Practice Test

7.EE.4a two-step equations

Question 1 of 39

2026 Practice Test A Math, question 27

Question

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

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

$24 = -0.5(x + 8)$

$24 = -0.5x +$

$= -0.5x$

$= x$

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. A tile can be used more than once. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: −4; blank 2: 28; blank 3: −56

Solution:

Answer for second equation: −4

Answer for third equation: 28

Answer for fourth equation: −56

$24 = -0.5(x + 8)$

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

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

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

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving equations of the form p(x + q) = r with rational numbers is named in the grade 7 standards. Here the steps are laid out, but the negative decimal factor forces a sign decision at every one of the three boxes.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer minor
Many steps minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
Two of the three boxes turn on a negative decimal: −0.5 times 8 must give −4, and dividing 28 by −0.5 means multiplying by −2, which doubles the size and makes it negative. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The pool pairs −4 with 4, −40 with 40 and −56 with 56, so a box filled with the right size and the wrong sign looks just as finished; estimating cannot separate −56 from 56. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
The three boxes form a chain and all three must be right, so a slip in the first carries through; the pool even holds the whole wrong chain 4, 20 and −40. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The second line prints a plus sign before its box, so the distributed −0.5x − 4 must be written as −0.5x + (−4), with the negative tile in the box. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

DEB Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Gives the minus of −0.5 to x but not to 8, writing −0.5x + 4, then correctly reaches 20 = −0.5x and x = −40 from that wrong start.
CFH Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Distributes and adds 4 correctly to reach 28 = −0.5x, then divides 28 by 0.5 instead of −0.5, dropping the negative and entering 56.
CEB A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Distributes correctly to −4 but moves it to the left side without changing its sign, so 24 − 4 = 20, and 20 ÷ (−0.5) = −40.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The distributive property: $a(b + c) = a \times b + a \times c$; a is the number in front of the parentheses, and b and c are the terms inside. Multiply a by each term inside. If a is negative, the minus sign goes into every product. It works because a groups of b + c is the same as a groups of b plus a groups of c.
Signs when you multiply or divide: $(-) \times (+) = (-) \qquad (-) \times (-) = (+)$; A negative number times a positive number gives a negative result. A negative times a negative gives a positive result. The same rules hold when you divide. Use them at every step where a negative number appears.
Doing the same to both sides: $\text{if } a = b \text{, then } a + c = b + c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number (not zero when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it level. That is how you get x by itself.

Sort out the question
You solve $24 = -0.5(x + 8)$ by filling three boxes:
- Box 1 comes after $-0.5x +$ , once the parentheses are gone.
- Box 2 is the number equal to $-0.5x$ .
- Box 3 is the value of x.
A tile can be used more than once, and some tiles are not used.
Distribute the −0.5
Multiply −0.5 by each term inside the parentheses. A negative times a positive is negative:
$-0.5 \times x = -0.5x$
$-0.5 \times 8 = -4$
So $24 = -0.5x + (-4)$ . The line already shows a plus sign, so box 1 gets −4.
Watch out: the −0.5 multiplies the 8 too. Writing 4 instead starts a wrong chain: 4, then 20, then −40. Every number in that chain is a tile.
Add 4 to both sides
You want −0.5x alone on one side, and the −4 is in the way. Adding 4 undoes subtracting 4, so add 4 to both sides:
$24 + 4 = -0.5x - 4 + 4$
$28 = -0.5x$
Box 2 gets 28.
Watch out: taking 4 away from 24 instead gives 20, another tile.
Divide both sides by −0.5
Now x is multiplied by −0.5. Dividing undoes multiplying, so divide both sides by −0.5:
$28 \div (-0.5) = -56$
Dividing by 0.5 is the same as multiplying by 2, since every whole holds two halves. A positive divided by a negative is negative. So $x = -56$ .
Watch out: dropping the minus sign gives 56, which is also a tile.
Name the three tiles
From top to bottom, the finished steps read:
$24 = -0.5x + (-4)$
$28 = -0.5x$
$-56 = x$
So the boxes get −4, 28 and −56, in that order. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

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

More questions like this

The original DOE question

Question 27 as shown on the DOE’s digital test, page 98 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 98, question 27.

Question 2 of 39

2026 Practice Test C Math, question 2

Question

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

A $\frac{1}{3}$
B $\frac{1}{2}$
C2
D3

Show the answer

Correct answer: B

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

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

Subtract 4 from both sides:

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

Multiply both sides of the equation by m:

$8 = 16m$

Solve for m:

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

Simplify:

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

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Undoing an addition and then a division is typical two-step grade 7 equation work. The unknown sitting in a denominator is less common than px + q = r, which makes the last step easier to get backward.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
The unknown is the divisor, so 8 ÷ m = 16 ends in m = 8 ÷ 16, a quotient less than 1; the familiar order, 16 ÷ 8, gives a whole number instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must see that (8 ÷ m) is the fraction 8/m and that freeing m from a denominator means multiplying both sides by it, a move grade 7 equations rarely need. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: weak
Adds 4 to 20 instead of subtracting it, so 8 ÷ m = 24 and m = 8/24, which simplifies to 1/3.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Gets 8 ÷ m = 16 correctly but then divides 16 by 8, putting the divisor and the dividend the wrong way round, so m comes out as 2.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Makes both mistakes, adding 4 to get 8 ÷ m = 24 and then dividing 24 by 8 instead of 8 by 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The equation is $(8 \div m) + 4 = 20$ . You need the value of m. Here m is the number you divide 8 by. In words: 8 divided by some number, plus 4, makes 20.
Undo the plus 4
A fraction bar means divide, so $8 \div m$ is the same as $\frac{8}{m}$ . Subtract 4 from both sides to undo the plus 4:
$\frac{8}{m} + 4 - 4 = 20 - 4$
$\frac{8}{m} = 16$
Now 8 divided by m equals 16.
Watch out: adding 4 instead gives 24 on the right. That leads to choice A, 1/3, or to D, 3.
Get m out of the bottom
m is on the bottom of the fraction. Multiply both sides by m to free it:
$8 = 16 \times m$
Now divide both sides by 16:
$m = \frac{8}{16} = \frac{8 \div 8}{16 \div 8} = \frac{1}{2}$
8 goes into both 8 and 16, so 8/16 simplifies to 1/2.
Choose the answer
Watch out: it is tempting to divide 16 by 8 and get 2, choice C. But ask yourself: 8 divided by what number makes 16? To get an answer bigger than 8, you must divide by a number less than 1.
So the answer is B, 1/2.

Check your answer

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

More questions like this

The original DOE question

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

Question 3 of 39

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 4 of 39

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 5 of 39

2026 Practice Test D Math, question 13

Question

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

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

A43°
B44°
C46°
D47°

Show the answer

Correct answer: A

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

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

$5x - 5 = 90$

$5x = 95$

$x = 19$

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

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

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

The smallest angle is 43°.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 6 of 39

2026 Practice Test D Math, question 44

Question

A bakery charges $1.25 for each cupcake and $2.50 for a serving tray. The equation describes p, the price of c cupcakes and a serving tray, not including tax.

$p = 2.50 + 1.25c$

If Samantha bought some cupcakes and a serving tray for a total of $22.50, not including tax, how many cupcakes did she buy?

Show the answer

Correct answer: C

Using the specified equation $p = 2.50 + 1.25c$ , determine the number of cupcakes Samantha bought by substituting the total paid ($22.50) for p, and solve the equation for the value of c, the number of cupcakes bought.

$22.50 = 2.50 + 1.25c$

$22.50 - 2.50 = 1.25c$

$20.00 \div 1.25 = c$

$16 = c$

Question analysis

A bakery's price for c cupcakes and a serving tray is given by the equation p = 2.50 + 1.25c, and Samantha paid $22.50 before tax. This equation question asks how many cupcakes she bought. Students substitute the total and solve a two-step equation with decimal money amounts, undoing the steps in the right order.

NYS standards: 7.EE.4a two-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step equation of the form px + q = r in a money context is core grade 7 algebra. The equation is given, so the work is substituting the total and dividing by a decimal, a routine example.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The equation is supplied, so the question is a two-step solve with money. The tray’s $2.50 has to come off the total before dividing, and students who divide the whole $22.50 by $1.25 get a whole number that matches a choice, so nothing warns them. Substituting the chosen number back into p = 2.50 + 1.25c is a quick, decisive check.

Error-prone arithmetic minor Working it out
After subtracting the tray, $20 must be divided by $1.25, a decimal divisor that is easiest handled as 2,000 ÷ 125 or by noting that 4 cupcakes cost $5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute the total for p, subtract the tray's cost from both sides, then divide by the price of one cupcake; check by putting your number back into the equation.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Notes that 4 cupcakes cost $5.00 and that the $20 left after the tray holds four such amounts, then gives the 4 groups instead of 16 cupcakes.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the $22.50 total by the $2.50 tray price, combining two numbers from the question without using the equation.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole $22.50 by $1.25 a cupcake, forgetting that $2.50 of the total paid for the serving tray.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } a - n = b - n \text{ and } \frac{a}{n} = \frac{b}{n}$; a and b are the two sides of an equation, and n is any number (not zero when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it level. That is how you get c, the number of cupcakes, by itself.

Sort out the question
The equation $p = 2.50 + 1.25c$ gives the price. p is the price in dollars, and c is the number of cupcakes. The 2.50 is the tray, and 1.25 is the price of one cupcake. Samantha paid $22.50. You need c.
Put in the total
Samantha paid $22.50 in all, so put 22.50 in place of p. Now the equation has only one unknown, c:
$22.50 = 2.50 + 1.25c$
Subtract the tray
Take 2.50 away from both sides. That removes the cost of the tray:
$22.50 - 2.50 = 1.25c$
$20.00 = 1.25c$
So $20.00 went to cupcakes alone.
Watch out: dividing all of $22.50 by $1.25 gives 18, choice D. That forgets the tray.
Divide by the price of one cupcake
Divide both sides by 1.25:
$c = 20.00 \div 1.25 = 16$
An easy way: 4 cupcakes cost $4 \times 1.25 = 5.00$ dollars. $20 is 4 groups of $5. So Samantha bought $4 \times 4 = 16$ cupcakes.
Watch out: 4 is the number of $5 groups, not cupcakes. That is choice A.
So the answer is C, 16.

Check your answer

Put 16 back into the equation. $1.25 \times 16 = 20.00$ , and $2.50 + 20.00 = 22.50$ . That is what Samantha paid, so 16 is right.

More questions like this

The original DOE question

Question 44 as shown on the DOE’s digital test, page 117 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 117, question 44.

Question 7 of 39

2026 Practice Test D Math, question 45

Question

Marcus and Isla are saving money to buy concert tickets.

They decide to save the same amount each week.
They already have $8 saved.
After 3 weeks, the two have $26 saved in total.

How much money, in dollars, is being saved by Marcus and Isla each week?

Enter your answer in the space.

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

Show the answer

Correct answer: 6

Define the variable:

Let w represent the amount of money Marcus and Isla save per week.

Construct an equation:

Start with $8.

Marcus and Isla save w each week for 3 weeks, which can be expressed as $3 \cdot w$ or $3w$ .

The total amount saved can be represented by the expression $8 + 3w$ .

After 3 weeks, Marcus and Isla have $26. This can be represented by the equation $8 + 3w = 26$ .

Solve the equation:

$8 + 3w = 26$

$3w = 18$ (subtracting 8 from both sides)

$w = 6$ (dividing both sides by 3)

Marcus and Isla are saving $6 per week.

Question analysis

Marcus and Isla already have $8 toward concert tickets, save the same amount each week, and have $26 in all after 3 weeks. Students type how many dollars the two of them save each week. It tests writing and solving a two-step equation, keeping the money saved beforehand apart from the weekly savings.

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r from a word problem is the grade 7 standard. This one uses small whole numbers and two steps, a typical classroom example, with no choices to compare.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 51 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two inverse steps solve it, subtracting the starting $8 and dividing by the 3 weeks, and most mistakes drop one of them. The $18 saved over three weeks is the natural stopping place, and dividing $26 by 3 without removing the $8 gives an awkward value that should prompt a second look. Adding three weeks of the answer to $8 and getting $26 confirms it.

Turning words into math minor Understanding the question
The bullet points hold a start amount, “already have $8 saved”, and a total after three weeks, “$26 saved in total”, which must become 8 + 3w = 26. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The amount is typed with no choices, so stopping at the 3-week total of $18 leaves a whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Take away the amount saved before the weeks began, then share what is left equally among the weeks; check by adding three weeks of savings to the starting amount.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

18 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Subtracts the $8 already saved from the $26 total and stops, giving the amount saved over all 3 weeks rather than in each week.
26/3 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $26 total by the 3 weeks, 26 ÷ 3 = 8 2/3, about $8.67, forgetting that $8 of it was saved before the weekly saving began.
3 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the weekly amount and then halves it for each person, though the question asks what Marcus and Isla save together each week.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total savings: $\text{total} = \text{start} + \text{weeks} \times \text{amount each week}$; The start is the money saved before the weeks begin. The amount each week is added once for every week, so you multiply it by the number of weeks. The total is all the money together.
Doing the same to both sides: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number (not zero when you divide). Doing the same thing to both sides keeps them equal. That is how you get the unknown by itself.

Sort out the question
You know:
- Marcus and Isla already have $8.
- They save the same amount each week.
- After 3 weeks they have $26 in all.
You need the amount the two of them save together each week.
Write an equation
Let w be the dollars saved each week. Three weeks add 3w dollars to the $8 they started with:
$8 + 3w = 26$
Subtract the starting money
Take 8 away from both sides:
$3w = 26 - 8 = 18$
They saved $18 over the 3 weeks.
Watch out: 18 is the amount for all 3 weeks, not for each week. Do not stop here.
Divide by the number of weeks
Divide both sides by 3:
$w = 18 \div 3 = 6$
They save $6 each week.
Watch out: dividing all $26 by 3 gives about 8.67. That forgets the $8 saved before. Halving 6 to get 3 is wrong too: the question asks what the two save together.
Enter the answer
Type only the number in the box, with no dollar sign and no units. So you would enter 6.

Check your answer

Count up week by week. Start at $8. After week 1: $8 + 6 = 14$ . After week 2: $14 + 6 = 20$ . After week 3: $20 + 6 = 26$ . That is the $26 in the question.

More questions like this

The original DOE question

Question 45 as shown on the DOE’s digital test, page 118 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 118, question 45.

Question 8 of 39

2026 Practice Test D Math, question 46

Question

A company charges $3.00 to ship a 4-pound package. The cost includes a base fee of $1.50 plus $0.50 for each pound over the first pound. The table below illustrates how the cost depends on the total weight of the package. Complete the table by matching each total weight to its correct total shipping cost.

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

SHIPPING COSTS

Total Weight (pounds)	Total Cost to Ship (dollars)
2
	2.50
6
	5.50

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. Each tile can be used once. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 2; blank 2: 3; blank 3: 4; blank 4: 9

Total Weight (pounds)	Total Cost to Ship (dollars)
2	2
3	2.50
6	4
9	5.50

Let x represent the weight of the package in pounds. The first pound is included in the base fee of $1.50, so $(x - 1)$ represents the number of pounds that cost $0.50 each.

The cost equation can be written as $\text{Cost} = \$1.50 + \$0.50\,(x - 1)$

The cost for a 2-pound package is $\$1.50 + \$0.50(2 - 1) = \$1.50 + \$0.50(1) = \$2.00$
The cost for a 6-pound package is $\$1.50 + \$0.50(6 - 1) = \$1.50 + \$0.50(5) = \$1.50 + \$2.50 = \$4.00$
The weight of a package that costs $2.50 to ship can be found by solving

$\$2.50 = \$1.50 + \$0.50(x - 1)$

$\$1.00 = \$0.50(x - 1)$

$2 = x - 1$

$x = 3 \text{ pounds}$
The weight of a package that costs $5.50 to ship can be found by solving

$\$5.50 = \$1.50 + \$0.50(x - 1)$

$\$4.00 = \$0.50(x - 1)$

$8 = x - 1$

$x = 9 \text{ pounds}$

To check:

Given that the total cost is $3.00 for the initial package

$\$3.00 = \$1.50 + \$0.50(x - 1)$

$\$1.50 = \$0.50(x - 1)$

$3 = x - 1$

$x = 4$ pounds fits the linear pattern of the table

Question analysis

A shipping company charges a base fee plus 50 cents for each pound over the first, and a 4-pound package costs $3.00. Students drag number tiles into a table to complete two costs and two weights, with some tiles left unused. It tests building and using a linear rule, and the wording about the first pound must be applied in every row.

NYS standards: 7.EE.4a two-step equations
6.EE.9 dependent and independent variables
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve word problems with equations like p(x + q) = r. This table needs that rule four times, twice run backward from a cost to a weight, and the phrase about the first pound must be applied every time.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 75 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The rule is stated in words and confirmed by the 4-pound example, so the work is applying it consistently in both directions. Charging every pound breaks the $3.00 example at once, since 4 pounds would cost $3.50, and models that ignore the wording leave costs or weights with no matching tile. A shortcut helps: each extra pound adds $0.50 to the $3.00, so the table can be filled by counting up or down from 4 pounds.

Turning words into math moderate Understanding the question
The phrase “$0.50 for each pound over the first pound” means the first pound is covered by the $1.50 fee, so a package of x pounds pays for only x − 1 of them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Four boxes must all be right, two found forward from a weight and two backward from a cost, and the backward rows need the first pound added back. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write the rule as the base fee plus $0.50 times the pounds after the first, check it against the 4-pound example, then use it forward for the costs and backward for the weights.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

BAEH Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Charges $0.50 for every pound, first included, so 2 pounds cost $2.50 and $2.50 ships 2 pounds; no tile fits its $4.50 or 8, so it settles for 4 and 9.
ACFH Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Triples the 2-pound cost for the 6-pound package, $2 × 3 = $6, treating the cost as proportional to weight though the base fee is paid only once.
ACBH Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Charges $0.50 for each of the 5 pounds over the first for the 6-pound package but stops there, leaving out the $1.50 base fee and entering 2.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Shipping cost: $C = 1.50 + 0.50 \times (w - 1)$; C is the total cost in dollars, and w is the weight in pounds. The $1.50 base fee is paid once, and it covers the first pound. Each pound after the first costs $0.50. There are w − 1 of those pounds.

Sort out the question
The base fee of $1.50 covers the first pound. Each pound over the first adds $0.50. A 4-pound package costs $3.00.
You fill four boxes: the costs for 2 and 6 pounds, and the weights that cost $2.50 and $5.50. Each tile is used once at most, and four tiles are not used.
Test the rule on 4 pounds
Check the rule with the example. A 4-pound package has 3 pounds over the first:
$C = 1.50 + 0.50 \times (4 - 1) = 1.50 + 1.50 = 3.00$
That is $3.00, so the rule fits.
Watch out: charging $0.50 for every pound, the first one too, gives $3.50 for 4 pounds. That does not match.
Find the two costs
For 2 pounds, 1 pound is over the first:
$1.50 + 0.50 \times 1 = 2.00$
For 6 pounds, 5 pounds are over the first:
$1.50 + 0.50 \times 5 = 1.50 + 2.50 = 4.00$
The costs are the tiles 2 and 4.
Watch out: 6 pounds is 3 times 2 pounds, but the cost is not 3 times $2. The base fee is paid once. Tripling gives the tile 6. Leaving out the base fee gives the tile 2.5.
Work back to the two weights
For $2.50, take off the base fee: $2.50 - 1.50 = 1.00$ . At $0.50 a pound, $1.00 pays for 2 extra pounds. Add the first pound: $2 + 1 = 3$ pounds.
For $5.50: $5.50 - 1.50 = 4.00$ . $4.00 pays for 8 extra pounds. Add the first pound: $8 + 1 = 9$ pounds.
Watch out: forgetting to add back the first pound gives 2 and 8.
Fill in the four boxes
From top to bottom, the boxes get 2, 3, 4 and 9. The tiles 2.5, 3.5, 6 and 7 are not used. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Each extra pound adds $0.50, so count from 4 pounds at $3.00. Going down: 3 pounds cost $2.50 and 2 pounds cost $2.00. Going up: 5 pounds $3.50, 6 pounds $4.00, 7 pounds $4.50, 8 pounds $5.00 and 9 pounds $5.50. All four boxes match.

More questions like this

The original DOE question

Question 46 as shown on the DOE’s digital test, page 119 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 119, question 46.

Question 9 of 39

2024 Test A Math, question 2

Question

Angle M and angle R are supplementary. The measure of angle R is 5 times the measure of angle M. What is the measure of angle R, in degrees?

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

Show the answer

Correct answer: 150

Angle M and angle R are supplementary, so the sum of the angles is 180°.

Let m equal the measure of angle M, and let r equal the measure of angle R.

$r + m = 180$

The measure of angle R is 5 times the measure of angle M.

$r = 5m$

$5m + m = 180$

$6m = 180$

$m = \frac{180}{6} = 30$

$r = 5m = 5(30) = 150$

Therefore, the measure of angle R is 150°.

Question analysis

A geometry grid-in about two supplementary angles, where one angle is five times the size of the other. Students must recall what supplementary means, write an equation linking the two angles, and solve it. The question asks for the larger angle, so stopping at the smaller one is the usual slip.

NYS standards: 7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students regularly write and solve simple equations for unknown angles using supplementary and complementary facts. This is a two-step version of exactly that task, with whole numbers and no figure to read.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once supplementary is read as a 180° total, the problem is a one-variable equation with whole numbers. Students lose the point in two ways: by stopping at the smaller angle, which the equation produces first, or by using 90° for the total. Checking that the gridded angle is the larger of the pair, and that the two angles add to 180°, catches both.

Something you must already know moderate Working it out
The stem never says what supplementary means; the 180° total has to be recalled, and a student who confuses it with complementary's 90° halves every angle. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The sentence "The measure of angle R is 5 times the measure of angle M" has to become R = 5M, so the 180° splits into six equal parts, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in there is nothing to compare against, so gridding angle M's 30°, the first value the equation gives, looks just as finished as angle R. (Enright, Morley & Sheehan, 2002)

Strategy: Call angle M one part and angle R five parts; together the six parts make the supplementary total, so find one part and then multiply for angle R.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

30 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Solves 6m = 180 correctly for angle M, 30°, and grids it instead of multiplying by 5 to get angle R, the angle the question asks for.
75 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses 90° for supplementary, the complementary total, so 6m = 90 gives m = 15 and angle R = 75, half its real size.
36 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides 180 by the 5 in the stem, combining the two numbers without an equation that gives angle M its own share of the 180°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Supplementary angles: $M + R = 180$; M and R are the sizes of two angles, in degrees. Supplementary means the two angles add up to 180 degrees. Put side by side, they make a straight line. Do not mix this up with complementary, which means a total of 90 degrees.

Sort out the question
Two angles, M and R, are supplementary. Angle R is 5 times as big as angle M. You need the measure of angle R, in degrees. Measure means its size. Notice that the question asks for R, not M.
Write the facts as equations
Supplementary angles add up to 180°.
$M + R = 180$
The sentence “The measure of angle R is 5 times the measure of angle M” becomes this:
$R = 5 \times M$
Watch out: supplementary means 180°, not 90°. Using 90° gives R = 75, which is wrong.
Find angle M
Put $5 \times M$ in place of R. Then the equation has only one unknown, M.
$M + 5 \times M = 180$
$6 \times M = 180$
$M = 180 \div 6 = 30$
Think of 6 equal parts. M is 1 part and R is 5 parts. So angle M is 30°.
Find angle R
Angle R is 5 times angle M.
$R = 5 \times 30 = 150$
Angle R is 150°. Watch out: do not stop at M. Gridding 30 gives the wrong angle. So you would grid 150.

Check your answer

Add the two angles: $30 + 150 = 180$ . So they are supplementary. Also, $150 \div 30 = 5$ , so R really is 5 times M. Both facts in the question check out.

More questions like this

The original DOE question

Question 59 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)

Question 10 of 39

2024 Test A Math, question 16

Question

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

Show the answer

Correct answer: D

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

$c = 40 + m$

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

$\frac{60}{80\%} = \frac{60}{0.80} = 75$

Use $75 as the total cost to find m.

$40 + m = 75$

$m = 35$

Dalia drove 35 miles.

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve two-step equations and work with discounts, usually separately. This combines them and makes students undo a 20% discount starting from the price paid, a reversal many grade 7 students have not practised.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Combines ideas moderate
A very close wrong answer moderate
Solution: 3 steps
Reading level: Question and choices: about grade 5, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
The sentence "After the coupon was applied, she spent a total of $60" makes 60 the price after 20% off, so it is 80% of the full cost, 40 + m. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Combines ideas moderate Working it out
A percent discount has to be undone before the fee-plus-mileage equation can be solved: 60 ÷ 0.8 = 75, then 75 − 40 = 35 miles. (Embretson, 1998; Daniel & Embretson, 2010)
A very close wrong answer moderate Choosing the answer
The choices 32 and 35 are three miles apart, too close to tell apart by estimate; only dividing 60 by 0.8, rather than adding 20% of 60, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Takes 20% off the $60 again, 60 × 0.8 = 48, instead of undoing the discount, then subtracts the $40 fee to get 8 miles.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Ignores the coupon and treats the $60 as the full cost, so it subtracts the $40 fee and gets 20 miles.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Undoes the coupon by adding 20% of $60 to get $72, though the 20% was taken of the larger pre-coupon cost, then subtracts the $40 fee.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The truck costs a $40 fee, plus $1 for each mile. Dalia’s coupon took 20% off the total. After the coupon, she paid $60. You need how many miles she drove. The $60 is the price after the discount, not the full cost.
Write what she paid
20% off means she paid what was left of the full cost.
$100\% - 20\% = 80\%$
As a decimal, 80% is 0.80.
$\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)$
$60 = \text{full cost} \times 0.80$
Undo the discount
To undo times 0.80, divide by 0.80. Move the decimal point one place in both numbers to make it easier.
$\text{full cost} = 60 \div 0.80 = 600 \div 8 = 75$
The full cost was $75. Watch out: the 20% was taken off the full cost, not off $60. Adding 20% of $60 gives $72 and leads to 32 miles, choice C. Taking 20% off $60 again leads to 8 miles, choice A.
Find the miles
$c = 40 + 1 \times m$
$75 = 40 + m$
$m = 75 - 40 = 35$
She drove 35 miles. Watch out: do not skip the coupon. Using $60 as the full cost gives 20 miles, choice B. So the answer is D, 35.

Check your answer

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

More questions like this

The original DOE question

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

Question 11 of 39

2024 Test A Math, question 18

Question

Each number in a sequence is formed by doubling the previous number and then adding 1. If the 9th number in the sequence is 63, what is the 10th number minus the 7th number?

A96
B111
C112
D127

Show the answer

Correct answer: C

First, write an equation based on the rule described in the item. Let x be the previous number in the sequence and y be the number in the sequence. The ninth number in the sequence is 63.

$2x + 1 = y$

Determine the 10th number in the sequence, y, by making $x = 63$ and solving:

$2(63) + 1 = y$

$126 + 1 = y$

$127 = y$

Determine the 8th number in the sequence, x, by making $y = 63$ and solving:

$2x + 1 = 63$

Subtract one from both sides of the equation.

$2x + (1 - 1) = 63 - 1$

$2x = 62$

$\frac{2x}{2} = \frac{62}{2}$

$x = 31$

Determine the 7th number in the sequence, x, by making $y = 31$ and solving:

$2x + 1 = 31$

$2x + (1 - 1) = 31 - 1$

$2x = 30$

$\frac{2x}{2} = \frac{30}{2}$

$x = 15$

Subtract the 7th number in the sequence from the 10th number:

$127 - 15 = 112$

The difference is 112.

Question analysis

A sequence question: each number is found by doubling the one before it and adding 1, and the 9th number is 63. Students find the 10th number minus the 7th. That means running the rule forward once and backward twice, while keeping track of which position each number holds.

NYS standards: 7.EE.4a two-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve two-step equations but rarely have to run a rule backward through a sequence. This mixes forward and backward steps with position counting, so it asks for more planning than a typical grade 7 task.
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: No obvious method moderate
Many steps moderate
Much to keep track of minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 32 words.
Corrected from the DOE handbook: A misplaced parenthesis in the explanation’s working was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Applying the rule forward is routine; running it backward is not, and the question needs both. Students must also count positions carefully, since the 9th number is given and the 7th is two steps back, not one. Each listed wrong answer stops or slips at a different stage, so writing the four numbers in a row, 15, 31, 63 and 127, before subtracting is the reliable way.

No obvious method moderate Working it out
The rule is stated only forward, double and add 1, but the 7th and 8th numbers come before the given 9th, so the rule must be undone: subtract 1, then halve. (Schoenfeld, 1985)
Many steps moderate Working it out
One forward step reaches the 10th number, 127, two backward steps reach the 8th and 7th, 31 and 15, and a final subtraction finishes. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Four positions and their values, 7th to 10th, have to be kept straight, and the 8th number is easy to mistake for the 7th. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Apply the rule to 63 to get the 10th number; to go back a position, subtract 1 and halve, doing it twice to reach the 7th; then subtract.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Steps back only once from the 9th number and subtracts the 8th, 127 − 31 = 96, one position short of the 7th.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Doubles 63 to 126 but forgets to add 1 on that step, then subtracts the correct 7th number: 126 − 15 = 111.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 10th number, 2 × 63 + 1 = 127, and stops without subtracting the 7th number from it.

What was corrected from the DOE handbook, and why

Explanation: Finding the 7th number, the handbook writes “2x + (1 − 1 = 31 − 1)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 1 − 1 = 31 − 1 is false). The site now reads 2x + (1 − 1) = 31 − 1, the same step the handbook writes correctly for the 8th number (2x + (1 − 1) = 63 − 1). It leads to 2x = 30 and x = 15 as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule: $y = 2x + 1$; A sequence is a list of numbers that follows a rule. x is a number in the list, and y is the next number. To get the next number, double x and then add 1.
Going back one number: $x = (y - 1) \div 2$; y is a number in the list, and x is the one just before it. To go back, undo the rule in reverse order. The rule doubles and then adds 1. So first subtract 1, then divide by 2.

Sort out the question
Each number is “formed by doubling the previous number and then adding 1”. Previous means the one just before. The 9th number is 63. You need the 10th number minus the 7th number. So you go forward one number and back two numbers.
Find the 10th number
Use the rule with x = 63.
$y = 2x + 1$
$y = 2 \times 63 + 1 = 126 + 1 = 127$
The 10th number is 127. Watch out: remember to add 1. Doubling only gives 126 and leads to 111, choice B.
Plan how to go back
The 7th number comes before the 9th, so you must undo the rule. The rule doubled, then added 1. Undo the last step first: subtract 1. Then undo the doubling: divide by 2.
$x = (y - 1) \div 2$
Find the 8th number
$x = (63 - 1) \div 2 = 62 \div 2 = 31$
The 8th number is 31. Watch out: 31 is the 8th number, not the 7th. Using it gives $127 - 31 = 96$ , choice A.
Find the 7th number
Step back once more, from the 8th number. Use the same rule with y = 31.
$x = (31 - 1) \div 2 = 30 \div 2 = 15$
The 7th number is 15.
Subtract the 7th from the 10th
$127 - 15 = 112$
Watch out: 127 alone is choice D. You still have to subtract the 7th number. So the answer is C, 112.

Check your answer

Go forward from 15 using the rule. $2 \times 15 + 1 = 31$ , then $2 \times 31 + 1 = 63$ , then $2 \times 63 + 1 = 127$ . You land on 63 as the 9th number, so the list is right. And $127 - 15 = 112$ .

More questions like this

The original DOE question

Question 75 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 75. Open page 80 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 12 of 39

2024 Test A Math, question 22

Question

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

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

A10.0° C
B15.0° C
C20.0° C
D32.4° C

Show the answer

Correct answer: A

The highest temperature was 86° F.

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

$86 = \frac{9}{5}C + 32$

Subtract 32 from both sides of the equation:

$86 - 32 = \frac{9}{5}C$

$54 = \frac{9}{5}C$

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

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

$30 = C$

The lowest temperature was 68° F.

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

$68 = \frac{9}{5}C + 32$

$68 - 32 = \frac{9}{5}C$

$36 = \frac{9}{5}C$

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

$20 = C$

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

$30 - 20 = 10$

Question analysis

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

NYS standards: 7.EE.4a two-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving equations of the form px + q = r with fraction coefficients is grade 7 work. Doing it twice and comparing the results is a typical multistep grade 7 task, with extra care over which scale the answer is in.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 29 words.
Corrected from the DOE handbook: A stray term in the explanation’s working was removed.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Many steps moderate Working it out
Each Fahrenheit reading needs its own two-step solve, subtracting 32 and multiplying by 5/9, and then the two Celsius values must be subtracted. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Multiplying 54 and 36 by 5/9 is easy only if the division by 9 comes first; multiplying by 5 first leaves 270/9 and 180/9 to reduce. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No conversion route gives 15; the gap is 18 × 5/9 by any method, so 15 needs a slip such as 54 ÷ 9 = 7, and it mostly draws guesses.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Converts the low reading, 68°F, to 20°C and gives that temperature instead of the difference between the two readings.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Converts in the wrong direction, multiplying the 18-degree gap by 9/5 instead of 5/9, the same as putting 86 and 68 in for C.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Fahrenheit and Celsius are two ways to measure temperature. The formula changes Celsius, C, into Fahrenheit, F. Yesterday’s high was 86° F and the low was 68° F. You need the difference between them in degrees Celsius. So change each one to Celsius, then subtract.
Undo the + 32 for the high
Put 86 in for F. Then subtract 32 from both sides. That keeps the equation balanced.
$F = \frac{9}{5}C + 32$
$86 = \frac{9}{5}C + 32$
$86 - 32 = \frac{9}{5}C$
$54 = \frac{9}{5}C$
Solve for the high in Celsius
To undo times $\frac{9}{5}$ , multiply both sides by $\frac{5}{9}$ . That is its reciprocal, the fraction flipped upside down.
$C = 54 \times \frac{5}{9} = 6 \times 5 = 30$
Dividing 54 by 9 first gives 6, which keeps the numbers small. The high was 30° C.
Change the low to Celsius
Do the same with 68.
$68 = \frac{9}{5}C + 32$
$68 - 32 = \frac{9}{5}C$
$36 = \frac{9}{5}C$
$C = 36 \times \frac{5}{9} = 4 \times 5 = 20$
The low was 20° C. Watch out: 20° C is the low, not the difference. That is choice C.
Subtract to find the difference
$30 - 20 = 10$
The difference is 10 degrees Celsius. Watch out: do not multiply the 18-degree Fahrenheit gap by $\frac{9}{5}$ . That goes the wrong way and gives 32.4, choice D. So the answer is A, 10.0° C.

Check your answer

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

More questions like this

The original DOE question

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

Question 13 of 39

2024 Test A Math, question 27

Question

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

A60
B80
C120
D160

Show the answer

Correct answer: D

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

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

$t + t + 100 + t - 50 = 230$

$t + t + t + 100 - 50 = 230$

$3t + 50 = 230$

$3t + (50 - 50) = 230 - 50$

$3t = 180$

$(3 \div 3)t = 180 \div 3$

$t = 60$

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

$t + 100 = 60 + 100 = 160$

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

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r from a word problem is a core grade 7 standard. This is a typical three-step version, with one negative adjustment and an extra step after solving.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 45 words.
Corrected from the DOE handbook: A misplaced parenthesis in the explanation’s working was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
Both comparisons are made with Tuesday, 100 more loaves on Wednesday and 50 fewer on Thursday, so the three days become t, t + 100 and t − 50. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
The equation's unknown is Tuesday's count, while the question asks for Wednesday's, so one more step remains after the equation is solved. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Solves 3t + 50 = 230 for Tuesday's loaves, t = 60, and gives that instead of Wednesday's t + 100.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts both 100 and 50 from the total, 230 − 100 − 50 = 80, combining the numbers without writing each day in terms of Tuesday.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds Tuesday's 60, then gets Wednesday by subtraction, 230 − 60 − 50, taking Thursday as 50 loaves instead of 60 − 50 = 10.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c$; a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you divide both sides by the same number that is not zero.

Sort out the question
The baker made bread on Tuesday, Wednesday and Thursday. Wednesday had 100 more loaves than Tuesday. Thursday had 50 fewer loaves than Tuesday. All three days add up to 230 loaves. You need the number of loaves made on Wednesday.
Write each day using Tuesday
Both other days are compared with Tuesday. So call Tuesday’s loaves t.
Day Loaves
Tuesday $t$
Wednesday $t + 100$
Thursday $t - 50$
100 more means add 100. 50 fewer means subtract 50.
Write the equation and combine
$t + (t + 100) + (t - 50) = 230$
Group the t’s together and the plain numbers together. Three t’s make 3t, and $100 - 50 = 50$ .
$3t + 50 = 230$
Watch out: do not just subtract 100 and 50 from 230. That gives 80, choice B, and skips writing each day with t.
Solve for Tuesday
Subtract 50 from both sides. Then divide both sides by 3.
$\text{if } a = b \text{, then } a - c = b - c$
$3t + 50 - 50 = 230 - 50$
$3t = 180$
$t = 180 \div 3 = 60$
Tuesday had 60 loaves. Watch out: 60 is Tuesday, choice A. The question asks about Wednesday.
Find Wednesday’s loaves
Wednesday had 100 more loaves than Tuesday, so add 100 to t.
$t + 100 = 60 + 100 = 160$
Wednesday had 160 loaves. So the answer is D, 160.

Day	Loaves
Tuesday	$t$
Wednesday	$t + 100$
Thursday	$t - 50$

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 39

2024 Test A Math, question 33

Question

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

A $\frac{11}{120}$
B $\frac{2}{7}$
C $\frac{5}{6}$
D $\frac{55}{6}$

Show the answer

Correct answer: D

$\left(\frac{3}{5} - \frac{1}{2}\right)x = \frac{1}{4} + \frac{2}{3}$

Rewrite with equivalent fractions that have like denominators.

$\left(\frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5}\right)x = \frac{1 \times 3}{4 \times 3} + \frac{2 \times 4}{3 \times 4}$

$\left(\frac{6}{10} - \frac{5}{10}\right)x = \frac{3}{12} + \frac{8}{12}$

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

Multiply both sides of the equation by 10.

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

$x = \frac{110}{12} = \frac{55}{6}$

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve equations such as (2/3)x = 4/5 with fractional coefficients. Here both sides must first be simplified over unlike denominators, so four fraction operations come before x is found: on level, but long.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps moderate
Solution: 4 steps
Reading level: Question and choices: about grade 8, 12 words.
Corrected from the DOE handbook: A minus sign printed in place of an equals sign in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
Both sides need common denominators, tenths on the left and twelfths on the right, and the coefficient comes out as 1/10, a fraction to divide by. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps moderate Working it out
Four operations come before x appears: 3/5 − 1/2, then 1/4 + 2/3, then 11/12 divided by 1/10, then reducing 110/12 to 55/6. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Simplifies both sides correctly but multiplies 11/12 by 1/10 instead of dividing by it, getting 11/120.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Combines each pair straight across, 3/5 − 1/2 as 2/3 and 1/4 + 2/3 as 3/7, then multiplies them: 2/3 × 3/7 = 2/7.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Adds 3/5 and 1/2 instead of subtracting, so the coefficient is 11/10 and x = 11/12 ÷ 11/10 = 5/6.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the equation
An equation says two things are equal. On the left, the subtraction in the parentheses multiplies x. On the right is a sum. You need the value of x. First make each side as simple as you can. Then get x by itself.
Simplify the left side
To subtract, the fractions need the same bottom. Both 5 and 2 go into 10, so use tenths.
$\frac{a}{b} = \frac{a \times n}{b \times n}$
$\frac{3}{5} - \frac{1}{2} = \frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5} = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}$
Watch out: this is a minus sign. Adding instead gives $\frac{11}{10}$ , and that leads to $\frac{5}{6}$ , choice C.
Simplify the right side
Both 4 and 3 go into 12, so use twelfths.
$\frac{1}{4} + \frac{2}{3} = \frac{1 \times 3}{4 \times 3} + \frac{2 \times 4}{3 \times 4} = \frac{3}{12} + \frac{8}{12} = \frac{11}{12}$
Now the equation is much simpler:
$\frac{1}{10}x = \frac{11}{12}$
Get x by itself
x is multiplied by $\frac{1}{10}$ . To undo that, multiply both sides by 10. That is the reciprocal of $\frac{1}{10}$ , the fraction flipped upside down.
$\text{if } a = b \text{, then } n \times a = n \times b$
$10 \times \frac{1}{10}x = 10 \times \frac{11}{12}$
$x = \frac{110}{12}$
Watch out: multiplying $\frac{11}{12}$ by $\frac{1}{10}$ does not undo anything. It gives $\frac{11}{120}$ , choice A.
Simplify the fraction
Both 110 and 12 are even, so divide the top and the bottom by 2.
$x = \frac{110 \div 2}{12 \div 2} = \frac{55}{6}$
So the answer is D, $\frac{55}{6}$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 15 of 39

2024 Test A Math, question 36

Question

CELL PHONE SALES BY COLOR

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

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

A18%
B20%
C22%
D28%

Show the answer

Correct answer: B

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

$8k + 30 + 30 - 2k + k + 5 = 100$

$8k - 2k + k + 30 + 30 + 5 = 100$

$7k + 65 = 100$

$7k + (65 - 65) = 100 - 65$

$7k = 35$

$k = 5$

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

$30 - 2k = 30 - 2(5) =$

$30 - 10 = 20$

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

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve two-step equations that are written out for them. Here the equation must be built from a table whose percentages total 100, with like terms to combine, and the value found must then be substituted back.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question hides an equation in a table, and the key insight is that the four percentages must add to 100. After that the algebra is routine, and no listed choice rewards stopping early, since neither k nor any other color's percentage appears. The wrong choices sit near the answer or use the blue expression with k ignored, so students who cannot set up the equation are left guessing.

Many steps moderate Working it out
The answer needs a chain: add the four expressions, set the sum to 100, collect 8k − 2k + k into 7k, solve for k, then substitute into 30 − 2k. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
No equation is printed; the table's Total row of 100 is what turns the four percentages written in k into one equation to solve. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Collecting like terms includes the negative coefficient in the blue row, −2k, and the constants 30, 30 and 5 have to be gathered too. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common slip gives k = 6 from 7k + 65 = 100, the value 18% would need; it sits 2 points below the answer and mostly draws guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No usual error in collecting terms or solving yields k = 4, which 22% requires, so this near neighbor of the answer mostly attracts guessers.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Never solves for k and reads the blue entry 30 − 2k as 30 − 2, as if k were 1, giving 28.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms: $ak + bk = (a + b)k$; k is the unknown number. a and b are the numbers in front of k, called coefficients. Terms with the same letter are called like terms. You can add their coefficients, because a groups of k plus b groups of k make a + b groups of k.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it balanced. Use it to undo each step and get k alone.

Sort out the table
The table gives the percent of phones sold in each color. Some entries use k, a letter that stands for an unknown number. 8k means 8 times k. You need the percent that were blue, $30 - 2k$ . To find it, you first need k.
Write an equation for the total
Every phone sold is one of the four colors. So the four percents add up to the total, 100, as the last row of the table shows.
$8k + 30 + (30 - 2k) + (k + 5) = 100$
Combine like terms
Terms with k in them are called like terms, so they can be combined. Group the k terms together and the plain numbers together. A k on its own means 1k.
$ak + bk = (a + b)k$
$8k - 2k + k = (8 - 2 + 1)k = 7k$
$30 + 30 + 5 = 65$
So the equation becomes $7k + 65 = 100$ . Watch out: the blue row takes away 2k. Keep its minus sign, or the k terms will be wrong.
Solve for k
Undo the adding first, then the multiplying.
$\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$
$7k + 65 - 65 = 100 - 65$
$7k = 35$
$k = 35 \div 7 = 5$
So k is 5. But k is not the answer. You still need the blue row.
Find the blue percent
Put 5 in for k in the blue row.
$30 - 2k = 30 - 2 \times 5 = 30 - 10 = 20$
So 20% of the phones sold were blue. Watch out: 2k means 2 times k, not 2. Reading $30 - 2k$ as $30 - 2$ gives 28%, choice D.
So the answer is B, 20%.

Check your answer

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

More questions like this

The original DOE question

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

Question 16 of 39

2023 Test A Math, question 3

Question

Solve for x:

$7x + 3 - 2(2x + 1) = 13$

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

Show the answer

Correct answer: 4

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

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students expand expressions with negative factors and solve equations such as p(x + q) = r. This equation needs both skills plus combining like terms, which is routine work with one negative sign that is easy to lose.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 4 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a standard multi-step linear equation, and the work is short once the parentheses are cleared. Everything rests on distributing −2 correctly; the later steps, combining 7x with −4x and 3 with −2, then undoing the + 1 and the × 3, are routine. Substituting 4 back into the left side, 28 + 3 − 18 = 13, is a fast check that catches a sign slip.

Error-prone arithmetic moderate Working it out
The −2 in front of (2x + 1) must reach both terms, giving −4x − 2; the second product, −2 × 1, is the one most often written as +2 or left as +1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in with no choices, a value such as 3 or 8/3 from a distributing slip looks just as finished as the right one; only substituting back exposes it. (Enright, Morley & Sheehan, 2002)

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

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

3 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies only the 2x by −2 and leaves the + 1 untouched, so the equation becomes 3x + 4 = 13 and x = 3.
8/3 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Gives the minus sign to 4x but not to the 2, writing −4x + 2, so the equation becomes 3x + 5 = 13 and x = 8/3.
12 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Simplifies correctly to 3x + 1 = 13, subtracts 1 to reach 3x = 12, and grids 12 without dividing by 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the equation
You need the value of x that makes $7x + 3 - 2(2x + 1) = 13$ true. x is the unknown number. The tricky part is the −2 in front of the parentheses. It multiplies everything inside them.
Multiply out the parentheses
Use the distributive property with a = −2, b = 2x and c = 1.
$-2(2x + 1) = (-2) \times 2x + (-2) \times 1 = -4x - 2$
So the equation becomes this:
$7x + 3 - 4x - 2 = 13$
Watch out: the −2 must multiply the 1 too. Leaving + 1 gives x = 3. Writing + 2 instead of − 2 gives $x = \frac{8}{3}$ .
Combine like terms
Like terms are terms of the same kind. Put the x terms together. Put the plain numbers together.
$7x - 4x = 3x$
$3 - 2 = 1$
$3x + 1 = 13$
Get x by itself
Undo the + 1 first. Subtract 1 from both sides, so the equation stays balanced.
$3x + 1 - 1 = 13 - 1$
$3x = 12$
3x means 3 times x. So divide both sides by 3.
$x = 12 \div 3 = 4$
Watch out: don’t stop at 3x = 12. 12 is three times the answer. So you would grid 4.

Check your answer

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

More questions like this

The original DOE question

Question 60 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 60. Open page 75 of the handbook (PDF)

Question 17 of 39

2023 Test A Math, question 8

Question

Jamel works at a computer store. He is paid an hourly rate plus a 15% commission on all computer products he sells. Last week, Jamel was paid $802.50 for working 30 hours and selling $1,250.00 worth of computer products. What is Jamel’s hourly rate?

A$20.50/hr
B$26.75/hr
C$33.00/hr
D$37.65/hr

Show the answer

Correct answer: A

First, determine the amount of Jamel’s weekly pay that is from commission:

$\$1{,}250.00 \times 15\% = \$187.50$

Next, subtract Jamel’s commission amount from his weekly pay:

$\$802.50 - \$187.50 = \$615.00$

Last, divide the result by 30 hours to find Jamel’s hourly rate:

$\$615.00 \div 30 = \$20.50$

Question analysis

Jamel earns an hourly wage plus a 15% commission on his sales, and one week he is paid $802.50 for 30 hours of work and $1,250 in sales. Students find his hourly rate. The question mixes a percent with a rate, and the commission has to be taken out of the total pay before dividing by the hours.

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 Commission is one of the percent applications named in the grade 7 standards, and the remaining work is a two-step equation with decimals. That matches a typical grade 7 multistep problem, with the order of the steps as the main 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
Many steps moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Jamel's pay has two parts, and the question asks for only one of them, so the commission has to be removed before dividing by the hours. Students who divide the whole $802.50 by 30 get a believable hourly figure that still contains the commission. Checking the answer against the story, 30 × $20.50 + $187.50 = $802.50, confirms both parts.

Turning words into math moderate Understanding the question
The phrase "paid an hourly rate plus a 15% commission on all computer products he sells" makes $802.50 the sum of two parts, 30 hours of pay and 15% of $1,250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Three operations in order: 15% of $1,250 is $187.50, $802.50 − $187.50 = $615.00, and $615.00 ÷ 30 = $20.50; dividing before subtracting gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Every amount carries cents, and 15% of $1,250 and $615 ÷ 30 are worked by hand, so a misplaced decimal point shifts the hourly rate. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Take 15% of the $1,250 in sales to find the commission, remove that from the week's pay, and divide what is left by the 30 hours worked.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole $802.50 by 30 hours, which gives Jamel's average pay per hour with the commission included, not his hourly rate.
C Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Follows the word plus and adds the $187.50 commission to the $802.50 instead of removing it, so (802.50 + 187.50) ÷ 30 = $33.00.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Swaps the two dollar amounts, taking 15% of the $802.50 pay and subtracting it from the $1,250 of sales: (1,250 − 120.38) ÷ 30 ≈ $37.65.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Pay with commission: $\text{pay} = \text{rate} \times \text{hours} + \text{commission}$; The rate is the money Jamel earns for each hour. Hours is how long he worked. The commission is extra money that is a percent of what he sells. His pay is these two amounts added together.

Sort out the question
Jamel’s pay has two parts. One part is an hourly rate, the money he earns for each hour. The other part is a commission. A commission is extra pay that is a percent of what he sells. He worked 30 hours and sold $1,250.00 of products. In all he got $802.50. You need his hourly rate.
Find the commission
His commission is 15% of what he sold. 15% means 15 out of 100, or 0.15.
$\text{part} = \frac{p}{100} \times \text{whole}$
$\text{part} = 0.15 \times 1{,}250 = 187.50$
His commission was $187.50. Watch out: take 15% of the $1,250 he sold, not of his $802.50 pay. Mixing them up leads to $37.65, choice D.
Take the commission out of his pay
The rest of his pay came from his hours.
$802.50 - 187.50 = 615.00$
He earned $615.00 for 30 hours of work. Watch out: the word plus does not mean add here. Adding the $187.50 to $802.50 counts the commission twice. That leads to $33.00, choice C.
Divide by the hours he worked
Share the $615.00 equally over the 30 hours.
$615 \div 30 = 20.50$
Watch out: don’t divide all $802.50 by 30. That gives $26.75, choice B. That amount still includes the commission. So the answer is A, $20.50 per hour.

Check your answer

Work forward from the answer. 30 hours at $20.50 is $30 \times 20.50 = 615$ dollars. Add the commission: $615 + 187.50 = 802.50$ . That is exactly Jamel’s pay for the week.

More questions like this

The original DOE question

Question 65 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 2024 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 18 of 39

2023 Test A Math, question 18

Question

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

A$212.55
B$262.55
C$302.55
D$312.55

Show the answer

Correct answer: A

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

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

First, subtract 90 from both sides:

$0.15 \times 12x = \$382.59$

$1.8x = \$382.59$

Last, divide both sides by 1.8:

$x = \$212.55$

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r from a word problem is core grade 7 work. Here the coefficient combines a percent with 12 weeks, and the decimals are long, which adds steps but no new idea.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The story has three layers, a starting balance, a weekly saving rate and a 12-week total, and the unknown sits under the percent. Students lose the point by forgetting the $90 was already there, or by moving it the wrong way. Testing a choice is a dependable alternative: 15% of $212.55 is about $31.88, and 12 of those plus $90 gives $472.59.

Turning words into math moderate Understanding the question
The total of $472.59 includes the $90.00 she started with, and only 15% of each week's pay is saved, so the equation is 90 + 12 × 0.15x = 472.59. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Dividing $382.59 by 1.8 by hand means shifting the decimal point to get 3,825.9 ÷ 18, which comes out to exactly $212.55 only if every digit is placed correctly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Subtract the $90, then divide by 12 weeks and by 0.15, or by their product 1.8; doing the division before the subtraction gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Overlooks the $90 she started with, treating the whole $472.59 as savings from earnings, and divides it by 1.8 to get $262.55.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No setup of the savings gives $302.55, the right weekly pay plus the $90 starting balance, a sum with no meaning here, so it mostly catches guesses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Adds the $90 to $472.59 instead of subtracting it, so the savings from earnings look like $562.59, and dividing by 1.8 gives $312.55.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Savings over time: $\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}$; The start is the money she had at the beginning. Weekly savings is what she adds each week. Weeks is how many weeks pass. Each week adds the same amount, so multiply. Then add what she started with.

Sort out the question
Sheila starts with $90.00. Each week she earns the same amount and saves 15% of it. After 12 weeks she has $472.59 in all. That total includes the $90 she started with. You need how much she earns each week, not how much she saves.
Write an equation
Call her weekly earnings x. She saves 15% of x each week. That is $0.15 \times x$ .
$\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}$
$472.59 = 90 + 0.15x \times 12$
Take away the starting $90
Only the money added during the 12 weeks came from her pay. Subtract 90 from both sides.
$0.15x \times 12 = 472.59 - 90 = 382.59$
She saved $382.59 from her pay. Watch out: subtract the $90, don’t add it. Adding leads to $312.55, choice D. Forgetting the $90 leads to $262.55, choice B.
Combine the numbers next to x
$0.15 \times 12 = 1.8$
$1.8x = 382.59$
Each week she saves 0.15 of one week’s pay. Over 12 weeks, that adds up to 1.8 times one week’s pay.
Divide by 1.8
Divide both sides by 1.8. First move the decimal point one place right in both numbers. Then you divide by the whole number 18.
$x = 382.59 \div 1.8 = 3{,}825.9 \div 18 = 212.55$
Sheila earns $212.55 each week. So the answer is A, $212.55.

Check your answer

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

More questions like this

The original DOE question

Question 19 of 39

2023 Test A Math, question 36

Question

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

A300
B800
C1,500
D2,500

Show the answer

Correct answer: C

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

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method moderate Working it out
No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)
Many steps minor Working it out
Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

Stops one step early: A choice is a number from partway through the solution.
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 a misconception Expected pull: moderate
Takes 60% of 500, applying the percent to the difference between the two shares as if the 500 stamps were the whole collection.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes 60% of 500 to get 300 and then adds the 500 extra stamps, combining the stem's numbers without a model of the two shares.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Sees that the 500-stamp gap is 20% of the total, finds the total of 2,500, and stops before taking Josef's 60% of it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.
Find Mai’s percent
All the stamps make 100%. Josef has 60%, so Mai has the rest.
$100\% - 60\% = 40\%$
Mai has 40% of the stamps.
Find what the 500 stamps are
Josef’s extra 500 stamps are the gap between his share and Mai’s share.
$60\% - 40\% = 20\%$
So 500 stamps are 20% of all the stamps.
Find the total number of stamps
$\text{part} = \frac{p}{100} \times \text{whole}$
$500 = \frac{20}{100} \times \text{whole}$
$\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500$
There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.
Find Josef’s share
$\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500$
Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

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

More questions like this

The original DOE question

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

Question 20 of 39

2023 Test A Math, question 55

Question

What is the least of four consecutive integers whose sum is 58?

Show the answer

Correct answer: D

Let n represent the least of the four integers. The other three integers are $(n + 1)$ , $(n + 2)$ , and $(n + 3)$ . Write an equation to solve for n:

$n + (n + 1) + (n + 2) + (n + 3) = 58$

$4n + 6 = 58$

$4n = 52$

$n = 13$

Question analysis

This number properties question asks for the least of four consecutive integers that add up to 58. Students write the integers in terms of one unknown and solve a two-step equation, or reason from the average of the four. Starting the list at n + 1 instead of n, which shifts every integer by one, is the main trap.

NYS standards: 7.EE.4a two-step equations
6.EE.6 variables representing unknown numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r is core grade 7 work. Expressing four consecutive integers in terms of the least one is a familiar setup, so this is a routine three-step grade 7 problem.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 11 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One equation, 4n + 6 = 58, solves the question, and dividing 58 by 4 to find the middle of the four, 14.5, is quicker still. The only real risk is the count: writing the integers as n + 1 through n + 4 and reporting n gives a least integer one too small. The two smallest choices match no reasonable working, so they draw only guesses. Adding the four integers back up to 58 is a ten-second check.

Turning words into math minor Understanding the question
The phrase “four consecutive integers” has to become n, n + 1, n + 2 and n + 3, with n the least, so the sum is 4n + 6, not 4n + 10. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choices 12 and 13 differ by one, the size of a fence-post error, so only checking that 13 + 14 + 15 + 16 = 58 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Call the least integer n, write the next three as n + 1, n + 2 and n + 3, add them and solve; or divide 58 by 4 to find the middle of the four.

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 equation for four consecutive integers gives 1; it may echo the step of 1 between them, but it mostly catches guesses.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with a sum of 58 leads to 2, which may echo the + 2 in n + 2, so it mostly draws guesses.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Writes the four integers as n + 1 through n + 4, solves 4n + 10 = 58 to get n = 12, and reports n instead of the least integer, n + 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers: $n,\; n + 1,\; n + 2,\; n + 3$; Integers are whole numbers and their negatives, like −2, 0 and 7. Consecutive integers come one right after another, like 4, 5, 6, 7. If n is the least one, each next one is 1 more than the one before.

Sort out the question
You need four consecutive integers, numbers in a row like 7, 8, 9, 10. Together they add up to 58. You need the least, or smallest, of the four.
Name the four numbers
Let n be the least number. Then the other three are 1, 2 and 3 more than n.
$n,\; n + 1,\; n + 2,\; n + 3$
Write and simplify the equation
Add the four numbers and set the sum equal to 58.
$n + (n + 1) + (n + 2) + (n + 3) = 58$
$4n + 6 = 58$
To combine, count the n’s: there are four of them. The plain numbers make 1 + 2 + 3 = 6.
Solve for n
Subtract 6 from both sides. Then divide both sides by 4.
$4n = 58 - 6 = 52$
$n = 52 \div 4 = 13$
Watch out: if you call the numbers n + 1 to n + 4, you get n = 12. Then 12 is not one of the four numbers; the least one is 13. Answering 12 is choice C.
So the answer is D, 13.

Check your answer

Add them up: $13 + 14 + 15 + 16 = 58$ . It works. Another way: $58 \div 4 = 14.5$ is their average. It falls halfway between the two middle numbers, 14 and 15. So the four numbers are 13, 14, 15 and 16.

More questions like this

The original DOE question

Question 112 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 112. 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 21 of 39

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 22 of 39

2023 Test B Math, question 12

Question

The perimeter of a 12-sided polygon is 100 centimeters. The length of one side is 25 centimeters. The length of another side is 35 centimeters. The remaining sides are equal in length to one another. What is the length of each of the remaining sides, in centimeters?

Show the answer

Correct answer: A

Let x represent the length of one of the 10 remaining sides:

$25 + 35 + 10x = 100$

$60 + 10x = 100$

$10x = 40$

$x = 4$

Question analysis

A 12-sided polygon has a perimeter of 100 centimeters: one side is 25 cm long, another is 35 cm, and the rest are all equal to one another. Students find the length of each remaining side. This perimeter problem turns into a two-step equation, and the tricky part is counting the sides that are left.

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving a two-step equation from a short description is standard grade 7 work. The perimeter setting is simple; the only extra care is counting the sides that remain after the two given ones.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 44 words.
Students, last 30 days: 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 perimeter and the two named sides make the subtraction obvious, so the real decision comes after it: what to divide the leftover 40 centimeters by. Students who stop at the leftover length, or who lose track of how many sides remain, reach a listed choice. Writing 25 + 35 + 10x = 100 keeps the count of 10 in view.

Turning words into math minor Understanding the question
“The remaining sides are equal in length to one another” hides the count: 12 sides minus the two given ones leaves 10 equal sides to share the rest of the perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Subtracting 25 and 35 from 100, counting the sides left, and dividing are three steps, and the first result, 40, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the two known sides from the perimeter, count how many sides are left, and share the remaining length equally among them.

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 Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 5; sharing the 40 centimeters among 8 sides would, but nothing in the question suggests 8, so it mostly draws guesses.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Shares the leftover 40 centimeters between two sides, as if the figure were four-sided, overlooking that it has 12 sides.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the total of the remaining sides, 100 − 25 − 35 = 40, and stops before sharing it among the 10 equal sides.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a polygon: $P = \text{the sum of all the side lengths}$; P is the perimeter, the total distance around a shape. A polygon is a flat shape with straight sides. A sum is the answer to an addition. So add the lengths of every side to get P.

Sort out the question
The shape is a polygon with 12 sides. Its perimeter is 100 centimeters. One side is 25 centimeters, and another is 35 centimeters. The other sides all have the same length. You need that length.
Count the equal sides
Two of the 12 sides are already known. The rest are the equal sides.
$12 - 2 = 10$
So there are 10 equal sides. Call the length of each one x.
Write an equation for the perimeter
Use the perimeter formula. All 12 sides add up to 100. Ten sides of length x add up to 10x, which means 10 times x.
$P = \text{the sum of all the side lengths}$
$25 + 35 + 10x = 100$
Solve for x
$60 + 10x = 100$
$10x = 100 - 60 = 40$
The 10 equal sides add up to 40 centimeters. Watch out: 40 is all ten sides together, not one side. Stopping here gives choice D. Now share the 40 among the 10 sides.
$x = 40 \div 10 = 4$
Give the answer
Each remaining side is 4 centimeters long. Watch out: sharing the 40 centimeters between only 2 sides gives 20, choice C. That treats the shape as if it had 4 sides, but it has 12. So the answer is A, 4.

Check your answer

Add all 12 sides: $25 + 35 + 10 \times 4 = 25 + 35 + 40 = 100$ . That is the perimeter the question gives, so 4 centimeters is right.

More questions like this

The original DOE question

Question 69 as printed on page 194 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 69. Open page 194 of the handbook (PDF)

Question 23 of 39

2023 Test B Math, question 14

Question

Each number in a sequence is 3 more than twice the number that comes just before it. If 93 is a number in the sequence, what number comes just before it?

Show the answer

Correct answer: D

Let n represent a number in the sequence, and let x represent the number that comes just before n:

$n = 3 + 2x$

$93 = 3 + 2x$

$90 = 2x$

$45 = x$

Question analysis

In a sequence, each number is 3 more than twice the number just before it, and 93 is one of the numbers. Students find the number that comes just before 93. That means running the rule backward, undoing the adding and the doubling in reverse order.

NYS standards: 7.EE.4a two-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students routinely solve two-step equations. The demand here is translating the wordy sequence rule into one, or running the rule backward in the right order, rather than any harder content.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is two short steps, so the question rests on reading a wordy rule correctly and running it in reverse. Each wrong choice marks a different place to go wrong: ignoring the rule, stopping after the subtraction, or undoing the adding with more adding. Checking the answer forward, twice the number plus 3, confirms it reaches 93.

Turning words into math moderate Understanding the question
“Each number … is 3 more than twice the number that comes just before it” must become 2x + 3 = 93, with x the earlier number, not the later one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method minor Working it out
The rule is stated forward, from a number to the next, but the question runs backward, so each operation has to be undone in reverse order: subtract 3, then halve. (Schoenfeld, 1985)

Strategy: Call the earlier number x, write the rule as an expression in x set equal to 93, and solve it by undoing each operation in reverse order.

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.
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 surface strategy Expected pull: weak
Reads “the number that comes just before it” as the whole number one less than 93, ignoring the sequence’s rule.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Undoes the adding of 3, 93 − 3 = 90, and stops before undoing the doubling.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds 3 instead of subtracting it before halving, (93 + 3) ÷ 2 = 48, undoing the rule in the wrong direction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule: $n = 2x + 3$; x is any number in the sequence, a list that follows a rule. n is the number right after it. Twice means times 2, so twice x is 2x. Then 3 more means add 3. The same rule links every number to the one after it.

Sort out the question
A sequence is a list of numbers that follows a rule. Here, each number is “3 more than twice the number that comes just before it.” 93 is in the list. You need the number right before 93. Call it x.
Write the rule as an equation
Twice x is 2x. Three more than that is 2x + 3. The rule says this equals the next number, which is 93.
$n = 2x + 3$
$93 = 2x + 3$
Undo the adding of 3
The rule doubles x first, then adds 3. To go backward, undo the steps in reverse order. First, subtract 3 from both sides.
$93 - 3 = 2x$
$90 = 2x$
Watch out: 90 is twice the number, not the number itself. Stopping here gives choice B. And don’t add 3, which leads to 48, choice C.
Undo the doubling
Now divide both sides by 2.
$x = 90 \div 2 = 45$
Watch out: “just before” means just before in this list. It does not mean one less than 93, so 92 is wrong. So the answer is D, 45.

Check your answer

Run the rule forward from 45. Twice 45 is 90, and 3 more is 93: $2 \times 45 + 3 = 93$ . So 45 really does come just before 93.

More questions like this

The original DOE question

Question 71 as printed on page 195 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 71. Open page 195 of the handbook (PDF)

Question 24 of 39

2023 Test B Math, question 19

Question

Orange juice costs $2 per gallon, and grapefruit juice costs $3 per gallon. A recent shipment of equal amounts of the two types of juice costs a total of $250. How many gallons of orange juice are in the shipment?

A50
B62.5
C100
D125

Show the answer

Correct answer: A

Let x represent the number of gallons of orange juice. Since the shipment contains equal amounts of the two types of juice, x also represents the number of gallons of grapefruit juice. Write an equation to solve for x:

$2x + 3x = 250$

$5x = 250$

$x = 50$

Question analysis

Orange juice costs $2 a gallon and grapefruit juice costs $3 a gallon, and a shipment with equal amounts of each costs $250. Students find how many gallons of orange juice are in the shipment. The word problem leads to a simple equation; the key phrase is equal amounts of juice, not equal amounts of money.

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation from a short word problem is standard grade 7 work. This one has a single unknown, since both juices come in the same number of gallons, and only two short steps.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The equation is short once the key phrase is read as equal gallons, since one gallon of each juice costs $5 together and $250 buys 50 such pairs. Read as equal spending, the same phrase leads to a listed answer that is not a whole number of gallons. The other trap is giving the gallons of both juices together when only orange juice is asked for.

Turning words into math moderate Understanding the question
“Equal amounts of the two types of juice” means the same number of gallons of each, x, so the total cost is 2x + 3x, not two equal halves of $250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Price one gallon of each juice together, then see how many such pairs $250 buys; that is the number of gallons of each kind.

Traps

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

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads “equal amounts” as equal spending, $125 on each juice, and divides $125 by $2 a gallon to get 62.5.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the right count for each juice but gives the whole shipment, 50 + 50 = 100 gallons, instead of the orange juice alone.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides $250 by 2, the orange juice price or the number of juices, combining numbers from the question without equal gallons of each.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost: $\text{cost} = \text{price per gallon} \times \text{number of gallons}$; The price per gallon is what one gallon costs. Multiply it by the number of gallons to get what you pay in all. For example, 3 gallons at $2 a gallon cost $6.

Sort out the question
Orange juice costs $2 a gallon. Grapefruit juice costs $3 a gallon. The shipment has equal amounts of each, so it has the same number of gallons of each juice. It costs $250 in all. You need the number of gallons of orange juice.
Write the cost of each juice
Let x be the number of gallons of each juice. Use the cost formula for each one.
$\text{cost} = \text{price per gallon} \times \text{number of gallons}$
$\text{orange juice cost} = 2 \times x = 2x$
$\text{grapefruit juice cost} = 3 \times x = 3x$
Add the costs and solve
The two costs add up to $250. Two x’s plus three x’s make five x’s.
$2x + 3x = 250$
$5x = 250$
$x = 250 \div 5 = 50$
There are 50 gallons of each juice.
Give the answer
Watch out: equal amounts means equal gallons, not equal dollars. Spending $125 on each juice gives 62.5 gallons, choice B. Also, 100 gallons is both juices together, choice C. So the answer is A, 50.

Check your answer

Find each cost. 50 gallons of orange juice cost $50 \times 2 = 100$ dollars. 50 gallons of grapefruit juice cost $50 \times 3 = 150$ dollars. Together, $100 + 150 = 250$ dollars, the total in the question.

More questions like this

The original DOE question

Question 76 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 76. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 25 of 39

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 26 of 39

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 27 of 39

2023 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

A−12
B−10
C−8
D−4

Show the answer

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

$n+(n+1)+(n+2)+(n+3)$

$=4n+6$

Set that expression equal to each of the options and solve for n:

Option A:

$4n+6=-12$

$4n=-18$

$n=-\frac{18}{4}$ is not an integer.

Option B:

$4n+6=-10$

$4n=-16$

$n=-4$ is an integer.

Option C:

$4n+6=-8$

$4n=-14$

$n=-\frac{14}{4}$ is not an integer.

Option D:

$4n+6=-4$

$4n=-10$

$n=-\frac{10}{4}$ is not an integer.

The only option for which n is an integer is −10.

Question analysis

A number properties question about consecutive integers: which of four negative numbers could be the sum of 4 consecutive integers? Students write the sum with a variable, or add runs of negative integers, and test each choice. Keeping signs straight with negative numbers, and remembering that the integers must be whole numbers, is the tricky part.

NYS standards: 7.EE.2 rewriting expressions to show relationships
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 consecutive-integer problems usually give the sum and ask for the integers. This one reverses the task: students test which negative total is possible at all, using the structure of the sum rather than one equation.
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: Every choice must be checked moderate
No obvious method minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 9 words.
Corrected from the DOE handbook: A stray word in the explanation’s last sentence was removed.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Three of the four choices are multiples of 4, and that is the whole trap: a student who pictures the sum as four times some integer finds three choices that seem to work. The sum of four consecutive integers is 4n + 6, so it is always even but never a multiple of 4. Trying the most natural run, −4, −3, −2 and −1, lands on −10 at once and is the fastest check.

Every choice must be checked moderate Choosing the answer
The stem asks which number “could be” the sum, so no single equation gives the answer; each of −12, −10, −8 and −4 has to be tested against the form a four-integer sum must take. (Embretson & Wetzel, 1987)
No obvious method minor Working it out
Nothing points to the shortcut that n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 is always 2 more than a multiple of 4, or to simply adding a run such as −1, −2, −3 and −4. (Schoenfeld, 1985)
Error-prone arithmetic minor Working it out
Every choice is negative, so solving 4n + 6 = −12 or adding a run of integers means subtracting 6 from a negative number and dividing a negative by 4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the integers as n, n + 1, n + 2 and n + 3, add them, and set the sum equal to each choice to see which one gives a whole-number n.

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

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
Assumes four consecutive integers add to a multiple of 4, as if they had a whole-number middle, and takes −12 = 4 × (−3), the first such choice.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Writes n + (n + 1) + (n + 2) + (n + 3) as 4n, dropping the 1 + 2 + 3, so −8 = 4 × (−2) seems to give a whole-number n.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Solves 4n + 6 = −4 to get n = −2.5 and accepts it, since −2.5, −1.5, −0.5 and 0.5 add to −4, though they are not integers.

What was corrected from the DOE handbook, and why

Explanation: The handbook ends with “The only option for which n is an integer answer is −10.” The word “answer” does not belong in that sentence; the site reads “The only option for which n is an integer is −10.” The working above it is unchanged: only 4n + 6 = −10 gives an integer, n = −4, and −4 + (−3) + (−2) + (−1) = −10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of 4 consecutive integers: $n + (n + 1) + (n + 2) + (n + 3) = 4n + 6$; n is the smallest of the four integers. Each next integer is 1 more than the one before, so they are n, n + 1, n + 2 and n + 3. Adding them gives four n’s, plus 1 + 2 + 3, which is 6. Use it to add any four integers in a row.

Sort out the question
Integers are whole numbers and their negatives, like −2, 0 and 5. They have no fractions or decimals. Consecutive integers come one right after another, like 7, 8, 9, 10 or −3, −2, −1, 0.
You need the choice that can be the sum of four consecutive integers. The sum is what you get when you add them.
Write the sum with a letter
You don’t know the integers yet. So call the smallest one n. Each next one is 1 more: n + 1, n + 2 and n + 3. Use the formula. Collect the four n’s, then add the numbers.
$n + n + n + n + 1 + 2 + 3 = 4n + 6$
Any four consecutive integers add up to 4n + 6.
Test each choice
Set 4n + 6 equal to each choice and solve for n. First subtract 6 from both sides. Then divide both sides by 4.
Choice 4n + 6 4n n
A −12 −18 −4.5
B −10 −16 −4
C −8 −14 −3.5
D −4 −10 −2.5
n must be an integer. Only choice B gives one: n = −4.
Name the four integers
Start at n = −4 and count up by 1. The integers are −4, −3, −2 and −1.
Watch out: Choices A, C and D are multiples of 4. A multiple of 4 is 4 times an integer, like 4 × (−3) = −12. It is tempting to think four integers must add to one of those. But the sum is 4n + 6, and the extra 6 gets in the way. Choice D works only with n = −2.5, and that is not an integer.
So the answer is B, −10.

Choice	4n + 6	4n	n
A	−12	−18	−4.5
B	−10	−16	−4
C	−8	−14	−3.5
D	−4	−10	−2.5

Check your answer

Add the integers: $-4 + (-3) + (-2) + (-1) = -10$ . That is choice B. Also, 4n + 6 is always 2 more than a multiple of 4. And −10 is 2 more than −12, which is 4 × (−3).

More questions like this

The original DOE question

Question 86 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 86. 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). This is the DOE’s version, before the corrections listed in the analysis.

Question 28 of 39

2023 Test B Math, question 40

Question

Angle A and angle B are complementary. If the measure of angle B is three times as great as the measure of angle A, what is the measure of angle B?

A45°
B67.5°
C75°
D135°

Show the answer

Correct answer: B

Let A represent the measure of angle A, and let B represent the measure of angle B, in degrees. Since the angles are complementary, $A+B=90$ . The question states that $B=3A$ , or $A=\frac{B}{3}$ . Substitute that expression for A in the first equation and solve for B:

$A+B=90$

$\frac{B}{3}+B=90$

$\frac{4B}{3}=90$

$4B=270$

$B=67.5$

Question analysis

A geometry question about two complementary angles, where angle B is three times as large as angle A. Students find the measure of angle B. They must recall what complementary means, since the question does not say, and set up an equation that splits the total in the right proportion.

NYS standards: 7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation from complementary angles is a grade 7 standard. This is the familiar version without a figure: one equation from the definition, then a division and a multiplication, typical of grade 7 classwork.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 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

Once complementary is read as a 90° total, the problem is one equation with a decimal answer, 4A = 90. The 22.5° for angle A is not listed, so stopping at the smaller angle is not rewarded here, but using 180° is: it gives 45° and 135°, and both appear among the choices. Checking that the chosen angle and one third of it add to 90° settles it.

Something you must already know moderate Working it out
The stem never says what complementary means; the 90° total has to be recalled, and a student who confuses it with the supplementary 180° doubles both angles. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The clause “the measure of angle B is three times as great as the measure of angle A” has to become B = 3A, so the 90° splits into four equal parts, not three. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Recall the total that complementary angles share, think of angle A as one part and angle B as three parts, find one part, then multiply for angle B.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Splits the 90° evenly, 90 ÷ 2 = 45, using the complementary total but ignoring that angle B is three times angle A.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 75°, which would need angle A to be 25° and the pair to add to 100°; it mostly draws guesses.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses 180°, the supplementary total, so 4A = 180 gives angle A = 45° and angle B = 135°, a pair that is not complementary.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Complementary angles: $A + B = 90$; A and B are the measures of the two angles, in degrees. Complementary angles are two angles that add up to 90 degrees, the size of a square corner. Use this whenever a problem says two angles are complementary.

Sort out the question
Complementary angles are two angles that add up to 90 degrees. Angle B is three times as big as angle A. You need the measure of angle B, in degrees.
Write the facts as equations
Let A and B stand for the two measures. The angles are complementary, so use the formula.
$A + B = 90$
Angle B is three times angle A, so:
$B = 3A$
Put 3A in place of B
Replace B in the sum with 3A. Then there is only one letter to find.
$A + 3A = 90$
$4A = 90$
Think of it as 4 equal parts. Angle A is 1 part, and angle B is 3 parts.
Find angle A
Divide both sides by 4.
$A = 90 \div 4 = 22.5$
Angle A measures 22.5 degrees. That is not the answer yet, because the question asks for angle B.
Find angle B
Angle B is three times angle A.
$B = 3 \times 22.5 = 67.5$
Angle B measures 67.5 degrees.
Watch out: Angles that add to 180 degrees are called supplementary. Using 180 by mistake gives 45 and 135, choice D. Splitting 90 in half gives 45, choice A. That ignores the fact that B is three times A.
So the answer is B, 67.5°.

Check your answer

Add the two angles: 22.5 + 67.5 = 90. So they are complementary. Now divide: 67.5 ÷ 22.5 = 3. So angle B is three times angle A. Both facts check out.

More questions like this

The original DOE question

Question 97 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 97. Open page 202 of the handbook (PDF)

Question 29 of 39

2022 Test A Math, question 15

Question

If $5(x - 6y) = 50y$ , what is the value of x in terms of y?

A $16y$
B $15y$
C $11\frac{2}{5}y$
D $11y$

Show the answer

Correct answer: A

To find the value of x in terms of y, solve $5(x - 6y) = 50y$ for x. Begin by distributing 5 on the left side of the equation:

$(5 \times x) - (5 \times 6y) = 50y$

$5x - 30y = 50y$

Solve for the value of x:

$5x = 80y$

$\frac{5x}{5} = \frac{80y}{5}$

$x = 16y$

Question analysis

This algebra question gives the equation 5(x − 6y) = 50y and asks for x in terms of y. Students use the distributive property, or divide both sides by 5, and then solve a literal equation with inverse operations. Forgetting that the 5 multiplies both terms inside the parentheses is the classic slip here.

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 equations such as 5(x − 6) = 50 with numbers. Solving for x in terms of another letter, through the distributive property, is grade 8 algebra, though the choices can be checked by substitution.
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
Solution: 3 steps
Reading level: Question and choices: about grade 6, 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 equation needs only three moves, distribute or divide by 5, add the y term, divide, so the risk lies in the first move. Multiplying only the x by 5 is the classic slip, and it gives 56/5 y, a value that is not listed but sits between two of the choices. Substituting y = 1 and the chosen x back into 5(x − 6) = 50 checks any choice in seconds.

Variables instead of numbers moderate Understanding the question
Both sides hold y and the result must be a multiple of y, so there is no number to check against unless the student substitutes, for instance y = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
The 5 has to multiply both terms in the parentheses, giving 5x − 30y, and then 80y ÷ 5 must come out exactly; dividing both sides by 5 first keeps the numbers small. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Distribute the 5 to both terms in the parentheses, or divide both sides by 5 first, then add the y term to both sides; substitute your choice back for x to check.

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

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 15y; it sits one y below the correct value and would need a slip such as 80 ÷ 5 = 15, so it mostly draws guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 57/5 y; multiplying only the x by 5 leads to 56/5 y, or 11 1/5 y, so this mostly draws students who land near it.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Multiplies only the x by 5, so 5x − 6y = 50y and 5x = 56y, then drops the remainder of 56 ÷ 5 = 11 R1, keeping 11y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. Multiply each term inside by a. Use it when a number multiplies a whole sum.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
The equation is $5(x - 6y) = 50y$ . It has two letters, x and y. In terms of y means get x alone on one side. The other side may still have y in it. So the answer will look like a number times y.
Multiply out the parentheses
Use the distributive property. The 5 multiplies both terms inside.
$5(x - 6y) = 5 \times x - 5 \times 6y = 5x - 30y$
$5x - 30y = 50y$
Watch out: the 5 must multiply the 6y too. Multiplying only the x gives $5x = 56y$ . Then x is $11\frac{1}{5}y$ . That is not a choice, but it sits close to C and D. Don’t pick one of them just because it is close.
Get the x term alone
Add 30y to both sides. This keeps the equation balanced.
$5x - 30y + 30y = 50y + 30y$
$5x = 80y$
Like terms add: 50 y’s plus 30 y’s is 80 y’s.
Divide by 5
5x means 5 times x. Divide both sides by 5 to get x by itself.
$x = \frac{80y}{5} = 16y$
So the answer is A, $16y$ .

Check your answer

Pick y = 1. Then x = 16. Put both into the first equation: $5(16 - 6 \times 1) = 5 \times 10 = 50$ . The right side is $50 \times 1 = 50$ . Both sides match. A faster route: divide both sides by 5 first. Then $x - 6y = 10y$ , so x = 16y.

More questions like this

The original DOE question

Question 72 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 2023 Admissions, page 83, question 72. 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 30 of 39

2022 Test A Math, question 50

Question

Nadia paid $1.00 to mail a letter. The cost to mail a letter is $0.55 for the first ounce of weight plus $0.15 for each additional ounce. What was the weight, in ounces, of the letter Nadia mailed?

A1.4
B1.5
C3
D4

Show the answer

Correct answer: D

Let x represent the number of ounces Nadia’s letter weighed. The first ounce costs $0.55, so $(x - 1)$ represents the number of ounces that cost $0.15 each.

Given that the total cost is $1.00, the equation that describes the situation can be written and solved as shown.

$0.55 + 0.15(x - 1) = 1.00$

$0.55 + 0.15x - 0.15 = 1.00$

$0.15x = 1.00 - 0.55 + 0.15$

$0.15x = 0.60$

$\frac{0.15x}{0.15} = \frac{0.60}{0.15}$

$x = 4$

So the letter Nadia mailed weighed 4 ounces.

Question analysis

Nadia pays $1.00 to mail a letter, and postage costs $0.55 for the first ounce plus $0.15 for each additional ounce. Students find the weight of the letter in ounces. It tests writing and solving a two-step equation from a pricing rule in which the two prices apply to different ounces.

NYS standards: 7.EE.4a two-step equations
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve word problems that lead to equations such as px + q = r. This one takes that form once each price is matched to the right ounces, a step that makes it a little more demanding than routine practice.
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
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Pricing the letter ounce by ounce, $0.55, $0.70, $0.85, $1.00, reaches the total at the fourth ounce without any algebra. The equation route works too, but the first ounce is easy to lose: dividing the $0.45 left by $0.15 counts only the ounces after the first. The two decimal choices come from merging the two prices or from treating dollars as ounces.

Turning words into math moderate Understanding the question
“$0.55 for the first ounce of weight plus $0.15 for each additional ounce” makes the cost 0.55 + 0.15(x − 1), so the $0.15 covers every ounce except the first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
The money is in decimals, $1.00 − $0.55 = $0.45 and then $0.45 ÷ $0.15 = 3, and dividing by a decimal invites a misplaced point. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the cost of a letter weighing x ounces, $0.55 plus $0.15 for each ounce after the first, set it equal to $1.00 and solve, then check by pricing the letter ounce by ounce.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Merges the two prices into $0.70 for every ounce and divides $1.00 by $0.70, about 1.43, which rounds to 1.4.
B Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Counts the first ounce, then treats the $0.45 left over as 0.45 ounce, getting 1.45 ounces, which rounds to 1.5.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 3 additional ounces that the $0.45 left over pays for, 0.45 ÷ 0.15 = 3, and stops before adding the first ounce.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost to mail a letter: $C = 0.55 + 0.15 \times (w - 1)$; C is the cost in dollars. w is the weight in ounces. The first ounce costs $0.55. Each ounce after the first costs $0.15, and there are w − 1 of them. Subtracting 1 leaves out the first ounce, which is already paid for.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
Nadia paid $1.00 to mail a letter. The first ounce costs $0.55. Each additional ounce, meaning each ounce after the first, costs $0.15. You need the weight of the letter in ounces.
Write the equation
Let w be the weight in ounces. The first ounce is 1 of the w ounces. So the extra ounces number w − 1. The cost is $1.00.
$0.55 + 0.15 \times (w - 1) = 1.00$
Take off the first ounce’s cost
Subtract 0.55 from both sides. This keeps the equation balanced.
$0.15 \times (w - 1) = 1.00 - 0.55 = 0.45$
So $0.45 pays for the extra ounces.
Find the extra ounces
Divide both sides by 0.15. Multiply both numbers by 100 to make them whole.
$w - 1 = 0.45 \div 0.15 = 45 \div 15 = 3$
There are 3 extra ounces after the first one.
Add back the first ounce
$w = 3 + 1 = 4$
The letter weighed 4 ounces. Watch out: 3 counts only the ounces after the first. It is choice C. Also, don’t join $0.55 and $0.15 into one price. $1.00 \div 0.70$ is about 1.4, choice A. So the answer is D, 4 ounces.

Check your answer

Price the letter one ounce at a time. 1 ounce: $0.55. 2 ounces: $0.70. 3 ounces: $0.85. 4 ounces: $1.00. The cost reaches $1.00 at 4 ounces. That matches.

More questions like this

The original DOE question

Question 107 as printed on page 95 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 95, question 107. Open page 95 of the handbook (PDF)

Question 31 of 39

2022 Test B Math, question 7

Question

If $\frac{x + 2y}{5} = 3y$ , what is the value of x in terms of y?

A $\frac{7y}{5}$
B $\frac{13y}{5}$
C13
D $13y$

Show the answer

Correct answer: D

The goal is to isolate the variable x by using inverse operations. First, multiply both sides by 5 and then subtract $2y$ from both sides.

$\frac{x + 2y}{5} = 3y$

$x + 2y = 15y$

$x = 13y$

Another approach is to make the right side of the equation a fraction and then use cross multiplication for the first step.

$\frac{x + 2y}{5} = \frac{3y}{1}$

$x + 2y = 15y$

$x = 13y$

Question analysis

An equation says that x plus 2y, all divided by 5, equals 3y, and students choose the expression for x in terms of y. This multiple-choice algebra question tests solving a literal equation with inverse operations. Because the 5 divides the whole left side, clearing the fraction first and keeping track of the y terms 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 (n + 4)/3 = 6. Solving for one letter in terms of another, with a fraction to clear first, is grade 8 work and a step beyond.
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
Solution: 2 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

Two inverse operations solve it, but the fraction bar spans the whole sum x + 2y, so multiplying both sides by 5 has to come first. Students who split the fraction and stop at x/5 find a matching choice, and those who divide 3y by 5 instead of multiplying end up with fifths of y as well. Substituting y = 1, which makes x + 2 equal 15, checks any choice in seconds.

Variables instead of numbers moderate Understanding the question
Both sides hold y, so the result must stay a multiple of y: after clearing the 5, the like terms 15y and 2y are combined, and there is no number to check against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Multiplying the whole of 3y by 5 gives 15y, but dividing it by 5 or splitting the fraction into x/5 + 2y/5 leaves fifths of y in the working. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply both sides by 5 to clear the fraction, then subtract 2y from both sides; check by putting y = 1 into both the equation and the chosen expression.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
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 3y by 5 instead of multiplying, so x + 2y = 3y/5 and x = −7y/5, then takes 7y/5, the choice that matches its size.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Splits the fraction and subtracts 2y/5 from 3y, reaching x/5 = 13y/5, then stops without multiplying both sides by 5.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reaches x = 15y − 2y but drops the letter, as if subtracting like terms cancelled the y, and gives the number 13 instead of 13y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Do the same to both sides: $\text{if } a = b, \text{ then } a \times c = b \times c$; a and b are two equal amounts, and c is any number. If you do the same thing to both sides of an equation, they stay equal, like a balanced scale. Use it to undo each step until the letter you want is alone.

Sort out the question
The equation says: take x plus 2y, divide the whole sum by 5, and you get 3y. You need x “in terms of y.” That means x alone on one side, with an expression using y on the other. An expression is a math phrase with no equal sign.
Clear the fraction first
The fraction bar divides the whole top, x + 2y, by 5. To undo dividing by 5, multiply both sides by 5.
$5 \times \frac{x + 2y}{5} = 5 \times 3y$
$x + 2y = 15y$
Watch out: multiply 3y by 5. Don’t divide it by 5. Dividing leaves fifths of y, as in choice A.
Get x alone
Now 2y is added to x. To undo adding, subtract 2y from both sides.
$x + 2y - 2y = 15y - 2y$
$x = 13y$
15y − 2y means 15 y’s take away 2 y’s. That leaves 13 y’s, so the y stays.
Give the answer
Watch out: you might split the fraction into x/5 + 2y/5 instead. Then you reach x/5 = 13y/5. Stopping there gives choice B, but you still have to multiply by 5. Dropping the y gives 13, choice C. So the answer is D, $13y$ .

Check your answer

Pick y = 1. Then x = 13 × 1 = 13. Put both into the equation: $\frac{13 + 2 \times 1}{5} = \frac{15}{5} = 3$ . And 3y = 3 × 1 = 3. Both sides match, so x = 13y works.

More questions like this

The original DOE question

Question 64 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 2023 Admissions, page 199, question 64. Open page 199 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 32 of 39

2018 Test A Math, question 5

Question

The sum of two consecutive integers is −15. If 1 is added to the smaller integer and 2 is subtracted from the larger integer, what is the product of the two resulting integers?

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

Show the answer

Correct answer: 63

If x is the smaller consecutive integer, then

$x+1$

is the larger consecutive integer. Use their sum (−15) to find x:

$x+(x+1)=-15$

$2x+1=-15$

$2x=-16$

$x=-8$

The two consecutive integers are −8 and −7. One is added to the smaller integer:

$-8+1=-7$

and 2 is subtracted from the larger integer:

$-7-2=-9$

Find the product:

$(-7)(-9)=63$

Question analysis

A grid-in problem about two consecutive integers whose sum is −15. Students find the two integers, add 1 to the smaller and subtract 2 from the larger, then multiply the results. Deciding which negative integer is the smaller one, and keeping signs straight through every step, are what make it tricky.

NYS standards: 7.EE.4a two-step equations
7.NS.2a rational number multiplication
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 A typical grade 7 task solves one two-step equation or multiplies signed numbers. This links both: set up a consecutive-integer equation, solve it with negatives, adjust each integer, and multiply, with no choices to check against.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math moderate
Many steps moderate
Solution: 4 steps
Reading level: Question and choices: about grade 9, 30 words.
Students, last 30 days: 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 the question collects most of the ways students lose points with negatives. The crucial judgment is which of −8 and −7 is the smaller; a student who picks −7 still produces a tidy, wrong product. Gridding the product of the original pair, or giving a negative product, are the other usual losses. Writing each intermediate pair down, rather than holding them in mind, protects against all three.

Error-prone arithmetic moderate Working it out
Every number in play is negative: −8 and −7, then −7 and −9, and a product of two negatives that must come out positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
"1 is added to the smaller integer and 2 is subtracted from the larger" depends on knowing that −8, not −7, is the smaller, since it lies farther below zero. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Four linked steps: set up x + (x + 1) = −15, solve for x, adjust each integer, then multiply; an error early in the chain carries through to the product. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Halve the sum to see which two integers sit on either side of it, remember that the smaller negative is farther from zero, then adjust each and multiply.

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

The wrong answers, classified

60 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes −7 as the smaller integer because 7 is less than 8, so adjusts to −7 + 1 = −6 and −8 − 2 = −10, and multiplies to 60.
56 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds −8 and −7 and multiplies them to get 56, skipping the sentence that adds 1 to one integer and subtracts 2 from the other.
-63 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Finds the adjusted integers −7 and −9 correctly but gives their product a negative sign, gridding −63.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers: $n, \; n + 1$; Integers are whole numbers and their opposites, like −2, −1, 0, 1 and 2. Consecutive integers come one right after another. If n is the smaller one, the next one is n + 1, because each step up the number line adds 1.
Multiplying two negative numbers: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. When you multiply two negative numbers, the answer is positive. The two minus signs cancel out.

Sort out the question
Two consecutive integers add up to −15. Integers are whole numbers and their negatives. Consecutive means one comes right after the other, like 4 and 5. You need to do three things:
1. Find the two integers.
2. Add 1 to the smaller one. Subtract 2 from the larger one.
3. Find the product of the two new numbers. The product is the answer when you multiply.
Find the two integers
Call the smaller integer n. Then the larger one is n + 1. Their sum is −15.
$n + (n + 1) = -15$
$2n + 1 = -15$
Subtract 1 from both sides. Then divide both sides by 2.
$2n = -16$
$n = -8$
So the integers are −8 and −7.
Decide which one is smaller
On a number line, −8 is to the left of −7. Numbers to the left are smaller. So −8 is the smaller integer, and −7 is the larger one. Watch out: 8 is bigger than 7, but −8 is smaller than −7. Mixing them up gives −6 and −10, and the wrong answer 60.
Change each integer
Add 1 to the smaller integer.
$-8 + 1 = -7$
Subtract 2 from the larger integer.
$-7 - 2 = -9$
The two new numbers are −7 and −9.
Multiply the new numbers
Both numbers are negative, so their product is positive.
$(-7) \times (-9) = 63$
Watch out: the answer is not −63, because two negatives make a positive. Also, don’t multiply the first pair, −8 and −7. That gives 56 and skips the changes. So you would grid 63.

Check your answer

Check the pair: $-8 + (-7) = -15$ , and they are one apart. Half of −15 is −7.5, and −8 and −7 sit on either side of it. Then $7 \times 9 = 63$ , and the sign is positive.

More questions like this

The original DOE question

Question 62 as printed on page 81 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 62. Open page 81 of the handbook (PDF)

Question 33 of 39

2018 Test A Math, question 13

Question

The perimeter of a rectangle is 510 centimeters. The ratio of the length to the width is 3:2. What are the dimensions of this rectangle?

A150 cm by 105 cm
B153 cm by 102 cm
C158 cm by 97 cm
D165 cm by 90 cm

Show the answer

Correct answer: B

Let $2x$ = the width and $3x$ = the length.

Draw the rectangle to help visualize.

$Rectangle with its top and bottom sides labeled 3x and its left and right sides labeled 2x.$

Since 2 times width + 2 times length = perimeter, we get

$2(2x)+2(3x)=510$

$4x+6x=510$

$10x=510$

$x=51$

$2x=102$ cm and $3x=153$ cm

Question analysis

A rectangle word problem that combines perimeter with a ratio: the perimeter is 510 centimeters and the length and width are in the ratio 3 to 2. Students find the rectangle's dimensions. Every choice has the correct perimeter, so only the ratio separates them.

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 Perimeter and ratio are each familiar from grades 6 and 7, and combining them with a five-part split is a typical grade 7 multistep ratio problem. It asks about the same as classroom 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: A very close wrong answer moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is built so that the perimeter, the easier condition, rules nothing out: every choice adds to 255. A student who tests the choices against 510 alone will accept whichever comes first, and the first is 150 by 105. Solving with ratio parts, or checking each pair's ratio, avoids that; 3:2 means the length is exactly one and a half times the width.

A very close wrong answer moderate Choosing the answer
153 by 102 and 150 by 105 differ by just 3 cm in each side and give the same 510 cm perimeter; only the ratio, exactly 3:2 against 10:7, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
"The ratio of the length to the width is 3:2" has to become 3 parts and 2 parts of the half-perimeter, 255, not of the full 510. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Halve the perimeter to get length plus width, split that sum into 5 equal parts, and give the length 3 parts and the width 2.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Checks only the perimeter, 2 × (150 + 105) = 510, and accepts the first pair that fits; 150 to 105 is 10:7, not 3:2.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Confirms that 158 + 97 = 255, half the perimeter, but never tests the ratio; 1.5 × 97 is 145.5, not 158.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Matches the perimeter, since 165 + 90 = 255, but a 3:2 rectangle 90 cm wide would be 135 cm long, not 165.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a rectangle: $P = 2 \times (l + w)$; P is the perimeter, the distance all the way around the shape. l is the length and w is the width. A rectangle has two lengths and two widths. So you add one length and one width, then double it.
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 perimeter of the rectangle is 510 cm. The perimeter is the distance all the way around. The ratio of length to width is 3:2. That means the length is 3 equal parts, and the width is 2 of the same parts. You need the length and the width.
Find length plus width
The perimeter counts each side twice, once for each matching side. So halve it to get one length plus one width.
$P = 2 \times (l + w)$
$510 = 2 \times (l + w)$
$l + w = 510 \div 2 = 255$
Length plus width is 255 cm.
Find the size of one part
Length and width together are $3 + 2 = 5$ parts. Share the 255 cm equally among them.
$\text{one part} = 255 \div 5 = 51$
Each part is 51 cm.
Find the length and width
The length is 3 parts and the width is 2 parts.
$l = 3 \times 51 = 153$
$w = 2 \times 51 = 102$
The length is 153 cm and the width is 102 cm.
Check the ratio, then answer
Watch out: every choice has a perimeter of 510 cm. For example, $150 + 105 = 255$ too. So checking only the perimeter can trick you into choice A. The ratio decides it. 153 is exactly 1.5 times 102, but 150 is not 1.5 times 105. So the answer is B, 153 cm by 102 cm.

Check your answer

Check both facts. Perimeter: $2 \times (153 + 102) = 2 \times 255 = 510$ . Ratio: $153 \div 51 = 3$ and $102 \div 51 = 2$ , so 153 to 102 is 3:2.

More questions like this

The original DOE question

Question 70 as printed on page 83 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 70. 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 34 of 39

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 35 of 39

2018 Test A Math, question 20

Question

Mrs. Cranston bought five bottles of water for $0.90 each and 8 pounds of meat. She paid a total of $26.90 for these items, not including tax. What was the price per pound of the meat?

A$2.80
B$3.25
C$14.40
D$22.40

Show the answer

Correct answer: A

Let $x$ be the price per pound for the meat. Set up an equation to show what Mrs. Cranston spent:

$5(0.90)+8x=26.90$

$4.50+8x=26.90$

$8x=22.40$

$x=2.80$

The price per pound for the meat is $2.80.

Question analysis

This money word problem has a shopper buying five bottles of water at a set price plus 8 pounds of meat, for a known total. Students write and solve a two-step equation to find the price per pound of the meat. The decimals call for care, and the question wants a price per pound rather than a total.

NYS standards: 7.EE.4a two-step equations
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Word problems that lead to an equation like px + q = r are the core of grade 7 algebra. This is a standard example with money decimals and no extra twist.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a two-step money problem: take away the water, then divide by the pounds. The listed wrong answers come from doing one of those steps wrongly or not at all, especially stopping at the total cost of the meat, which looks like a finished dollar amount. Checking that 8 pounds at the chosen price plus $4.50 makes $26.90 confirms the answer.

Turning words into math minor Understanding the question
"five bottles of water for $0.90 each" has to become 5 × 0.90 before the total is split, and the unknown is a price per pound, not the meat's total cost. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Every amount is a decimal of dollars, so 26.90 − 4.50 and 22.40 ÷ 8 must be lined up and worked carefully without a calculator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the cost of all five bottles from the total, then divide what is left by the 8 pounds of meat.

Traps

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

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Misses the word each and subtracts one bottle's $0.90, then divides: (26.90 − 0.90) ÷ 8 = 3.25.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Finds the $22.40 spent on meat and subtracts the 8 pounds from it instead of dividing, getting 14.40.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Subtracts the five bottles from the total, 26.90 − 4.50 = 22.40, and stops at the meat's total cost without dividing by 8 pounds.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost: $C = p \times q$; C is the total cost. p is the price of one item, or of one pound. q is the quantity, how many items or pounds you buy. Each one costs p, so q of them cost p times q.

Sort out the question
Mrs. Cranston bought five bottles of water “for $0.90 each”. She also bought 8 pounds of meat. Everything cost $26.90 in all. You need the price per pound of the meat, the cost of just 1 pound.
Find the cost of the water
Each bottle costs $0.90, and there are 5 bottles. Use the cost formula.
$C = p \times q$
$C = 0.90 \times 5 = 4.50$
The water cost $4.50. Watch out: each means every bottle costs $0.90. Taking away just one $0.90 leads to $3.25, choice B.
Find the cost of the meat
Take the water away from the total. What is left paid for the meat.
$26.90 - 4.50 = 22.40$
All 8 pounds of meat cost $22.40. Watch out: this is the cost of all the meat, not of one pound. Stopping here gives choice D.
Find the price per pound
Use the cost formula again. Now C is 22.40 and q is 8 pounds. Divide to find p.
$22.40 = p \times 8$
$p = 22.40 \div 8 = 2.80$
Each pound costs $2.80. So the answer is A, $2.80.

Check your answer

Put it all back together. Meat: $8 \times 2.80 = 22.40$ . Water: 4.50. Total: $22.40 + 4.50 = 26.90$ . That is exactly what she paid.

More questions like this

The original DOE question

Question 77 as printed on page 85 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 77. Open page 85 of the handbook (PDF)

Question 36 of 39

2018 Test B Math, question 2

Question

$\frac{147-x}{12}=12$

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

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

Show the answer

Correct answer: 3

$\frac{147-x}{12}=12$

$147-x=144$

$x=3$

Question analysis

This grid-in asks students to solve a two-step linear equation in which an expression containing x is divided by 12 and set equal to 12. The algebra is short, but x is being subtracted from 147, so undoing the operations in the right order and keeping the minus sign straight is where students slip.

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 This is exactly the two-step equation practice of grade 7: undo a division, then undo a subtraction. The only twist is that the unknown is being subtracted, so the last step needs care with its sign.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a two-move equation that most students finish quickly: multiply both sides by 12, then find what must be taken from 147 to leave 144. Points are lost only at the last move, where the minus sign in front of x invites 144 − 147 or 147 + 144. Substituting the result back into the fraction catches either slip in a few seconds.

Error-prone arithmetic minor Working it out
After multiplying by 12, the unknown sits behind a minus sign in 147 − x = 144, so students must ask what is taken from 147 rather than subtract in the order the numbers appear. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply both sides by 12 to clear the fraction, then ask what number must be taken away from 147 to leave that product.

Traps

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

The wrong answers, classified

-3 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches 147 − x = 144 but subtracts in the order the numbers appear, writing x = 144 − 147.
146 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Divides the right side by 12 instead of multiplying, so sets 147 − x equal to 1 and gets 146.
291 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves 147 across the equals sign without changing its sign and drops the minus on x, so x = 144 + 147.

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. Division and multiplication undo each other. So if a divided by b gives c, then b times c gives back a. This is the same as multiplying both sides by b.

Read the equation in words
The equation says: start with 147, take away a mystery number x, then divide by 12. The result is 12.
You need to find x. To get x alone, undo the steps in reverse order. Dividing by 12 was done last, so undo it first.
Undo the division by 12
Multiply both sides by 12. Whatever you do to one side, you must do to the other. That keeps the equation balanced.
$\frac{147 - x}{12} = 12$
$147 - x = 12 \times 12 = 144$
Watch out: Do not divide 12 by 12. That gives 147 − x = 1 and the wrong answer 146.
Find what was taken away
Now ask: 147 minus what number leaves 144? The gap between the two numbers is x.
$x = 147 - 144 = 3$
Watch out: Do not work out 144 − 147. That gives −3. Since 147 is bigger than 144, x must be positive.
So you would grid 3.

Check your answer

Put 3 back in for x. First, $147 - 3 = 144$ . Then $144 \div 12 = 12$ . That matches the right side of the equation, so x = 3 is correct.

More questions like this

The original DOE question

Question 59 as printed on page 153 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 59. Open page 153 of the handbook (PDF)

Question 37 of 39

2018 Test B Math, question 18

Question

PRICES FOR NEWSPAPER ADVERTISING

Page Space	Price
$\frac{1}{4}$ page	$200
$\frac{1}{2}$ page	$350
full page	$600

The table above shows prices for newspaper advertising. A store purchased $\frac{1}{4}$ pages, $\frac{1}{2}$ pages, and full pages of page space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?

A $1\frac{3}{4}$ pages
B10 pages
C $16\frac{1}{2}$ pages
D $17\frac{1}{2}$ pages

Show the answer

Correct answer: D

The question says that an equal number ( $x$ ) of each type of page space was purchased. To find the number of each type of page space that was purchased, multiply the price per type by $x$ and set it equal to the total amount spent, then solve for $x$ :

$200x+350x+600x=11,500$

$1,150x=11,500$

$x=10$

Thus, the store purchased 10 units of each type of page space. To find the total amount of page space purchased, multiply each type of page space by 10, and add:

$\left(10\times\frac{1}{4}\text{ page}\right)+\left(10\times\frac{1}{2}\text{ page}\right)+(10\times 1\text{ page})$

$=17\frac{1}{2}\text{ pages}$

Question analysis

This word problem gives a table of newspaper advertising prices for quarter-page, half-page and full-page ads, and says a store bought equal numbers of each for a set total. Students write and solve an equation for how many ads of each size were bought. The question wants total page space, which takes a second calculation with fractions.

NYS standards: 7.EE.4a two-step equations
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students write and solve equations like px = r from word problems. This adds a table, combines three prices into one coefficient and needs a second step with fractions after solving, a longer chain than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps major
Turning words into math moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 59 words.
Corrected from the DOE handbook: An arithmetic slip in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Solving for the number of ads is the obvious goal, and finding it feels like the finish, but the question asks for page space, which needs one more pass through the table. The awkward wording about buying pages in equal numbers slows students before any arithmetic starts. The wrong choices mark each place to stop: one set of ads, the number of each ad, or a slip in adding the fractional pages.

Many steps major Working it out
The work has two phases: turn the money into a count, 1,150x = 11,500 so x = 10, then turn the count into space by multiplying 10 by 1/4, 1/2 and 1 and adding. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math moderate Understanding the question
The phrase “purchased 1/4 pages, 1/2 pages, and full pages of page space in equal numbers” has to be read as the same number of ads of each size, so the three prices add to one coefficient. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
Both columns of the table are needed: the price column builds the equation, and the page-space column turns the count of ads into the final total. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Let x be the number of each ad size, add the three prices to write one equation for the total, solve it, then multiply x by the page space in one set of ads.

Traps

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 partial knowledge Expected pull: moderate
Adds 1/4 + 1/2 + 1, the page space in one set of the three ads, without using the $11,500 to find how many sets were bought.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Solves 1,150x = 11,500 for the number of ads of each size and stops there instead of converting the count to page space.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Takes ten quarter pages as 1 1/2 pages instead of 2 1/2, so adds 1 1/2 + 5 + 10.

What was corrected from the DOE handbook, and why

Explanation: The handbook adds 200x + 350x + 600x and writes 1,500x = 11,500. The sum is 1,150x, and the handbook’s next line, x = 10, only follows from 1,150x = 11,500. The site shows 1,150x = 11,500.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal sets: $\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}}$; A set here means one ad of each size. The cost of one set is what one of each costs together. Dividing the total money by it tells how many sets were bought, because every set costs the same.

Sort out the question
The store bought the same number of each ad size: quarter page, half page and full page. It spent $11,500 in all.
You need the total page space. That is not the number of ads, and not the money.
Find the cost of one set
Think of one quarter page, one half page and one full page as a set. The store bought equal numbers of each, so it bought whole sets.
$200 + 350 + 600 = 1{,}150$
One set costs $1,150.
Find how many sets
$\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}} = \frac{11{,}500}{1{,}150} = 10$
The store bought 10 sets. That means 10 ads of each size.
Watch out: 10 is the number of each ad, not the page space. Stopping here gives choice B.
Find the page space in one set
Add the sizes in one set. Write them in quarters so they have the same bottom number.
$\frac{1}{4} + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}$
One set is $\frac{7}{4}$ , or $1\frac{3}{4}$ , pages.
Watch out: $1\frac{3}{4}$ pages is only one set. That is choice A.
Multiply by the number of sets
There are 10 sets, so multiply the space in one set by 10.
$10 \times \frac{7}{4} = \frac{70}{4} = 17\frac{1}{2}$
So the answer is D, $17\frac{1}{2}$ pages.

Check your answer

Add the sizes another way. 10 quarter pages make 2.5 pages, not 1.5. 10 half pages make 5 pages. 10 full pages make 10 pages. Then $2.5 + 5 + 10 = 17.5$ , the same answer.

More questions like this

The original DOE question

Question 75 as printed on page 156 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 75. Open page 156 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 38 of 39

2018 Test B Math, question 33

Question

A large container is partially filled with $n$ liters of water. Ito adds 10 liters of water to the container, making it 60% full. If Ignacio adds 6 more liters of water, the container will be 75% full. What is the value of $n$ ?

Show the answer

Correct answer: A

Let $x$ be the total number of liters the container can hold.

$\frac{n+10}{x}=60\%\text{ and }\frac{n+16}{x}=75\%$

First, solve each equation for $x$ :

Equation 1:

$\frac{n+10}{x}=\frac{60}{100}$

$\frac{n+10}{x}=\frac{3}{5}$

$3x=5(n+10)$

$x=\frac{5n+50}{3}$

Equation 2:

$\frac{n+16}{x}=\frac{75}{100}$

$\frac{n+16}{x}=\frac{3}{4}$

$3x=4(n+16)$

$x=\frac{4n+64}{3}$

Now, set the two equations equal to each other and solve for $n$ .

$\frac{5n+50}{3}=\frac{4n+64}{3}$

$5n+50=4n+64$

$n+50=64$

$n=14\text{ liters}$

Question analysis

Water is poured into a large container in two stages: 10 liters make it 60% full, and 6 more liters bring it to 75% full. Students must work out how much water was in it to begin with. This percent word problem hides two unknowns, the starting amount and the capacity, and several in-between numbers look like answers.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here the whole is unknown and must be found from the change between two percents before a second step reaches the answer.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 3, strategic thinking · mainly analyze
What makes it hard: Turning words into math major
No obvious method moderate
Many steps moderate
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The container's size is the hidden key, and nothing in the stem names it. Students who write an equation for each fill level, or who notice that 6 liters equal 15 percent, get there; others grab a number from the story or stop at the 75% level. Because every wrong choice is a number that appears in the problem or arises while solving it, a careless finish is rewarded with a match.

Turning words into math major Understanding the question
Two percent-full statements about a container of unknown size have to become equations, n + 10 = 60% of the capacity and n + 16 = 75% of it, and the capacity is never given. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method moderate Working it out
The quick route is to see that Ignacio's 6 liters raise the level by 15 percentage points, which fixes the capacity at 40 liters; the wording does not point to it. (Schoenfeld, 1985)
Many steps moderate Working it out
After the capacity, the solution still needs the 60% amount, 24 liters, and a subtraction of Ito's 10 liters, so two unknowns are chased through four steps. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
15 sits one unit from the right 14, so a rough estimate of the starting amount cannot choose between them; only solving exactly, 24 − 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compare the two fill levels: the extra 6 liters account for the change in percent, which gives the capacity. Then use the 60% level to work back to the starting amount.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts the percents, 75 − 60 = 15, and gives a gap in percentage points as a number of liters.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: weak
Works back from the 30 liters at 75% full but adds Ignacio's 6 liters instead of removing them: 30 + 6 − 10 = 26.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the 40-liter capacity and its 75% level, 30 liters, then gives the amount after both additions instead of the starting amount n.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the full amount, here how much the container holds. The part is p percent of it. Percent means out of 100, so you divide p by 100 and then multiply.

Sort out what you know
The container starts with $n$ liters of water. Ito adds 10 liters, and it is 60% full. Then 6 more liters make it 75% full. You do not know how much the container holds when full. Call that its capacity. You need $n$ , the starting amount.
See what the 6 liters did
The extra 6 liters raised the level from 60% to 75%.
$75 - 60 = 15$
So 6 liters is 15% of the capacity.
Watch out: 15 is a percent, not a number of liters. Picking 15 gives choice B.
Find the full capacity
If 15% is 6 liters, then 1% is $6 \div 15 = 0.4$ liters. A full container is 100%.
$100 \times 0.4 = 40$
The container holds 40 liters when full.
Find the water at 60 percent
Use the percent formula with 60 for p and 40 liters for the whole.
$\text{part} = \frac{60}{100} \times 40 = 24$
After Ito adds his water, there are 24 liters.
Undo Ito’s 10 liters
Ito added 10 liters to the starting amount. To go back, subtract them.
$n = 24 - 10 = 14$
The container started with 14 liters.
Watch out: 75% of 40 is 30 liters. That is the amount after both people add water, not at the start. It is choice D.
Give the answer
The question asks for $n$ , the amount before anyone added water. So the answer is A, 14.

Check your answer

Run the story forward. Start with 14 liters. Add 10 to get 24 liters, and 24 is 60% of 40. Add 6 more to get 30 liters, and 30 is 75% of 40. Both facts match.

More questions like this

The original DOE question

Question 90 as printed on page 159 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 90. Open page 159 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 39 of 39

2018 Test B Math, question 45

Question

Phan chose an Internet service that charges $18.00 per month plus $0.024 per minute. Deion chose an Internet service that charges $30.00 per month for unlimited usage. At the end of the month, Phan’s and Deion’s charges were identical. For how many minutes did Phan use the Internet service that month?

A50
B60
C100
D500

Show the answer

Correct answer: D

Let x be the number of minutes Phan used his internet service in the month. Phan’s monthly charges were

18 + 0.024x

Since Deion’s charges were the same as Phan’s, set the expression equal to 30 and solve for x:

18 + 0.024x = 30

0.024x = 12

x = 500

Phan used his service for 500 minutes.

Question analysis

Phan pays $18 a month plus 2.4 cents per minute for Internet service, Deion pays a flat $30, and their bills come out the same. Students find how many minutes Phan used. It tests writing and solving a two-step linear equation, and the small decimal rate makes the division error-prone.

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving equations of the form px + q = r from a word problem is central grade 7 work. This is a typical example; only the three-decimal-place rate adds some care to the division.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Setting up the equation is routine, so the question is mostly a test of decimal division. Dividing 12 by 0.024 invites a misplaced decimal point, and one such slip gives a choice exactly ten times too small. A quick check, 500 × 0.024 = 12, confirms the answer in seconds.

Error-prone arithmetic moderate Working it out
The last step divides 12 by 0.024, a three-place decimal, and moving the decimal point one place too few gives 50 instead of 500. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
"Phan's and Deion's charges were identical" has to become 18 + 0.024m = 30, with the flat $30 on one side and Phan's fee plus per-minute cost on the other. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Set Phan's cost, the monthly fee plus the per-minute charge times the minutes, equal to Deion's flat fee, then subtract and divide, checking the decimal places.

Traps

An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Divides 12 by 0.024 but shifts the decimal point one place too few, as if the rate were 24 cents a minute: 12 ÷ 0.24 = 50.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No step with the $12 gap and the $0.024 rate gives 60; it most likely attracts students thinking of the 60 minutes in an hour.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No computation from the given numbers yields 100, and 100 minutes would add only $2.40 to Phan's $18 fee.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost with a monthly fee: $C = f + r \times m$; C is the total cost for the month. f is the fixed fee you pay no matter what. r is the price for each minute, and m is the number of minutes. Use it when a bill has a set fee plus a charge for each unit.

Sort out what you know
Phan pays $18.00 a month plus $0.024 for each minute. Deion pays $30.00 a month, however much he uses. Their bills were the same. You need the number of minutes Phan used. Call it $m$ .
Write an equation
Use the cost formula for Phan. His bill equals Deion’s $30.
$18 + 0.024 \times m = 30$
An equation says two amounts are equal. Here it says the two bills are equal.
Take away the monthly fee
Subtract 18 from both sides. That leaves the minutes part alone.
$0.024 \times m = 30 - 18 = 12$
So Phan paid $12 for his minutes.
Divide to find the minutes
Divide 12 by 0.024. To get rid of the decimal, move the point 3 places right in both numbers.
$m = 12 \div 0.024 = 12{,}000 \div 24 = 500$
Phan used 500 minutes.
Watch out: 0.024 is not 0.24. Using 0.24 gives $12 \div 0.24 = 50$ , choice A. Move the point the same number of places in both numbers.
Give the answer
Phan used 500 minutes that month. That is the only choice that makes the two bills equal. So the answer is D, 500.

Check your answer

Put 500 back in. The minutes cost $500 \times 0.024 = 12$ dollars. Add the fee: $18 + 12 = 30$ dollars. That is Deion’s bill, so the bills match.

More questions like this

The original DOE question

Question 102 as printed on page 161 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 102. Open page 161 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

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