p - f)$ .

Check your answer

Try a book with 5,000 words, 250 words a page and 4 front pages. The first formula gives $p = 5{,}000 \div 250 + 4 = 24$ pages. Choice A gives back the words: $250 \times (24 - 4) = 5{,}000$ . Choice C gives $4 + 250 \times 24 = 6{,}004$ , which is wrong.

More questions like this

The original DOE question

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

Question 7 of 51

2026 Practice Test C Math, question 43

Question

A softball player bought bottles of water for her team to have at practice. The total number of bottles she bought, b, is proportional to n, the number of players at practice. Which equation represents the relationship between b and n? The constant of proportionality is p.

A $b = pn$
B $b = p + n$
C $b + n = p$
D $b + n + p = 1$

Show the answer

Correct answer: A

In this context, the constant of proportionality, p, represents the relationship between b, the number of bottles of water, and n, the number of players on the team.

The number of bottles of water is the product of the constant of proportionality and the number of players.

The equation that represents this relationship is $b = pn$ .

Question analysis

A softball player buys bottles of water for practice, and the number of bottles, b, is proportional to the number of players, n, with constant of proportionality p. Students choose the equation that represents the relationship. It tests writing an equation for a proportional relationship using letters only, so the meaning of proportional has to be clear.

NYS standards: 7.RP.2c proportional relationship equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students write equations for proportional relationships from tables, graphs and descriptions, usually with a number to find. Here the constant is already given as a letter, so only the general form must be recalled, a short task.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly remember
What makes it hard: Something you must already know minor
Variables instead of numbers minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 51 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a recall question dressed in a short story: the softball team adds nothing but names for the letters. The one idea needed is that proportional means multiplied by a constant, so doubling the players doubles the bottles. Students who think of a constant as something added, or as a fixed total, are drawn to the two sums that use the same letters.

Something you must already know minor Working it out
The general form of a proportional relationship, one amount equal to the constant times the other, has to be recalled; the stem names p but never says it multiplies. (Anderson & Krathwohl, 2001)
Variables instead of numbers minor Understanding the question
No numbers are given, only b, n and p, so a choice cannot be checked by computing unless students make up a case, such as 2 bottles per player for 10 players. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Recall how two proportional quantities are linked through the constant of proportionality, then test each choice with sample numbers: if the players double, the bottles should double too.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats proportional as adding a fixed amount, so the bottles would be the players plus p, rather than the players multiplied by a constant.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes the constant of proportionality as the total of bottles and players, confusing a constant sum with a constant ratio.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Puts all three letters into one sum equal to 1, an equation that uses the right letters but matches no idea of proportion, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx$; y and x are two amounts that change together. k is the constant of proportionality, the number you always multiply x by to get y. Because it is a multiplication, doubling x doubles y.

Sort out the question
The letter b is the number of bottles, and n is the number of players. The bottles are proportional to the players. The constant of proportionality is p.
You need the equation that links b and n. There are no numbers, only letters.
Recall what proportional means
Proportional means one amount is always a fixed number times the other. For example, say there are 2 bottles for each player. Then 10 players need 20 bottles, and 20 players need 40 bottles. Doubling the players doubles the bottles.
The fixed number, here 2, is the constant of proportionality.
Write it with these letters
Start with the rule $y = kx$ . The bottles take the place of y, the players take the place of x, and p is the constant:
$b = pn$
Here pn means p times n.
Check the other choices
Watch out: choice B, $b = p + n$ , adds instead of multiplying. With p = 2, 10 players give 12 bottles. Then 20 players give 22, which is not double.
Choice C makes p a total of bottles and players, not a multiplier. Choice D makes all three letters add up to 1, which does not fit at all.
Give the answer
Only $b = pn$ multiplies the players by a constant. So it is the only choice where doubling the players doubles the bottles.
So the answer is A, $b = pn$ .

Check your answer

Try p = 3 bottles per player. With 5 players, $b = 3 \times 5 = 15$ . With 10 players, $b = 3 \times 10 = 30$ . Twice the players, twice the bottles. Also, $15 \div 5 = 3$ , so b ÷ n gives back the constant.

More questions like this

The original DOE question

Question 43 as shown on the DOE’s digital test, page 116 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 116, question 43.

Question 8 of 51

2026 Practice Test D Math, question 2

Question

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

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

Show the answer

Correct answer: D

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

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

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

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

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

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

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

$x + 2y - 2x = z$

$2y - x = z$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 9 of 51

2026 Practice Test D Math, question 3

Question

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

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

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

What is the value of $x - y$ ?

A−7
B−5
C5
D7

Show the answer

Correct answer: B

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

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

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

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

$x + 16 = y + 11$

$x = y - 5$

$x - y = -5$

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 10 of 51

2026 Practice Test D Math, question 6

Question

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

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

Show the answer

Correct answer: B

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

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

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

4 pounds is

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

5 pounds is

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

p pounds is

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 11 of 51

2026 Practice Test D Math, question 9

Question

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

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

Select the three correct answers.

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

A $-m + k$
B $-m \cdot k$
C $m - k$
D $-k \cdot m$
E $\frac{m}{k}$

Show the answer

Correct answer: A, B, and D

Each of the expressions must be considered:

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

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

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

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

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

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

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

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

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

A negative divided by a positive remains negative.

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 12 of 51

2026 Practice Test D Math, question 48

Question

The perimeter of square W is twice the perimeter of equilateral triangle X. What is the ratio of the length of one side of square W to the length of one side of equilateral triangle X?

A3:8
B2:3
C3:2
D2:1

Show the answer

Correct answer: C

First, set up the ratio of the perimeter of the square to the triangle as 2:1 because the question states that the perimeter of the square is twice the perimeter of the triangle.

Recognizing that the square has 4 equal sides and that an equilateral triangle has 3 sides, use division to find the ratio of the sides.

Perimeter: $2 : 1 = \frac{2}{1}$

Sides:

$\frac{\frac{2}{4}}{\frac{1}{3}} = \frac{2}{4} \div \frac{1}{3} =$

$\frac{2}{4} \times \frac{3}{1} = \frac{6}{4} =$

$\frac{3}{2} = 3 : 2$

Question analysis

The perimeter of square W is twice the perimeter of equilateral triangle X, and students find the ratio of a side of the square to a side of the triangle. This ratio question uses perimeters of regular shapes with no lengths given. Turning a comparison of perimeters into one of sides, and keeping the ratio in the order asked, are the tricky parts.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4 equations and inequalities from problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 ratio problems usually supply numbers to work with. This one gives only a relationship between two perimeters, so students must bring in the number of sides of each shape themselves and keep the ratio in the order asked.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question turns a statement about perimeters into a ratio of sides, and the natural guess, that the sides keep the perimeters’ 2 to 1, ignores that the square shares its perimeter among 4 sides and the triangle among 3. Choosing a triangle perimeter such as 12 makes the comparison concrete. The order of the ratio still has to be checked at the end, because its reverse is listed too.

Turning words into math moderate Understanding the question
“The perimeter of square W is twice the perimeter of equilateral triangle X” must become 4s = 2 × 3t, with the 2 on the triangle’s side; written backward, it gives the triangle the larger perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers moderate Understanding the question
No side length is given, so students must either work with letters for the two sides or choose a convenient perimeter themselves, such as 12 for the triangle. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The ratios 3:2 and 2:3 use the same two numbers, so only keeping the square’s side first, as the question orders it, separates the answer from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Pick an easy side length for the triangle, find its perimeter, double that for the square, divide by the square's number of sides, and write square to triangle.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

A Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: weak
Writes the comparison backward, 2 × 4s = 3t, giving the triangle the larger perimeter, so the ratio of the sides comes out 3:8.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Finds the right relationship but turns the ratio upside down, giving the triangle’s side to the square’s instead of the square’s side first.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Assumes the sides are in the same ratio as the perimeters, 2:1, ignoring that the square has 4 sides and the triangle only 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a square: $P = 4s$; P is the perimeter, the distance around the shape. s is the length of one side. A square has 4 equal sides, so its perimeter is 4 times one side.
Perimeter of an equilateral triangle: $P = 3t$; An equilateral triangle has 3 equal sides. t is the length of one side. So its perimeter is 3 times one side.
Ratio: $a : b$; A ratio compares two amounts in order. a : b means a compared with b. The order matters, so write the amounts in the order the question names them.

Sort out the question
Square W has a perimeter twice as long as the perimeter of triangle X. Equilateral means all 3 sides of X are equal.
You need the ratio of one side of the square to one side of the triangle, in that order. No lengths are given, so you can pick a size to try.
Pick a size for the triangle
Choose an easy side length for the triangle, such as 2 units. All 3 sides are 2, so its perimeter is:
$P = 3 \times 2 = 6$
Find the square’s side
The square’s perimeter is twice the triangle’s:
$2 \times 6 = 12$
The square has 4 equal sides, so divide by 4:
$s = 12 \div 4 = 3$
Each side of the square is 3 units.
Write the ratio in order
The square comes first, then the triangle. Their sides are 3 and 2, so the ratio is 3 : 2.
Watch out: 2 : 1, choice D, compares the perimeters, not the sides. The square splits its perimeter into 4 sides, the triangle into only 3. Choice B, 2 : 3, puts the triangle first. Choice A, 3 : 8, comes from giving the triangle the bigger perimeter.
So the answer is C, 3:2.

Check your answer

Try a different size. Give the triangle sides of 4. Its perimeter is 12, so the square’s is 24. Each side of the square is $24 \div 4 = 6$ . The ratio 6 : 4 simplifies to 3 : 2. The size does not change the ratio.

More questions like this

The original DOE question

Question 48 as shown on the DOE’s digital test, page 121 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 121, question 48.

Question 13 of 51

2024 Test A Math, question 7

Question

If $\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}$ and $x \neq 0$ , what expression is represented by $\square$ ?

A $y$
B $2y$
C $3y$
D $4y$

Show the answer

Correct answer: C

Use properties of equations to determine the missing expression:

$\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}$

$\frac{2}{2} \bullet \frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}$

$\frac{4y}{2x} - \frac{y}{2x} = \frac{\square}{2x}$

$\frac{4y - y}{2x} = \frac{\square}{2x}$

$\frac{3y}{2x} = \frac{\square}{2x}$

The missing expression is $3y$ .

Question analysis

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

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students subtract fractions with unlike denominators and combine like terms, but rarely see a variable in a denominator. The steps are the same two as with numbers, so the demand is typical once the form is familiar.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
The denominators are x and 2x, so the common denominator is built with a letter, and there is no number to check unless students substitute values such as x = 1 and y = 2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Only the first fraction changes: 2y/x becomes 4y/2x with top and bottom both doubled, while y/2x already has the denominator the box sits over. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Subtracts the numerators straight away, 2y − y = y, and puts the result over 2x as if the two fractions already shared a denominator.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts the numerators over x, (2y − y)/x = y/x, then doubles top and bottom to reach the denominator 2x, giving 2y/2x.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Rewrites 2y/x as 4y/2x correctly to match the box's denominator, then stops before subtracting the y of the second fraction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The box, $\square$ , stands for a missing expression. An expression is a math phrase made of numbers and letters, like 5y. You need what goes in the box to make the two sides equal. $x \neq 0$ means x is not zero, so no fraction divides by zero.
Pick a shared bottom
You can only subtract fractions when their bottoms match. The bottoms here are x and 2x. The box sits over 2x, so use 2x for both. The second fraction, $\frac{y}{2x}$ , already has it.
Rewrite the first fraction over 2x
To turn x into 2x, multiply by 2. Multiply the top by 2 as well, so the value stays the same.
$\frac{a}{b} = \frac{a \times k}{b \times k}$
$\frac{2y}{x} = \frac{2y \times 2}{x \times 2} = \frac{4y}{2x}$
Watch out: this is only the first fraction. Stopping here and choosing 4y gives choice D.
Subtract the tops
$\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}$
$\frac{4y}{2x} - \frac{y}{2x} = \frac{4y - y}{2x} = \frac{3y}{2x}$
4y take away y is like 4 apples take away 1 apple. That leaves 3y. Watch out: do not subtract the tops, 2y − y, before the bottoms match. That gives y, choice A.
Match the box
The left side equals $\frac{3y}{2x}$ . The right side is $\frac{\square}{2x}$ . The bottoms are the same, so the tops must be equal. The box is 3y. So the answer is C, $3y$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 51

2024 Test A Math, question 9

Question

A list of consecutive integers begins with m and ends with n. If $n - m = 66$ , how many integers are in the list?

Show the answer

Correct answer: D

Consider the case where $m = 1$ .

$n - m = 66$

$n - 1 = 66$

$n - 1 + 1 = 66 + 1$

$n = 67$

The list of integers from 1 to 67 contains 67 numbers.

Question analysis

A number question about a list of consecutive integers that starts at m and ends at n, where n minus m is 66. Students decide how many integers the list contains. The tricky part is relating the difference between the two ends to the number of integers in the list.

NYS standards: 7.NS.1c rational number subtraction
6.EE.6 variables representing unknown numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Consecutive integers and differences are familiar grade 7 ideas, and the arithmetic is trivial. What keeps this at grade 7 rather than below it is the counting trap, which needs a moment of careful reasoning.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers moderate
Solution: 2 steps
Reading level: Question and choices: about grade 7, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is trivial, so the question tests one idea: the number of integers from m to n is one more than their difference, because both ends belong to the list. The stem supplies 66 ready-made, and taking it as the count is the natural first reaction. A five-number list, 1 to 5, whose ends differ by 4, shows the extra one in seconds.

A very close wrong answer moderate Choosing the answer
The choices 66 and 67 differ by one, and 66 is the difference the stem hands over; only counting both ends of the list separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers moderate Understanding the question
The list is given only by its ends, m and n, so the count has to be reasoned out in general, or checked on a small list such as 1 through 5. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Try a small list you can count, such as 1 through 5, and compare its count with the difference between its first and last numbers.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working counts only the two named ends, m and n, in a list said to hold consecutive integers, so 2 mostly draws guesses.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Halves 66, as if consecutive integers went up by 2 like consecutive odd or even numbers, so the list held only every other integer.
C Counting the ends (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes the difference of the ends as the count, forgetting that a list from m to n includes both m and n.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting consecutive integers: $\text{count} = n - m + 1$; m is the first number in the list and n is the last. n − m counts the steps from each number to the next. A list always has one more number than it has steps, so you add 1. This works for any list that goes up by 1.

Sort out the question
Integers are whole numbers and their negatives, like −2, 0 and 5. Consecutive integers come one after another, going up by 1, like 4, 5, 6. The list starts at m and ends at n. The difference $n - m$ is 66. You need how many integers are in the list.
Try a small list
Letters are hard to count, so try a list you can count: 1, 2, 3, 4, 5. Here m = 1 and n = 5.
$n - m = 5 - 1 = 4$
But the list has 5 numbers, not 4. The difference counts the steps: 1 to 2, 2 to 3, 3 to 4 and 4 to 5. Both ends are in the list, so there is one more number than steps.
Use the counting rule
$\text{count} = n - m + 1$
$\text{count} = 66 + 1 = 67$
Watch out: 66 is the difference, not the count. Choosing 66, choice C, leaves out one end of the list. Also, consecutive integers go up by 1, not 2, so halving 66 to get 33 is wrong. So the answer is D, 67.

Check your answer

Try m = 1. Then $n = 1 + 66 = 67$ . The list 1, 2, 3, and so on up to 67, plainly has 67 numbers. Any other start gives the same count, because the list just slides along.

More questions like this

The original DOE question

Question 66 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 66. Open page 78 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 51

2024 Test A Math, question 12

Question

If n is a whole number, and 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ , what is the value of n?

Show the answer

Correct answer: D

$0.01 = \frac{1}{100}$ , which is between $\frac{1}{101}$ and $\frac{1}{99}$ .

Therefore, $n + 2 = 101$ and $n = 99$ .

Question analysis

A number sense question: 0.01 lies between 1 over n and 1 over n + 2, where n is a whole number, and students find n. It tests comparing a decimal with unit fractions and knowing how a fraction changes as its denominator grows. The unusual form, with n in two denominators, is what slows students down.

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students compare fractions and decimals routinely, but not with an unknown in the denominators. The comparison is simple once 0.01 is written as a fraction; the unfamiliar setup is the main demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 18 words.
Students, last 30 days: 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 setup looks unfamiliar, with n in two denominators, but one rewrite makes it routine: 0.01 is 1/100, so the two denominators must sit on either side of 100. The three small choices fail a quick check, since every fraction from 1/1 to 1/4 is far above 0.01, and 0 cannot be a denominator at all. Students who test the choices instead of rewriting the decimal spend a little longer but reach the same place.

Variables instead of numbers moderate Understanding the question
The unknown sits in two denominators, n and n + 2, so students must reason about which unit fractions lie on either side of 0.01 rather than solve an equation. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Two facts are assumed: 0.01 equals 1/100, and a unit fraction shrinks as its denominator grows, so 1/99 is just above 0.01 and 1/101 just below. (Anderson & Krathwohl, 2001)

Strategy: Rewrite 0.01 as a unit fraction, recall that a larger denominator gives a smaller fraction, and test each choice in both 1/n and 1/(n + 2).

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats 1/0 as 0, so 0.01 seems to lie between 0 and 1/2; but division by zero is undefined, so n cannot be 0.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 1 from 0.01 or from 1/n as the value of n, without checking that 0.01 is far below both 1 and 1/3.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 2 from n + 2, or the two decimal places of 0.01, without testing 1/2 and 1/4, which are both far above 0.01.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; A unit fraction has 1 on top, like one fifth. a and b are positive whole numbers on the bottom. A bigger bottom cuts the whole into more pieces, so each piece is smaller. So the fraction with the bigger bottom is the smaller number.

Sort out the question
n is a whole number: 0, 1, 2, 3 and so on. The number 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ . Between means one fraction is bigger than 0.01 and the other is smaller. You need the value of n.
Write 0.01 as a fraction
0.01 means one hundredth.
$0.01 = \frac{1}{100}$
Now all three numbers are fractions with 1 on top. That makes them easy to compare.
Find fractions on each side
With 1 on top, a bigger bottom makes a smaller fraction.
$\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$
So $\frac{1}{99}$ is a little more than $\frac{1}{100}$ , and $\frac{1}{101}$ is a little less.
$\frac{1}{101} < \frac{1}{100} < \frac{1}{99}$
The bottoms 99 and 101 are 2 apart, just like n and n + 2.
Match the fractions to n
The smaller bottom is n, and the larger bottom is n + 2.
$n = 99 \text{ and } n + 2 = 101$
Both fit. Watch out: small choices do not work. For n = 1, the fractions are 1 and $\frac{1}{3}$ . Both are far bigger than 0.01. For n = 0, $\frac{1}{0}$ is not a number, because you can’t divide by zero. So the answer is D, 99.

Check your answer

Write the fractions as decimals. $\frac{1}{99} \approx 0.0101$ , a bit more than 0.01. $\frac{1}{101} \approx 0.0099$ , a bit less. The sign ≈ means about. So 0.01 sits right between them.

More questions like this

The original DOE question

Question 69 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 69. Open page 78 of the handbook (PDF)

Question 16 of 51

2024 Test A Math, question 20

Question

A piece of wood that is $4\frac{1}{2}$ feet long is cut into 2 pieces of different lengths. The shorter piece has a length of x feet. Which inequality expresses all possible values of x?

A $0 < x < 2\frac{1}{4}$
B $0 \leq x \leq 2\frac{1}{4}$
C $0 < x < 4\frac{1}{2}$
D $2\frac{1}{4} < x < 4\frac{1}{2}$

Show the answer

Correct answer: A

The original piece of wood is $4\frac{1}{2}$ feet long.

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

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

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

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

Question analysis

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

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve and graph simple inequalities. Writing a compound inequality from a description, where the words decide whether each end is strict, is less familiar and calls for careful reading of every condition.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
Two phrases set the two ends: "cut into 2 pieces" means x is more than 0, and "of different lengths" with "the shorter piece" means x is less than half of 4 1/2. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
The inequalities 0 < x < 2 1/4 and 0 ≤ x ≤ 2 1/4 share both bounds and differ only in their symbols; only checking whether each end can happen separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
The question asks for every possible length of the shorter piece, a range rather than one length, so students must picture the cut moving along the board. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Finds the right bounds but includes both ends, allowing a piece of 0 feet or two equal halves, though the stem says 2 pieces of different lengths.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Ignores that x is the shorter piece and allows any length up to the whole 4 1/2-foot board.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Describes the longer piece, whose length lies between half the board and the whole board, instead of the shorter piece the stem calls x.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
A board is $4\frac{1}{2}$ feet long. It is cut into 2 pieces of different lengths. The shorter piece is x feet long. You need an inequality for every possible value of x. An inequality uses signs like <, less than, to show a range of numbers.
Find the top end for x
If you cut exactly in the middle, each piece is half the board.
$4\frac{1}{2} \div 2 = \frac{9}{2} \times \frac{1}{2} = \frac{9}{4} = 2\frac{1}{4}$
But the pieces have different lengths, so the cut is not in the middle. The shorter piece must be less than half.
$x < 2\frac{1}{4}$
Find the bottom end for x
The shorter piece can be very small, like 0.1 foot. But it can’t be 0 feet. Then nothing would be cut off, and there would not be 2 pieces.
$0 < x$
Put the two ends together
$0 < x < 2\frac{1}{4}$
This says x is more than 0 and less than $2\frac{1}{4}$ . Watch out: choice B uses ≤, which allows x = 0 or $x = 2\frac{1}{4}$ . Neither can happen. Choice C lets x be more than half, so it would not be the shorter piece. Choice D describes the longer piece. So the answer is A, $0 < x < 2\frac{1}{4}$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 17 of 51

2024 Test A Math, question 23

Question

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

A $x$
B $2x$
C $2x - 2$
D $2x - 4$

Show the answer

Correct answer: B

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

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

Question analysis

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

NYS standards: 6.EE.2a writing algebraic expressions
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students write and combine linear expressions; here the expressions describe neighboring integers, a familiar idea from consecutive-integer problems. Two short steps and a quick number check put it at typical grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
Here x stands for any odd number, so the two even neighbors must be written as x − 1 and x + 1; trying x = 7, with neighbors 6 and 8, makes it concrete. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The phrase "the two even numbers closest to x" means one just below and one just above x, not two in a row on the same side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the number midway between the two even neighbors, x itself, which is their mean rather than their sum.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Treats x as if it were even and adds it to the even number below it, x + (x − 2), ignoring that x is odd.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Takes the two even numbers just below x, x − 1 and x − 3, instead of the closest one on each side.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
x is an odd number. You need the sum of the two even numbers closest to x. Sum means the result of adding. In terms of x means write the answer using the letter x.
Try a real number first
Pick an odd number, like 7. The even numbers closest to 7 are 6 and 8, one on each side.
$6 + 8 = 14$
Notice that 14 is $2 \times 7$ .
Write the neighbors with x
Odd and even numbers take turns. So both numbers right next to an odd x are even.
$x - 1 \text{ and } x + 1$
For x = 7, these are 6 and 8. Watch out: take one even number below x and one above. Using two even numbers below x, $x - 1$ and $x - 3$ , gives $2x - 4$ , choice D.
Add the two neighbors
$(x - 1) + (x + 1) = x + x - 1 + 1 = 2x$
The −1 and the +1 cancel, and $x + x = 2x$ . Watch out: x is the number in the middle, not the sum, so choice A is wrong. So the answer is B, $2x$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 18 of 51

2024 Test A Math, question 25

Question

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

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

A $\overline{PQ}$
B $\overline{NP}$
C $\overline{MQ}$
D $\overline{MP}$

Show the answer

Correct answer: A

These facts are given:

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

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

Question analysis

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

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

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
Every length is a letter: MN is x, MP is y, and the target is the expression 2x − y, so the matching segment cannot be found by measuring. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph moderate Understanding the question
The figure marks only MN and MP with arrows; the length of MQ comes from the stem's midpoint fact, not from the drawing, and has to be written as 2x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

Traps

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

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Leaves out the midpoint fact, so the only difference it can read from the arrows is NP = y − x, which it takes for 2x − y.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Uses the midpoint to find that MQ is 2x centimeters and picks that segment, stopping before subtracting the y of MP.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Sees y in the expression and picks MP, the segment labeled y, without working out the rest of 2x − y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the figure
Points M, N, P and Q lie in that order on a line. Segment MN is x cm long, and segment MP is y cm long. N is the midpoint of $\overline{MQ}$ , the point exactly halfway from M to Q. You need the segment that is $2x - y$ cm long.
Find the length of MQ
MN is half of MQ, so MQ is twice as long.
$MQ = 2 \times MN = 2x$
MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.
Take MP away from MQ
Subtracting y from 2x means taking MP away from MQ. MP covers the part from M to P. What is left of MQ is the piece from P to Q.
$MP + PQ = MQ$
$PQ = MQ - MP = 2x - y$
So PQ is $2x - y$ cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is $y - x$ , choice B. So the answer is A, $\overline{PQ}$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 19 of 51

2024 Test A Math, question 46

Question

A graph shows the proportional relationship between the number of test questions a student gets correct, x, and the student’s test score, y. The ordered pair $\left(1, \frac{5}{4}\right)$ is on the graph. What does the y-coordinate of the ordered pair represent in this relationship?

AThe test will last $1\frac{1}{4}$ hours.
BEach test question is worth $1\frac{1}{4}$ points.
CAn average student can answer 5 questions in 4 minutes.
DA student who answers 5 questions correctly will earn 4 points.

Show the answer

Correct answer: B

The graph represents a proportional relationship, which means the y-intercept of the graph is (0, 0).

This means that the point $\left(1, \frac{5}{4}\right)$ represents the unit rate for this situation.

In this situation, x represents the number of questions answered correctly, and y represents the number of points received.

Therefore, for each question answered correctly, the student receives $\frac{5}{4}$ , or $1\frac{1}{4}$ , points.

Question analysis

A proportional relationship links the number of test questions a student gets right, x, to the test score, y, and the point (1, 5/4) lies on its graph. Students decide what the y-coordinate of that point means in the situation. The choices reuse the numbers 5, 4 and 1 1/4 in different ways, so the meaning of each coordinate must be kept straight.

NYS standards: 7.RP.2d points on proportional graphs
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on the graph of a proportional relationship means is named in the grade 7 standards. This item is a direct case, with the graph described in words rather than drawn and only one point to interpret.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Variables instead of numbers minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 73 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question checks one idea: when x is 1, the y-coordinate is the score for one correct answer, the unit rate. The pitfall is reading 5/4 as two separate numbers, 5 and 4, and attaching them to questions and points the wrong way round. Ruling out the choices about hours and minutes is quick, since the relationship never mentions time.

Variables instead of numbers minor Understanding the question
No graph is drawn: the relationship and the point (1, 5/4) exist only in words, so students must say for themselves that x counts correct answers and y counts points. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The point gives 5/4, but two choices write it as 1 1/4 and two split it into 5 and 4, so the right reading must be recognized in another form. (Enright, Morley & Sheehan, 2002)

Strategy: Say in words what x and y stand for, then read the ordered pair as a sentence: when x has the first value, y has the second.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Attaches 1 1/4 to hours of testing, matching the number while ignoring that y measures the test score, not time.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Reads 5/4 as 5 questions in 4 minutes, using the numbers 5 and 4 with units that the relationship never measures.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Reads 5/4 as 5 correct answers for 4 points, turning the rate upside down; the point means 1 correct answer earns 5/4 points.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = k \times x$; x and y are two amounts that change together, here right answers and the test score. k is the constant of proportionality: how much y goes with 1 of x. In a proportional relationship, y is always the same number times x. So when x is 1, y equals k.

Sort out the question
x is the number of questions a student gets right. y is the student’s test score, in points. An ordered pair such as $\left(1, \frac{5}{4}\right)$ lists x first and y second. The y-coordinate is the second number, $\frac{5}{4}$ . You need what that number means.
Read the point as a sentence
In the point, x is 1 and y is $\frac{5}{4}$ . Put those into words: a student who gets 1 question right scores $\frac{5}{4}$ points.
Use the proportional rule
The relationship is proportional. So the score is always the same number times the questions right.
$y = k \times x$
$\frac{5}{4} = k \times 1$
So k is $\frac{5}{4}$ . Every right answer adds the same $\frac{5}{4}$ points. That makes $\frac{5}{4}$ the points for each question.
Write it as a mixed number
A mixed number is a whole number plus a fraction. Four quarters make 1 whole, so five quarters are 1 whole and 1 quarter.
$\frac{5}{4} = 1\frac{1}{4}$
Each question right is worth $1\frac{1}{4}$ points.
Match the choices
Choice B says each question is worth $1\frac{1}{4}$ points. That matches. Watch out: choice D splits $\frac{5}{4}$ into 5 and 4. But 5 right answers earn $5 \times \frac{5}{4} = 6\frac{1}{4}$ points, not 4. Choices A and C talk about time, which this relationship does not measure.
So the answer is B, each test question is worth $1\frac{1}{4}$ points.

Check your answer

Try another point. A student with 4 right answers scores $4 \times \frac{5}{4} = 5$ points. That fits B, since four questions at $1\frac{1}{4}$ points each make 5 points. It does not fit D, which says 5 questions earn 4 points.

More questions like this

The original DOE question

Question 103 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 88, question 103. Open page 88 of the handbook (PDF)

Question 20 of 51

2024 Test B Math, question 8

Question

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

Show the answer

Correct answer: C

$\frac{9}{2x}=\frac{3y}{8}$

$9\bullet 8=72$

$2x\bullet 3y=6xy$

$72=6xy$

$12=xy$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
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.

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)
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

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Matches the denominators, 2x = 8, to get x = 4 and gives the value of x instead of the product of x and y.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 6 in 72 = 6xy, the product of the coefficients 2 and 3, as the answer without dividing 72 by it.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 2 and the 8 from the two denominators, combining numbers in the equation instead of cross-multiplying.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; 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.

Sort out the question
The equation says that two fractions are equal:
$\frac{9}{2x} = \frac{3y}{8}$
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 $x \times y$ , written xy. You do not need x or y alone.
Cross-multiply the fractions
An equation without fractions is easier to work with. So cross-multiply.
$\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$
$9 \times 8 = 2x \times 3y$
Watch out: do not just set the bottoms equal. Writing 2x = 8 gives x = 4, choice A. That is x, not the product.
Multiply out both sides
On the left, $9 \times 8 = 72$ . On the right, multiply the numbers and keep the letters together.
$2x \times 3y = 2 \times 3 \times x \times y = 6xy$
So the equation is now:
$72 = 6xy$
The xy is the product you want, all in one piece.
Divide both sides by 6
xy is being multiplied by 6. To undo that, divide both sides by 6.
$xy = 72 \div 6 = 12$
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: $\frac{9}{2 \times 4} = \frac{9}{8}$ . Right side: $\frac{3 \times 3}{8} = \frac{9}{8}$ . 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 21 of 51

2024 Test B Math, question 36

Question

The integers $(n-1)$ , n, and $(n+1)$ are factors of 168. What is the greatest possible value of n?

Show the answer

Correct answer: B

First, determine all possible factors of 168. One method of doing this is to use a table to list pairs of factors.

Factors of 168
1	168
2	84
3	56
4	42
6	28
7	24
8	21
12	14

Then determine the group of three sequential factors $(n-1)$ , n, and $(n+1)$ where n is as large as possible. This group is 6, 7, and 8, where $n=7$ .

Question analysis

Three consecutive integers, written as n − 1, n and n + 1, must all be factors of 168, and students find the greatest possible value of n. It tests factors and consecutive integers. The catch is that a number can be a factor while its neighbors are not, so every member of the run needs checking.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Listing factors is grade 6 work, but here students must read n − 1, n and n + 1 as three consecutive integers and search the factor list for the largest such run. Pairing algebraic notation with factors matches grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the factors of 168 are listed in order, 1, 2, 3, 4, 6, 7, 8, 12 and up, the runs of three in a row stand out, and the last of them is 6, 7, 8. Students lose points by stopping at the first run they see or by checking only n itself. Testing each choice with its two neighbors, starting from the largest, settles it in under a minute.

Variables instead of numbers minor Understanding the question
The stem names the numbers only as (n − 1), n and (n + 1), which has to be read as three whole numbers in a row that all divide 168. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Each candidate must pass three divisibility tests at once, for itself and for both neighbors, while the search continues for the greatest such n. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: List the factor pairs of 168 in order, then look along the list for three whole numbers in a row; or test each choice together with its two neighbors.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the run 2, 3, 4 among the small factors and stops searching, missing the later run 6, 7, 8 that 168 also contains.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Checks only that 14 divides 168, since 12 × 14 = 168, without testing its neighbors; neither 13 nor 15 is a factor.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Takes the largest factor below 168 itself, 84, for the greatest n, ignoring that 83 and 85 would also have to divide 168.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair: $a \times b = 168$; a and b are whole numbers. If a times b equals 168, then a and b are both factors of 168. A factor divides a number with no remainder. Listing every pair, from small a upward, finds every factor.

Sort out the question
A factor of 168 divides 168 with no remainder. The numbers n − 1, n and n + 1 are three whole numbers in a row, such as 6, 7 and 8. The middle one is n.
All three must be factors of 168. You need the greatest n that works.
List the factor pairs of 168
Try each whole number from 1 up. When it divides 168, write it with its partner. Stop when the pairs meet, after 12 and 14.
Small factor Partner
1 168
2 84
3 56
4 42
6 28
7 24
8 21
12 14
Look for three in a row
Here are all the factors in order:
1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168
The runs of three in a row are 1, 2, 3, then 2, 3, 4, then 6, 7, 8. No other three factors sit side by side.
Pick the greatest middle number
The middle numbers of the runs are 2, 3 and 7. The greatest is 7, so n = 7.
Watch out: Stopping at the run 2, 3, 4 gives 3, choice A. Keep looking. Also, 14 and 84 are factors, but their neighbors are not. Neither 13 nor 15 divides 168, so 14, choice C, fails. For 84, choice D, neither 83 nor 85 is a factor.
So the answer is B, 7.

Check your answer

Divide 168 by each number in the run: $168 \div 6 = 28$ , $168 \div 7 = 24$ and $168 \div 8 = 21$ . All three are whole numbers, so 6, 7 and 8 are all factors.

More questions like this

The original DOE question

Question 93 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 93. Open page 209 of the handbook (PDF)

Question 22 of 51

2024 Test B Math, question 49

Question

Suppose that w, x, y, and z are integers, none of which are zero. Under which set of conditions would the product $x^{2}yz^{3}|w|$ always be positive?

Ax is positive and w is positive.
Bx is negative and w is negative.
Cy is negative and z is negative.
Dx and y are negative, and z and w are positive.

Show the answer

Correct answer: C

First, analyze the role of each variable in the expression $x^{2}yz^{3}|w|$ .

$x^{2}$ will always be a positive number.

y can be a positive or a negative number.

$z^{3}$ can be a positive or a negative number.

$|w|$ will always be a positive number.

The expressions $x^{2}$ and $|w|$ will always be positive. Therefore, making the values of y and z negative would result in the expression $yz^{3}$ being positive, since a negative multiplied by a negative is positive. Thus the given product would always be positive.

Question analysis

Four nonzero integers appear in a product that includes a square, a cube and an absolute value. Students choose the set of conditions that makes the product positive every time. It tests the sign rules for multiplication together with exponents and absolute value, and the trap is judging the sign by counting negative variables without looking at what is done to each.

NYS standards: 7.NS.2a rational number multiplication
6.EE.1 whole-number exponent expressions
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students learn sign rules for products of signed numbers. Here they must reason about four unknown integers under an even power, an odd power and an absolute value, and decide what is always true, which is more abstract than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 58 words.
Students, last 30 days: 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 insight decides the question: x² and |w| are positive whatever x and w are, so only the signs of y and z³ matter. Students who count negative letters instead of negative factors pick conditions that look balanced but leave y as the only negative factor. Students who test a single set of values can be fooled by a choice that leaves y and z unspecified.

Variables instead of numbers moderate Understanding the question
There are four variables and no numbers, and always positive must hold for every allowed value, so one test case can rule a choice out but never confirm it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
Each set of conditions has to be checked against all four factors; the first two fix only x and w and leave y and z free to make the product negative. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
Students must know that a nonzero square and an absolute value are always positive, while an odd power such as z³ keeps the sign of z. (Anderson & Krathwohl, 2001)

Strategy: Work out, factor by factor, which signs each part of the product can take, then test each set of conditions against every factor rather than counting negatives.

Traps

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

The wrong answers, classified

A Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Tests the choice with every variable positive, so the product is positive and it seems to work; y = −1 with z = 1 would make it negative.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Pairs the negatives x and w as if they must cancel, not seeing that x² and |w| are positive anyway, so y and z still decide.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Counts two negative letters, x and y, as an even number of negatives, but x² is positive, so y is the only negative factor.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.

Sort out the question
w, x, y and z are integers, meaning whole numbers or their negatives. None of them is 0. The product $x^2 y z^3 |w|$ means $x^2 \times y \times z^3 \times |w|$ .
You need the conditions that make the product positive every time, for every number that fits them.
Look at x squared and |w|
$x^2$ means x × x. Two positives or two negatives multiply to a positive. So $x^2$ is always positive.
$|w|$ is the absolute value of w, its distance from 0. It is always positive, since w is not 0.
So x and w never change the sign of the product. Their signs do not matter.
Look at y and z cubed
y can be positive or negative. $z^3$ means z × z × z, and it has the same sign as z. For example, $(-2)^3 = -8$ .
So the sign of the whole product is the sign of $y \times z^3$ . It depends only on y and z.
Test each choice
- A and B say nothing about y and z. If y = −1 and z = 1, the product is negative. So they do not always work.
- C: y is negative, and $z^3$ is negative. A negative times a negative is positive. This always works.
- D: y is negative and z is positive. So the product is always negative.
Watch out: D has two negative letters, x and y. But $x^2$ is positive anyway, so y is the only negative factor.
So the answer is C, y is negative and z is negative.

Check your answer

Try numbers for C. Let x = −2, y = −3, z = −1 and w = −5. Then $x^2 \times y \times z^3 \times |w| = 4 \times (-3) \times (-1) \times 5 = 60$ , which is positive. Even with x and w negative, the product stays positive.

More questions like this

The original DOE question

Question 106 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 106. Open page 213 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 23 of 51

2024 Test B Math, question 57

Question

If x and y are both negative integers with $x<y$ , which expression has the least value?

A $x^{2}y$
B $xy^{2}$
C $-x^{2}y$
D $-xy^{2}$

Show the answer

Correct answer: A

Since x and y are both negative integers, both $x^{2}$ and $y^{2}$ are positive.

Therefore, both $x^{2}y$ and $xy^{2}$ are negative, and both $-x^{2}y$ and $-xy^{2}$ are positive.

Because $x<y$ and both x and y are negative, $|y|<|x|$ and $y^{2}<x^{2}$ . So $|y^{2}|<|x^{2}|$ .

Since $|y|<|x|$ and x and y are integers, $|xy^{2}|<|x^{2}y|$ .

So $x^{2}y<xy^{2}$ .

$-xy^{2}=|xy^{2}|\text{ and }-x^{2}y=|x^{2}y|$

Therefore, $x^{2}y<xy^{2}<-xy^{2}<-x^{2}y$ .

Question analysis

Two negative integers x and y, with x less than y, appear in four expressions built from squares and minus signs. Students find the expression with the least value. It tests multiplying negative numbers and comparing their sizes, and the difficulty is tracking how each minus sign and each square changes both the sign and the size of the result.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7d absolute value versus order
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students multiply negative numbers, but this asks them to compare four variable expressions for any pair of negative integers. Deciding how the squares compare when x is less than y goes beyond typical grade 7 sign practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two facts settle the question: squares of negatives are positive, and the more negative number has the larger square. Students who track the minus signs carelessly think a leading minus makes an expression negative, and students who assume the larger y has the larger square choose the wrong negative expression. Substituting x = −3 and y = −2 gives −18, −12, 18 and 12, which orders all four at once.

Variables instead of numbers moderate Understanding the question
No values are given, only that x and y are negative integers with x < y, so students must reason about signs or invent a pair such as −3 and −2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
x²y and xy² use the same letters with the exponent moved; both are negative, and only comparing x² with y² separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
All four expressions have to be placed in order, because the minus signs in front of the last two turn those expressions positive. (Embretson & Wetzel, 1987)

Strategy: Pick two negative integers with x less than y, such as −3 and −2, evaluate all four expressions, and confirm the order with a second pair.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Knows x²y and xy² are both negative but takes y² as the larger square because y > x, though the more negative x has the bigger square.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads the leading minus as making −x²y negative, missing that y is negative, so −x²y is in fact positive.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Assumes the leading minus makes −xy² negative, missing that x is negative, so −xy² is in fact positive.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.
Squaring a number: $x^2 = x \times x$; x is any number. x² (x squared) means x times itself. The square of a number other than 0 is always positive, because two negatives or two positives multiply to a positive.

Sort out the question
x and y are negative integers, such as −3 and −2. Since x < y, x is to the left of y on the number line. So x is more negative, farther from 0.
You need the expression with the least value. That is the most negative one.
Find the sign of each expression
Squares are positive, so $x^2$ and $y^2$ are positive.
- $x^2 y$ is a positive times a negative, so it is negative.
- $x y^2$ is a negative times a positive, so it is negative.
- $-x^2 y$ and $-x y^2$ are the opposites of these, so they are positive.
So the least value must be A or B. Watch out: The minus sign in front of C and D does not make them negative. It turns a negative into a positive.
Compare A and B with numbers
Try x = −3 and y = −2, since −3 < −2.
$x^2 y = (-3)^2 \times (-2) = 9 \times (-2) = -18$
$x y^2 = (-3) \times (-2)^2 = (-3) \times 4 = -12$
Since −18 < −12, A is smaller.
See why it always works
Both expressions contain x × y, which is positive. $x^2 y = (xy) \times x$ and $x y^2 = (xy) \times y$ . Multiplying the same positive number by the more negative x gives the more negative result.
Watch out: y is bigger than x, but $y^2$ is smaller than $x^2$ : 4 is less than 9. Thinking y² is bigger leads to choice B.
So the answer is A, $x^2 y$ .

Check your answer

Try another pair, x = −5 and y = −1. Then A is $25 \times (-1) = -25$ , B is $(-5) \times 1 = -5$ , C is 25 and D is 5. A is still the least.

More questions like this

The original DOE question

Question 114 as printed on page 215 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 114. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 24 of 51

2023 Test A Math, question 12

Question

If n is an integer and $3n + 3$ is an even number, which expression must also represent an even number?

A $5n + 1$
B $4n + 5$
C $2n + 3$
D $n + 2$

Show the answer

Correct answer: A

The expression $3n + 3$ is equal to $3(n + 1)$ , and it is given that this is an even number. The expression $(n + 1)$ must also represent an even number, because 3 is an odd number: if $(n + 1)$ were odd, $3(n + 1)$ would be the product of two odd numbers, which is an odd number. If $(n + 1)$ is an even number, then n must be an odd number. Choose an odd number to substitute for n in the answer options to find which expression represents an even number:

Let $n = 3$ .

Option A:

$5n + 1 = 5(3) + 1 = 16$

Option B:

$4n + 5 = 4(3) + 5 = 17$

Option C:

$2n + 3 = 2(3) + 3 = 9$

Option D:

$n + 2 = 3 + 2 = 5$

Since 16 is the only even number, the correct answer is option A.

Question analysis

This number properties question says that 3n + 3 is even for an integer n and asks which of four expressions must also be even. Students reason about odd and even numbers written with a variable. The hidden step is deciding what the given fact says about n itself before testing the expressions.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students rewrite and evaluate expressions, but rarely reason about whether an expression is odd or even for every value of a variable. Here that reasoning comes in two layers: first about n, then about each choice.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 23 words.
Corrected from the DOE handbook: The explanation gave the wrong reason why n + 1 must be even; the reason was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question hides its key step: 3n + 3 = 3(n + 1), and since 3 is odd, n + 1 must be even, so n is odd. A student who skips that step and tests a convenient even value never checks the condition and lands on a wrong expression. Once n is known to be odd, any odd test value, such as 1, picks out 5n + 1 at once.

Variables instead of numbers moderate Understanding the question
No value of n is given; the only fact is that 3n + 3 is even, and it has to be turned into a fact about n itself, that n is odd, before any choice is judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
The parity rules are assumed, not stated: an odd number times an odd number is odd, and an even number plus an odd number is odd. (Anderson & Krathwohl, 2001)
Every choice must be checked minor Choosing the answer
Each expression has to be tested with an odd n or by parity rules; three of the four come out odd whenever n is odd, so none can be dismissed on sight. (Embretson & Wetzel, 1987)

Strategy: First work out what 3n + 3 being even says about n, by factoring out 3 or testing small integers, then check each expression with a value of n that fits.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Sees the even term 4n and treats the whole expression as even, overlooking that adding the odd 5 makes 4n + 5 odd for every integer n.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Assumes an expression that starts with 2n must be even, though 2n + 3 is an even number plus 3 and so is always odd.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Tests n = 2 without checking the condition, though 3(2) + 3 = 9 is odd; with n = 2, only n + 2 = 4 comes out even.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation said that n + 1 must be even “because 3 is an odd number and the product of an odd number and an even number is an even number.” That fact does not show that n + 1 is even; the reason is that if n + 1 were odd, 3(n + 1) would be the product of two odd numbers, which is odd, and it is given as even. The site now gives that reason. The answer, A, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Pulling out a common factor: $a \times b + a \times c = a(b + c)$; a is a factor, a number being multiplied, that both terms share. b and c are what is left. This is the distributive property run backward. It turns a sum into a product, which is easier to reason about.
Multiplying odd numbers: $\text{odd} \times \text{odd} = \text{odd}$; An even number splits into pairs, like 2, 4 or 6. An odd number has one left over, like 1, 3 or 5. An odd number of odd groups always leaves one extra. So the product is odd.
Adding odd and even numbers: $\text{odd} + \text{odd} = \text{even}, \quad \text{even} + \text{odd} = \text{odd}$; Two odd numbers each have one left over. Together the two extras make a pair, so the sum is even. An even number plus an odd number still has one left over, so the sum is odd.

Sort out the question
n is an integer, a whole number that can be positive, negative or 0. You are told that $3n + 3$ is even. You need the expression that must be even too. Must means it is even every time. First find out what the fact tells you about n.
Pull out the 3
Both terms in $3n + 3$ have a factor of 3. Pull it out.
$3n + 3 = 3(n + 1)$
So 3 times $(n + 1)$ is an even number.
Decide whether n is odd or even
3 is odd. If $n + 1$ were odd, $3(n + 1)$ would be odd times odd. That is always odd. But we are told it is even. So $n + 1$ must be even. One less than an even number is odd. So n is odd. Watch out: don’t test n = 2. Then $3n + 3 = 9$ , which is odd. With n = 2, only $n + 2$ comes out even, choice D.
Test each choice with an odd n
Pick an odd number, n = 3. It fits, since $3 \times 3 + 3 = 12$ is even.
- A: $5n + 1 = 5 \times 3 + 1 = 16$ , even
- B: $4n + 5 = 4 \times 3 + 5 = 17$ , odd
- C: $2n + 3 = 2 \times 3 + 3 = 9$ , odd
- D: $n + 2 = 3 + 2 = 5$ , odd
Only A is even.
Make sure A always works
Must means every odd n, not just 3. If n is odd, 5n is odd times odd, so it is odd. Odd plus 1 is even. So $5n + 1$ is always even. Watch out: 4n is even, but adding the odd 5 makes $4n + 5$ odd, choice B. In the same way, $2n + 3$ is always odd, choice C. So the answer is A, $5n + 1$ .

Check your answer

Try another odd number, n = 1. Then $3n + 3 = 6$ , which is even, so n = 1 fits. And $5n + 1 = 6$ is even too. The other three choices give 9, 5 and 3, all odd. Choice A wins again.

More questions like this

The original DOE question

Question 25 of 51

2023 Test A Math, question 22

Question

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

Show the answer

Correct answer: C

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

Option A:

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

Option B:

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

Option C:

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

Option D:

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

Question analysis

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

NYS standards: 6.EE.5 solutions and substitution
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 The equations involve only doubling, halving and multiplying by 6, all grade 6 skills. What goes beyond grade 7 practice is the chain of four equal expressions, with every letter limited to a short list of different numbers.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Solution: 3 steps
Reading level: Question and choices: about grade 5, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The chain looks forbidding, but it gives each letter as a simple multiple of p: m is half of p, q a sixth of it, and n twice it. Seen that way, q = p/6 forces p to be 6 or 12, and n = 2p rules out 12, so only one value fits. Students who check one or two of the letters, but not all three, can accept 2 or 12.

Variables instead of numbers moderate Understanding the question
Four letters are tied by one chain of equalities, 2m = 6q = ½n = p, and the only numbers are the five allowed values, so each letter has to be written in terms of p. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
Each choice for p must be tested against three letters, m = p/2, q = p/6 and n = 2p, and all of them must land on different numbers from 1, 2, 3, 6 and 12. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Checks only m, finding that p = 2 gives m = 1 on the list, and misses that q would be 1/3 and n would be 4.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds the solution, m = 3, q = 1, n = 12 and p = 6, then answers with m's value, 3, instead of the value of p.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Tests p = 12 against 2m and 6q, getting m = 6 and q = 2, both on the list, and stops before seeing that ½n = 12 needs n = 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
m, n, p and q are four different numbers. Each one is 1, 2, 3, 6 or 12. The chain $2m = 6q = \frac{1}{2}n = p$ says that four amounts are all equal to p. You need the value of p.
Write each letter using p
Each part of the chain equals p. Undo each multiplication.
$2m = p \text{, so } m = p \div 2$
$6q = p \text{, so } q = p \div 6$
$\frac{1}{2}n = p \text{, so } n = 2 \times p$
Half of n is p, so n is twice p.
Test each choice for p
p m q n All on the list?
2 1 $\frac{1}{3}$ 4 No
3 1.5 0.5 6 No
6 3 1 12 Yes
12 6 2 24 No
Only p = 6 puts every letter on the list.
Check that the letters are different
With p = 6, you get m = 3, q = 1 and n = 12. All four numbers are on the list, and no two are the same. Watch out: p = 12 makes m = 6 and q = 2, which look fine. But n = 24 is not on the list, so choice D fails. Also, the question asks for p, not m. Picking 3 is choice B. So the answer is C, 6.

p	m	q	n	All on the list?
2	1	$\frac{1}{3}$	4	No
3	1.5	0.5	6	No
6	3	1	12	Yes
12	6	2	24	No

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 51

2023 Test A Math, question 26

Question

Which statement must be true if x is a whole number greater than or equal to 1?

A $\frac{1}{x + 1} > \frac{1}{x + 2}$
B $\frac{1}{x + 1} < \frac{1}{x + 2}$
C $\frac{1}{x + 1} - \frac{1}{x + 2} > 1$
D $\frac{1}{x + 1} - \frac{1}{x + 2} > \frac{1}{x}$

Show the answer

Correct answer: A

Let $x = 2$ , and substitute that value for x in each inequality to determine which one is correct:

Option A:

$\frac{1}{3} > \frac{1}{4}$ . This is true.

Option B:

$\frac{1}{3} < \frac{1}{4}$ . This is false.

Option C:

$\frac{1}{3} - \frac{1}{4} > 1$ , or $\frac{1}{12} > 1$ . This is false.

Option D:

$\frac{1}{3} - \frac{1}{4} > \frac{1}{2}$ , or $\frac{1}{12} > \frac{1}{2}$ . This is false.

Option A is the only true inequality.

Question analysis

This question asks which of four inequalities must be true when x is a whole number of at least 1. The inequalities compare fractions with denominators x + 1 and x + 2, or their difference. Students reason about how the size of a fraction changes as its denominator grows, and must be sure a statement holds for every allowed value.

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Comparing fractions is familiar, but these fractions have variable denominators and the question asks what must be true for every whole number from 1 up. That abstraction makes it as demanding as typical grade 7 work, though the arithmetic is light.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Choosing a small value, such as x = 1 or 2, and testing every statement settles this quickly, because three of the four fail for any value tried. The trap is the belief that a bigger denominator makes a bigger fraction, which turns the true comparison around. Students unsure of fraction subtraction lose time on the two difference statements, though both differences are tiny, never more than 1/6.

Variables instead of numbers moderate Understanding the question
All four statements are about the unknown x, and the bold must means each has to hold for every whole number from 1 up, not just for a single test value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
The statements are four separate inequalities, two comparisons and two differences, so each has to be tested; one counterexample is enough to rule a statement out. (Embretson & Wetzel, 1987)
Error-prone arithmetic minor Working it out
Testing the differences means subtracting unlike fractions, such as 1/3 − 1/4 = 1/12 when x = 2, and comparing the result with 1 and with 1/x. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute a small whole number such as 1 or 2 for x in all four statements, rule out any that fail, and confirm the survivor by thinking about larger denominators.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes a bigger denominator makes a bigger fraction, so 1/(x + 2) seems larger than 1/(x + 1), the comparison turned around.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Both fractions are at most 1/2, so their difference can never exceed 1, and no common working makes it do so; this choice mostly draws guesses.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts over a common denominator but keeps only the numerator, (x + 2) − (x + 1) = 1, so the difference looks like 1, larger than 1/x once x is 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } 0 < a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; a and b are positive numbers, and b is bigger. One over a means one whole cut into a equal pieces. Cutting the same whole into more pieces makes each piece smaller. So the fraction with the bigger bottom is smaller.
Subtracting fractions: $\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$; a and b are the bottoms of the two fractions. Rewrite both fractions over the common bottom a × b. Then subtract the tops. The bottom stays a × b.

Sort out the question
x is a whole number that is 1 or more, so x can be 1, 2, 3 and so on. You need the statement that must be true. That means it is true for every such x. One x that makes a statement false rules it out.
Test x = 1 in each statement
With x = 1, x + 1 is 2 and x + 2 is 3.
- A: $\frac{1}{2} > \frac{1}{3}$ . True.
- B: $\frac{1}{2} < \frac{1}{3}$ . False.
- C: $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$ , which is not more than 1. False.
- D: $\frac{1}{6}$ is not more than $\frac{1}{1}$ . False.
Only A is true when x = 1. So B, C and D are out.
Make sure A is always true
Must means A has to work for every x, not just 1. In $\frac{1}{x + 1}$ and $\frac{1}{x + 2}$ , both tops are 1. The bottom x + 2 is always 1 more than x + 1. Cutting a whole into more pieces makes each piece smaller. So A is always true. Watch out: a bigger bottom makes a smaller fraction, not a bigger one. Thinking the other way gives choice B.
See why C and D always fail
Use the subtracting rule with a = x + 1 and b = x + 2.
$\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$
$\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}$
The bottom is at least 2 × 3 = 6, so the difference is at most $\frac{1}{6}$ . That is less than 1. Its bottom is also bigger than x, so it is less than $\frac{1}{x}$ . Watch out: keep the bottom when you subtract. Keeping only the top, 1, leads to choice D. So the answer is A.

Check your answer

Try a bigger number, x = 9. A tenth of a pizza is more than an eleventh, so $\frac{1}{10} > \frac{1}{11}$ . The difference, $\frac{1}{110}$ , is tiny. It is less than 1 and less than $\frac{1}{9}$ . A still holds, and C and D still fail.

More questions like this

The original DOE question

Question 83 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 27 of 51

2023 Test A Math, question 29

Question

Integer x is evenly divisible by 3. Which expression below is also evenly divisible by 3?

A $2x + 1$
B $3x - 5$
C $4x - 1$
D $4x + 6$

Show the answer

Correct answer: D

Since integer x is evenly divisible by 3, substitute 3 for x in the answer options to determine which expression is also divisible by 3:

Let $x = 3$ .

Option A:

$2x + 1 = 2(3) + 1 = 7$

Option B:

$3x - 5 = 3(3) - 5 = 4$

Option C:

$4x - 1 = 4(3) - 1 = 11$

Option D:

$4x + 6 = 4(3) + 6 = 18$

Option D is the only value that is divisible by 3.

Question analysis

This number properties question says an integer x is divisible by 3 and asks which of four linear expressions in x is also divisible by 3, whatever multiple of 3 the integer is. Students can test a multiple of 3 in each expression or think about remainders. Testing a value that is not a multiple of 3 can make a wrong expression look right.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite and evaluate linear expressions, but they seldom ask whether one is divisible by 3 for every allowed value. One well-chosen substitution settles it, so the demand is about typical for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The algebra is light; the question turns on choosing a test value that obeys the condition. The habit of trying x = 1 backfires here, because 1 is not a multiple of 3 and it makes two wrong expressions come out divisible. Students who substitute 3, or who see that 4x and 6 are both multiples of 3 whenever x is, settle it quickly, and a second multiple such as 6 confirms it.

Variables instead of numbers moderate Understanding the question
Nothing pins x down except that it is a multiple of 3, so each expression must be judged for every such x, and any test value has to be a multiple of 3 itself, such as 3 or 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked minor Choosing the answer
The four expressions differ only in their coefficients and constants, from 2x + 1 to 4x + 6, so none can be dismissed at a glance; each needs a test or a look at its terms. (Embretson & Wetzel, 1987)

Strategy: Substitute a multiple of 3, such as 3 or 6, for x in every expression and keep the one whose value divides evenly by 3; a second multiple confirms it.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Tests x = 1, which is not a multiple of 3, finds that 2(1) + 1 = 3 divides evenly, and stops at the first choice; with x = 3 it gives 7.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Sees the 3 in 3x and takes the whole expression to be a multiple of 3, overlooking that subtracting 5 spoils it: with x = 3 it gives 4.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Tries a value that is not a multiple of 3, such as x = 4, where 4x − 1 = 15 divides evenly, though x = 3 gives 11.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding multiples of 3: $3a + 3b = 3(a + b)$; a and b are whole numbers, so 3a and 3b are multiples of 3. A multiple of 3 is a number that 3 divides evenly, with nothing left over. Adding two multiples of 3 always gives another multiple of 3, because you can pull the 3 out of both parts.

Sort out the question
“Evenly divisible by 3” means 3 goes into the number with nothing left over. Such a number is called a multiple of 3, like 3, 6, 9 or 12.
You know that x is a multiple of 3. You need the expression that is a multiple of 3 every time, whatever multiple of 3 x is.
Test x = 3 in every choice
Pick a number for x that follows the rule. The easiest one is 3. Put it into each expression.
Choice Work Divisible by 3?
A $2(3) + 1 = 7$ No
B $3(3) - 5 = 4$ No
C $4(3) - 1 = 11$ No
D $4(3) + 6 = 18$ Yes
Watch out: your test number must be a multiple of 3. With x = 1, choice A gives 3 and looks right. That leads to the wrong choice, A.
See why D always works
One test could be lucky, so see why D works for every x. Since x is a multiple of 3, 4x is one too. For example, 4 × 3 = 12 and 4 × 6 = 24. The number 6 is also a multiple of 3, since 6 = 3 × 2.
$3a + 3b = 3(a + b)$
So 4x + 6 is a multiple of 3 plus a multiple of 3. That is always a multiple of 3.
Rule out the other choices
Watch out: choice B starts with 3x, but then it subtracts 5. The 5 is not a multiple of 3, so it spoils the result, and 3(3) − 5 = 4. Choices A and C also add or take away a number that is not a multiple of 3.
So the answer is D, $4x + 6$ .

Choice	Work	Divisible by 3?
A	$2(3) + 1 = 7$	No
B	$3(3) - 5 = 4$	No
C	$4(3) - 1 = 11$	No
D	$4(3) + 6 = 18$	Yes

Check your answer

Try two more multiples of 3. With x = 6, choice D gives $4(6) + 6 = 30$ , and 30 ÷ 3 = 10. With x = 9, it gives $4(9) + 6 = 42$ , and 42 ÷ 3 = 14. Both divide evenly, so D keeps working.

More questions like this

The original DOE question

Question 86 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 86. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 28 of 51

2023 Test A Math, question 39

Question

A scientist mixed three chemicals, R, S, and T, in a glass container. The amount of R is 3 times the amount of S, and the amount of T is $\frac{1}{6}$ the amount of S. What is the ratio of the amount of R to the amount of T?

A1:18
B2:1
C3:1
D18:1

Show the answer

Correct answer: D

First, set up the equations based on the information in the question.

$\mathrm{R} = 3\mathrm{S}$

$\mathrm{T} = \frac{1}{6}\mathrm{S}$

So, the ratio of R:T is

$3\mathrm{S}:\frac{1}{6}\mathrm{S}$

Divide both sides by S to eliminate the variable. Then, multiply both sides by 6 to eliminate the fraction. The resulting ratio is 18:1.

Question analysis

This ratio question describes three chemicals mixed in a container: R is 3 times the amount of S, and T is one sixth of S. Students must combine the two relationships through S to find the ratio of R to T. Writing the ratio in the wrong order, or using only one of the relationships, produces the tempting wrong choices.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 ratio problems often link two quantities through a third. Here R and T are each described relative to S, one as a multiple and one as a fraction, and combining them into one ratio is typical grade 7 work with an order trap.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question chains two comparisons through S, and the 1/6 makes the chain awkward to do with letters. Choosing S = 6 turns it into whole numbers at once. The listed wrong ratios come from writing the right numbers in the wrong order, combining 3 and 6 by division, or stopping at the first comparison. Rereading which amount the question names first settles the order.

Turning words into math moderate Understanding the question
Both sentences compare a chemical with S, R as 3 times S and T as 1/6 of S, so S must serve as the link before R can be compared with T. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
The ratios 1:18 and 18:1 use the same two numbers in opposite order, so only checking that R, the larger amount, comes first separates the right ratio from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
No amounts are given, only relationships, so students either reason with S as a letter or choose a number for it, such as 6, which makes R 18 and T 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Pick a convenient amount for S, such as 6, so that both R and T come out as whole numbers, then write R to T in that order 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).
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Works out the amounts correctly but writes the ratio upside down, T to R, putting the smaller amount first.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides the 6 by the 3 instead of multiplying them, combining the two comparisons with S the wrong way to get 2:1.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Uses only the first sentence, R is 3 times S, and gives the ratio of R to S instead of R to T.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios: $a : b = (a \times n) : (b \times n)$; a and b are two amounts being compared, and n is any number except 0. Multiplying or dividing both parts of a ratio by the same number gives an equal ratio. For example, 2 : 1 is the same as 4 : 2.

Sort out the question
There are three chemicals, R, S and T. R is 3 times S. T is $\frac{1}{6}$ of S. You need the ratio of R to T, written R : T. Both facts compare a chemical with S, so S links R and T.
Pick an easy amount for S
No amounts are given, so you may choose one. Pick 6 units for S, because 6 divides evenly by 6.
$\mathrm{R} = 3 \times 6 = 18$
$\mathrm{T} = \frac{1}{6} \times 6 = 1$
So R is 18 units and T is 1 unit.
Write the ratio of R to T
$\mathrm{R} : \mathrm{T} = 18 : 1$
R comes first because the question asks for R to T. Watch out: writing it the other way round gives 1:18, choice A. Using only the first fact gives R : S = 3 : 1, choice C.
Check with letters instead
Without picking a number, R = 3S and T is $\frac{1}{6}\mathrm{S}$ . Divide both parts by S, then multiply both parts by 6.
$a : b = (a \times n) : (b \times n)$
$3\mathrm{S} : \frac{1}{6}\mathrm{S} = 3 : \frac{1}{6} = 18 : 1$
So the answer is D, 18:1.

Check your answer

R should be much bigger than T. R is 3 times S, while T is only a small piece of S. Try another amount, S = 12. Then R = 36 and T = 2, and 36 : 2 is also 18 : 1.

More questions like this

The original DOE question

Question 96 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 96. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 29 of 51

2023 Test A Math, question 40

Question

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

A $k - 6 = 2(j + 6)$
B $k - 6 = 2j + 6$
C $2(k - 6) = j$
D $2(k - 6) = j + 6$

Show the answer

Correct answer: A

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

$k - 6 = 2(j + 6)$

Question analysis

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

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.2a writing algebraic expressions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students write equations from word problems. This one tracks two people's counts after a transfer and states a times-as-many comparison, so both counts change and the factor of 2 must go on the right side: typical but error-prone.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is all setup: no solving is needed, just an equation that says what the story says. The trade moves games out of one count and into the other, and the comparison has to be written with the 2 on the smaller side. The wrong equations reverse the comparison, forget Jeff's gain, or drop the parentheses, and all four look alike. Plugging in sample counts is the surest check.

Turning words into math moderate Understanding the question
After the gift both counts change, Ken's to k − 6 and Jeff's to j + 6, and “twice as many video games as Jeff” must double Jeff's whole new count, putting the 2 on his side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
All four equations share the pieces k − 6, j + 6 and a 2; the right one differs from 2(k − 6) = j + 6 only in where the 2 sits, and from k − 6 = 2j + 6 by parentheses. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
There are no numbers to check against, only k and j, so testing an equation means inventing counts that fit the story, such as Jeff with 4 games and Ken with 26. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

Traps

Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Leaves out the parentheses, doubling only Jeff's old count j and then adding 6, instead of doubling his whole new count j + 6.
C Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: weak
Puts the 2 on Ken's side, writing the comparison backwards, and also forgets that Jeff gains the 6 games, two errors in one equation.
D Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: strong
Writes the comparison backwards word for word, doubling Ken's new count, which would mean Jeff ends with twice as many games as Ken.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Ken has k video games, and Jeff has j. Then Ken gives 6 of his games to Jeff. After that, Ken has twice as many games as Jeff. You need the equation that links k and j.
Write each new amount
Ken gives away 6 games, so he has k − 6 left. Jeff gets those 6 games, so he has j + 6. Both amounts change, so use both new amounts.
Write “twice as many”
After the gift, Ken has more games. So Jeff’s new amount is the one that gets doubled.
$A = 2 \times B$
$k - 6 = 2(j + 6)$
Watch out: putting the 2 on Ken’s side, as in $2(k - 6) = j + 6$ , would mean Jeff has twice as many. That is choice D.
Keep the parentheses
$a(b + c) = ab + ac$
$2(j + 6) = 2j + 12$
Twice Jeff’s new amount is 2j + 12, not 2j + 6. Watch out: dropping the parentheses gives choice B. So the answer is A, $k - 6 = 2(j + 6)$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 30 of 51

2023 Test A Math, question 56

Question

$Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. The line is shaded from the unlabeled tick halfway between 0 and 1 to the unlabeled tick halfway between 1 and 2, with closed dots at both ends.$

If all possible values of x are indicated by the shaded part of the number line above, which number line best shows all possible values of $\frac{1}{x}$ ?

A $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 1 and 2 to the tick halfway between 2 and 3, with closed dots at both ends.$
B $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from 1 to 2, with closed dots at both ends.$
C $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 0 and 1 to 2, with closed dots at both ends.$
D $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from a closed dot between the tick halfway between 0 and 1 and the tick at 1, about two-thirds of the way from 0 to 1, to a closed dot at 2.$

Show the answer

Correct answer: D

The shaded region goes from $\frac{1}{2}$ to $\frac{3}{2}$ . So, the possible values of $\frac{1}{x}$ would be between the reciprocal of $\frac{1}{2}$ and the reciprocal of $\frac{3}{2}$ , which are 2 and $\frac{2}{3}$ , respectively. Find the number line in the answer options that has a shaded region that ranges from $\frac{2}{3}$ to 2, which is option D.

Question analysis

This number line question shades all possible values of x from one half to three halves and asks which number line shows all possible values of 1/x. Students find reciprocals of fractions and work out how a whole range of values changes. Keeping track of what happens to each end of the range, and placing fractions accurately on the line, makes it tricky.

NYS standards: 6.NS.1 dividing fractions by fractions
6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Reciprocals of single fractions are grade 6 work. Applying them to every value in a shaded range at once, and working out where the new range begins and ends, is reasoning about a whole set of numbers that grade 7 classes rarely practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 12, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Few students have flipped an interval before, so the question rewards testing its two ends and noticing that the order reverses. The harder part comes at the end: 2/3 falls between the tick marks, and a graph ending at the nearby 1/2 tick looks right at a glance. Students who believe a reciprocal is always at least 1, or who shift the interval instead of flipping it, reach the other graphs.

Reading a figure, table or graph moderate Understanding the question
The shaded segment's ends sit on unlabeled half-unit ticks, 1/2 and 3/2, and each choice must be read the same way, with 2/3 falling between ticks rather than on one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer moderate Choosing the answer
The graph from 1/2 to 2 matches the right graph's right end and differs at the left end by only a sixth of a unit, 1/2 against 2/3. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
The question asks about 1/x for every x in a range, so a whole interval must be transformed, and its ends swap: the smallest x gives the largest 1/x. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Take the reciprocal of each end of the shaded part, ask whether a bigger x gives a bigger or a smaller 1/x, and look for the graph whose endpoints are exactly those values.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds 1 to each endpoint instead of taking reciprocals, sliding the shaded part from 1/2–3/2 one unit right, to 3/2–5/2.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gets 2 from 1/2 but assumes a reciprocal cannot be less than 1, as flipping a proper fraction suggests, so it starts the shading at 1 instead of 2/3.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Finds the reciprocals 2/3 and 2 but places 2/3 at the half-unit tick, 1/2, though 2/3 lies a third of the way from that tick toward 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal: $\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$; a and b are the top and bottom of a fraction. The reciprocal flips the fraction upside down. The number 1 over x is the reciprocal of x. A number times its reciprocal always makes 1.

Sort out the question
x can be any number in the shaded part of the first number line. You need the number line that shows every possible value of $\frac{1}{x}$ , the reciprocal of x. The reciprocal is the number flipped upside down, like $\frac{1}{2}$ for 2.
Read the shaded part
The ticks come every half unit. The shading starts at the tick halfway between 0 and 1, which is $\frac{1}{2}$ . It ends halfway between 1 and 2, which is $\frac{3}{2}$ , or $1\frac{1}{2}$ . Both dots are filled in, so both ends count.
Find the reciprocals of the ends
Flip each end of the shaded part.
$\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$
$\text{reciprocal of } \frac{1}{2} = \frac{2}{1} = 2$
$\text{reciprocal of } \frac{3}{2} = \frac{2}{3}$
So the ends of the new shading are 2 and $\frac{2}{3}$ .
Notice that the ends swap
When x gets bigger, $\frac{1}{x}$ gets smaller. So the smallest x, $\frac{1}{2}$ , gives the biggest value, 2. The biggest x, $\frac{3}{2}$ , gives the smallest, $\frac{2}{3}$ . So $\frac{1}{x}$ runs from $\frac{2}{3}$ to 2, ends included.
Find the matching number line
$\frac{2}{3}$ lies between the half tick and 1, a third of the way from $\frac{1}{2}$ to 1. Line D starts there and ends at 2. Watch out: line C starts at the half tick, $\frac{1}{2}$ , not at $\frac{2}{3}$ . Line B starts at 1.
So the answer is D, the line shaded from $\frac{2}{3}$ to 2.

Check your answer

Test a value in the middle. x = 1 is shaded, and $\frac{1}{1} = 1$ . On line D, 1 is shaded too. Test an end: $\frac{3}{2} \times \frac{2}{3} = 1$ , so $\frac{2}{3}$ really is the reciprocal of $\frac{3}{2}$ .

More questions like this

The original DOE question

Question 113 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 113. Open page 91 of the handbook (PDF)

Question 31 of 51

2023 Test B Math, question 8

Question

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

A $x = \frac{y - 4}{2}$
B $x = 2(y + 4)$
C $x = \frac{y + 4}{2}$
D $x = 2(4 - y)$

Show the answer

Correct answer: A

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

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers minor
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.

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)
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

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies by 2 instead of dividing and adds 4 instead of subtracting, applying the operations on x again rather than undoing them.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Moves the 4 across the equals sign without changing its sign, getting 2x = y + 4, then divides by 2 correctly.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two unrelated errors at once, the subtraction written backward as 4 − y and a multiplication by 2 in place of division, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. 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.

Sort out the question
The equation is $2x + 4 = y$ . “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.
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.
$2x + 4 - 4 = y - 4$
$2x = y - 4$
Watch out: the 4 must be subtracted from y. Writing $2x = y + 4$ leads to choice C.
Undo the doubling
Next, divide both sides by 2. Divide the whole right side, not only the y.
$\frac{2x}{2} = \frac{y - 4}{2}$
$x = \frac{y - 4}{2}$
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, $x = \frac{y - 4}{2}$ .

Check your answer

Pick a value, such as x = 1. Then $y = 2 \times 1 + 4 = 6$ . Put y = 6 into choice A: $\frac{6 - 4}{2} = 1$ . That gives back x = 1. Choice C would give $\frac{6 + 4}{2} = 5$ , 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 32 of 51

2023 Test B Math, question 20

Question

Five consecutive numbers have a sum of 110. What is the least of these consecutive numbers?

Show the answer

Correct answer: C

Let n represent the least of the five numbers. Then the next four consecutive numbers are $(n + 1)$ , $(n + 2)$ , $(n + 3)$ , $(n + 4)$ . Write an equation to solve for n:

$n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110$

$5n + 10 = 110$

$5n = 100$

$n = 20$

Question analysis

Five consecutive numbers add up to 110, and students find the least of them. This number-properties question can be solved with an equation in one unknown, or by noting that the middle of an odd number of consecutive numbers equals their average. The bold word least matters, because the middle number is easier to find.

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 A sum of consecutive numbers leads to a two-step equation, standard grade 7 work. The average shortcut makes it quicker, and the only extra demand is answering for the least number rather than the middle one.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Dividing 110 by 5 is the natural first move, and it gives the middle number, which is a listed choice. The bold word least signals that one more step is needed, counting down two from the middle. The equation route, 5n + 10 = 110, goes straight to the least number but takes longer to set up.

Turning words into math minor Understanding the question
“Consecutive numbers” means each is 1 more than the last, and the bold “least” asks for the first of the five, not the middle one that dividing by 5 gives. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
None of the five numbers is given, so they must be written as n through n + 4, or pictured as two on each side of a middle value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Divide 110 by 5 to find the middle of the five numbers, then count down to the least of them.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 11, which is 110 ÷ 10; five consecutive numbers starting at 11 add to only 65, so it mostly draws guesses.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Steps by 2 as if the numbers were consecutive even numbers, n through n + 8, so 5n + 20 = 110 gives 18.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides 110 by 5 to get 22, the middle number, and gives it without counting down to the least number.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive numbers: $n,\ n + 1,\ n + 2,\ n + 3,\ n + 4$; Consecutive numbers come one right after another when you count, like 7, 8, 9. Each is 1 more than the one before. If n is the least number, the next ones are n + 1, n + 2, and so on.

Sort out the question
Five consecutive numbers add up to 110. Consecutive means each number is 1 more than the one before. You need the least, or smallest, of the five.
Name the five numbers
Let n be the least number. Then the five numbers are:
$n,\ n + 1,\ n + 2,\ n + 3,\ n + 4$
Each one is 1 more than the one before it.
Write the sum as an equation
Add all five and set the sum equal to 110.
$n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110$
There are five n’s, which make 5n. The extra numbers add up to 1 + 2 + 3 + 4 = 10.
$5n + 10 = 110$
Solve for n
Subtract 10 from both sides. Then divide both sides by 5.
$5n = 100$
$n = 100 \div 5 = 20$
So the least number is 20.
Give the answer
Watch out: $110 \div 5 = 22$ is the middle number, not the least. Picking 22 is choice D. The question asks for the least of the five. So the answer is C, 20.

Check your answer

Write out the five numbers and add them: $20 + 21 + 22 + 23 + 24 = 110$ . The sum matches. Also, the middle number, 22, is 110 ÷ 5, and the least is 2 less than the middle.

More questions like this

The original DOE question

Question 77 as printed on page 197 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 77. Open page 197 of the handbook (PDF)

Question 33 of 51

2023 Test B Math, question 21

Question

$-mn$

$(-m)(-n)$

$|mn|$

$|-mn|$

$|(-m)(-n)|$

If m and n are positive integers, how many of the five expressions above are equivalent to $mn$ ?

Show the answer

Correct answer: D

Since m and n are positive, the value of $mn$ is also positive. Of the five options given:

$-mn$ is negative.

$(-m)(-n)$ is positive.

$|mn|$ is positive.

$|-mn|$ is positive.

$|(-m)(-n)|$ is positive.

There are 4 expressions that are positive, and each one is equal to $mn$ .

Question analysis

Five expressions are listed: −mn, (−m)(−n), |mn|, |−mn| and |(−m)(−n)|. If m and n are positive integers, students count how many are equivalent to mn. It tests the sign rules for multiplying negatives together with what absolute value does, and each expression has to be judged on its own.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multiplying signed numbers and absolute value are grade 6 and 7 content. Checking five expressions one by one takes longer than a typical single computation, but each check uses a familiar rule.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Much to keep track of minor
Something you must already know minor
Solution: 5 steps
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each expression takes only a moment, but the answer is a count, so one misjudged expression changes it. The minus signs are designed to unsettle: four of the five expressions show at least one, and only one of those actually stays negative. Substituting small positive numbers and comparing each value with mn turns the question into arithmetic.

Variables instead of numbers moderate Understanding the question
The five expressions use only m and n, so each sign rule has to be applied in general unless the student substitutes values such as m = 2 and n = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Five expressions are judged one at a time and a count kept, and two of them combine the product rule for negatives with absolute value bars. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The rules that two negative factors give a positive product and that absolute value removes a minus sign are assumed, not stated. (Anderson & Krathwohl, 2001)

Strategy: Pick small values such as m = 2 and n = 3, evaluate each of the five expressions, and count the ones equal to 6.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Treats any minus sign as making the value negative, even inside bars or in a product of two negatives, so counts only |mn|.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Accepts |mn| and (−m)(−n) but believes bars cannot cancel a minus sign written inside them, rejecting |−mn| and |(−m)(−n)|.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Thinks a negative times a negative stays negative, so rejects (−m)(−n) along with −mn and counts only the three absolute values.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying two negatives: $(-a) \times (-b) = a \times b$; a and b are positive numbers. A negative number times a negative number gives a positive number. The two minus signs cancel out. One negative and one positive give a negative.
Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so |−6| = 6 and |6| = 6. The bars remove any minus sign.

Sort out the question
m and n are positive integers. That means whole numbers greater than 0. Writing mn means m times n, so mn is positive. You need to count how many of the five expressions always equal mn.
Pick numbers to test
Pick easy values: $m = 2$ and $n = 3$ . Then $mn = 2 \times 3 = 6$ . Any expression that equals mn must come out to 6.
Test the first two expressions
$-mn = -(2 \times 3) = -6$
This is negative, so it does not equal 6.
$(-m)(-n) = (-2)(-3) = 6$
A negative times a negative is positive, so this one equals 6. Watch out: if you think a negative times a negative stays negative, you reject this one and get 3, choice C.
Test the three absolute values
$|mn| = |6| = 6$
$|-mn| = |-6| = 6$
$|(-m)(-n)| = |6| = 6$
All three equal 6, because absolute value bars remove a minus sign. Watch out: the bars work even when the minus sign is inside them. Rejecting the last two gives 2, choice B.
Count the matches
Four expressions equal mn: $(-m)(-n)$ , $|mn|$ , $|-mn|$ and $|(-m)(-n)|$ . Only $-mn$ does not. The sign rules work the same for any positive m and n, not just 2 and 3. So the answer is D, 4.

Check your answer

Try other numbers, such as $m = 1$ and $n = 5$ . Then $mn = 5$ . The five expressions give −5, 5, 5, 5 and 5. Again, four of them match mn.

More questions like this

The original DOE question

Question 78 as printed on page 197 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 78. Open page 197 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 34 of 51

2023 Test B Math, question 28

Question

The sum of the numbers x, y, and z is 40. The ratio of x to y is 1:2, and the ratio of y to z is 2:5. What is the value of y?

Show the answer

Correct answer: D

The ratio of x to y is 1:2, so $\frac{x}{y} = \frac{1}{2}$ or $x = \frac{1}{2}y$ . Similarly, the ratio of y to z is 2:5, so $\frac{y}{z} = \frac{2}{5}$ . Solve for z in terms of y:

$5y = 2z$

$\frac{5}{2}y = z$

Use the given sum of x, y, and z to find the value of y:

$x + y + z = 40$

$\frac{1}{2}y + y + \frac{5}{2}y = 40$

$4y = 40$

$y = 10$

Question analysis

This ratio question gives two ratios that share y: x to y is 1:2 and y to z is 2:5, and x, y and z total 40. Students find the value of y. The numbers in a ratio count parts, not amounts, so the work is to combine the ratios and find what one part is worth.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Most grade 7 ratio tasks split a total using one ratio. Here two ratios linked through y must first become a single three-part ratio; since y's term already matches, that step is short and the problem stays at grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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 question rewards seeing that the two ratios share y’s term, which makes one three-part ratio. From there, one part is worth 5 and y is two parts. The wrong choices are numbers that live inside the ratios, and adding all four ratio numbers, which counts y twice, also leads to the part count, so that choice draws students along two paths.

No obvious method moderate Working it out
Nothing says to merge the ratios, but y is 2 in both, so x : y : z becomes 1 : 2 : 5 at once; the explanation’s route through fractions of y is slower. (Schoenfeld, 1985)
Variables instead of numbers minor Understanding the question
x, y and z appear only in two ratios and a sum of 40, so the ratio numbers 1, 2 and 5 are easy to mistake for the values themselves. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Adding the 8 parts, dividing 40 by 8 to get 5 per part, and taking y’s 2 parts are three steps, and the part count is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write x : y : z as one ratio using y's matching term, add the parts, divide 40 by that sum to get one part, and multiply by y's parts.

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

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes y’s term in the ratios, 2, as y’s actual value, never using the sum of 40.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two 2s that stand for y in the two ratios, 2 + 2 = 4, treating ratio terms as amounts.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Combines the ratios into 1 : 2 : 5 and gives the total number of parts, 1 + 2 + 5 = 8, instead of y’s share of the 40.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Splitting a total by a ratio: $\text{one part} = \text{total} \div \text{number of parts}$; Each number in a ratio counts parts that are all the same size. Add the numbers in the ratio to find how many parts make up the total. Dividing the total by that number gives the size of one part. Each amount is its number of parts times that size.

Sort out the question
x, y and z add up to 40. A ratio compares two amounts. The ratio of x to y is 1:2, so y is twice x. The ratio of y to z is 2:5. You need the actual value of y, not its number in a ratio.
Combine the two ratios
y is 2 in both ratios. So both ratios count the same size of part, and they join into one.
$x : y : z = 1 : 2 : 5$
So x is 1 part, y is 2 parts and z is 5 parts.
Find the total number of parts
$1 + 2 + 5 = 8$
The 40 is split into 8 equal parts. Watch out: 8 counts parts. It is not y, so choice C is wrong.
Find the size of one part
$\text{one part} = \text{total} \div \text{number of parts}$
$\text{one part} = 40 \div 8 = 5$
The 40 is shared equally among the 8 parts. So each part is worth 5.
Find y
y is 2 parts, so multiply the size of one part by 2.
$y = 2 \times 5 = 10$
Watch out: the 2 in the ratio counts parts, so y is not 2, choice A. Adding the two 2s to get 4 is wrong too, choice B. So the answer is D, 10.

Check your answer

Find all three: x = 5, y = 10 and z = 25. They add to $5 + 10 + 25 = 40$ . x to y is 5 to 10, which is 1:2. y to z is 10 to 25, which is 2:5.

More questions like this

The original DOE question

Question 85 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 85. Open page 199 of the handbook (PDF)

Question 35 of 51

2023 Test B Math, question 31

Question

If $xy=30$ , what is the value of $10\div x$ in terms of y?

A $\frac{1}{10}y$
B $\frac{1}{3}y$
C $3y$
D $10y$

Show the answer

Correct answer: B

Solve $xy=30$ , the given equation, for x:

$xy=30$

$x=\frac{30}{y}$

Then:

$10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y$

Question analysis

An algebra question: given that x times y equals 30, students express 10 divided by x using only y. It tests solving for one variable and dividing by an algebraic fraction. Working with letters instead of numbers, and dividing in the right order, are what trip students up.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.1 dividing fractions by fractions
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The algebra is three short moves, but each is a place to turn a fraction upside down: solving xy = 30 for x, dividing 10 by that fraction, and reducing 10/30. Students who reach for the bigger-over-smaller division they know from whole numbers land on 3y. Choosing two numbers with a product of 30, such as 3 and 10, and testing every choice settles the question in seconds.

Variables instead of numbers moderate Understanding the question
x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Solve xy = 30 for x, substitute that fraction into 10 ÷ x, and divide by multiplying by its reciprocal; or choose values of x and y with a product of 30 and test each choice.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Writes the 10 from 10 ÷ x under y as one tenth of y, never using the product xy = 30 to link the two letters.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Replaces x with 30/y correctly but divides 30 by 10 instead of 10 by 30, taking the larger number over the smaller, and gets 3y.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats dividing by x as multiplying by y, as if x and y were reciprocals, though their product is 30, not 1, which gives 10y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is the number you divide. b over c is the fraction you divide by: b is its top and c is its bottom. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. For example, 6 ÷ 1/2 = 6 × 2 = 12, because 12 halves fit in 6.

Sort out the question
$xy$ means x times y. So x times y is 30. You need $10 \div x$ “in terms of y.” That means your answer can use y and numbers, but no x.
Solve for x
To swap x out, first get x by itself. Divide both sides of $xy = 30$ by y.
$x = \frac{30}{y}$
This says x is 30 divided by y.
Put the fraction in place of x
Replace x in $10 \div x$ with this fraction.
$10 \div x = 10 \div \frac{30}{y}$
Now use the rule for dividing by a fraction. Flip the fraction and multiply.
$10 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}$
Simplify the fraction
Both 10 and 30 can be divided by 10.
$\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y$
Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.
So the answer is B, $\frac{1}{3}y$ .

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then $10 \div x = 10 \div 3 = \frac{10}{3}$ . Choice B gives $\frac{1}{3} \times 10 = \frac{10}{3}$ . They match. Choice C would give 30, and choice D would give 100.

More questions like this

The original DOE question

Question 88 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 88. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 51

2023 Test B Math, question 38

Question

How is $\frac{1}{x}-\frac{1}{2x}$ written as a single fraction? (Assume $x\neq 0$ .)

A $\frac{1}{2x}$
B $\frac{1}{3x}$
C $-\frac{1}{x}$
D $-\frac{1}{2x}$

Show the answer

Correct answer: A

$\frac{1}{x}-\frac{1}{2x}=\frac{2}{2x}-\frac{1}{2x}=\frac{1}{2x}$

Question analysis

This algebra question asks students to write 1/x minus 1/(2x) as a single fraction, assuming x is not zero. It tests finding a common denominator when the denominators contain a variable. Getting both fractions over the same denominator before subtracting, and keeping track of the sign, are what matter.

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students subtract fractions with unlike denominators and work with linear expressions, but seldom see a variable in a denominator. The steps match those for numerical fractions, so it is a short task once the common denominator is found.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 11, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is subtraction of unlike fractions with a letter in the bottom, and every wrong choice is a familiar fraction error: combining the denominators straight across by adding or subtracting, or subtracting the numerators the wrong way round. Substituting x = 1 makes it 1 − 1/2 = 1/2, which rules out the negative choices and 1/3 at once.

Variables instead of numbers moderate Understanding the question
The denominators are x and 2x, so the common denominator is built with a letter, and no number checks the result unless students substitute a value such as x = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Only the first fraction changes: 1/x becomes 2/2x, and then the subtraction 2 − 1 must be done in the order written, first numerator minus second. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite 1/x with the denominator 2x by multiplying its top and bottom by 2, then subtract the numerators and keep the common denominator.

Traps

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

The wrong answers, classified

B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Adds the denominators, x + 2x = 3x, and keeps a numerator of 1, as if two unit fractions combined by adding their bottoms.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the denominators straight across, x − 2x = −x, and keeps 1 on top, so the result 1/(−x) is written −1/x.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Rewrites 1/x as 2/2x correctly but subtracts the numerators in reverse order, 1 − 2 = −1, giving a negative result.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; a is the top, or numerator. b is the bottom, or denominator. k is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, because k over k is 1.
Subtracting fractions with the same bottom: $\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$; a and b are the tops. c is the shared bottom. When the bottoms match, the pieces are the same size. So subtract the tops and keep the bottom.

Sort out the question
You need to write $\frac{1}{x} - \frac{1}{2x}$ as one fraction. The problem says x is not 0. That matters because you can’t divide by 0.
Find a common bottom
To subtract fractions, their bottoms must match. 2x is 2 times x. So multiply the top and bottom of $\frac{1}{x}$ by 2.
$\frac{1}{x} = \frac{1 \times 2}{x \times 2} = \frac{2}{2x}$
Now both fractions have the bottom 2x.
Subtract the tops
Subtract the tops in order. Keep the bottom.
$\frac{2}{2x} - \frac{1}{2x} = \frac{2 - 1}{2x} = \frac{1}{2x}$
Watch out: Never add or subtract the bottoms. Adding them gives $\frac{1}{3x}$ , choice B. Subtracting them gives $-\frac{1}{x}$ , choice C. And keep the order: 2 − 1, not 1 − 2. The wrong order gives choice D.
So the answer is A, $\frac{1}{2x}$ .

Check your answer

Try x = 1. The problem becomes $1 - \frac{1}{2} = \frac{1}{2}$ . Choice A gives $\frac{1}{2 \times 1} = \frac{1}{2}$ . It matches. Choices B, C and D give $\frac{1}{3}$ , −1 and $-\frac{1}{2}$ .

More questions like this

The original DOE question

Question 95 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 95. Open page 202 of the handbook (PDF)

Question 37 of 51

2023 Test B Math, question 41

Question

The height of a tree in 2013 is 300% of the height it was in 2003. By what percentage did its height increase over this ten-year period?

A $33\frac{1}{3}\%$
B100%
C200%
D300%

Show the answer

Correct answer: C

Let h represent the height of the tree in 2003. Then the height of the tree in 2013 is $3h$ . Calculate the percent increase:

$\frac{3h-h}{h}=\frac{2h}{h}=2=200\%$

Question analysis

A percent question about a tree whose height in 2013 is 300% of its height in 2003. Students find the percent increase over the ten years. No heights are given, so the key is telling the difference between a percent of an amount and a percent change.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent-increase problems give two amounts to compare. This gives only a percent of the earlier height and no heights at all, so students must work out what that percent means for the change with nothing concrete to calculate with.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The stem hands students a percent and asks for a different percent, and the given 300% looks like a finished answer. The real step is separating “300% of” from “increased by”: a height that triples has grown by twice its original size. Picking a 2003 height of 100 units makes it concrete, since 300 − 100 = 200 units of growth on a base of 100.

Turning words into math moderate Understanding the question
“300% of the height it was in 2003” describes the new height, three times the old, not the increase; the question then asks for the change, which is a different percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers moderate Understanding the question
No heights are given, only a percent, so students must invent a height such as 100 units or work with h and 3h to measure the change. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
A percent increase divides the change by the starting amount, (3h − h) ÷ h, a rule the question assumes without stating. (Anderson & Krathwohl, 2001)

Strategy: Call the 2003 height 100 units, find the 2013 height from the percent given, then divide the change by the 2003 height and write the result as a percent.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the 2003 height by the 2013 height, h ÷ 3h = 1/3, turning the comparison upside down and reporting 33 1/3%.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 100%, which would mean the height only doubled; it echoes the 100% that the 2003 height stands for, so it mostly draws guesses.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Reads “300% of” as an increase of 300%, giving the 2013 height as a percent of the 2003 height instead of the percent increase.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; Original is the starting amount. New is the amount after the change. New minus original is how much it grew. Dividing by the original compares the growth with the start. Multiplying by 100 turns that into a percent.

Sort out the wording
The 2013 height is 300% of the 2003 height. That means it is 3 times the 2003 height. You need the percent increase. That is how much the tree grew, compared with its 2003 height.
Pick an easy height
The question gives no real height, so choose one. Say the tree was 100 feet tall in 2003. Then the 2013 height is 300% of 100 feet.
$\frac{300}{100} \times 100 = 300$
In 2013, the tree was 300 feet tall.
Find how much it grew
The growth is how much taller the tree got. Subtract the old height from the new one.
$300 - 100 = 200$
The tree grew 200 feet in ten years.
Find the percent increase
Use the formula. The original height is the 2003 height.
$\text{percent increase} = \frac{300 - 100}{100} \times 100 = 2 \times 100 = 200$
The height went up by 200%.
Watch out: 300% compares the new height with the old one. It is not the increase. That trap is choice D. Dividing the old height by the new one gives $33\frac{1}{3}\%$ , choice A.
So the answer is C, 200%.

Check your answer

Think about doubling. A tree that doubles adds 100% of its height. A tree that triples adds 200% of its height. So 100 feet plus 200% of 100 feet is 100 + 200 = 300 feet. That is 300% of 100 feet.

More questions like this

The original DOE question

Question 98 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 98. 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 38 of 51

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is $\frac{3}{5}$ ?

Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to $\frac{3}{5}$ and solve the equation for w:

$\frac{16+w}{14+10+16+w}=\frac{3}{5}$

$\frac{16+w}{40+w}=\frac{3}{5}$

$5(16+w)=3(40+w)$

$80+5w=120+3w$

$2w=40$

$w=20$

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards: 6.SP.8a uniform probability models
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

Turning words into math moderate Understanding the question
“How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals partial knowledge Expected pull: strong
Takes 3/5 of the 40 marbles now in the box, 24, and subtracts the 16 white ones, forgetting that added marbles raise the total.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 10, the number of black marbles in the stem, without setting up the new probability or testing it.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 14, the number of red marbles in the stem, without checking that adding 14 white marbles gives 30 of 54, not 3/5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}$; P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

Sort out the box
The box has 14 red, 10 black and 16 white marbles.
$14 + 10 + 16 = 40$
That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is $\frac{3}{5}$ .
Write the new probability
Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.
$\frac{16 + w}{40 + w} = \frac{3}{5}$
Watch out: Taking $\frac{3}{5}$ of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.
Cross multiply
Fractions are hard to solve with. Clear them with cross multiplication.
$5 \times (16 + w) = 3 \times (40 + w)$
Then multiply each outside number by both terms inside the parentheses.
$80 + 5w = 120 + 3w$
Solve for w
Subtract 3w from both sides. Then subtract 80 from both sides.
$2w = 40$
Divide both sides by 2.
$w = 20$
You need to add 20 white marbles.
So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: $\frac{36}{60} = \frac{3}{5}$ . It works.

More questions like this

The original DOE question

Question 114 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 39 of 51

2018 Test A Math, question 6

Question

The set of possible values of $m$ is $\left \{ 5,7,9 \right \}$ . What is the set of possible values of $k$ if $2k=m+3$ ?

A $\left \{ 3,4,5 \right \}$
B $\left \{ 4,5,6 \right \}$
C $\left \{ 8,10,12 \right \}$
D $\left \{ 10,14,18 \right \}$

Show the answer

Correct answer: B

$2k=m+3$

$k=\cfrac{m+3}{2}$

Substitute each value of m to find the values of k:

$k=\cfrac{5+3}{2}=\cfrac{8}{2}=4$

$k=\cfrac{7+3}{2}=\cfrac{10}{2}=5$

$k=\cfrac{9+3}{2}=\cfrac{12}{2}=6$

The set k is { 4, 5, 6}.

Question analysis

An algebra question about substitution: m can be 5, 7 or 9, and an equation links k to m. Students find every possible value of k by substituting each value of m and solving. The curly-brace set notation can look unfamiliar, and each value has to be carried all the way through the solve.

NYS standards: 6.EE.7 one-step equations
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Each substitution leads to a one-step grade 6 equation, easier than a typical grade 7 two-step problem. What adds demand is doing it three times and reading the answer as a set of values, which grade 7 rarely uses.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The equation is simple, and the main hazard is stopping one step short: m + 3 gives 8, 10 and 12, a listed choice even though those are values of 2k. Checking a single value of m is the other risk, since k = 4 and k = 5 both appear in the first two sets. Students who substitute all three values and halve each result finish quickly.

Variables instead of numbers minor Understanding the question
There is no single unknown: m stands for the set {5, 7, 9}, and the answer is the matching set for k, so the equation has to be applied to three values. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the equation so k equals m plus 3, all divided by 2, then evaluate it for each of the three values of m.

Traps

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

The wrong answers, classified

A Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Checks only m = 5, gets k = 4, and takes the first set containing 4; m = 9, giving k = 6, would rule it out.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Adds 3 to each value of m to get 8, 10 and 12, which are the values of 2k, and forgets to divide by 2.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Doubles each value of m to get 10, 14 and 18, moving the 2 onto m instead of dividing by it, and drops the + 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Getting k by itself: $k = \frac{m + 3}{2}$; k and m are the letters from the equation 2k = m + 3. The fraction bar means divide. If 2 times k equals m + 3, then k is half of m + 3. You get this by dividing both sides of the equation by 2.

Sort out the question
The letter m can be 5, 7 or 9. The curly braces { } show a set, which is a list of numbers. The equation $2k = m + 3$ links k to m. You need every value k can have, one for each value of m.
Get k by itself
2k means 2 times k. To undo times 2, divide both sides by 2. Doing the same thing to both sides keeps the equation true.
$k = \frac{m + 3}{2}$
Try each value of m
Put each value of m into the formula. Add 3 first, then divide by 2.
$k = \frac{5 + 3}{2} = \frac{8}{2} = 4$
$k = \frac{7 + 3}{2} = \frac{10}{2} = 5$
$k = \frac{9 + 3}{2} = \frac{12}{2} = 6$
Watch out: 8, 10 and 12 are the values of m + 3, which equal 2k, not k. Stopping before you divide by 2 gives choice C.
Pick the matching set
The values of k are 4, 5 and 6. Watch out: try all three values of m. If you only try m = 5, you get k = 4, and choice A has a 4 too. So the answer is B, {4, 5, 6}.

Check your answer

Put each k back into the equation. $2 \times 4 = 8 = 5 + 3$ . $2 \times 5 = 10 = 7 + 3$ . $2 \times 6 = 12 = 9 + 3$ . All three work, so the set is right.

More questions like this

The original DOE question

Question 63 as printed on page 82 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 63. Open page 82 of the handbook (PDF)

Question 40 of 51

2018 Test A Math, question 17

Question

The sum of the numbers x, y, and z is 50. The ratio of x to y is 1:4, and the ratio of y to z is 4:5. What is the value of y ?

Show the answer

Correct answer: D

Since both ratios have $y$ in common, solve for $x$ and $z$ in terms of $y$ in both equations. Using $x:y=1:4$ , solve for $x$ in terms of $y$ .

$\frac{x}{y}=\frac{1}{4}$

$x=\frac{y}{4}$

Using the ratio $y:z=4:5$ , solve for $z$ in terms of $y$ :

$\frac{y}{z}=\frac{4}{5}$

$z=\frac{5}{4}y$

The question states $x+y+z=50$ . Substitute from the two equations above and solve for $y$ .

$\frac{1}{4}y+y+\frac{5}{4}y=50$

$\frac{10}{4}y=50$

$10y=200$

$y=20$

Question analysis

A ratio problem with three unknowns: x, y and z add to 50, x to y is 1 to 4, and y to z is 4 to 5. Students merge the two ratios into one three-part ratio and use the sum to find y. The step that is easy to skip is turning ratio parts into actual values.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 ratio problems usually involve one ratio. Here two ratios share a term and must be merged before the total is used, though the shared term already matches, which keeps it near typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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 question rewards noticing that y has the same term in both ratios, which turns two ratios into one three-part ratio in a single move. Students who miss this often write x and z in terms of y, as the explanation does, which works but is slower. The wrong choices are numbers taken from the ratios themselves, so a student who never scales up to the total of 50 finds an answer waiting.

No obvious method moderate Working it out
The two ratios have to be merged into one, x : y : z = 1 : 4 : 5, which works only because y is 4 in both; nothing in the wording says to combine them. (Schoenfeld, 1985)
Variables instead of numbers minor Understanding the question
x, y and z are given only through two ratios and their sum of 50, so there is no actual count to anchor the work until the parts are scaled. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Adding the parts, dividing 50 by their total and scaling y's 4 parts are three steps, and the count of parts after the first step is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Since y is 4 in both ratios, write x : y : z as one combined ratio, find the value of one part from the total of 50, and scale up.

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

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes y's term in the ratios, 4, as y's actual value, never using the sum of 50.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two 4s that stand for y in the two ratios, 4 + 4 = 8, treating ratio terms as amounts.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Combines the ratios into 1 : 4 : 5 and gives the total number of parts, 1 + 4 + 5 = 10, instead of y's share of the 50.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sharing a total in a ratio: $\text{one part} = \text{total} \div \text{number of parts}$; In a ratio like 3:2, the amounts are made of equal-sized parts. Add the numbers in the ratio to count the parts. Divide the total by that count to get the size of one part. Then multiply by how many parts each amount has.

Sort out the question
The three numbers x, y and z add up to 50. The ratio of x to y is 1:4. The ratio of y to z is 4:5. A ratio compares sizes. You need the actual value of y.
Join the two ratios
y is 4 in both ratios. So the two ratios use the same size parts, and you can join them into one.
$x : y : z = 1 : 4 : 5$
This means x is 1 part, y is 4 parts, and z is 5 parts.
Count the parts
$1 + 4 + 5 = 10$
There are 10 parts in all. Watch out: 10 is the number of parts, not the value of y. Picking it is choice C.
Find the size of one part
Use the sharing formula. The 10 parts share the total of 50.
$\text{one part} = 50 \div 10 = 5$
Each part is worth 5.
Find y
y is 4 parts.
$y = 4 \times 5 = 20$
Watch out: the 4 in the ratio counts parts. It is not the value of y. Picking 4 is choice A. So the answer is D, 20.

Check your answer

Find all three: x = 5, y = 20, z = 25. They add up: $5 + 20 + 25 = 50$ . x to y is 5 to 20, which is 1:4. y to z is 20 to 25, which is 4:5. Everything fits.

More questions like this

The original DOE question

Question 74 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 74. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 41 of 51

2018 Test A Math, question 32

Question

The least of 5 consecutive integers is $l$ , and the greatest is $g$ . What is the value of $\frac{l+g}{2}$ in terms of $l$ ?

A $2l$
B $3l$
C $l+2$
D $l+5$

Show the answer

Correct answer: C

The first integer is $l$ , so the second is $l+1$ , the third is $l+2$ , then $l+3$ , and finally $l+4$ .

Since $g$ is the fifth and greatest of the integers,

$g=l+4$

Substitute $l+4$ for $g$ and simplify:

$\frac{l+g}{2}=\frac{l+l+4}{2}=\frac{2l+4}{2}=l+2$

Question analysis

This algebra question about five consecutive integers calls the least one l and the greatest g, then asks for (l + g)/2 written in terms of l. Students must express g using l and simplify a fraction that contains a variable. Counting how far the fifth integer is from the first is where many students slip.

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students combine like terms in expressions such as 3(x + 4) − x. This item first asks them to build the expression from a description of consecutive integers, then simplify it, with only letters to work with.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Turning words into math minor
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The algebra is short once g is written as l + 4, so most errors come from the fence-post count or from testing numbers carelessly. The wrong choices in front of the answer are exactly the ones a single test value cannot rule out, which punishes students who stop at the first match. Using two test values, or simplifying (2l + 4)/2 directly, avoids both traps.

Variables instead of numbers moderate Understanding the question
Every expression is in l, so the simplification of (l + l + 4)/2 has to be done with letters and cannot be checked against a single number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The description "the least of 5 consecutive integers is l, and the greatest is g" must become g = l + 4, four steps up, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
A single test value fits two choices: l = 1 makes 3l and l + 2 both 3, and l = 2 makes 2l and l + 2 both 4. (Embretson & Wetzel, 1987)

Strategy: Write the integers as l, l + 1, and so on up to the fifth, substitute for g and simplify; then test two different starting values, since one value can match a wrong choice by coincidence.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: weak
Tests only l = 2, where 2 through 6 give (2 + 6)/2 = 4, and takes 2l, the first choice that also equals 4.
B Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Tests only l = 1, where 1 through 5 give (1 + 5)/2 = 3, and stops at 3l, which matches before l + 2 is checked.
D Counting the ends (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Writes the greatest integer as l + 5, counting five steps for five integers, then halves only the 2l in (2l + 5)/2, leaving l + 5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest of 5 consecutive integers: $g = l + 4$; l is the least (smallest) integer and g is the greatest. Consecutive integers go up by 1 each time. Five integers have only 4 jumps from the first to the last, so you add 4, not 5.
Dividing a sum: $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$; a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. You must divide every part of the top, not just one.

Sort out the words
Integers are whole numbers, like −2, 0 and 7. Consecutive integers come one right after another, like 8, 9, 10. There are 5 of them here.
The smallest is called l and the biggest is called g. You must write $\frac{l + g}{2}$ using only the letter l.
Write all five integers
Each integer is 1 more than the one before:
$l,\ l+1,\ l+2,\ l+3,\ l+4$
The fifth one is the greatest, so g = l + 4. Watch out: five integers have only four jumps between them. Writing g = l + 5 leads to choice D.
Put l + 4 in place of g
Swap g for l + 4 in the fraction:
$\frac{l + g}{2} = \frac{l + l + 4}{2} = \frac{2l + 4}{2}$
Here l + l is 2l, which means 2 times l.
Split the fraction
Divide each part of the top by 2:
$\frac{2l}{2} + \frac{4}{2} = l + 2$
So the answer is C, $l + 2$ .

Check your answer

Test two starting numbers. If l = 1, the integers are 1 to 5, and (1 + 5) ÷ 2 = 3, which is 1 + 2. If l = 10, they are 10 to 14, and (10 + 14) ÷ 2 = 12, which is 10 + 2. Use two tests, because l = 1 also makes 3l equal 3, choice B.

More questions like this

The original DOE question

Question 89 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 89. Open page 87 of the handbook (PDF)

Question 42 of 51

2018 Test A Math, question 39

Question

Suppose $M=\frac{w}{x}$ , $N=\frac{y}{z}$ , and $w,x,y$ , and $z$ do not equal 0. What is $\frac{M}{N}$ in terms of $w,x,y$ , and $z$ ?

A $\frac{wx}{yz}$
B $\frac{wy}{xz}$
C $\frac{wz}{xy}$
D $\frac{xy}{wz}$

Show the answer

Correct answer: C

It is easier to rewrite

$\frac{M}{N}$

$M\div N$

since they are both fractions.

$M\div N=\frac{w}{x}\div \frac{y}{z}=\frac{w}{x}\times \frac{z}{y}=\frac{wz}{xy}$

Question analysis

This algebra question defines M and N as fractions built from the letters w, x, y and z, and asks for M divided by N in terms of those letters. Students must divide one fraction by another using only variables. Without numbers to check against, it is easy to multiply instead of divide or flip the wrong fraction.

NYS standards: 6.NS.1 dividing fractions by fractions
7.NS.2c properties for multiplying and dividing
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students divide numeric fractions such as 2/3 ÷ 4/5. This item applies the same rule with four letters instead of numbers, which adds abstraction but no new steps.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is fraction division in disguise, and students who recognize M/N as M ÷ N need one rule. The difficulty is that every choice is built from the same four letters, so a wrong rule still produces something that looks like a reasonable answer. Substituting small numbers, such as w = 1, x = 2, y = 3 and z = 4, checks the result in seconds.

Variables instead of numbers moderate Understanding the question
All four letters stand for unknown nonzero numbers, so dividing w/x by y/z must be done symbolically, with no value to check the result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
All four choices arrange w, x, y and z as one fraction, and xy/wz is wz/xy turned over, so only carrying out the division in the right order tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite M/N as M ÷ N, then multiply by the reciprocal of the second fraction; substituting small numbers for the letters checks the result.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Stacks M's letters over N's letters, writing wx over yz, without carrying out any fraction division.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies M by N straight across, top times top and bottom times bottom, instead of dividing.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Flips the first fraction instead of the second, computing x/w × y/z, which gives N divided by M.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a, b, c and d are numbers that are not 0. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing by a number is the same as multiplying by 1 over that number.
Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops and b and d are the bottoms. Multiply the tops together and the bottoms together. Use it after you flip the second fraction.

Sort out what you know
M is the fraction w over x. N is the fraction y over z. The letters stand for numbers that are not 0.
The fraction bar in $\frac{M}{N}$ means M divided by N. So you must divide one fraction by another.
Write it as a division
Rewrite the big fraction as a division. Then put in the fractions for M and N. This lets you use the rule for dividing fractions:
$\frac{M}{N} = M \div N = \frac{w}{x} \div \frac{y}{z}$
Flip the second fraction
Use the rule for dividing fractions. Keep the first fraction. Flip only the second one, and change divide to multiply:
$\frac{w}{x} \times \frac{z}{y}$
Watch out: flipping the first fraction instead gives choice D, which is N ÷ M. Multiplying without flipping at all gives choice B.
Multiply across
Multiply the tops and the bottoms:
$\frac{w \times z}{x \times y} = \frac{wz}{xy}$
Here wz means w times z. So the answer is C, $\frac{wz}{xy}$ .

Check your answer

Try small numbers. Let w = 1, x = 2, y = 3 and z = 4. Then M = 1/2 and N = 3/4. So M ÷ N = 1/2 × 4/3 = 4/6. Choice C gives (1 × 4) ÷ (2 × 3) = 4/6 too. They match.

More questions like this

The original DOE question

Question 96 as printed on page 88 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 96. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 43 of 51

2018 Test A Math, question 43

Question

$\frac{w}{x}=\frac{y}{z}$

In the equation above, $w,x,y$ , and $z$ are positive numbers. Which of these is equal to $z$ ?

A $x$
B $xy$
C $\frac{w}{xy}$
D $\frac{xy}{w}$

Show the answer

Correct answer: D

Solve the equation for $z$ :

$\frac{w}{x}=\frac{y}{z}$

$wz=xy$

$z=\frac{xy}{w}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
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.

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)
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

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches z with x because both sit in a denominator, picking a letter from the equation without solving for z.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Cross-multiplies to wz = xy but stops there, giving xy without dividing both sides by w.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Cross-multiplies correctly but divides the wrong way round, putting w over xy instead of xy over w.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; 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.

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.
Cross-multiply the fractions
Multiply each top by the other fraction’s bottom:
$w \times z = x \times y$
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.
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:
$\frac{w \times z}{w} = \frac{x \times y}{w}$
$z = \frac{xy}{w}$
Watch out: dividing the wrong way round gives w over xy, choice C. So the answer is D, $\frac{xy}{w}$ .

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 44 of 51

2018 Test A Math, question 47

Question

If $3n$ is a positive even number, how many odd numbers are in the range from $3n$ up to and including $3n+5$ ?

Show the answer

Correct answer: B

Since $3n$ is even, then $3n+1$ must be odd.

Thus $3n+3$ and $3n+5$ are also odd.

So there are a total of 3 numbers in this range that are odd.

Question analysis

This number properties question says 3n is a positive even number and asks how many odd numbers lie from 3n up to and including 3n + 5. Students must decide which integers in that span are odd. The coefficient in 3n is a distraction; only the fact that 3n is even matters.

NYS standards: 6.EE.2b parts of an expression
6.EE.6 variables representing unknown numbers
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students work with expressions like 3n + 5 but rarely reason about whether they are even or odd. The idea is simple once the integers are written out, so the demand is about typical for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 12, 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

Writing the six integers from 3n to 3n + 5 and marking them even, odd, even, odd, even, odd answers the question in seconds. The traps come from not writing them: dropping the last one, which "up to and including" keeps in, or reading the 5 in 3n + 5 as a count. Picking any even value for 3n, such as 6, makes the check concrete.

Variables instead of numbers moderate Understanding the question
No value of n is given, so students must reason about parity: since 3n is even, 3n + 1, 3n + 3 and 3n + 5 are the odd ones. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
The parity rules, even plus odd is odd and even plus even is even, are assumed; the coefficient 3 is a distraction once 3n is said to be even. (Anderson & Krathwohl, 2001)
A very close wrong answer minor Choosing the answer
The choices 2, 3, 4 and 5 are consecutive counts, so 2 and 3 cannot be told apart by sizing up; only checking whether 3n + 5 is inside the range separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write out every integer from 3n to 3n + 5, mark 3n as even, alternate even and odd from there, and count the odd ones.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Leaves out 3n + 5, stopping before the last number although "up to and including" keeps it, and counts only 3n + 1 and 3n + 3.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Counts 3n itself as odd because its coefficient 3 is odd, ignoring the stem's statement that 3n is even, and adds it to the three.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 5 in 3n + 5 as the count, subtracting the ends instead of listing the six integers and picking out the odd ones.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Even plus odd: $\text{even} + \text{odd} = \text{odd}$; An even number splits into pairs. An odd number has one left over. Adding them keeps that one left over, so the sum is odd. So adding 1, 3 or 5 to an even number gives an odd number.
Even plus even: $\text{even} + \text{even} = \text{even}$; Two numbers that each split into pairs still split into pairs when you add them. So adding 2 or 4 to an even number gives an even number.

Sort out what you know
You are told that 3n is even. You do not need to know n itself. You need to count the odd numbers from 3n to 3n + 5.
“Up to and including” means 3n + 5 counts too.
List all the numbers
Go up by 1 each time, from 3n to 3n + 5. That gives six numbers:
$3n,\ 3n+1,\ 3n+2,\ 3n+3,\ 3n+4,\ 3n+5$
Mark each as even or odd
3n is even. Adding 1, 3 or 5 makes it odd. Adding 2 or 4 keeps it even.
Number Even or odd
$3n$ even
$3n+1$ odd
$3n+2$ even
$3n+3$ odd
$3n+4$ even
$3n+5$ odd
Watch out: 3n is even because the question says so. The 3 being odd does not matter. Counting 3n as odd gives choice C, 4.
Count the odd numbers
The odd ones are 3n + 1, 3n + 3 and 3n + 5. That is 3 numbers.
Watch out: leaving out 3n + 5 gives 2, choice A. It is the last number, and it is included. So the answer is B, 3.

Number	Even or odd
$3n$	even
$3n+1$	odd
$3n+2$	even
$3n+3$	odd
$3n+4$	even
$3n+5$	odd

Check your answer

Try a real number. If n = 2, then 3n = 6, which is even. The numbers from 6 to 11 are 6, 7, 8, 9, 10 and 11. The odd ones are 7, 9 and 11. That is 3.

More questions like this

The original DOE question

Question 104 as printed on page 90 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 104. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 45 of 51

2018 Test A Math, question 55

Question

If $4x-3y=12$ , what is $x$ in terms of $y$ ?

A $x=\frac{3}{4}y+12$
B $x=-\frac{3}{4}y+12$
C $x=\frac{3}{4}y+3$
D $x=-\frac{3}{4}y+3$

Show the answer

Correct answer: C

$4x-3y=12$

$4x=12+3y$

$x=\frac{3}{4}y+\frac{12}{4}$

$x=\frac{3}{4}y+3$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Error-prone arithmetic minor
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.

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)
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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Moves 3y correctly but divides only the y term by 4, leaving the constant 12 undivided.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Keeps the minus sign on the 3y term when moving it and also leaves the 12 undivided, two errors at once.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Divides every term by 4 correctly but keeps the minus on 3y, as if a term kept its sign when moved across.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $a = b \text{ means } a + c = b + c$; 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: $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$; 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.

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 $4x - 3y = 12$ .
Move the y term
The 3y is being subtracted on the left. Add 3y to both sides to clear it:
$4x - 3y + 3y = 12 + 3y$
$4x = 12 + 3y$
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.
Divide both sides by 4
x is being multiplied by 4. Divide both sides by 4 to undo it:
$x = \frac{12 + 3y}{4}$
Split the fraction
Divide each part of the top by 4:
$x = \frac{12}{4} + \frac{3y}{4} = 3 + \frac{3}{4}y$
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, $x = \frac{3}{4}y + 3$ .

Check your answer

Pick y = 4. Choice C gives x = 3 + 3 = 6. Put x = 6 and y = 4 into the start: 4 × 6 − 3 × 4 = 24 − 12 = 12. It works. Choice D gives x = 0, and 4 × 0 − 12 = −12, not 12.

More questions like this

The original DOE question

Question 112 as printed on page 91 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 112. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 46 of 51

2018 Test B Math, question 8

Question

$M=3N=\frac{P}{4}=Q+5=\frac{R}{7}> 0$

Based on the statement above, which variable has the greatest value?

A $M$
B $N$
C $P$
D $R$

Show the answer

Correct answer: D

Break the equations apart to each equal $M$ :

$M=3N$

$M=\frac{P}{4}$

$M=Q+5$

$M=\frac{R}{7}$

Pick a number to substitute into the equations, and solve the equations to find the values of $M$ , $N$ , $P$ , $Q$ , and $R$ .

Let $M=2$ . Since all the equations are equal to 2, substitute 2 to find each variable.

$M=3N$

$2=3N$

$\frac{2}{3}=N$

$M=\frac{P}{4}$

$2=\frac{P}{4}$

$8=P$

$M=Q+5$

$2=Q+5$

$-3=Q$

$M=\frac{R}{7}$

$2=\frac{R}{7}$

$14=R$

Variable $R$ has the greatest value.

Question analysis

This algebra question sets five expressions equal to one another and to a positive number, then asks which variable has the greatest value. It tests whether students can rewrite each variable in terms of one shared value and compare them. What makes it tricky is that there are no numbers to compute, so each variable's size has to be reasoned out.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students rewrite expressions to see how quantities relate, usually one equation at a time. This chain of five linked expressions, with no numbers to compute and a positivity condition to use, is a reasoning task beyond routine grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 3, strategic thinking · mainly analyze
What makes it hard: Variables instead of numbers major
No obvious method moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards a strategy, not a calculation: once the shared value is named, each variable follows in one line and R is clearly largest. Without that step, students judge by how the expressions look, and the look is backwards, since the 3 in 3N makes N small and the 7 under R makes R large. The positivity condition at the end is easy to skip but is what fixes the order.

Variables instead of numbers major Understanding the question
There are no numbers to compute: five variables are tied together in the one chain M = 3N = P/4 = Q + 5 = R/7, and each must be compared through the shared value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
No obvious method moderate Working it out
Nothing says how to compare; the efficient move is to name the common value, or pick one such as 12, and rewrite each variable, which students must think of themselves. (Schoenfeld, 1985)
Much to keep track of minor Working it out
Five relations and the condition > 0 must be held at once; the condition matters, since with a negative common value the order of M, N, P and R would reverse. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Call the shared value k, a positive number, and rewrite each variable as an expression in k; or substitute a convenient positive value such as 12 and compare the results.

Traps

Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

A Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: weak
Reads P/4 and R/7 as a quarter and a seventh of M, the reverse of the truth, so M looks largest of all.
B Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: strong
Reads the 3 in 3N as making N three times as large, when 3N = M actually makes N one third of M.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Correctly sees that P is four times M but stops there without checking R, which at seven times M is larger.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing division: $\frac{a}{b} = c \text{ means } a = b \times c$; a is the top of the fraction, b is the bottom, and c is the result. Multiplying undoes dividing. So if a divided by b is c, then a is b times c.
Undoing multiplication: $b \times a = c \text{ means } a = c \div b$; b is the number multiplied by a, and c is the result. Dividing undoes multiplying. So to get a alone, divide c by b.

Understand the chain
The statement is a chain of equal signs. It says five expressions all have the same value. The last part says that value is greater than 0, so it is positive.
You need to find which letter has the largest value.
Pick a value to test
All five expressions are equal, so give them one shared value. Pick a positive number that 3 divides evenly. That way N will be a whole number.
Let the shared value be 12. Then $M = 12$ , and each of the other four expressions also equals 12.
Solve for each letter
Undo each operation to find each letter. Divide to undo times. Multiply to undo divide. Subtract to undo plus.
$3N = 12 \text{, so } N = 12 \div 3 = 4$
$\frac{P}{4} = 12 \text{, so } P = 12 \times 4 = 48$
$Q + 5 = 12 \text{, so } Q = 12 - 5 = 7$
$\frac{R}{7} = 12 \text{, so } R = 12 \times 7 = 84$
Compare the values
The values are M = 12, N = 4, P = 48, Q = 7 and R = 84. R is the largest.
Watch out: 3N does not make N big. It means 3 times N equals M, so N is only one third of M. That trap points to choice B. Also, P is big, but R is bigger. Check every letter before you pick choice C.
So the answer is D, R.

Check your answer

Think without numbers. R divided by 7 is M, so R is 7 times M. P is only 4 times M. N is a third of M, and Q is 5 less than M. Since M is positive, 7 times M is the biggest.

More questions like this

The original DOE question

Question 65 as printed on page 154 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 65. Open page 154 of the handbook (PDF)

Question 47 of 51

2018 Test B Math, question 20

Question

If $\frac{36}{y}=4x$ , what is the value of $x$ when $y=3$ ?

Show the answer

Correct answer: A

Substitute 3 for $y$ and solve for $x$ :

$\frac{36}{y}=4x$

$\frac{36}{3}=4x$

$12=4x$

$3=x$

Question analysis

This equation question gives 36 divided by y equal to 4x and asks for the value of x at one given value of y. Students substitute, simplify the fraction and solve a one-step equation. It is short, so the risk lies in stopping one step early or dividing by the wrong number.

NYS standards: 6.EE.7 one-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Substituting a value and solving a one-step equation are grade 6 skills. With two quick steps and whole numbers, this is lighter than the multistep equations typical of grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is substitution followed by a one-step equation, with small whole numbers throughout, so most students answer correctly. The only real risk is stopping at 12, which is the value of 4x rather than of x. Writing 12 = 4x as its own line before reading the choices makes the last division hard to skip.

Variables instead of numbers minor Understanding the question
The variable y sits in a denominator, 36/y, which looks less familiar than an equation in one unknown, though substituting y = 3 removes it at once. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Replace y with its value, simplify the left side to a single number, and then divide by the coefficient of x.

Traps

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

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks the coefficient 4 from 4x, a number that appears in the equation, without solving for x.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Clears the fraction to get 36 = 4xy, divides by 4 to reach xy = 9, and stops before dividing by y.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Substitutes y = 3 to get 36/3 = 12, which is the value of 4x, and stops before dividing by 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing multiplication: $a \times x = b \text{ means } x = b \div a$; a is the number multiplied by x, and b is the result. Division undoes multiplication. So to get x alone, divide both sides by a.

Write down what you know
The equation is $\frac{36}{y} = 4x$ . You are told y = 3. You need the value of x.
The letter y is on the bottom of a fraction. Putting in its value turns the fraction into a plain number.
Put in the value of y
Replace y with 3.
$\frac{36}{3} = 4x$
The fraction means 36 divided by 3, which is 12.
$12 = 4x$
Watch out: 12 is the value of 4x, not x. Stopping here gives choice D.
Divide both sides by 4
4x means 4 times x. To undo multiplying by 4, divide both sides by 4.
$x = 12 \div 4 = 3$
So the answer is A, 3.

Check your answer

Put x = 3 and y = 3 back in. The left side is $36 \div 3 = 12$ . The right side is $4 \times 3 = 12$ . Both sides match, so x = 3 is right.

More questions like this

The original DOE question

Question 77 as printed on page 157 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 77. Open page 157 of the handbook (PDF)

Question 48 of 51

2018 Test B Math, question 28

Question

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

A40%
B60%
C64%
D80%

Show the answer

Correct answer: C

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

$0.80x$

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

$0.80(0.80x)=0.64x\text{ or }64\%$

Question analysis

This percent problem describes a chemical that loses 20% of what is present each week and asks what percent of the original amount remains after two weeks. Students apply a percent decrease twice in a row. Deciding which amount each week's 20 percent is taken from is the heart of the problem.

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually apply a single increase or decrease. Applying a decrease twice, each time to the new amount, is an early form of exponential decay that goes beyond typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer major
Turning words into math moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a two-step percent decrease, and the intuitive answer, 20% plus 20%, is wrong for a reason students rarely check: the second loss is taken from what is left after the first. The stem states this rule in its own words, but quickly. Starting with 100 units and working week by week makes the difference visible, since the second week removes 16 units, not 20.

A very close wrong answer major Choosing the answer
60% and 64% are four points apart, too close for a rough estimate; only seeing that week two's 20% comes from 80, not 100, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math moderate Understanding the question
The stem measures each week's loss against “the amount at the beginning of that same week”, and students must see that this resets the base to 80% for the second week. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
No starting amount is given, so students must supply one, such as 100 units, or reason directly with a percent of a percent. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Start with 100 units, take away 20 percent of the current amount for the first week, then take 20 percent of what is left for the second week.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Adds the two 20% losses to get 40% and reports that amount lost, when the question asks how much is left.
B Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes both weeks' 20% from the original amount, 100 − 20 − 20, missing that the second week's loss comes from the smaller 80%.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 80% left after one week and stops before applying the second week's decrease.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent decrease: $\text{amount left} = \text{amount} \times \left(1 - \frac{p}{100}\right)$; p is the percent taken away, here 20. Taking away 20% leaves 80% of the amount, so you multiply by 0.80. Each week starts from the amount left at the end of the week before.

Understand the decay
Each week, the chemical loses 20% of what it had at the start of that week. The amount at the start of each week keeps changing.
So the second week’s 20% is taken from a smaller amount. You need the percent left after two weeks.
Start with 100 units
No amount is given, so pick an easy one. Start with 100 units. Then the number of units left is also the percent left.
Work out week one
20% of 100 is 20, so 20 units are lost.
$100 - 20 = 80$
After week one, 80 units are left. That is 80% of the start.
Watch out: 80% is only one week. Stopping here gives choice D.
Work out week two
Now the week starts with 80 units, not 100. Take 20% of 80.
$\frac{20}{100} \times 80 = 16$
$80 - 16 = 64$
After two weeks, 64 units are left. That is 64% of the 100 units you started with.
Watch out: Taking 20 away twice from 100 gives 60%, choice B. That uses the old amount in week two.
So the answer is C, 64%.

Check your answer

Use the percent decrease formula. Each week keeps 80%, or 0.80, of the amount. Two weeks gives $0.80 \times 0.80 = 0.64$ , which is 64%. The two methods agree.

More questions like this

The original DOE question

Question 85 as printed on page 158 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 85. Open page 158 of the handbook (PDF)

Question 49 of 51

2018 Test B Math, question 29

Question

If $w-1$ is an odd integer, which one of the following must be an even integer?

A $w+1$
B $2w-1$
C $2w-2$
D $2w+1$

Show the answer

Correct answer: C

One more than an odd integer must be even. One more than $w-1$ is $w$ , therefore $w$ must be even.

Two times an even integer must be even, therefore $2w$ is even. An even integer decreased by 2 must be even.

Therefore, $2w-2$ must be even.

Question analysis

This algebra question says that w − 1 is an odd integer and asks which of four expressions must be even. It tests reasoning about odd and even numbers written with a variable. The difficulty is that "must be" means even for every allowed value of w, so one example that happens to work is not enough.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 work evaluates and rewrites expressions but rarely asks whether one is odd or even for every value. The algebra here is light; the demand is reasoning about parity and reading "must be" carefully.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question turns on one early inference: since w − 1 is odd, w itself is even. Students who skip that step, or quietly treat w as the odd number, carry the wrong parity into every choice. After that the work is mechanical, because each expression adds an even or an odd amount to an even number. Trying one value, such as w = 4, confirms the choice in seconds.

Variables instead of numbers moderate Understanding the question
The stem gives no number at all: w − 1 is only known to be odd, so the parity of w itself has to be worked out before any choice can be judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
All four choices are built from w or 2w plus or minus 1 or 2, so each one has to be tested for evenness; three of them come out odd for every allowed w. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
The parity rules are assumed, not stated: one more than an odd number is even, any multiple of 2 is even, and an even number plus or minus 1 is odd. (Anderson & Krathwohl, 2001)

Strategy: Choose a value of w that makes w − 1 odd, evaluate all four expressions, and keep the one that comes out even; then confirm it with a second value.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Treats w itself as the odd number, so w + 1 looks even; in fact w − 1 being odd makes w even and w + 1 odd.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Doubles the odd number w − 1 to make it even but distributes the 2 to w only, writing 2(w − 1) as 2w − 1, which is odd.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Assumes any expression that starts with 2w is even, overlooking that 2w + 1 is one more than an even number and so always odd.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding 1 to an odd number: $\text{odd} + 1 = \text{even}$; Odd and even numbers take turns on the number line: 1, 2, 3, 4 and so on. Adding 1 moves you to the next whole number. So the number right after an odd number is always even, and the one after an even number is always odd.
Doubling makes an even number: $2 \times n = \text{even}$; n is any whole number. An even number is one you can split into two equal whole groups. Two times n is always two equal groups of n, so it is always even.

Sort out what you know
You know that $w-1$ is an odd integer. An integer is a whole number, like 3, 0 or −5. An even number can be split into two equal groups, like 4 or 10. An odd number is 1 more than an even number, like 3 or 11.
You need the choice that must be even. Must means it is even for every possible value of $w$ , not just one.
Decide whether w is odd or even
The number $w$ is 1 more than $w-1$ . One more than an odd number is always even. For example, if $w-1$ is 3, then $w$ is 4. So $w$ is even.
Watch out: do not treat $w$ itself as the odd number. That mistake makes $w+1$ look even and leads to choice A.
Show that 2w − 2 is even
Two times any whole number is even, so $2w$ is even. Taking 2 away from an even number moves you back two spots. You land on another even number. So $2w-2$ is even.
Here is another way to see it. $2w-2$ is the same as $2 \times (w-1)$ . That is 2 times a whole number, so it must be even.
Check the other three choices
Each one adds or takes away 1 from an even number. That always gives an odd number.
- A: $w+1$ is even plus 1, so it is odd.
- B: $2w-1$ is even minus 1, so it is odd.
- D: $2w+1$ is even plus 1, so it is odd.
Watch out: starting with $2w$ does not make a number even. The +1 in choice D makes it odd.
Give the answer
Only one choice is even for every possible $w$ . So the answer is C, $2w-2$ .

Check your answer

Pick a number. Let $w = 4$ , so $w-1 = 3$ , which is odd. Then A is 5, B is 7, C is 6 and D is 9. Only C is even. Try $w = 10$ too. You get 11, 19, 18 and 21. C is even again.

More questions like this

The original DOE question

Question 86 as printed on page 158 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 86. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 51

2018 Test B Math, question 43

Question

$A quarter circle whose center, marked with a dot, is at the bottom left corner. A segment labeled x + 1 runs from the center to a point on the arc, and from that point a vertical segment drops to the bottom edge, meeting it at a right angle. Along the bottom edge, the part from the center to the foot of that segment is labeled y, and the part from the foot to the end of the arc is labeled 2.$

In the quarter circle above, what is $y$ in terms of $x$ ?

A $x-1$
B $x+1$
C $\frac{x+1}{2}$
D $\sqrt{\frac{\left ( x+1 \right )^{2}}{2}}$

Show the answer

Correct answer: A

Both $x+1$ and $y+2$ are radii.

So, set them equal to each other and solve for $y$ .

$y+2=x+1$

$y=x-1$

Question analysis

In a quarter circle, one segment is labeled x + 1 and the bottom edge is split into lengths y and 2, and students must write y in terms of x. It tests basic circle facts together with solving a simple equation. The challenge is seeing which marked lengths must be equal.

NYS standards: 7.G.4 circle area and circumference
7.EE.4 equations and inequalities from problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the definition of a radius and a one-step equation, but spotting the relationship in a figure is less routine than plugging into a formula.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Reading a figure, table or graph major
Variables instead of numbers minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The figure looks like a right-triangle problem, and that is the trap: the right angle and the slanted radius pull students toward the Pythagorean theorem, which cannot work without the vertical leg. The solution needs only the fact that all radii of a circle are equal, applied to the bottom edge. Students who mark every segment that starts at the center see it at once.

Reading a figure, table or graph major Understanding the question
The labels sit on two different radii: x + 1 on the slanted one, and y + 2 along the bottom edge, which runs from the center all the way to the arc. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Variables instead of numbers minor Understanding the question
The answer is an expression in x, so there is no number to check; trying x = 5, which makes the radius 6 and y = 4, is the fallback. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
x + 1 differs from the right x − 1 by a single sign; only seeing that y plus the 2-unit piece makes a radius, so the 2 is subtracted, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Look for two lengths in the figure that are both radii of the quarter circle, set their expressions equal, and solve for y.

Traps

Trusting the picture: The figure suggests something it does not state; work from the given facts.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Sets y equal to the whole radius x + 1, ignoring the 2-unit piece that completes the bottom radius beyond y.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Treats y and the 2 as factors, y × 2 = x + 1, and divides by 2 where it should subtract 2.
D Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a misconception Expected pull: strong
Trusts the drawing's nearly 45° radius, assumes the triangle's two legs are equal, and solves y² + y² = (x + 1)² for y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radii of a circle are equal: $r_1 = r_2$; A radius is a segment from the center of a circle to its curved edge. r₁ and r₂ stand for any two radii of the same circle. They are always equal, because every point on the curve is the same distance from the center.

Read the figure
The shape is a quarter circle, one fourth of a whole circle. Its center is the corner with the dot. A radius is a segment from the center to the curved edge. The slanted segment, labeled $x+1$ , goes from the center to the curve. The bottom edge is split into two parts, $y$ and 2.
Spot the two radii
The slanted segment is a radius, so the radius is $x+1$ . The whole bottom edge also runs from the center to the curve. Its length is $y+2$ , so it is a radius too.
Watch out: the slanted radius only looks like it is at 45 degrees. Do not assume the triangle’s two short sides are equal. That idea leads to choice D.
Set the radii equal
Use the rule that all radii of a circle are equal.
$y + 2 = x + 1$
Now you have one equation with $x$ and $y$ .
Solve for y
Get $y$ alone by taking 2 from both sides.
$y = x + 1 - 2 = x - 1$
Watch out: $y$ is only part of the bottom radius. Setting $y$ equal to $x+1$ gives choice B. Dividing by 2 instead of subtracting 2 gives choice C.
Give the answer
$y$ is 2 less than the radius, and the radius is $x+1$ . So the answer is A, $x-1$ .

Check your answer

Pick a number. If $x = 5$ , the radius is 6. Then $y = 5 - 1 = 4$ , and $4 + 2 = 6$ , which matches the radius. Also, $y$ is a short side of the right triangle. It should be shorter than the slanted side, and 4 is less than 6.

More questions like this

The original DOE question

Question 100 as printed on page 161 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 100. Open page 161 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 51

2018 Test B Math, question 48

Question

$Coordinate axes with origin O. A shaded triangle stands on the x-axis: its base runs from the point marked m to the point marked n, both to the right of O, and its top vertex touches a dashed horizontal line drawn at 4 on the y-axis.$

What is the area of the shaded triangle shown above?

A $m+n$
B $n-m$
C $2\left ( n-m \right )$
D $4\left ( n-m \right )$

Show the answer

Correct answer: C

The height of the triangle is 4 units. The length of the base is $n-m$ . So the area is

$A=\frac{1}{2}(n-m)(4)=2(n-m)$

Question analysis

Students find the area of a shaded triangle drawn on coordinate axes, with its base between the points m and n on the x-axis and its top vertex at a height of 4. It tests the triangle area formula with an algebraic side length. Writing the base as an expression in m and n is the step that needs care.

NYS standards: 6.G.1 area of polygons
6.EE.2a writing algebraic expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Triangle area is grade 6 content, but a base written with two variables is less familiar. With only two steps, the demand matches typical grade 7 work that mixes geometry with algebraic expressions.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Reading a figure, table or graph moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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 arithmetic is trivial, but students must treat a length as a difference of coordinates and keep a formula's one half when the answer is a letter expression. Those who forget the half, or who stop at the base length, find their result among the choices. Trying m = 1 and n = 4 gives a triangle of area 6 that checks the expressions quickly.

Variables instead of numbers moderate Understanding the question
The base runs from m to n on the x-axis, so its length is the expression n − m, and the area stays in letters with no number to check. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph moderate Understanding the question
The height, 4, is read from the dashed line through the apex rather than labeled on the triangle, and the base endpoints appear only as tick labels. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer moderate Choosing the answer
4(n − m) is the right 2(n − m) doubled, written the same way except for the coefficient; only remembering the one half in a triangle's area separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the base as the difference between its endpoint values, read the height from the figure, and apply one-half times base times height.

Traps

A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two labels, m + n, as if the base were the sum of the endpoints, and goes no further with the height.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the base length n − m correctly and gives it as the area, never multiplying by the height or taking one half.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies the base n − m by the height 4 but forgets the one half in the triangle area formula.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.
Length along the x-axis: $\text{length} = \text{right end} - \text{left end}$; The right end and the left end are the numbers where a segment starts and stops on the x-axis. Subtracting the smaller from the larger tells how long the segment is.

Read the figure
The shaded triangle sits on the x-axis, the flat number line. Its base runs from $m$ to $n$ . Its top corner touches the dashed line at height 4. You need the area, written with the letters $m$ and $n$ .
Find the length of the base
The base starts at $m$ and ends at $n$ . Use the length rule: right end minus left end.
$\text{base} = n - m$
Watch out: adding the ends, $m+n$ , does not give a length. That is choice A.
Find the height
The height is the straight-up distance from the base to the top corner. The top corner is on the dashed line, which is at 4 on the y-axis. So the height is 4.
Find the area
Use the triangle area formula.
$A = \frac{1}{2} \times (n - m) \times 4 = 2(n - m)$
Half of 4 is 2, so the area is $2(n-m)$ .
Watch out: forgetting the one half gives $4(n-m)$ , choice D. Stopping at the base gives $n-m$ , choice B.
Give the answer
The area is 2 times the length of the base. So the answer is C, $2(n-m)$ .

Check your answer

Try numbers. Say $m = 1$ and $n = 4$ . The base is 3 and the height is 4, so the area is $\frac{1}{2} \times 3 \times 4 = 6$ . Choice C gives $2(4-1) = 6$ . It matches. Choice D would give 12.

More questions like this

The original DOE question

Question 105 as printed on page 162 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 105. Open page 162 of the handbook (PDF)

Small factor	Partner
1	168
2	84
3	56
4	42
6	28
7	24
8	21
12	14