Practice Test

Linear equations

Question 1 of 43

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 43

2026 Practice Test A Math, question 32

Question

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

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

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

x =

Show the answer

Correct answer: x = -13/3

Solution:

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

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

$6x - 9x = 13$

Combine like terms.

$-3x = 13$

Divide both sides by −3 to solve for x.

$x = -\frac{13}{3}$

$x = -4\frac{1}{3}$

Question analysis

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

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

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
Collecting the x terms gives 6x − 9x = −3x, a negative coefficient, and 13 ÷ (−3) is a negative fraction that does not come out whole, so both the sign and the fraction must survive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The answer is typed after the printed x =, with no choices to compare, and its decimal form, −4.333…, repeats forever, so only a fraction or mixed number is exact. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

13/3 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches −3x = 13 correctly but drops the negative when dividing 13 by −3, entering a positive 13/3.
-4.33 Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Solves correctly but turns −13/3 into a decimal and rounds it to −4.33, which is close to the exact value but not equal to it.
-3/13 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Gets to −3x = 13 but divides the wrong way round, putting the coefficient −3 over the constant 13 instead of 13 over −3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You need the value of x that makes $6x = 9x + 13$ true. The letter x appears on both sides, so first gather the x terms on one side. There are no choices to check against, so work carefully.
Move the x terms to one side
Subtract 9x from both sides. This clears the x term off the right side:
$6x - 9x = 9x + 13 - 9x$
$-3x = 13$
Since 6 − 9 = −3, the left side is −3 times x.
Divide both sides by −3
x is multiplied by −3, so divide both sides by −3:
$x = \frac{13}{-3} = -\frac{13}{3}$
A positive divided by a negative is negative, so x is negative.
Watch out: dropping the minus sign gives 13/3, which is wrong. Also, the number you divide by goes on the bottom. Writing −3/13 turns the fraction upside down.
Write it as a mixed number
A mixed number is a whole number with a fraction, like 4 1/3. Divide 13 by 3. The 3 fits 4 times, with 1 left over.
$\frac{13}{3} = 4\frac{1}{3}$
So $x = -4\frac{1}{3}$ .
Watch out: as a decimal, −13/3 is −4.333…, and the 3s never end. Rounding to −4.33 is close, but it is not exact. Keep the fraction.
Enter the answer
The box already shows x =, so you enter only the number. On the test, press the minus key first. Then press the fraction key, a small box over a small box. Type 13 in the top box and 3 in the bottom box. On this site, type −13/3 or −4 1/3.
So the finished answer reads x = −13/3. You type only the −13/3, which is the same as −4 1/3.

Check your answer

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

More questions like this

The original DOE question

Question 32 as shown on the DOE’s digital test, page 103 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 103, question 32.

Question 3 of 43

2026 Practice Test B Math, question 46

Question

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

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

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

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

Show the answer

Correct answer: y = 20x + 50

Solution: $y = 20x + 50$

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

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

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

Question analysis

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

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.9 dependent and independent variables
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an equation for a fixed fee plus a steady monthly charge is a familiar grade 7 task, and the numbers are whole. Typing the full equation, with both variables and no choices to compare, makes it less forgiving.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 51 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
Only the words “onetime setup fee” and “per month” tell which amount is multiplied by x and which is added once, and the $50 comes first in the sentence. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The whole equation, with y, an equals sign and x, is built on a keypad with no choices, so a reversed or incomplete equation looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

y = 20x Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Multiplies the $20 monthly charge by x months but never adds the one-time $50 setup fee, so the equation gives only the cost of service.
y = 50x + 20 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Attaches x to the $50 fee and adds the $20 once, putting the numbers into y = mx + b in the order they appear rather than by which one repeats.
y = 70x Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the $50 fee and the $20 charge into $70 and multiplies by the months, which would charge the setup fee again every month.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Amiya pays two kinds of charges:
- a setup fee of $50, paid one time only;
- a charge of $20 for every month of service.
y is her total cost in dollars, and x is the number of months. You need an equation that links them. The answer is an equation, not a number.
Work out a few totals
Find the total for a few months by hand. Each month adds one more $20:
Months Monthly charges ($) Total ($)
1 $1 \times 20 = 20$ $50 + 20 = 70$
2 $2 \times 20 = 40$ $50 + 40 = 90$
3 $3 \times 20 = 60$ $50 + 60 = 110$
The $50 shows up once in every total. The $20 shows up once for each month.
Match each number to its job
Use the fee-and-rate model, $y = (\text{rate})x + (\text{fee})$ . The rate is the amount that repeats each month, $20. It multiplies the months, so it becomes 20x. The fee, $50, is paid once, so it is added once.
Watch out: the $50 comes first in the question, but it is not the rate. $y = 50x + 20$ charges $50 a month. $y = 70x$ charges the fee every month. $y = 20x$ forgets the fee.
Write the equation
Put the rate and the fee into the model:
$y = 20x + 50$
Here 20x means 20 times x. An equal form is fine too, such as $y = 50 + 20x$ , because you can add in either order.
Enter it on the keypad
Type the whole equation, including y =. Without it, you have an expression, not an equation. On the test, the keypad has x, y and = keys. Press y, =, 2, 0, x, +, 5, 0. You do not need a times sign or dollar signs. On this site, type it the same way.
So you would enter y = 20x + 50.

Months	Monthly charges ($)	Total ($)
1	$1 \times 20 = 20$	$50 + 20 = 70$
2	$2 \times 20 = 40$	$50 + 40 = 90$
3	$3 \times 20 = 60$	$50 + 60 = 110$

Check your answer

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

More questions like this

The original DOE question

Question 46 as shown on the DOE’s digital test, page 118 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 118, question 46.

Question 4 of 43

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 5 of 43

2026 Practice Test C Math, question 3

Question

$6x = x - 1{,}680$

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

A336
B240
C−240
D−336

Show the answer

Correct answer: D

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

$6x - x = x - x - 1{,}680$

$5x = -1{,}680$

$\frac{5x}{5} = \frac{-1{,}680}{5}$

$x = -336$

Question analysis

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

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

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
Subtracting x from both sides leaves 5x = −1,680, and 1,680 ÷ 5 = 336 has to be worked out by hand while the negative sign is carried through to the result. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
The choices are two sizes, 336 and 240, each offered with both signs, so 336 and −336 are written the same except for the minus; only tracking the sign of 1,680 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches 5x = −1,680 correctly but drops the negative sign when dividing by 5, giving a positive 336 of the right size.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves x to the left without changing its sign, getting 7x, and also loses the minus on 1,680, so 1,680 ÷ 7 = 240.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Moves x to the left side without changing its sign, writing 6x + x = −1,680, so divides by 7 and gets −240.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The equation is $6x = x - 1{,}680$ . You need the value of x. Here 6x means 6 times x. The unknown x is on both sides. So first gather it on one side.
Subtract x from both sides
Take x away from both sides. On the right, x − x is 0, so only −1,680 is left:
$6x - x = x - x - 1{,}680$
$5x = -1{,}680$
Six x’s take away one x leaves five x’s.
Watch out: moving x to the left without changing its sign gives 7x. Dividing by 7 then leads to choice C, −240, or B, 240.
Divide both sides by 5
5x means 5 times x. Divide both sides by 5 to undo it:
$\frac{5x}{5} = \frac{-1{,}680}{5}$
Find the size first. Split 1,680 into 1,500 + 180:
$1{,}500 \div 5 + 180 \div 5 = 300 + 36 = 336$
A negative divided by a positive is negative:
$x = -336$
Choose the answer
Watch out: the minus sign is easy to drop. Positive 336 is choice A. Carry the sign of −1,680 all the way to the end.
So the answer is D, −336.

Check your answer

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

More questions like this

The original DOE question

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

Question 6 of 43

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 7 of 43

2026 Practice Test C Math, question 35

Question

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

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

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

a =

Show the answer

Correct answer: a = 10b

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

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

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

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

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

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

$2a = 20b$

$a = 10b$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 8 of 43

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 9 of 43

2026 Practice Test D Math, question 2

Question

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

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

Show the answer

Correct answer: D

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

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

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

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

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

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

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

$x + 2y - 2x = z$

$2y - x = z$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 10 of 43

2026 Practice Test D Math, question 3

Question

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

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

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

What is the value of $x - y$ ?

A−7
B−5
C5
D7

Show the answer

Correct answer: B

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

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

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

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

$x + 16 = y + 11$

$x = y - 5$

$x - y = -5$

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 11 of 43

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 12 of 43

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 13 of 43

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 14 of 43

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 15 of 43

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 16 of 43

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 17 of 43

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 18 of 43

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 19 of 43

2024 Test B Math, question 8

Question

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

Show the answer

Correct answer: C

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

$9\bullet 8=72$

$2x\bullet 3y=6xy$

$72=6xy$

$12=xy$

Question analysis

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

NYS standards: 8.EE.7 linear equations in one variable
6.EE.2b parts of an expression
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students cross-multiply numeric proportions to find one unknown. Here two variables sit in the proportion and the question wants their product, a literal-equation idea from grade 8 that many grade 7 students have not met.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question looks harder than it is, because it asks for a product rather than for x or y. Cross-multiplying turns the proportion into 72 = 6xy, and dividing by 6 finishes it, so the only real hurdle is trusting that xy can be solved for as one quantity. The wrong choices come from grabbing numbers along the way, not from a nearly complete solution.

No obvious method moderate Working it out
Neither x nor y can be found alone, so students must see that cross-multiplying gives 72 = 6xy, where the product xy appears as one block to solve for. (Schoenfeld, 1985)
Variables instead of numbers minor Understanding the question
Both unknowns stay letters and only their product is asked for, so there is no value of x or y to check a result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 20 of 43

2024 Test B Math, question 13

Question

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

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

Show the answer

Correct answer: B

First substitute 2 for r in the equation.

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

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

Then solve for p.

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

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

$p+4=2p+3$

$p+(4-3)=2p+(3-3)$

$p+1=2p$

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

$1=p$

The value of p is 1.

Question analysis

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

NYS standards: 8.EE.7b linear equations with rational coefficients
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 Equations with the variable on both sides are formally grade 8, but after substituting, this is a short solve with small whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Adds the outer 1 inside the parentheses before multiplying, writing 2(p + 2) = 2p + 4, so p + 4 = 2p + 4 gives 0.
C Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Multiplies only the p by 2, writing 2p + 1 + 1 = 2p + 2, so p + 4 = 2p + 2 and p comes out as 2.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Drops the 1 inside the parentheses, reading 2(p + 1) as 2p, so the right side is 2p + 1 and p + 4 = 2p + 1 gives 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. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c$; a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. Adding the same thing, or dividing both sides by the same number that is not zero, works too.

Sort out the question
The equation has two letters, p and r. You are told that r = 2. You need the value of p. Here 2r means 2 times r, and r(p + 1) means r times the group p + 1.
Put 2 in for r
Replace every r with 2. Keep the parentheses so you can see what gets multiplied.
$p + 2(2) = 2(p + 1) + 1$
On the left, $2(2) = 4$ , so the left side is $p + 4$ .
Multiply out the parentheses
The 2 multiplies both the p and the 1 inside the parentheses. Then add the 1 that is outside.
$a(b + c) = a \times b + a \times c$
$2(p + 1) + 1 = 2p + 2 + 1 = 2p + 3$
Watch out: multiply the 1 inside by 2 as well. Writing 2p + 1 + 1 gives p = 2, choice C. Adding the outside 1 inside first gives p = 0, choice A.
Get p by itself
Now the equation is $p + 4 = 2p + 3$ . Subtract p from both sides. Then subtract 3 from both sides.
$\text{if } a = b \text{, then } a - c = b - c$
$p + 4 - p = 2p + 3 - p$
$4 = p + 3$
$4 - 3 = p + 3 - 3$
$1 = p$
So p = 1, and the answer is B, 1.

Check your answer

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

More questions like this

The original DOE question

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

Question 21 of 43

2024 Test B Math, question 55

Question

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

A2,000
B2,500
C3,000
D4,285

Show the answer

Correct answer: A

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

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

$30{,}000=5x$

$6{,}000=x$

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

$6{,}000=3y$

$2{,}000=y$

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

Question analysis

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

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving a one-variable equation from a word problem is grade 7 work. Two linked multiplicative comparisons, with Saturday measured against Sunday rather than a weekday, add a layer of translation, but the equation itself stays simple.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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 12, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is two comparisons chained through Sunday, and most wrong answers come from linking them incorrectly. Students who read Saturday's 4 times as 4 weekdays, or who leave Sunday out of the 30,000, find listed answers, one of them from a division that does not come out even. Letting a weekday be one share, so Sunday is 3 and Saturday 12, makes the weekend 15 shares and the check easy.

Turning words into math moderate Understanding the question
"4 times as many tickets on Saturday as on Sunday" compares Saturday with Sunday, not with a weekday, so Saturday is 4 × 3 = 12 weekdays' worth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Three days have to be held in relation at once, a weekday, Sunday and Saturday, with the 30,000 covering only the last two. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Makes Saturday 3 × 4 = 12 weekdays' worth but divides all 30,000 by 12, leaving Sunday's 3 shares out of the weekend total.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Adds the multipliers, so Saturday is 3 + 4 = 7 weekdays' worth; with Sunday's 3 the weekend is 10 shares, and 30,000 ÷ 10 = 3,000.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Reads Saturday as 4 times a weekday rather than 4 times Sunday, so splits 30,000 into 3 + 4 = 7 shares of about 4,285.7.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Sunday sells 3 times as many tickets as a weekday. Saturday sells 4 times as many as Sunday. Saturday and Sunday together sell 30,000.
Monday is a weekday. You need Monday’s tickets. Notice that Saturday is compared with Sunday, not with a weekday.
Write an equation for the weekend
Let x be Sunday’s tickets. Then Saturday’s tickets are 4x. Together they make 30,000.
$4x + x = 30{,}000$
$5x = 30{,}000$
Solve for Sunday
Use the rule for undoing multiplication. Divide both sides by 5.
$x = 30{,}000 \div 5 = 6{,}000$
Sunday sells 6,000 tickets. Saturday sells $4 \times 6{,}000 = 24{,}000$ .
Find Monday’s tickets
Sunday sells 3 times as many as a weekday. Let y be Monday’s tickets.
$3 \times y = 6{,}000 \text{ means } y = 6{,}000 \div 3 = 2{,}000$
Watch out: If you read Saturday as 4 times a weekday, you split 30,000 into 7 shares. That gives about 4,285, choice D. Dividing 30,000 by 12 leaves out Sunday and gives 2,500, choice B.
So the answer is A, 2,000.

Check your answer

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

More questions like this

The original DOE question

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

Question 22 of 43

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 23 of 43

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

Question 24 of 43

2023 Test A Math, question 22

Question

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

Show the answer

Correct answer: C

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

Option A:

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

Option B:

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

Option C:

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

Option D:

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

Question 25 of 43

2023 Test A Math, question 35

Question

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

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

A−2
B0
C2
D4

Show the answer

Correct answer: B

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

$x = \frac{2}{3}(10 - x)$

Solve for x:

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

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 2 in 2/3 as the length of JL and moves 2 units right from J at −4, without relating JL to LK or to JK.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Turns the ratio upside down, making LK two thirds of JL, so JL is 6 and L lands at 2, nearer K than J.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds JL = 4 correctly but gives that length as L's position, stopping before counting the 4 units from J at −4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is $\frac{2}{3}$ of LK. You need the position of L.
Find the length of JK
$\text{distance} = \text{right point} - \text{left point}$
K is on the right, so subtract J’s position from K’s.
$\mathrm{JK} = 6 - (-4) = 6 + 4 = 10$
Taking away −4 is the same as adding 4. J and K are 10 units apart.
Turn the rule into equal parts
JL is $\frac{2}{3}$ of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.
$\text{one part} = \frac{\text{whole length}}{a + b}$
$\text{one part} = \frac{10}{2 + 3} = 2$
Each part is 2 units long.
Find the two lengths
$\mathrm{JL} = 2 \times 2 = 4$
$\mathrm{LK} = 3 \times 2 = 6$
Watch out: JL is $\frac{2}{3}$ of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.
Find the position of L
L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.
$-4 + 4 = 0$
Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 43

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 27 of 43

2023 Test A Math, question 44

Question

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

A $\frac{75}{11}$
B $\frac{75}{31}$
C $\frac{15}{7}$
D $\frac{65}{31}$

Show the answer

Correct answer: B

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

Question analysis

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

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 equations have the form px + q = r. This one has x on both sides, a fractional coefficient and a product to expand, which are grade 8 equation skills, and every wrong choice matches a plausible slip along the way.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps moderate
A very close wrong answer minor
Solution: 5 steps
Reading level: Question and choices: about grade 8, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
The term x/5 sits beside whole-number terms and a negative −6x, so clearing the fraction means multiplying every term by 5, and the answer is the unfriendly 75/31. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps moderate Working it out
Five moves in a row, distributing the 3, combining 12 and −1, collecting the x terms, clearing the fifth and dividing by 31, each give a slip a place to enter. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The fraction 75/11 shares its numerator with 75/31 and differs by one digit of the denominator, so only the coefficient of x, 6 1/5 rather than 2 1/5, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Distributes the 3 only to the 4, writing 12 − 2x, so the equation becomes x/5 + 2x = 15 and then 11x = 75.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Multiplies only x/5 by 5 to clear the fraction, leaving 6x and 15 unchanged, so it solves x + 6x = 15 and gets 7x = 15.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Applies the 3 to the −1 as well, as if it sat inside the parentheses, so the right side becomes 9 − 6x and the constant 13 instead of 15.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The distributive property: $a(b - c) = ab - ac$; a, b and c are numbers or terms. Multiplying a by a difference means multiplying a by each part. The number outside must reach every term inside the parentheses.
Doing the same to both sides: $\text{if } a = b\text{, then } a + c = b + c \text{ and } a \times c = b \times c$; a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. Adding the same amount to both sides, or multiplying both sides by the same number, keeps it balanced.

Sort out the question
You need the value of x that makes both sides of the equation equal. The plan: clear the parentheses, gather the x terms on one side, and get rid of the fraction. Then divide.
Clear the parentheses
$a(b - c) = ab - ac$
$3(4 - 2x) = 12 - 6x$
$\frac{x}{5} - 4 = 12 - 6x - 1 = 11 - 6x$
Watch out: the 3 multiplies both 4 and 2x. Writing 12 − 2x leads to $\frac{75}{11}$ , choice A. The −1 is outside the parentheses, so the 3 does not touch it.
Gather the x terms
Add 6x to both sides to move the x terms to the left. Add 4 to both sides to move the plain numbers to the right.
$\text{if } a = b\text{, then } a + c = b + c \text{ and } a \times c = b \times c$
$\frac{x}{5} + 6x = 11 + 4 = 15$
Clear the fraction
Multiply both sides by 5. Every term gets multiplied, not only the fraction.
$5 \times \frac{x}{5} + 5 \times 6x = 5 \times 15$
$x + 30x = 75$
$31x = 75$
Watch out: multiplying only the fraction by 5 gives x + 6x = 15. That leads to $\frac{15}{7}$ , choice C.
Divide by 31
$x = \frac{75}{31}$
75 and 31 share no factor except 1, so the fraction is already simplified. So the answer is B, $\frac{75}{31}$ .

Check your answer

Put $x = \frac{75}{31}$ back in. Then $\frac{x}{5} = \frac{15}{31}$ and $2x = \frac{150}{31}$ . Write 4 as $\frac{124}{31}$ . The left side is $\frac{15}{31} - \frac{124}{31} = -\frac{109}{31}$ . On the right, $4 - \frac{150}{31} = -\frac{26}{31}$ , and $3 \times \left(-\frac{26}{31}\right) - 1 = -\frac{78}{31} - \frac{31}{31} = -\frac{109}{31}$ . Both sides match.

More questions like this

The original DOE question

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

Question 28 of 43

2023 Test B Math, question 5

Question

Solve the equation for x:

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

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

Show the answer

Correct answer: 7

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

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

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

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

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

Question analysis

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

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.2c properties for multiplying and dividing
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
No obvious method minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 6 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
No obvious method minor Working it out
The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)
An answer in an unexpected form minor Choosing the answer
The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

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

Traps

Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

8 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Simplifies the coefficient correctly to 1/8 but enters the 8 from its denominator, never multiplying it by 0.875.
0.109375 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 0.875 by 8 instead of multiplying, undoing the coefficient 1/8 the wrong way, so x comes out as 0.109375.
3.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Slips in reducing 14/48, taking it as 7/12 instead of 7/24, so the coefficient becomes 1/4 and x = 0.875 × 4 = 3.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

Sort out the question
You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.
Multiplying 49 by 48 makes a big number. It is easier to simplify first.
Split the fraction into two
Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.
$\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}$
Now the equation says these two fractions times x equal 0.875.
Simplify each fraction
To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.
$\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}$
Watch out: 48 ÷ 2 is 24, not 12. Writing $\frac{7}{12}$ leads to 3.5.
Multiply the simple fractions
$\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}$
The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get $\frac{1}{8}$ . So the equation becomes:
$\frac{1}{8} \times x = 0.875$
Solve for x
One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.
$x = 0.875 \times 8 = 7$
Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

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

More questions like this

The original DOE question

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

Question 29 of 43

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 30 of 43

2023 Test B Math, question 17

Question

$m \bullet t = 72$

For the equation above, what is the sum of the values of m when $t = 1$ , when $t = 2$ , and when $t = 3$ ?

A12
B72
C132
D135

Show the answer

Correct answer: C

First, find the value of m for each value of t:

For $t = 1$ :

$1m = 72$

$m = 72$

For $t = 2$ :

$2m = 72$

$m = 36$

For $t = 3$ :

$3m = 72$

$m = 24$

Then, add those three values together:

$72 + 36 + 24 = 132$

Question analysis

The equation m • t = 72 links two variables, and the raised dot means multiplication. Students find m when t is 1, 2 and 3, and then add those three values of m. Each value is a one-step division, and the trap is combining the t values before dividing.

NYS standards: 6.EE.7 one-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Solving equations like 3m = 72 is grade 6 work. This item repeats that step three times and adds the results, more bookkeeping than difficulty, so it is below typical grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each value of m comes from one division, so the work is bookkeeping rather than algebra. The trap is structural: adding 1, 2 and 3 first and dividing once feels efficient but answers a different question, since m does not change in step with t. Writing the three values of m in a row before adding them keeps the steps separate.

Turning words into math minor Understanding the question
“The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Divide 72 by each value of t in turn to get three separate values of m, then add the three quotients.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds the three t values to get 6 and divides 72 by 6, as if the sum of the m values were m at the sum of the t values.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives back 72, the constant in the equation and the value of m only when t = 1, without the values for t = 2 and t = 3.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 135, which is 3 more than the right sum; it would need a slip such as 72 ÷ 3 = 27, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division undoes multiplication: $\text{if } a \times b = c \text{, then } a = c \div b$; a, b and c are numbers, and b is not 0. If you know a product and one of the numbers that made it, divide to find the other one. For example, 4 × 5 = 20, so 4 = 20 ÷ 5.

Sort out the question
The equation is $m \bullet t = 72$ . The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.
Get m by itself
Since m times t is 72, m is 72 divided by t.
$m = 72 \div t$
Now put in each value of t, one at a time.
Find m for each value of t
Each time, divide 72 by the value of t. A table keeps the three answers in order.
Value of t Value of m
1 $m = 72 \div 1 = 72$
2 $m = 72 \div 2 = 36$
3 $m = 72 \div 3 = 24$
Add the three values
$72 + 36 + 24 = 132$
Watch out: don’t add the t values first. $72 \div (1 + 2 + 3) = 12$ is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.

Value of t	Value of m
1	$m = 72 \div 1 = 72$
2	$m = 72 \div 2 = 36$
3	$m = 72 \div 3 = 24$

Check your answer

Check each pair by multiplying: $72 \times 1 = 72$ , $36 \times 2 = 72$ and $24 \times 3 = 72$ . All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.

More questions like this

The original DOE question

Question 74 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 74. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 31 of 43

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 32 of 43

2023 Test B Math, question 57

Question

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

Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

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

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

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

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

$80+5w=120+3w$

$2w=40$

$w=20$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 33 of 43

2018 Test A Math, question 3

Question

For what value of $w$ is $4w=2w-8$ ?

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

Show the answer

Correct answer: -4

$4w=2w-8$

$2w=-8$

$w=-4$

Question analysis

A one-line algebra grid-in: find the value of w that makes 4w equal to 2w minus 8. Students solve a linear equation with the variable on both sides by collecting the w terms and dividing. Errors come from handling the subtraction sign or stopping before w stands alone.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.NS.2b rational number division
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 Equations with the variable on both sides are formally a grade 8 topic, but this one takes two short steps with whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 7 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One of the easiest questions on the test: collect the w terms on one side and divide. Errors come from carelessness rather than method, such as dropping the negative sign or gridding −8 after the first step because it already looks like a finished answer. Substituting the result back into both sides of the equation takes seconds and catches both.

Error-prone arithmetic minor Working it out
The only constant is −8, so the answer is negative; dividing −8 by the coefficient 2 is the step where the sign is most often lost. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The grid-in needs the negative sign filled in its own column, and there are no choices to show that the answer lies below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract 2w from both sides so the variable appears only once, divide by the remaining coefficient, and check by substituting into both sides.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

4 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Gets 2w = −8 correctly but drops the sign when dividing by 2, gridding 4 instead of a negative value.
-8 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Subtracts 2w from both sides to reach 2w = −8 and grids −8 without dividing by the coefficient 2.

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

Sort out the equation
An equation says two things are equal. You need the value of w that makes $4w$ equal to $2w - 8$ . Here 4w means 4 times w, and 2w means 2 times w. The answer may be a negative number, so keep track of the signs.
Get w on one side
w is on both sides. Subtract 2w from both sides. Then w is only on the left, and the equation stays balanced.
$4w - 2w = 2w - 8 - 2w$
$2w = -8$
Now 2 times w is −8.
Divide both sides by 2
To undo times 2, divide both sides by 2. A negative number divided by a positive number is negative.
$w = -8 \div 2 = -4$
Watch out: keep the minus sign. Dropping it gives 4. Also, don’t stop at $2w = -8$ and grid −8. So you would grid -4, with the minus sign in its own column.

Check your answer

Put −4 back in for w. Left side: $4 \times (-4) = -16$ . Right side: $2 \times (-4) - 8 = -8 - 8 = -16$ . Both sides are −16, so −4 works.

More questions like this

The original DOE question

Question 60 as printed on page 81 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 60. Open page 81 of the handbook (PDF)

Question 34 of 43

2018 Test A Math, question 6

Question

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

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

Show the answer

Correct answer: B

$2k=m+3$

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

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

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 35 of 43

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 43

2018 Test A Math, question 43

Question

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

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

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

Show the answer

Correct answer: D

Solve the equation for $z$ :

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

$wz=xy$

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

Question analysis

This algebra question gives the proportion w/x = y/z, with all four letters positive, and asks which expression equals z. Students must solve the equation for one letter in terms of the others. With only letters on the page, it is easy to multiply when you should divide or to stop one step early.

NYS standards: 8.EE.7 linear equations in one variable
6.EE.7 one-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students cross-multiply numeric proportions such as 3/4 = 9/x. Solving a proportion made only of letters for one of them is a literal equation, a grade 8 algebra skill, though it takes just two steps.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 10, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method, cross-multiply and divide, is the same one students use on numeric proportions, but letters hide whether each step went the right way. The two most attractive wrong choices are both one step from the answer: one never divides, and the other divides upside down. A quick check with 2/3 = 4/6, where z = 6, settles it.

Variables instead of numbers moderate Understanding the question
The proportion has only letters, so cross-multiplying to wz = xy and dividing by w must be done without numbers to show whether w goes on top or bottom. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
The choices w/(xy) and xy/w use the same three letters with the fraction turned over, so only dividing in the right direction after wz = xy separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cross-multiply to clear both fractions, then divide both sides by whatever multiplies z; check with a true proportion such as 2/3 = 4/6.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a, b, c and d are numbers that are not 0. When two fractions are equal, each top times the other bottom gives the same answer. It works because you multiply both sides by both bottoms, which clears the fractions.

Sort out what you need
The equation says w over x equals y over z. All four letters are positive numbers. You need to get z alone on one side. Then the other side shows what z equals.
Cross-multiply the fractions
Multiply each top by the other fraction’s bottom:
$w \times z = x \times y$
So wz = xy, and z is no longer on the bottom. Watch out: stopping here and picking xy gives choice B. z is not alone yet.
Get z alone
z is being multiplied by w. To undo that, divide both sides by w. Doing the same to both sides keeps them equal:
$\frac{w \times z}{w} = \frac{x \times y}{w}$
$z = \frac{xy}{w}$
Watch out: dividing the wrong way round gives w over xy, choice C. So the answer is D, $\frac{xy}{w}$ .

Check your answer

Use a true pair of equal fractions: 2/3 = 4/6. So w = 2, x = 3, y = 4 and z = 6. Choice D gives 3 × 4 ÷ 2 = 6, which is z. Choice C gives 2 ÷ 12, which is not 6.

More questions like this

The original DOE question

Question 100 as printed on page 89 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 100. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 37 of 43

2018 Test A Math, question 55

Question

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

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

Show the answer

Correct answer: C

$4x-3y=12$

$4x=12+3y$

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

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

Question analysis

This algebra question gives the linear equation 4x − 3y = 12 and asks for x in terms of y. Students must isolate the x term and then divide to solve for x. With two variables and a negative coefficient, small sign and division slips each lead to a different wrong choice.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve one-variable equations like 4x − 3 = 12. Rearranging an equation with two variables to express one in terms of the other is grade 8 algebra, with sign and division slips to avoid.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The solution is two moves, add 3y to both sides and divide everything by 4, yet each move has a classic slip attached. The four choices are the four combinations of getting the sign right or wrong and dividing the 12 or not, so every partial mistake has a home. Checking one point that satisfies the original equation catches any of them.

Variables instead of numbers moderate Understanding the question
With two variables there is no number to check against unless students invent one; the point x = 6, y = 4 satisfies the equation and separates the choices. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
The choices x = 3/4 y + 3 and x = −3/4 y + 3 are identical except for the sign of the y term, which only correctly moving −3y across the equals sign settles. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Dividing 12 + 3y by 4 must reach both terms, and the negative coefficient −3 has to change sign when it moves, two slips in a short line. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Move the y term to the right side, watching its sign, so only the x term remains on the left, then divide the entire right side, term by term, by 4.

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $a = b \text{ means } a + c = b + c$; a and b are the two sides of an equation, and c is any number. An equation is like a balance. If you do the same thing to both sides, such as adding 3y or dividing by 4, they stay equal.
Dividing a sum: $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$; a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. It works because sharing a + b into c groups shares a and b each into c groups.

Sort out what you need
The question asks for x in terms of y. That means get x alone on one side. The other side may have y and numbers, but no x. Start with $4x - 3y = 12$ .
Move the y term
The 3y is being subtracted on the left. Add 3y to both sides to clear it:
$4x - 3y + 3y = 12 + 3y$
$4x = 12 + 3y$
Watch out: the term was minus 3y. Adding 3y to both sides makes it plus 3y on the right. Keeping the minus sign leads to choice D.
Divide both sides by 4
x is being multiplied by 4. Divide both sides by 4 to undo it:
$x = \frac{12 + 3y}{4}$
Split the fraction
Divide each part of the top by 4:
$x = \frac{12}{4} + \frac{3y}{4} = 3 + \frac{3}{4}y$
Watch out: the 12 must be divided by 4 too. Leaving it as 12 gives choice A. The choices put the y term first. So the answer is C, $x = \frac{3}{4}y + 3$ .

Check your answer

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

More questions like this

The original DOE question

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

Question 38 of 43

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 39 of 43

2018 Test B Math, question 8

Question

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

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

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

Show the answer

Correct answer: D

Break the equations apart to each equal $M$ :

$M=3N$

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

$M=Q+5$

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

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

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

$M=3N$

$2=3N$

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

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

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

$8=P$

$M=Q+5$

$2=Q+5$

$-3=Q$

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

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

$14=R$

Variable $R$ has the greatest value.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 40 of 43

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 41 of 43

2018 Test B Math, question 20

Question

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

Show the answer

Correct answer: A

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

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

$\frac{36}{3}=4x$

$12=4x$

$3=x$

Question analysis

This equation question gives 36 divided by y equal to 4x and asks for the value of x at one given value of y. Students substitute, simplify the fraction and solve a one-step equation. It is short, so the risk lies in stopping one step early or dividing by the wrong number.

NYS standards: 6.EE.7 one-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Substituting a value and solving a one-step equation are grade 6 skills. With two quick steps and whole numbers, this is lighter than the multistep equations typical of grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is substitution followed by a one-step equation, with small whole numbers throughout, so most students answer correctly. The only real risk is stopping at 12, which is the value of 4x rather than of x. Writing 12 = 4x as its own line before reading the choices makes the last division hard to skip.

Variables instead of numbers minor Understanding the question
The variable y sits in a denominator, 36/y, which looks less familiar than an equation in one unknown, though substituting y = 3 removes it at once. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Replace y with its value, simplify the left side to a single number, and then divide by the coefficient of x.

Traps

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

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks the coefficient 4 from 4x, a number that appears in the equation, without solving for x.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Clears the fraction to get 36 = 4xy, divides by 4 to reach xy = 9, and stops before dividing by y.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Substitutes y = 3 to get 36/3 = 12, which is the value of 4x, and stops before dividing by 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing multiplication: $a \times x = b \text{ means } x = b \div a$; a is the number multiplied by x, and b is the result. Division undoes multiplication. So to get x alone, divide both sides by a.

Write down what you know
The equation is $\frac{36}{y} = 4x$ . You are told y = 3. You need the value of x.
The letter y is on the bottom of a fraction. Putting in its value turns the fraction into a plain number.
Put in the value of y
Replace y with 3.
$\frac{36}{3} = 4x$
The fraction means 36 divided by 3, which is 12.
$12 = 4x$
Watch out: 12 is the value of 4x, not x. Stopping here gives choice D.
Divide both sides by 4
4x means 4 times x. To undo multiplying by 4, divide both sides by 4.
$x = 12 \div 4 = 3$
So the answer is A, 3.

Check your answer

Put x = 3 and y = 3 back in. The left side is $36 \div 3 = 12$ . The right side is $4 \times 3 = 12$ . Both sides match, so x = 3 is right.

More questions like this

The original DOE question

Question 77 as printed on page 157 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 77. Open page 157 of the handbook (PDF)

Question 42 of 43

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).

Question 43 of 43

2018 Test B Math, question 56

Question

There are now $x$ cans stacked on a shelf that holds 36 cans when full. If 4 of these cans were removed, the shelf would be half full. What is the value of $x$ ?

Show the answer

Correct answer: D

The shelf, when full, holds 36 cans. When the shelf is half full, it holds 18 cans.

x − 4 = 18

x = 22

Question analysis

A shelf holds 36 cans when full, and removing 4 of the cans now on it would leave it half full. Students find how many cans are on the shelf now. It tests writing and solving a one-step equation from words, and the risk is undoing the removal in the wrong direction.

NYS standards: 6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 6 students solve equations like x − p = q, and this one needs only that and halving 36. It is a quick, routine word problem, easier than the two-step equations typical of grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
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: 2 steps
Reading level: Question and choices: about grade 3, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Short and routine, this question still punishes a quick reading. Students must keep three amounts apart, the 36-can capacity, the 18 cans on a half-full shelf and the x cans there now, and then undo the removal by adding. Writing x − 4 = 18 before computing prevents every wrong choice.

Turning words into math minor Understanding the question
The 36 cans are the shelf's capacity, not its contents, and "if 4 of these cans were removed" describes a hypothetical, so the equation is x − 4 = 18. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
18 sits just below the right 22, and both are near half of 36; only remembering that x counts the cans before the 4 were removed separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out how many cans make a half-full shelf, then undo the removal of the 4 cans; write it as an equation in x if that helps.

Traps

Stops one step early: A choice is a number from partway through the solution.
Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.

The wrong answers, classified

A Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Halves 36 to get 18 and then subtracts the 4 removed cans, following the word removed, instead of adding them back to undo it.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Reads the 36 cans as what is on the shelf now, removes 4 to get 32, and halves that to 16.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 18 cans on a half-full shelf and stops, forgetting that x counts the cans before the 4 were removed.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing subtraction: $x - a = b \text{ means } x = b + a$; x is the number you want. a is the amount taken away, and b is what is left. To undo taking away a, add a back. Adding the same number to both sides keeps the equation balanced.

Sort out what you know
The shelf holds 36 cans when it is full. That is its size, not how many cans are on it now. Right now it has $x$ cans. If 4 cans were taken off, the shelf would be half full. You need $x$ .
Find half of a full shelf
A full shelf holds 36 cans, so half full is half of that.
$36 \div 2 = 18$
A half-full shelf has 18 cans.
Write an equation
Taking 4 cans away from $x$ leaves 18 cans.
$x - 4 = 18$
Watch out: 18 is the number after the cans are taken off. Picking 18 gives choice C.
Add the 4 cans back
Undo the subtraction by adding 4 to both sides.
$x = 18 + 4 = 22$
Watch out: subtracting the 4 again gives $18 - 4 = 14$ , choice A. The cans were taken from $x$ , so $x$ is more than 18.
Give the answer
There are 22 cans on the shelf now, before any are taken off. So the answer is D, 22.

Check your answer

Start with 22 cans and take off 4. That leaves $22 - 4 = 18$ cans. Half of 36 is 18, so the shelf is half full. The answer works.

More questions like this

The original DOE question

Question 113 as printed on page 164 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 113. Open page 164 of the handbook (PDF)