Practice Test

An answer in an unexpected form

Question 1 of 69

2026 Practice Test A Math, question 10

Question

Three expressions are given:

$-2(2.5n + 4)$

$5 - n$

$3(2n + 7)$

Which of these are equivalent to the sum of these three expressions?

Select the two correct answers.

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

A $5 - n + 6n + 21 - 5n - 8$
B $5 - n + 6n + 21 - 5n + 8$
C $5 - n + 5n + 21 - 5n - 8$
D $18 - n$
E18
F34

Show the answer

Correct answer: A and E

Add the three expressions: $5 - n + 3(2n + 7) + (-2)(2.5n + 4)$ .

Distribute 3 and −2: $5 - n + 6n + 21 - 5n - 8$ .

Combine the variable terms and combine the constants: $(-n + 6n - 5n) + (5 + 21 - 8) = 18$ .

Question analysis

This algebra question gives three expressions, one with a decimal coefficient and a negative factor, and asks which two of six choices are equivalent to their sum. It tests distributing, combining like terms and recognizing equivalent expressions. Signs from the negative factor are easy to lose, and several choices are built from exactly those slips.

NYS standards: 7.EE.1 linear expressions with rational coefficients
6.EE.4 identifying equivalent expressions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Expanding and combining linear expressions with a negative factor and a decimal coefficient is standard grade 7 work. Choosing the two equivalent ones out of six, where each common slip appears as a choice, takes more care than one simplification.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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
Many steps minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 8, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The six choices form three consistent pairs, each an expansion with its simplified total: the right pair, a pair built on a sign slip with −2 × 4 that totals 34, and a pair built on 3 × 2n read as 5n that totals 18 − n. A student who makes either slip finds two choices that agree with each other and seem to confirm the work. The question rewards one careful simplification and the confidence to accept a sum with no n left.

Error-prone arithmetic moderate Working it out
The first expression puts a negative factor in front of a decimal term: −2 times 2.5n is −5n, and the −2 must reach the 4 as well, giving −8, not +8. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
The first two expansions are identical except for their last sign, − 8 against + 8, and the third differs from the first only in 5n for 6n, so only a term-by-term check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
Two distributions come first, then the n terms and the constants are collected separately, and six choices are compared with the result; both right ones must be picked to score. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
One right answer is the expansion before any terms are combined and the other is the plain number 18, which has to be accepted as equal to a sum in which every expression held n. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute each factor across its parentheses, keeping the sign with every product, then combine the n terms and the constants separately and test each of the six choices against the result.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Gives the minus of −2 to 2.5n but not to 4, expanding −2(2.5n + 4) as −5n + 8 instead of −5n − 8, so this expansion totals 34.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: weak
Expands 3(2n + 7) as 5n + 21, adding 3 and 2 instead of multiplying them, so the n terms no longer cancel and leave −n.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Expects a sum of expressions in n to keep an n term, so it doubts the plain 18 and takes 18 − n, which the 5n expansion also simplifies to.
F Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes −2 × 4 as +8, so the constants add to 5 + 21 + 8 = 34 while −n + 6n − 5n still cancels, leaving no n term.

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. You multiply a by every term inside, not only the first one. If a is negative, its minus sign goes into every product. It works because a groups of b + c is a groups of b plus a groups of c.
Multiplying signed numbers: $(-a) \times b = -(a \times b)$; a and b are positive numbers, so −a is negative. A negative number times a positive number is negative. Use it whenever a minus sign sits in front of the parentheses: −2 times 4 is −8, not 8.
Combining like terms: $an + bn = (a + b)n$; Like terms have the same letter part, such as 6n and −5n. a and b are their coefficients, the numbers in front of n. Add the coefficients and keep the n. Constants, the plain numbers, combine only with other constants.

Sort out the question
An expression is a math phrase made of numbers, letters and signs, with no equal sign. The sum of the three expressions means you add all of them:
$-2(2.5n + 4) + (5 - n) + 3(2n + 7)$
You need the two choices that are equivalent to this sum. Equivalent means equal for every value of n. Some choices show the sum before it is simplified. Others show the result.
Distribute the 3
To distribute means to multiply the number outside the parentheses by each term inside.
$3(2n + 7) = 3 \times 2n + 3 \times 7 = 6n + 21$
Watch out: 3 times 2n is 6n. Adding 3 and 2 instead gives 5n. That slip leads to choice C, and then to 18 − n, choice D.
Distribute the −2
Now multiply −2 by both terms. A negative times a positive is negative, so both products are negative.
$-2(2.5n + 4) = -2 \times 2.5n + (-2) \times 4 = -5n - 8$
Watch out: the −2 multiplies the 4 too. Writing +8 instead of −8 leads to choice B. That sum works out to 34, choice F.
Write the whole sum
Put the three pieces together, starting with 5 − n:
$5 - n + 6n + 21 - 5n - 8$
This matches choice A exactly, term by term. So A is one of the answers.
Combine like terms
Gather the n terms first. Remember that −n means −1 times n.
$-n + 6n - 5n = (-1 + 6 - 5)n = 0n = 0$
Then gather the constants:
$5 + 21 - 8 = 18$
The n terms cancel out, so the whole sum is just 18.
Watch out: a sum with no n left is fine. Keeping an extra −n leads to choice D.
Pick the two answers
Choice A is the sum before you combine like terms. Choice E, 18, is the sum after you combine them. Both equal the sum for every value of n. So the answers are A and E.

Check your answer

Try a value, such as n = 2, in each expression. $-2(2.5 \times 2 + 4) = -2 \times 9 = -18$ . $5 - 2 = 3$ . $3(2 \times 2 + 7) = 3 \times 11 = 33$ . The sum is $-18 + 3 + 33 = 18$ , which matches E. Choice D would give 18 − 2 = 16 instead.

More questions like this

The original DOE question

Question 10 as shown on the DOE’s digital test, page 81 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 81, question 10.

Question 2 of 69

2026 Practice Test A Math, question 17

Question

Consider this expression:

$|(-8) - (-3) + 6| - |6 - 11|$

Create an equivalent expression.

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

|| − ||

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: 1; blank 2: −5

To create an expression equivalent to $|(-8) - (-3) + 6| - |6 - 11|$ :

Step 1: Simplify the first absolute value term.

$(-8) - (-3) = -8 + 3 = -5$

$-5 + 6 = 1$

Simplify the second absolute value term.

$6 - 11 = -5$

Therefore the equivalent expression is

$|1| - |-5|$ .

Question analysis

This absolute value question shows an expression with two pairs of bars and asks students to build an equivalent one by placing number tiles in two boxes. It tests subtracting negative integers and understanding what absolute value bars do. Not every tile is used, and several tiles match the results of common sign mistakes.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Simplifying signed integers inside absolute value bars, including subtracting a negative, is core grade 7 work. This asks only for the values inside the bars, not the final number, but one sign slip leads straight to a wrong tile.
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
Many steps minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 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

Every move here involves a sign, and the tiles are built from the usual misreadings of absolute value: 11 for making each number positive before combining, 17 for erasing every minus sign. A slip on − (−3) gives −5 for the first pair, but the second pair needs the only −5 tile, so that slip leaves a box with no tile to fit. Students who simplify each inside completely, and leave the bars untouched, finish quickly.

Error-prone arithmetic moderate Working it out
Inside the first bars, subtracting −3 must become adding 3, so −8 + 3 = −5, and adding 6 then crosses zero to reach 1; reading − (−3) as − 3 gives −5 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Two pairs of bars are simplified separately, and both boxes need the right tile, kept in the original order, before the answer is right. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
Each box takes what is left inside a pair of bars, not its absolute value, so the second box needs the tile −5 rather than a 5, and the expression's value, −4, is never entered. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −3 as adding 3, then simplify the inside of each pair of bars separately, left to right; each box replaces the inside of one pair, in the same order.

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.
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

EC A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Takes the absolute value of each number before combining, so the first bars become 8 − 3 + 6 = 11; the same rule still gives 6 − 11 = −5 in the second.
BC Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Believes absolute value bars make everything inside positive, so it drops every minus sign and adds 8 + 3 + 6 = 17, then works the short pair 6 − 11 normally.
CA Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Simplifies both insides correctly but places −5 first and 1 second, reversing the subtraction so the expression equals 5 − 1 = 4 instead of 1 − 5 = −4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a is any number, and b is a positive number, so −b is negative. Subtracting a negative number is the same as adding its opposite, the same number with the other sign. Think of money: taking away a debt of $3 leaves you $3 richer.
Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; The bars around a number stand for its absolute value. a is any number. The absolute value is how far a is from 0 on a number line, so it is never negative. Both |5| and |−5| equal 5. Always simplify inside the bars first.

Sort out the question
The expression has two pairs of absolute value bars. The new expression keeps both pairs, with a box inside each one:
$|\square| - |\square|$
Each box gets the number you reach when you simplify inside that pair of bars. The bars stay, so you do not take the absolute value yet. Some tiles are not used.
Simplify inside the first bars
Inside the first bars is $(-8) - (-3) + 6$ . Subtracting −3 is the same as adding 3. Work from left to right:
$(-8) - (-3) = -8 + 3 = -5$
$-5 + 6 = 1$
The first box gets the tile 1.
Watch out: the bars do not make each number positive before you combine. Changing −8 and −3 to 8 and 3 first gives 8 − 3 + 6 = 11. Dropping every minus sign gives 8 + 3 + 6 = 17. Both are tiles, and both are wrong.
Simplify inside the second bars
Inside the second bars is $6 - 11$ . You take a bigger number from a smaller one, so the result is negative:
$6 - 11 = -5$
The second box gets the tile −5.
Watch out: do not change −5 to 5. The bars are still there in the new expression, so the box holds what is inside them.
Keep the boxes in order
The first bars came first, so 1 goes in the first box and −5 in the second. The new expression has the same value as the old one:
$|1| - |-5| = 1 - 5 = -4$
Swapping the tiles gives $|-5| - |1| = 5 - 1 = 4$ , which is not equal. So the boxes get 1 and −5, making |1| − |−5|.
On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Use a number line for the first pair. Start at −8 and move 3 to the right to reach −5. Then move 6 more to the right to reach 1. For the second pair, start at 6 and move 11 to the left. You land on −5. Both boxes check out.

More questions like this

The original DOE question

Question 17 as shown on the DOE’s digital test, page 88 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 88, question 17.

Question 3 of 69

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 4 of 69

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 5 of 69

2026 Practice Test A Math, question 43

Question

Graph the solution set for the inequality $-3 \leq 2x - 5$ .

Select a ray. Then, move the endpoint of the ray to the correct position on the number line.

On the test you draw the ray on the number line; here, choose its endpoint, whether the endpoint is closed or open, and its direction. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Endpoint closed (filled dot) open (empty dot) pointing left pointing right

Show the answer

Correct answer: a ray with a closed endpoint at 1, pointing right

Solution:

$A number line from −5 to 5 with a tick at every integer, labeled at −5, 0 and 5. A solid dot at 1, with the line drawn thick from 1 to the right and ending in an arrow.$

Solve the inequality for x, then graph the solution on the number line:

$-3 \leq 2x - 5$

Add 5 to both sides of the inequality: $2 \leq 2x$

Divide both sides by 2: $1 \leq x$

Rewrite the inequality with x on the left side: $x \geq 1$

The graph of $x \geq 1$ is a closed ray (solid point) beginning at 1 and extending to the right on the number line.

Question analysis

Students graph the solution set of a two-step inequality on a number line by choosing a ray: where it starts, whether its endpoint is open or closed, and which way it points. It tests solving a linear inequality and reading its solution. The variable sits on the right of the inequality symbol, so reading the result the right way round is part of the challenge.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step inequality and graphing its solution set is a named grade 7 skill. This one puts the variable on the right of the symbol and asks students to build the graph themselves rather than pick it from four.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Many steps minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Solving takes two routine moves; the points are lost in turning 1 ≤ x into a picture. Because x sits on the right, the symbol seems to point left, and a student who shades the way it points draws the values below 1, which all fail. The dot must also be solid, since x = 1 makes both sides −3. Testing x = 2, where 2x − 5 = −1 and −3 ≤ −1 holds, confirms the direction.

Reading a figure, table or graph moderate Understanding the question
The solution comes out as 1 ≤ x, with x on the right, so the ≤ sign points away from the shaded side; it has to be read as x ≥ 1 before the ray is drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Two solving moves, adding 5 and then halving, come before three graph choices, the endpoint, a solid dot and a direction, and a mistake in any of the five loses the item. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The ray is built, by picking one of four ray shapes and dragging its endpoint, rather than chosen from four finished graphs, so nothing on screen flags a ray that starts at the right number but points the wrong way. (Enright, Morley & Sheehan, 2002)

Strategy: Undo the subtraction and then the multiplication on both sides, rewrite the result with x first, decide whether the endpoint is included, and test one value on your ray in the original inequality.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

1|closed|left Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Solves to 1 ≤ x but shades the way the ≤ symbol points, a rule that holds only with x on the left, so the ray covers x ≤ 1.
1|open|right Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Finds the right endpoint and direction but draws an open dot, leaving out 1 itself, though the "or equal to" in ≤ includes it.
-4|closed|right A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves the −5 to the left side without changing its sign, writing −3 − 5 = −8 ≤ 2x, so the ray starts at −4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a \leq b \text{, then } a + c \leq b + c$; a and b are the two sides of an inequality, and c is any number. Adding the same number to both sides keeps the inequality true. So does dividing both sides by the same positive number. Dividing by a negative number would flip the symbol, but that does not happen here.
Turning an inequality around: $a \leq b \text{ means the same as } b \geq a$; a and b are any two numbers or terms. You may swap the two sides if you also turn the symbol around. The pointed end of the symbol always faces the smaller side. So 1 ≤ x and x ≥ 1 say the same thing.

Sort out the question
The symbol ≤ means is less than or equal to. The solution set is every value of x that makes $-3 \leq 2x - 5$ true. You show it as a ray. A ray is a dot, called the endpoint, with an arrow going one way forever. You need three things: where the dot goes, whether it is closed or open, and which way the ray points.
Add 5 to both sides
The right side is 2x − 5. Adding 5 undoes subtracting 5, so add 5 to both sides:
$-3 + 5 \leq 2x - 5 + 5$
$2 \leq 2x$
Watch out: taking 5 away from −3 instead gives −8 ≤ 2x. That wrong move puts the dot at −4.
Divide both sides by 2
Now x is multiplied by 2, so divide both sides by 2. You divide by a positive number, so the symbol stays the same:
$\frac{2}{2} \leq \frac{2x}{2}$
$1 \leq x$
Read it with x first
$1 \leq x$ says 1 is less than or equal to x. So x is 1 or any number bigger than 1. With x first, that is:
$x \geq 1$
Bigger numbers are to the right on a number line, so the ray points right.
Watch out: in 1 ≤ x, the symbol seems to point left. Do not follow it. A ray pointing left from 1 shows numbers that do not work.
Choose a closed dot
The words or equal to mean 1 itself is a solution. Check it: $2 \times 1 - 5 = -3$ , and −3 ≤ −3 is true. So the dot at 1 is closed, which means filled in. An open dot, an empty circle, leaves the endpoint out. Use it only for < or >.
Watch out: an open dot at 1 would leave out a number that works.
Draw the ray
On the test, choose the ray with a filled dot and an arrow pointing right. Then drag its dot to 1. On this site, choose 1 for the endpoint, then closed (filled dot) and pointing right.
So the answer is a closed dot at 1, with the ray pointing right.

Check your answer

Test a number on your ray, such as 3. $2 \times 3 - 5 = 1$ , and −3 ≤ 1 is true. Now test a number off the ray, such as 0. $2 \times 0 - 5 = -5$ , and −3 ≤ −5 is false. So the ray points the right way.

More questions like this

The original DOE question

Question 43 as shown on the DOE’s digital test, page 114 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 114, question 43.

Question 6 of 69

2026 Practice Test A Math, question 50

Question

Let $b = 3$ , $c = 4$ , and $d = 5$ . Determine the value of $\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}$ .

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

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

Show the answer

Correct answer: 7/5

Solution:

$\frac{7}{5}$

Substitute $b = 3$ , $c = 4$ , and $d = 5$ in $\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}$ : $\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5}$ .

Then simplify the complex fraction by multiplying the complex fraction by a form of 1, $\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right)$ : $\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5} = \frac{\frac{4}{5}}{\frac{4}{3}}\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right) + \frac{4}{5}$ .

Simplify after multiplying by a form of 1: $\frac{\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)}{\left(\frac{4}{3}\right)\left(\frac{3}{4}\right)} + \frac{4}{5} = \frac{\frac{3}{5}}{1} + \frac{4}{5}$ .

Then simplify again and add the two fractions: $\frac{\frac{3}{5}}{1} + \frac{4}{5} = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}$ .

Question analysis

This substitution question gives values for b, c and d and asks for the value of an expression made of a fraction divided by a fraction, plus another fraction. Students substitute, simplify the complex fraction and add, then type the result. Dividing fractions in the stacked form is the main hurdle, and the item wants a fraction, not a decimal.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.NS.1 dividing fractions by fractions
7.NS.2c properties for multiplying and dividing
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students evaluate expressions and divide and add fractions. A fraction stacked over another fraction looks unfamiliar and there are no choices to check, but with these values the work is three short steps.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the values are in, the work is one fraction division and one addition, but the division is printed as a fraction over a fraction, in small type. Multiplying the two fractions, or flipping the top one, gives a sum in fifteenths that looks as finished as the right one. Noticing that the 4s cancel, so the stacked part is simply 3/5, turns the last step into adding fifths.

Error-prone arithmetic moderate Working it out
Dividing 4/5 by 4/3 means multiplying by 3/4, giving 3/5, which is then added to 4/5; multiplying straight across or flipping the top fraction instead both lead to fifteenths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Nothing on the page says that a fraction over a fraction is a division, or that dividing by 4/3 means multiplying by its reciprocal, and the stacked form is rare in grade 7 work. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The value is typed into a fraction box with no choices to check against, and the item asks for a fraction, so the equal decimal 1.4 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Substitute the values, rewrite the stacked fraction as the top fraction times the reciprocal of the bottom one, where the 4s cancel, then add the last fraction and leave the result as a fraction.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

28/15 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Multiplies 4/5 by 4/3 instead of dividing, getting 16/15, then adds 4/5, or 12/15, to reach 28/15.
37/15 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Flips the top fraction instead of the bottom one, so the stacked part becomes 5/4 × 4/3 = 5/3, and 5/3 + 4/5 = 37/15.
1.4 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the right value but works in decimals, 0.6 + 0.8 = 1.4, and enters that, though the item asks for a fraction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by. It works because a fraction times its reciprocal is 1, so the reciprocal undoes the fraction, just as dividing does.
Adding fractions with the same bottom: $\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}$; d is the denominator, the bottom number, which tells the size of each piece. a and b are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

Sort out the question
You know $b = 3$ , $c = 4$ and $d = 5$ . You need the value of:
$\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}$
The first part is a fraction on top of a fraction. A fraction bar means divide, so this part is $\frac{c}{d} \div \frac{c}{b}$ . The answer must be typed as a fraction.
Put in the numbers
Replace each letter with its value. Look closely: the bottom fraction has b under the c, not d.
$\frac{c}{d} = \frac{4}{5} \qquad \frac{c}{b} = \frac{4}{3}$
So the expression becomes:
$\frac{4}{5} \div \frac{4}{3} + \frac{4}{5}$
Divide the two fractions
Use the rule for dividing fractions. The reciprocal of 4/3 is 3/4. To multiply fractions, multiply the tops and multiply the bottoms:
$\frac{4}{5} \div \frac{4}{3} = \frac{4}{5} \times \frac{3}{4} = \frac{12}{20} = \frac{3}{5}$
To simplify 12/20, divide its top and bottom by 4.
Watch out: flip only the bottom fraction. Multiplying without flipping gives 16/15 and a total of 28/15. Flipping the top one gives 5/3 and a total of 37/15.
Add the last fraction
Both fractions are now fifths, so add the tops and keep the bottom:
$\frac{3}{5} + \frac{4}{5} = \frac{3 + 4}{5} = \frac{7}{5}$
The value is 7/5.
Enter it as a fraction
The item asks for a fraction. 7/5 equals the decimal 1.4, but 1.4 is not accepted. On the test, the answer box already holds a fraction template, a small box over a small box. Type 7 in the top box. Then click the bottom box and type 5. On this site, type 7/5.
So you would enter 7/5.

Check your answer

Check with decimals. $\frac{4}{5} = 0.8$ , and 4/3 is about 1.33. Then $0.8 \div 1.33$ is about 0.6, which is 3/5. Last, $0.6 + 0.8 = 1.4$ , and 1.4 is the same as 7/5.

More questions like this

The original DOE question

Question 50 as shown on the DOE’s digital test, page 121 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 121, question 50.

Question 7 of 69

2026 Practice Test B Math, question 3

Question

At a diner, 18 dozen buns were used yesterday for 150 lunch customers. Today, there were 250 lunch customers. Assuming that the bun usage rate is the same for both days, which proportions can be used to determine how many dozen buns are needed for 250 lunch customers?

Select the three correct answers.

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

A $\frac{18}{150} = \frac{x}{250}$
B $\frac{18}{250} = \frac{x}{150}$
C $\frac{150}{18} = \frac{250}{x}$
D $\frac{x}{18} = \frac{250}{150}$
E $\frac{x}{150} = \frac{18}{250}$

Show the answer

Correct answer: A, C, and D

The three correct answers are:

$\frac{18}{150} = \frac{x}{250}$

$\frac{150}{18} = \frac{250}{x}$

$\frac{x}{18} = \frac{250}{150}$

Set up a proportion to compare the number of buns per customer. These are all equivalent:

$\frac{18\text{ dozen buns}}{150\text{ customers}} = \frac{x\text{ dozen buns}}{250\text{ customers}}$

To check for equivalency, solve each proportion for x.

Each of the three correct proportions simplifies to $150x = 18(250)$ .

Then solve for x to determine how many dozen buns are needed for 250 lunch customers: $150x = 4{,}500$

$x = \frac{4{,}500}{150}$

$x = 30$

Question analysis

A diner used 18 dozen buns for 150 lunch customers yesterday, and students select the three proportions, out of five, that can find how many dozen buns 250 customers need at the same rate. It tests setting up proportions and recognizing equivalent ones. All five equations use the same numbers, so each has to be checked for which quantities it pairs.

NYS standards: 7.RP.2c proportional relationship equations
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Setting up a proportion from a rate is everyday grade 7 work. Here nothing is solved: five setups built from the same numbers must each be judged, some arranged in unfamiliar ways, and all three usable ones found.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Every choice must be checked moderate
A very close wrong answer moderate
An answer in an unexpected form minor
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 54 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

Nothing needs solving here; the question is which pairings keep the usage rate fixed. Students who check only for buns over customers on both sides find exactly three such setups, two of them wrong, and pass over the two right ones written another way. The wrong pair gives 10.8 dozen, fewer buns for more customers, which a quick sense check would catch, and cross-multiplying every setup settles all five.

Every choice must be checked moderate Choosing the answer
All five setups are built from 18, 150, 250 and x, so none can be dismissed on sight; each has to be cross-multiplied and compared with 150x = 18 × 250. (Embretson & Wetzel, 1987)
A very close wrong answer moderate Choosing the answer
The first two setups differ only in where 150 and 250 sit, 18/150 = x/250 against 18/250 = x/150, and both show buns over customers on each side, so a units check passes both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
An answer in an unexpected form minor Choosing the answer
One right setup puts customers over dozens of buns, and another sets today’s x over yesterday’s 18 equal to 250 customers over 150, forms that look wrong to students who learned only rate equals rate. (Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Three of the five must be picked and all three must be right to score, so the verdict on each setup has to be kept until the last one is tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write one proportion you trust by matching both units and days, cross-multiply it, then cross-multiply each choice and keep only those that give the same equation.

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.
Answer in another form: The right answer is written differently from the way you worked it out.

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
Checks only that both fractions show dozens of buns over customers, missing that 18 sits over today’s 250 customers; cross-multiplying gives 250x = 2,700, so x = 10.8.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Left out by students who expect each fraction to compare buns with customers, not seeing that today over yesterday for buns and for customers also gives 150x = 4,500.
E Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Accepts x over 150 because it shows buns over customers like the trusted setup, but today’s buns sit over yesterday’s customers, which yields only 10.8 dozen for more people.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A rate: $\text{rate} = \frac{\text{dozens of buns}}{\text{customers}}$; The rate is how many dozen buns are used for each customer. Yesterday it was 18 dozen over 150 customers. Today it is x dozen over 250 customers. The question says the rate is the same both days, so those two fractions are equal.
Cross-multiplying a proportion: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; A proportion says two fractions are equal. a and c are the tops, and b and d are the bottoms. Multiply each top by the other bottom, and the two products are equal. This turns any proportion into a plain equation, so you can compare setups that look different.

Sort out the question
Yesterday the diner used 18 dozen buns for 150 customers. Today it has 250 customers. Let x be the number of dozen buns needed today. The bun usage rate is the same both days.
A proportion is two equal fractions. You must pick the three proportions that give the right value of x.
Write one proportion you trust
Put each day’s buns over that same day’s customers. The rates are equal, so the fractions are equal:
$\frac{18}{150} = \frac{x}{250}$
This is choice A. Cross-multiply it:
$150 \times x = 18 \times 250$
$150x = 4{,}500$
Any choice that leads to 150x = 4,500 is a right one.
Solve for x
Divide both sides by 150 to get x by itself:
$x = 4{,}500 \div 150 = 30$
So the diner needs 30 dozen buns today. A wrong setup gives a different number.
Test choices C and D
Cross-multiply each one.
- C: $\frac{150}{18} = \frac{250}{x}$ gives $150 \times x = 18 \times 250$ . That is the same equation, so C works. It is choice A with both fractions flipped.
- D: $\frac{x}{18} = \frac{250}{150}$ gives $x \times 150 = 18 \times 250$ . The same equation again, so D works too.
D compares today with yesterday. Today’s buns over yesterday’s buns equals today’s customers over yesterday’s customers.
Watch out: D looks odd because it puts buns over buns. Do not skip it. It still keeps the rate the same.
Test choices B and E
- B: $\frac{18}{250} = \frac{x}{150}$ gives $250 \times x = 18 \times 150$ , so 250x = 2,700.
- E: $\frac{x}{150} = \frac{18}{250}$ gives $x \times 250 = 150 \times 18$ , the same 250x = 2,700.
Both give $x = 2{,}700 \div 250 = 10.8$ . That is fewer buns for more customers, which makes no sense.
Watch out: B and E show buns over customers, so they look right. But each one mixes up the days. B puts yesterday’s 18 dozen over today’s 250 customers. E puts today’s buns over yesterday’s 150 customers.
Name the three answers
Choices A, C and D all give 150x = 4,500, so x = 30. On the test, you check the box next to each one. So the answers are A, C and D.

Check your answer

Find the buns a second way. Today has $\frac{250}{150} = \frac{5}{3}$ times as many customers as yesterday. So it needs 5/3 times as many buns: $18 \times \frac{5}{3} = 30$ dozen. That matches the 30 from choices A, C and D.

More questions like this

The original DOE question

Question 3 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 B Math, page 75, question 3.

Question 8 of 69

2026 Practice Test B Math, question 4

Question

Maria is playing a game in which players earn points called minks, zogs, and barts.

$1\text{ mink} = 3\text{ zogs}$

$1\text{ mink} = \frac{1}{4}\text{ bart}$

Maria has 60 zogs and 24 barts. She wants to exchange her zogs and barts for minks.

How many minks will Maria receive from the exchange?

Enter your answer in the space.

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

Show the answer

Correct answer: 116

Step 1: Convert the zogs to minks.
Since $1\text{ mink} = 3\text{ zogs}$ , divide the number of zogs Maria has by 3: $60 \div 3 = 20$ .

So, $60\text{ zogs} = 20\text{ minks}$ .

Step 2: Convert the barts to minks.
Since $1\text{ mink} = \frac{1}{4}\text{ bart}$ , multiply the number of barts by 4 (because 1 bart is equal to 4 minks): $24 \times 4 = 96$ .

So, $24\text{ barts} = 96\text{ minks}$ .

Step 3: Add the mink conversions.
Add the results from Steps 1 and 2 to find the total number of minks Maria will receive: $20 + 96 = 116$ .

Question analysis

In a game with made-up points called minks, zogs and barts, Maria trades 60 zogs and 24 barts for minks, and students type how many minks she receives. It tests unit conversion with invented units and a fraction. One rate says a mink equals a fraction of a bart, so deciding whether to multiply or divide is the tricky part.

NYS standards: 6.RP.3d converting measurement units
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with a given rate is grade 6 ratio reasoning. Invented units remove any real-world sense check, and the two conversions run in opposite directions, one through a fraction, which matches the care grade 7 problems need.
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
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 47 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 step is one multiplication, division or addition; the risk is direction. The two rates point opposite ways, three zogs to a mink but four minks to a bart, and the fraction in the second rate invites dividing by 4. Asking of each unit whether one of it is worth more or less than a mink fixes the direction before any arithmetic, and it also shows why the 60 and the 24 can never simply be added.

Turning words into math moderate Understanding the question
Both rates put the mink on the left, so “1 mink = 1/4 bart” has to be turned around into 1 bart = 4 minks, and the invented units give no sense of which is worth more. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two conversions run in opposite directions, dividing the zogs by 3 and multiplying the barts by 4, and their results are added, so a slip in either step changes the total. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The total is typed into a box with no choices to compare, so a conversion done the wrong way still produces a believable whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: For each kind of point, decide whether one of it is worth more or less than a mink, then multiply or divide to turn it into minks, and add the two results.

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).
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

26 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Gets 20 minks from the zogs but divides the barts by 4, 24 ÷ 4 = 6, as if a bart were worth a quarter of a mink, for 20 + 6 = 26.
276 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies the 60 zogs by 3, as if each zog were worth 3 minks, then adds the 96 minks from the barts correctly: 180 + 96 = 276.
84 Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the 60 zogs and 24 barts as they stand, 60 + 24 = 84, without converting either kind of point into minks first.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Converting with a rate: $\text{minks} = \text{number of points} \times \text{minks for one point}$; Number of points is how many zogs or barts Maria has. Minks for one point is what one zog, or one bart, is worth in minks. Find that worth first for each kind of point. Then multiply, because every point of the same kind trades for the same number of minks.

Sort out the question
Maria trades two kinds of points for minks. You know:
- 3 zogs trade for 1 mink.
- 1 mink is worth $\frac{1}{4}$ of a bart.
- Maria has 60 zogs and 24 barts.
You need the total number of minks. Zogs and barts are different units, like dimes and dollars. So you cannot just add them.
Watch out: 60 + 24 = 84 adds zogs to barts. That is not a number of minks.
Find what one zog is worth
It takes 3 zogs to make 1 mink. So one zog is only part of a mink. One zog is worth $\frac{1}{3}$ of a mink.
A zog is smaller than a mink. So Maria will get fewer minks than she has zogs.
Turn the bart rule around
The rule says 1 mink is $\frac{1}{4}$ of a bart. So a mink is a quarter of a bart. It takes 4 quarters to make a whole:
$4 \times \frac{1}{4} = 1$
So 4 minks make 1 bart. One bart is worth 4 minks. A bart is bigger than a mink.
Watch out: the fraction tempts you to divide by 4. Then 24 barts become only 6 minks, and the total is 26.
Change the zogs into minks
Use the rate. Each of the 60 zogs is worth 1/3 of a mink:
$60 \times \frac{1}{3} = 60 \div 3 = 20$
Multiplying by 1/3 is the same as dividing by 3. So 60 zogs trade for 20 minks.
Watch out: multiplying 60 by 3 gives 180. That treats a zog as bigger than a mink, and the total becomes 276.
Change the barts into minks
Each of the 24 barts is worth 4 minks:
$24 \times 4 = 96$
So 24 barts trade for 96 minks.
Add the minks
Both amounts are now in minks, so you can add them:
$20 + 96 = 116$
Maria receives 116 minks. Type only the number in the box, with no units. So you would enter 116.

Check your answer

Work backward. 20 minks are worth $20 \times 3 = 60$ zogs. 96 minks are worth $96 \times \frac{1}{4} = 24$ barts. Both match what Maria started with. Also, a bart is worth more than a mink, so 24 barts should give many more than 24 minks. They do.

More questions like this

The original DOE question

Question 4 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 B Math, page 76, question 4.

Question 9 of 69

2026 Practice Test B Math, question 13

Question

A basket contains 3 apples, 4 oranges, 5 bananas, and 8 pears.

If one piece of fruit is randomly pulled from the basket, what is the probability that it will be a pear?

Select the place on the number line to plot the point.

On the test you click a point on the number line; here, choose it from the menu.

Point

Show the answer

Correct answer: the point 0.4

The total number of fruits is:

$3(\text{apples}) + 4(\text{oranges}) + 5(\text{bananas}) + 8(\text{pears}) = 20\text{ fruits}$

The probability of picking a pear is:

$\frac{8\ (\text{pears})}{20\ (\text{total})} = \frac{8}{20} \text{ or } 0.4$

Question analysis

A basket holds apples, oranges, bananas and pears, and students find the probability that a piece of fruit pulled at random is a pear, then plot it as a point on a number line from 0 to 1. It tests simple probability as a part of a whole. With no choices to pick from, the fraction has to be turned into tenths to find its place.

NYS standards: 6.SP.8a uniform probability models
6.SP.6 probability from 0 to 1
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of a simple event from counts is grade 6 work in New York. This one adds four whole numbers, forms one fraction and plots it among marked tenths, a shorter and more routine task than grade 7 probability.
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: Something you must already know minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Adding the four counts to 20 and writing 8 over 20 is routine, so most students succeed. The losses come from the number line: its decimal labels make it easy to read the 8 pears as 0.8, and a part-to-part fraction such as 8/12 falls between marks and has to be rounded, which should prompt a second look. Simplifying 8/20 to 2/5 and then to 4/10 makes the mark plain.

Something you must already know minor Working it out
The question does not say that a probability is the favorable count over the whole count, here the 8 pears over all 20 pieces of fruit rather than over the other 12. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The number line is labeled only in tenths from 0 to 1, so 8/20 has to be rewritten as a decimal before the right mark can be chosen. (Enright, Morley & Sheehan, 2002)

Strategy: Add up all the fruit for the total, write the number of pears over that total, rewrite the fraction in tenths, and select that mark on the line.

Traps

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

The wrong answers, classified

0.8 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Takes the 8 pears straight to the mark labeled 0.8, never dividing by the 20 pieces of fruit in the basket.
0.7 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Compares the 8 pears with the 12 other pieces of fruit, 8/12 or about 0.67, a part-to-part ratio, and plots the nearest tenth.
0.6 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Puts the 12 apples, oranges and bananas over the 20 pieces of fruit, plotting the chance of not getting a pear instead of getting one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of results you want}}{\text{total number of results}}$; P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want, here pears. The bottom counts every possible result, here every piece of fruit. It works when each result is equally likely, like grabbing fruit without looking.
Equal fractions: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top of a fraction, and b is the bottom. n is any number except 0. Dividing the top and the bottom by the same number gives an equal fraction. Use it to get 10 on the bottom, which is easy to write as a decimal.

Sort out the question
The basket has 3 apples, 4 oranges, 5 bananas and 8 pears. You pull out one piece without looking. You need the probability that it is a pear. Then you plot that number as a point. The number line runs from 0 to 1, with a mark at every tenth.
Count all the fruit
The bottom of the fraction is every piece of fruit you could pull out. Add all four kinds:
$3 + 4 + 5 + 8 = 20$
There are 20 pieces of fruit in all.
Write the probability as a fraction
Use the probability formula. There are 8 pears out of 20 pieces of fruit:
$P = \frac{8}{20}$
Watch out: the bottom is all the fruit, not just the other fruit. $\frac{8}{12}$ compares the pears with the 12 fruits that are not pears. It is about 0.67, and it leads you to the mark at 0.7. And 12 over 20 is the chance of not getting a pear.
Change it to tenths
The number line is marked in tenths. So rewrite the fraction with 10 on the bottom. Divide the top and the bottom by 2:
$\frac{8}{20} = \frac{8 \div 2}{20 \div 2} = \frac{4}{10} = 0.4$
Four tenths is written 0.4.
Watch out: the 8 pears do not make 0.8. That skips dividing by the 20 pieces of fruit.
Plot the point
Start at 0 and count four marks to the right: 0.1, 0.2, 0.3, 0.4. On the test, you click the mark labeled 0.4. On this site, you choose 0.4 from the menu.
So the answer is the point at 0.4.

Check your answer

The chance of not getting a pear is $\frac{12}{20} = 0.6$ . A pear and not a pear cover every result, so the two must add to 1: $0.4 + 0.6 = 1$ . Also, pears are less than half of the fruit, so the point should be left of 0.5. It is.

More questions like this

The original DOE question

Question 13 as shown on the DOE’s digital test, page 85 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 85, question 13.

Question 10 of 69

2026 Practice Test B Math, question 17

Question

How many positive integers satisfy the inequality $x + 7 < 23$ ?

Enter your answer in the space.

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

Show the answer

Correct answer: 15

Solve the inequality for x.

$x + 7 < 23$

$x < 16$

There are 15 positive integers that are less than 16.

Question analysis

This inequality question asks how many positive integers make x + 7 < 23 true, and the count is typed into a box. Students solve a one-step inequality, then count its whole-number solutions. The strict less-than sign and the word positive decide where the count starts and stops, so both ends need checking.

NYS standards: 7.EE.4b two-step inequalities
6.EE.5 solutions and substitution
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 inequality problems usually take two steps, such as 2x − 5 > 9. This one needs a single subtraction; the care goes into counting the whole-number solutions and deciding whether the boundary itself counts.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 12, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One subtraction solves the inequality, so the points are lost in the count that follows. The boundary 16 is the last number the solving produces and is easy to type in, or to count up to; checking that 16 + 7 is 23, which is not less than 23, rules it out. Listing 1 through 15, the positive integers below the boundary, settles the count.

An answer in an unexpected form minor Choosing the answer
Solving gives x < 16, but the box asks how many positive integers make the inequality true, a count that still has to be made after the solving is done. (Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Two conditions set the ends of the count at once: the word positive starts it at 1 rather than 0, and the strict < stops it below 16. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Undo the addition to find the boundary for x, then count the positive whole numbers below it, checking the largest one and the boundary in the original inequality.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

16 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Solves to x < 16 and then enters 16, either the boundary itself or a count of 1 through 16, though x = 16 makes x + 7 equal 23, not less than it.
29 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves the 7 to the right side without changing its sign, 23 + 7 = 30, and counts the 29 positive integers below 30.

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 the two sides of an inequality, and c is any number. Taking the same number away from both sides keeps the inequality true. Think of two people who each walk down the same number of stairs: the lower one stays lower. Use it to get x by itself.

Sort out the question
An inequality compares two amounts. The sign < means is less than. A positive integer is a whole number bigger than 0: 1, 2, 3, and so on. You need to count how many of them make $x + 7 < 23$ true. The answer is a count, not a value of x.
Get x by itself
The 7 is added to x. Subtracting 7 undoes adding 7, so subtract 7 from both sides:
$x + 7 - 7 < 23 - 7$
$x < 16$
So x must be less than 16.
Watch out: adding 7 to 23 instead gives x < 30, and a count of 29.
Check the boundary number
The number 16 is the boundary, where the solutions stop. Does 16 itself work? Try it:
$16 + 7 = 23$
But 23 is not less than 23. It is equal. So 16 does not count. The largest number that works is 15, since $15 + 7 = 22$ , and 22 is less than 23.
Watch out: entering 16 counts the boundary, which fails.
Count the positive integers
The numbers that work start at 1, the smallest positive integer. Zero is not positive. They stop at 15:
$1, 2, 3, \ldots, 14, 15$
Counting from 1 up to 15 gives 15 numbers.
Enter the count
There are 15 positive integers that make the inequality true. Type the count in the box as a plain whole number. So you would enter 15.

Check your answer

Test both ends. For 1: $1 + 7 = 8$ , and 8 is less than 23, so 1 works. For 15: 22 is less than 23, so 15 works. For 16: 23 is not less than 23, so 16 fails. That leaves exactly the numbers 1 through 15.

More questions like this

The original DOE question

Question 17 as shown on the DOE’s digital test, page 89 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 89, question 17.

Question 11 of 69

2026 Practice Test B Math, question 28

Question

A circle has a radius of 5 inches. What is the area of the circle in terms of $\pi$ ?

Use $\pi$ to express your answer.

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

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

Show the answer

Correct answer: 25π

Solution:

$25\pi$

The area of a circle is its radius squared times $\pi$ .

$A = \pi r^2$

$r = 5$

$A = \pi(5)^2$

$A = 25\pi$

Question analysis

This circle question gives a radius of 5 inches and asks for the area in terms of π, typed into a box rather than chosen. Students recall and apply the formula for the area of a circle. The formula is not printed on the test, and the area and circumference formulas are easy to mix up.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the area of a circle from its radius is the grade 7 circle standard applied directly, in one routine calculation. Typing the answer in terms of π, with no choices to compare, adds a little 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: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is one formula and one square, so the question comes down to recall and to the form of the answer. The area and circumference formulas both pair π with the radius, and 2πr gives a tidy value with π in it that looks just as right. The other losses are about form: replacing π with 3.14, which the instructions rule out twice, or dropping π after squaring 5.

Something you must already know moderate Working it out
The formula A = πr² is not printed with the question, and the circumference formula 2πr uses the same r and π, so recalling the wrong one gives a value that looks just as finished. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The answer is built on a keypad with a π key and must keep π as a symbol; a decimal worked out with 3.14 is not accepted, and no choices show the expected form. (Enright, Morley & Sheehan, 2002)

Strategy: Recall the area formula rather than the circumference formula, square the radius before multiplying, and leave π as a symbol instead of replacing it with 3.14.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

10π Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses the circumference formula 2πr, or doubles the radius instead of squaring it, so 2 × 5 gives 10π, the distance around the circle.
78.5 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the area correctly but replaces π with 3.14, 25 × 3.14 = 78.5, though the question asks for the area in terms of π.
25 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Squares the radius, 5 × 5 = 25, and stops before multiplying by π, so the entry leaves out the π the instructions ask for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle: $A = \pi r^2$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. r² means r times r. Square the radius first, then multiply by π. The area is in square units.
Circumference of a circle: $C = 2 \pi r$; C is the circumference, the distance around the circle. r is the radius, and π is the same number as above. This formula measures a length, not the space inside. It is here so you can tell the two formulas apart.

Sort out the question
The circle has a radius of 5 inches. The radius is the distance from the center to the edge. You need the area, the space inside the circle. In terms of π means π stays in the answer as a symbol. Do not change it to 3.14.
Choose the area formula
The area formula is $A = \pi r^2$ . It uses the radius squared.
Watch out: the circumference formula, $C = 2\pi r$ , also uses π and the radius. It gives $2 \times \pi \times 5 = 10\pi$ , the distance around the circle in inches. That is not the area.
Square the radius
Squaring a number means multiplying it by itself:
$r^2 = 5^2 = 5 \times 5 = 25$
Watch out: squaring is not doubling. $5 \times 2 = 10$ would lead to 10π again.
Multiply by π
Put 25 into the formula in place of r²:
$A = \pi \times 25 = 25\pi$
The area is 25π square inches. Write the number first and then π, the way you write 3x for 3 times x.
Watch out: leave π as π. Using 3.14 gives 78.5, which is not accepted. Dropping π leaves 25, which is not the area.
Enter it with the π key
On the test, the keypad has a π key in its bottom left corner. Press 2, then 5, then the π key. You do not need a times sign. Enter only the answer, with no A = and no units. On this site, type 25π or 25pi.
So you would enter 25π.

Check your answer

Estimate. π is a little more than 3, so 25π is a little more than 75 square inches. A square drawn around the circle is 10 inches on each side, so its area is 100 square inches. The circle fits inside that square, so its area must be less than 100. It is.

More questions like this

The original DOE question

Question 28 as shown on the DOE’s digital test, page 100 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 100, question 28.

Question 12 of 69

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 13 of 69

2026 Practice Test C Math, question 5

Question

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

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

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

Show the answer

Correct answer: 4n – 12

Let n represent the number of pages that Maya read.

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

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

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

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

Ali’s pages $= 4n - 12$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 69

2026 Practice Test C Math, question 15

Question

A grocery store sells 1-gallon containers of milk for $3.99. The store also sells orange juice in a 6-pack of 5.5-fluid-ounce bottles for $1.79. Suppose the store wants to sell its orange juice in gallon containers instead. To the nearest dollar, how much more would a gallon of orange juice cost than a gallon of milk?

Enter your answer in the space.

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

Show the answer

Correct answer: 3

To determine the cost of a gallon of orange juice, begin by finding the total number of ounces in the 6-pack of bottles:

$5.5\text{ oz} \times 6 = 33\text{ oz}$

So, 33 oz of orange juice costs $1.79.

There are 128 oz in 1 gallon. Set up a proportion using the $1.79 cost per 33 oz of orange juice and the unknown cost (x) per 1 gallon of orange juice:

$\frac{33\text{ oz}}{\$1.79} = \frac{128\text{ oz}}{x}$

Cross multiply to solve for x in the proportion:

$(33) \times (x) = (\$1.79) \times (128)$

$33x = \$229.12$

$x \approx \$6.94$

So, the orange juice costs $6.94 per gallon, and the milk costs $3.99 per gallon.

Subtract the cost per gallon of milk from the cost per gallon of orange juice.

$\$6.94 - \$3.99 = \$2.95$

Round the difference to the nearest dollar.

$2.95 rounds to $3.

Question analysis

A store sells milk by the gallon for $3.99 and orange juice in six-packs of 5.5-fluid-ounce bottles for $1.79. Students work out what a gallon of the juice would cost at the same price per ounce and how much more that is than the milk, to the nearest dollar. The ounces in a gallon must be known, and several steps involve decimals.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3d converting measurement units
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Unit prices are grade 6 and 7 work, but this chains five steps with decimals and needs a fact the question does not give, the number of fluid ounces in a gallon. Typing a rounded difference, with no choices, adds risk.
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: Something you must already know moderate
Many steps moderate
Error-prone arithmetic minor
An answer in an unexpected form minor
Solution: 5 steps
Reading level: Question and choices: about grade 5, 60 words.
Corrected from the DOE handbook: A rounded value in the explanation is now marked as approximate (≈ instead of =).
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 calculation is long, and its key fact is missing from the page. Students who know the 128 fluid ounces still face a division by 33 that does not come out evenly, which tempts them to round the price per ounce to 5 cents and lose about 50 cents a gallon. Keeping the exact quotient to the end, or scaling 33 ounces up by about 3.9, keeps the difference near $2.95, which rounds to 3.

Something you must already know moderate Working it out
The question never says that a gallon holds 128 fluid ounces, yet the whole comparison depends on it; the 6-pack holds only 6 × 5.5 = 33 ounces. (Anderson & Krathwohl, 2001)
Many steps moderate Working it out
Five moves follow one another: the ounces in a pack, the scale-up to 128 ounces, the division by 33, the subtraction of $3.99 and the rounding to a whole dollar. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
1.79 × 128 = 229.12 and 229.12 ÷ 33 are long decimal calculations without a calculator, and a slip in either moves the price of a gallon. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The rounded difference is typed with no choices, so an unrounded $2.95, or a figure rounded too soon, looks just as finished as the whole-dollar answer. (Enright, Morley & Sheehan, 2002)

Strategy: Find the ounces in one six-pack, scale its price up to the 128 fluid ounces in a gallon, subtract the milk's price, and round only at the end.

Traps

Stops one step early: A choice is a number from partway through the solution.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

7 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the juice's cost per gallon, about $6.94, rounds it to $7, and enters that without subtracting the $3.99 price of the milk.
2 Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds the price per ounce, 1.79 ÷ 33 ≈ 0.054, to $0.05, so a gallon costs $6.40 and the difference, $2.41, rounds to 2.
2.95 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Works out the difference of $2.95 correctly but enters it as it stands, though the question asks for it to the nearest dollar.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation goes from 33x = $229.12 to “x = $6.94”. But 229.12 ÷ 33 = 6.9430…, so $6.94 is the cost per gallon rounded to the cent. The site writes x ≈ $6.94. Nothing else changes: $6.94 − $3.99 = $2.95, which rounds to $3, and the unrounded cost gives the same answer.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fluid ounces in a gallon: $\text{fluid ounces} = 128 \times \text{gallons}$; A gallon holds 128 fluid ounces. This is a fact you must know, because the question does not give it. A fluid ounce measures an amount of liquid. Multiply gallons by 128 to change them into fluid ounces.
Cross-multiplying a proportion: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; A proportion says two ratios are equal. Here each ratio is ounces over dollars. a and c are the tops, and b and d are the bottoms. Multiply each top by the other bottom, and the two products are equal. That turns the proportion into an equation.

Sort out the question
Milk costs $3.99 for 1 gallon. Orange juice comes in a 6-pack of 5.5-fluid-ounce bottles for $1.79. You need how much more a gallon of juice would cost than a gallon of milk. Round to the nearest dollar.
The question doesn’t say how big a gallon is. You must know this fact: 1 gallon is 128 fluid ounces.
Find the ounces in a 6-pack
Each of the 6 bottles holds 5.5 fluid ounces:
$6 \times 5.5 = 33$
So $1.79 buys 33 fluid ounces of juice.
Set up a proportion
Let x be the cost in dollars of a gallon, 128 fluid ounces, of juice. The price per ounce stays the same. So ounces over dollars is the same for the pack and the gallon:
$\frac{33}{1.79} = \frac{128}{x}$
Cross-multiply:
$33 \times x = 1.79 \times 128 = 229.12$
Find the cost of a gallon
Divide both sides by 33:
$x = 229.12 \div 33 \approx 6.94$
The sign ≈ means about equal. The exact value is 6.943…, so a gallon of juice costs about $6.94.
Watch out: don’t round too early. One ounce costs about $0.054. Rounding that to $0.05 makes a gallon $6.40, and your answer becomes 2.
Subtract the milk’s price
Find how much more the juice costs:
$6.94 - 3.99 = 2.95$
A gallon of juice costs $2.95 more than a gallon of milk.
Watch out: entering 7, the juice’s price, forgets to subtract the milk.
Round and enter it
The question asks for the nearest dollar. $2.95 is between $2 and $3. Its cents, 95, are 50 or more, so it rounds up to $3. Do not enter 2.95.
Type only the number in the box. So you would enter 3.

Check your answer

Estimate. 128 ounces is almost 4 packs, since $4 \times 33 = 132$ . Four packs cost $4 \times 1.79 = 7.16$ dollars, so a gallon costs a bit less than that. Milk costs about $4. About $7 minus $4 is $3, which matches.

More questions like this

The original DOE question

Question 15 as shown on the DOE’s digital test, page 88 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 88, question 15. This is the DOE’s version, before the corrections listed in the analysis.

Question 15 of 69

2026 Practice Test C Math, question 22

Question

During a hiking trip, a group of students recorded the elevation (in feet) at several points along the trail:

$-120\text{ ft},\ -80\text{ ft},\ 0\text{ ft},\ 45\text{ ft},\ -30\text{ ft},\ 90\text{ ft}$

What is the difference between the highest and lowest elevations recorded on the trail?

Enter your answer in the space.

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

Show the answer

Correct answer: 210

Determine the difference between the highest and lowest elevations:

Determine the highest elevation:

Of −120, −80, 0, 45, −30, and 90, the greatest value is 90 because its location is farthest to the right on a number line.

Highest elevation $= 90\text{ ft}$

Determine the lowest elevation:

Of −120, −80, 0, 45, −30, and 90, the lowest value is −120 because its location is farthest to the left on a number line.

Lowest elevation $= -120\text{ ft}$

Determine the difference between the two elevations:

$90\text{ ft} - (-120)\text{ ft} = 90\text{ ft} + 120\text{ ft} = 210\text{ ft}$

Question analysis

A group of students on a hiking trip records six elevations in feet, some below zero and some above, and finds the difference between the highest and the lowest, typing the result. It tests ordering positive and negative numbers and subtracting integers. Spotting the lowest reading among several negatives and subtracting a negative number correctly are where errors creep in.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7b order statements in context
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the gap between a positive and a negative value by subtracting is core grade 7 work with integers. Picking the extremes from six readings is quick, so the demand is the subtraction of a negative number.
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
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Picking the highest and lowest of six readings is quick, though −120 must be recognized as lower than −80 and −30. The points are lost in the subtraction: the span runs 120 feet up to zero and 90 more above it, and students who treat the readings as plain sizes find only 30. Picturing the two readings on a vertical number line shows the whole span.

Error-prone arithmetic moderate Working it out
The difference is 90 − (−120), and subtracting a negative means adding 120, so the part of the trail below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The difference is typed with no choices, so 30 or −30, the results of mishandling the negative, gets no warning that the two readings lie on opposite sides of zero. (Enright, Morley & Sheehan, 2002)

Strategy: Pick the greatest and the least elevation by their places on a number line, then subtract the least from the greatest, turning the subtraction of a negative into an addition.

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).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

30 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 120 − 90 = 30, as if both readings were on the same side of zero, leaving out the stretch below zero.
-30 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Combines the two extremes by adding, 90 + (−120) = −30, the net of the readings, instead of subtracting the lowest from the highest.
-210 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −120 − 90 = −210, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between highest and lowest: $\text{difference} = \text{highest} - \text{lowest}$; The difference between the highest and lowest values is how far apart they are. Subtract the lowest from the highest, so the answer is never negative.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding a positive. On a number line, it means moving to the right.

Sort out the question
Elevation is height above or below a starting level, here 0 feet. Negative elevations are below 0. The six readings are −120, −80, 0, 45, −30 and 90 feet. You need the difference between the highest and the lowest.
Find the highest and the lowest
Picture the readings on a number line. The highest is farthest right: 90 feet. The lowest is farthest left: −120 feet.
For negative numbers, the bigger the number after the minus sign, the lower it is. So −120 is lower than −80.
Subtract the lowest from the highest
Use the difference formula. Subtracting a negative is the same as adding:
$90 - (-120) = 90 + 120 = 210$
So the two readings are 210 feet apart.
Watch out: $120 - 90 = 30$ treats both readings as if they were above 0. Adding them, $90 + (-120) = -30$ , is wrong too.
Enter the answer
A difference like this is a distance, so it is positive. Subtracting in the other order gives −210, which is wrong. Type only the number, with no units.
So you would enter 210.

Check your answer

Count the distance in two parts. From −120 feet up to 0 is 120 feet. From 0 up to 90 is 90 more feet. Together, $120 + 90 = 210$ feet. That matches.

More questions like this

The original DOE question

Question 22 as shown on the DOE’s digital test, page 95 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 95, question 22.

Question 16 of 69

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 17 of 69

2026 Practice Test C Math, question 29

Question

A smoothie shop uses a consistent ratio of fruit to yogurt in its recipes. The table shows the proportional relationship between the number of cups of fruit (x) and the number of cups of yogurt (y) in different smoothie batches.

SMOOTHIE RECIPE

Cups of Fruit x	Cups of Yogurt y
6	1.5
8	2.0
10	2.5
12	3.0

What is the constant of proportionality of the number of cups of yogurt to the number of cups of fruit?

Enter your answer in the space.

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

Show the answer

Correct answer: 0.25

To find p, the constant of proportionality of the number of cups of yogurt, y, to the number of cups of fruit, x, use the equation $y = px$ .

The table shows that there are 1.5 cups of yogurt for 6 cups of fruit.

$1.5 = 6p$

$\frac{1.5}{6} = \frac{6p}{6}$

$\frac{1}{4} = 0.25 = p$

The constant of proportionality is 0.25.

Question analysis

A smoothie shop's recipe table lists cups of fruit and cups of yogurt for four batches in a proportional relationship. Students type the constant of proportionality of yogurt to fruit. It tests finding a unit rate from a table, and since fruit is listed first while the question names yogurt first, the ratio is easy to take upside down.

NYS standards: 7.RP.2b constant of proportionality
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the constant of proportionality in a table is a core grade 7 skill, and these values are friendly decimals. Typing the answer with no choices, and taking the ratio in the order the question gives, are the added demands.
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: 1 step
Reading level: Question and choices: about grade 6, 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

Any row gives the answer with one division, and the values divide evenly, so the points are lost on direction. Fruit fills the first column and comes first in the stem’s own ratio, which invites dividing it by the yogurt. The table also climbs by a steady half cup of yogurt, a number that looks like a constant but ignores the 2 cups of fruit added each time. Checking that 0.25 times each fruit value gives its yogurt value settles it.

Turning words into math moderate Understanding the question
The opening sentence speaks of a ratio of fruit to yogurt, but the question asks for the constant of yogurt to fruit, so the division is y ÷ x, against the table’s column order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed into a box with no choices, so 4, the ratio taken the other way round, is a believable entry that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Check which quantity the question names first, divide that column by the other in any row, and confirm the quotient is the same in a second row.

Traps

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

The wrong answers, classified

4 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides cups of fruit by cups of yogurt, 6 ÷ 1.5 = 4, following the stem’s “ratio of fruit to yogurt” and turning the requested constant upside down.
0.5 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the steady rise of 0.5 cup of yogurt from row to row as the constant, ignoring that the fruit rises by 2 cups each time.
1.5 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives 1.5, the first yogurt value in the table, as if the constant were an amount in one batch rather than a ratio of two amounts.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; k is the constant of proportionality. y and x are two matching amounts from the same row of the table. In a proportional relationship, y = kx, so y is always k times x. That is why dividing y by x gives the same k in every row.

Sort out the question
The table lists cups of fruit (x) and cups of yogurt (y) for four batches. You need the constant of proportionality of yogurt to fruit.
That is the number of cups of yogurt for each 1 cup of fruit. It is also the number you multiply x by to get y.
Divide yogurt by fruit
Use the first row: 6 cups of fruit and 1.5 cups of yogurt. Yogurt goes on top, because the question says yogurt to fruit:
$k = \frac{y}{x} = \frac{1.5}{6} = 0.25$
So there is 0.25 cup, or one quarter of a cup, of yogurt for each cup of fruit.
Watch out: the first sentence says fruit to yogurt, but the question asks the other way. Dividing $6 \div 1.5 = 4$ turns the answer upside down.
Check two more rows
The constant must be the same in every row. Test two more:
$\frac{2.0}{8} = 0.25 \qquad \frac{3.0}{12} = 0.25$
They match. Watch out: the yogurt goes up by 0.5 cup from row to row. That 0.5 is not the constant. The fruit goes up by 2 cups at the same time, and $0.5 \div 2 = 0.25$ .
Enter the number
Type only the number in the box, with no units. You can type 0.25 or .25. On this site, 1/4 counts too.
So you would enter 0.25.

Check your answer

Multiply each amount of fruit by 0.25. $6 \times 0.25 = 1.5$ , $8 \times 0.25 = 2$ , $10 \times 0.25 = 2.5$ and $12 \times 0.25 = 3$ . Every result matches the yogurt column. Also, each batch has less yogurt than fruit, so the constant should be less than 1.

More questions like this

The original DOE question

Question 29 as shown on the DOE’s digital test, page 102 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 102, question 29.

Question 18 of 69

2026 Practice Test C Math, question 32

Question

$A box plot titled MONTHLY HIGH TEMPERATURES IN NEW YORK above a number line labeled Temperature (°F), from 36 to 88 with a labeled tick every 4. The left whisker starts at about 39, just left of the tick for 40. The box runs from 48 to about 78, between the ticks for 76 and 80, with the median line at 64. The right whisker ends at 84.$

The box plot shows the average monthly high temperatures in New York City for 12 months. What is the difference between the range and the interquartile range of the temperature data?

Enter your answer in the space.

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

Show the answer

Correct answer: 15

Determine the range by subtracting the minimum value on the box plot from the maximum value on the box plot:

Maximum value $= 84^{\circ}\text{ F}$ , Minimum value $= 39^{\circ}\text{ F}$

$84^{\circ}\text{ F} - 39^{\circ}\text{ F} = 45^{\circ}\text{ F}$

Determine the interquartile range by subtracting the lower quartile value on the box plot from the upper quartile value on the box plot:

Upper Quartile Value $= 78^{\circ}\text{ F}$ , Lower Quartile Value $= 48^{\circ}\text{ F}$

$78^{\circ}\text{ F} - 48^{\circ}\text{ F} = 30^{\circ}\text{ F}$

Determine the difference between the range (45° F) and the interquartile range (30° F):

$45^{\circ}\text{ F} - 30^{\circ}\text{ F} = 15^{\circ}\text{ F}$

Question analysis

A box plot of New York City's average monthly high temperatures for 12 months sits above a number line labeled every 4 degrees. Students type the difference between the range and the interquartile range of the data. Two of the plot's values fall between labeled ticks, so careful reading of the scale decides the answer.

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Reading a box plot and finding its interquartile range are grade 7 skills. Here two of the five values fall between ticks spaced 4 apart, and the range, the IQR and their difference must all be found and typed with no choices.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Many steps minor
Something you must already know minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 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 method is standard, but the key rests on reading the scale between its labels. The axis is ticked only every 4 degrees, and the minimum and the upper quartile land off the ticks, so rounding either one to the nearest label moves the result by a degree or two with nothing to flag it. Students who find the range correctly can also stop there, entering 45, the larger of the two spreads, instead of the difference the question asks for.

Reading a figure, table or graph major Understanding the question
Two of the five values fall between ticks spaced 4 degrees apart: the left whisker ends just short of the 40 tick, at 39, and the box’s right edge sits midway between 76 and 80, at 78. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Four values are read and combined in three subtractions, the range, the interquartile range and then their difference, so one misread value changes the final entry. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Neither measure is defined on the page: the range spans the whiskers, and the interquartile range is the width of the box, the upper quartile minus the lower. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The result is typed with no choices to compare, so an entry one or two degrees off, from rounding a value to the nearest tick, looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Read the minimum, both quartiles and the maximum, judging where each mark falls between the ticks, then subtract for the range and for the IQR, and subtract those two results.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

14 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Reads the left whisker’s end as 40, the nearest labeled tick, instead of 39, so the range becomes 84 − 40 = 44 and 44 − 30 = 14.
45 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the range, 84 − 39 = 45 degrees, and enters it, stopping before the interquartile range of 30 is subtracted from it.
17 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads the box’s right edge as 76, the tick just left of it, instead of 78, so the IQR becomes 76 − 48 = 28 and 45 − 28 = 17.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range: $\text{range} = \text{maximum} - \text{minimum}$; The maximum is the greatest value, at the right end of the right whisker. The minimum is the least value, at the left end of the left whisker. The range tells how far the whole set of data spreads.
Interquartile range (IQR): $\text{IQR} = Q_3 - Q_1$; Q3 is the upper quartile, the right edge of the box. Q1 is the lower quartile, the left edge of the box. The IQR is the width of the box. It tells how far the middle half of the data spreads.

Sort out the question
A box plot shows data with five numbers. The two whiskers are the lines on each side of the box. Their ends are the minimum, the least value, and the maximum, the greatest. The box runs from the lower quartile to the upper quartile. These are the middles of the lower half and the upper half of the data. The line inside the box is the median, the middle value.
You need the range minus the interquartile range (IQR), in degrees.
Read the values between the ticks
The ticks on the number line are 4 degrees apart. So each degree is one quarter of the space between two ticks. Three values sit right on ticks: the box’s left edge at 48, the median at 64, and the right whisker’s end at 84.
- The left whisker starts just left of the 40 tick. It sits about one quarter of the way back toward 36. So the minimum is $40 - 1 = 39$ degrees.
- The box’s right edge is halfway between 76 and 80. Halfway is 78 degrees.
Find the range
Use the range formula:
$\text{range} = 84 - 39 = 45$
The data spread over 45 degrees.
Watch out: rounding the minimum to the 40 tick gives a range of 44. Then the final answer comes out 14, which is wrong.
Find the IQR
Use the IQR formula with the two edges of the box:
$\text{IQR} = 78 - 48 = 30$
The middle half of the data spreads over 30 degrees.
Watch out: reading the box’s right edge as 76 gives an IQR of 28. Then the final answer comes out 17.
Subtract the IQR from the range
The question asks for the difference between the two:
$45 - 30 = 15$
Watch out: 45 is only the range. Do not stop there. Type only the number, with no degree sign.
So you would enter 15.

Check your answer

The range covers the whole plot, and the IQR covers only the box. So their difference is the two whiskers added together. The left whisker is $48 - 39 = 9$ degrees long. The right whisker is $84 - 78 = 6$ degrees long. $9 + 6 = 15$ , which matches.

More questions like this

The original DOE question

Question 32 as shown on the DOE’s digital test, page 105 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 105, question 32.

Question 19 of 69

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 20 of 69

2026 Practice Test C Math, question 40

Question

In basketball, a player can earn 1 point for making a free throw and 2 points for making another basket. Marcus makes a mean of 8 free throws per game, and he scores a total of at least 25 points per game. The inequality below shows this relationship, where b represents the number of 2-point baskets Marcus makes.

$2b + 8 \geq 25$

What is the number of 2-point baskets Marcus needs to make if he wants to score at least his minimum total points for a game and make his mean number of free throws?

A $b \leq 8$
B $b \geq 9$
C $b \geq 16$
D $b \leq 25$

Show the answer

Correct answer: B

Solve the inequality $2b + 8 \geq 25$ .

$2b + 8 - 8 \geq 25 - 8$

$2b \geq 17$

$\frac{2b}{2} \geq \frac{17}{2}$

$b \geq 8.5$

Since the number of baskets made must be a whole number, the fewest 2-point baskets, b, he must make is 9.

Question analysis

In basketball, Marcus makes a mean of 8 one-point free throws a game and wants at least 25 points, shown by the inequality 2b + 8 ≥ 25. Students find how many 2-point baskets he needs. It tests solving a two-step inequality and interpreting the solution in context, since a player cannot make part of a basket.

NYS standards: 7.EE.4b two-step inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step inequality from a word problem is a grade 7 standard. Here the inequality is already written, so the extra demand is interpreting its solution as a count of baskets, typical of grade 7 tasks.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: An answer in an unexpected form minor
Complex sentences minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 92 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The inequality is handed over, so the solving is two routine moves; the judgment comes after. Dividing 17 by 2 leaves a half, and since a basket cannot be split, the boundary has to move up to 9: eight baskets give 2(8) + 8 = 24 points, one short of 25. Students who flip the symbol while dividing, or who move the 8 without changing its sign, still find a choice to match.

An answer in an unexpected form minor Choosing the answer
Solving gives b ≥ 8.5, which no choice shows; it has to be recognized as b ≥ 9, because Marcus cannot make half a basket. (Enright, Morley & Sheehan, 2002)
Complex sentences minor Understanding the question
The 33-word question restates the given inequality as “at least his minimum total points” and “make his mean number of free throws”, wording harder than the math it describes. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Undo the added 8 and then the 2 to solve for b, then ask which whole numbers of baskets actually reach 25 points, testing the boundary value in the original inequality.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reverses the inequality when dividing by 2, as if dividing always flipped it, so b ≤ 8.5, and whole baskets give b ≤ 8.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves the 8 to the right without changing its sign, so 2b ≥ 33 and b ≥ 16.5, then drops the half to match b ≥ 16.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 25-point total from the inequality as a limit on baskets and writes at most, a number copied from the question rather than solved for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality: $\text{if } A \geq B, \text{ then } A - C \geq B - C \text{ and } \frac{A}{2} \geq \frac{B}{2}$; A and B are the two sides. The sign ≥ means greater than or equal to. Taking the same amount C from both sides keeps the sign the same. Dividing both sides by a positive number, like 2, also keeps it the same. Only a negative number flips it.

Sort out the question
A free throw is worth 1 point, and each other basket is worth 2 points. Marcus makes 8 free throws, which give 8 points. The letter b is the number of 2-point baskets, so they give 2b points.
He wants at least 25 points in all. That is the inequality $2b + 8 \geq 25$ . An inequality compares two amounts. The sign ≥ means greater than or equal to. You need the values of b that work.
Subtract 8 from both sides
Undo the added 8 first:
$2b + 8 - 8 \geq 25 - 8$
$2b \geq 17$
Watch out: adding 8 instead gives $2b \geq 33$ . That leads to choice C, which is wrong.
Divide both sides by 2
2b means 2 times b. Divide both sides by 2. Since 2 is positive, the sign stays the same:
$\frac{2b}{2} \geq \frac{17}{2}$
$b \geq 8.5$
Watch out: do not flip the sign here. Flipping it gives $b \leq 8.5$ , which leads to choice A.
Think about whole baskets
Marcus cannot make half a basket. So test the whole numbers near 8.5:
$2 \times 8 + 8 = 24$
$2 \times 9 + 8 = 26$
Eight baskets give 24 points, one short of 25. Nine baskets give 26 points, which is enough. So he needs 9 or more baskets.
Give the answer
The number of baskets is 9 or more, written $b \geq 9$ . Choice D, $b \leq 25$ , just copies the 25 from the points.
So the answer is B, $b \geq 9$ .

Check your answer

Try a bigger number, such as b = 12. Then $2 \times 12 + 8 = 32$ points, which is at least 25. Any number of baskets above 9 also works, and that is what ≥ says. Below 9, the total stays at 24 points or fewer.

More questions like this

The original DOE question

Question 40 as shown on the DOE’s digital test, page 113 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 113, question 40.

Question 21 of 69

2026 Practice Test C Math, question 41

Question

Elijah is at the school bookstore. He wants to buy a water bottle for $12.50 and some notebooks that cost $4.75 each. He has at most $35.00 to spend. Write an inequality that represents the number of notebooks, n, Elijah can purchase.

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

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

Show the answer

Correct answer: 12.50 + 4.75n ≤ 35.00

The question uses the phrase “at most,” which means “less than or equal to” (≤).

Since notebooks cost $4.75 each and Elijah will buy n notebooks, the total cost just for the notebooks is found by multiplying $4.75 by n, or $4.75n$ .

Combining the cost of a water bottle ($12.50) with the cost of n notebooks, the inequality is $12.50 + 4.75n \leq 35.00$ .

This may also be written as $4.75n + 12.50 \leq 35.00$ .

Question analysis

Elijah has at most $35.00 to buy one water bottle and some notebooks at the school bookstore, with a price given for each. Students type an inequality for the number of notebooks, n, he can buy. It tests writing an inequality from a word problem. The two prices play different roles, and the phrase "at most" decides the inequality symbol.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an inequality for a budget with a one-time cost and a cost per item is typical grade 7 work. The answer is typed, so both the symbol for "at most" and the role of each price must be right.
Estimated difficulty: Straightforward: two steps or one careful reading.
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 5, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The prices are given in the order they are used, so the expression for the total is the easier half. The points are lost on the symbol: “at most” has to be heard as no more than, with equality allowed, and the keypad offers all four inequality signs side by side. Pricing one and two notebooks with the bottle, $17.25 and $22.00, shows which amount repeats and confirms the total.

Turning words into math moderate Understanding the question
The $12.50 bottle is bought once and the $4.75 is paid for each of n notebooks, and “at most $35.00” has to become ≤, which allows a total of exactly $35.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The whole inequality, with n, its symbol and both prices, is built on a keypad offering ≤, <, ≥ and >, so a wrong symbol looks as complete as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Price a few cases by hand, such as 1 and 2 notebooks with the water bottle, to see which amount repeats, then decide whether a total of exactly $35.00 is allowed.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

12.50 + 4.75n ≥ 35.00 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Builds the total correctly but reads “at most” as at least, writing ≥, which would mean Elijah must spend $35.00 or more.
12.50 + 4.75n < 35.00 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Builds the total correctly but uses a strict less-than sign, leaving out a total of exactly $35.00, which “at most” allows.
17.25n ≤ 35.00 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the bottle’s price to each notebook’s, $12.50 + $4.75 = $17.25, and multiplies by n, as if a bottle came with every notebook.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a one-time cost: $\text{total} = \text{one-time cost} + (\text{price each}) \times n$; The one-time cost is paid once, no matter what. The price each is paid once for every item. n is the number of items. You multiply the price each by n because you pay it n times.
At most: $\text{total} \leq \text{limit}$; The sign ≤ means less than or equal to. The total can be under the limit or exactly equal to it, but never over it. Use it for words like at most and no more than.

Sort out the question
Elijah buys one water bottle for $12.50. He also buys n notebooks at $4.75 each. He can spend at most $35.00.
You need an inequality for n. An inequality compares two amounts with a sign such as ≤ instead of =.
Price a few cases
Work out the cost for a few numbers of notebooks:
Notebooks Total ($)
1 $12.50 + 4.75 = 17.25$
2 $12.50 + 9.50 = 22.00$
3 $12.50 + 14.25 = 26.75$
The $12.50 shows up once. The $4.75 shows up once for each notebook.
Write the total cost
Use the total cost formula. The bottle is the one-time cost, and each notebook costs $4.75:
$12.50 + 4.75n$
Here 4.75n means 4.75 times n.
Watch out: do not add the two prices first. $17.25n$ charges for a new bottle with every notebook.
Choose the sign
At most $35.00 means the total can be less than $35.00 or exactly $35.00. It cannot be more. So use ≤:
$12.50 + 4.75n \leq 35.00$
Watch out: ≥ means at least, the opposite. And < leaves out a total of exactly $35.00, which Elijah is allowed to spend.
Enter it on the keypad
Type the whole inequality. Nothing is printed before the box. On the test, press 1, 2, ., 5, 0, +, 4, ., 7, 5, the n key, the ≤ key, then 3, 5, ., 0, 0. The ≤ key is in the fourth row. Skip the dollar signs. On this site, type <= for ≤.
So you would enter 12.50 + 4.75n ≤ 35.00. An equal form, such as 4.75n + 12.50 ≤ 35.00, counts too.

Notebooks	Total ($)
1	$12.50 + 4.75 = 17.25$
2	$12.50 + 9.50 = 22.00$
3	$12.50 + 14.25 = 26.75$

Check your answer

Test some numbers. For 4 notebooks, $12.50 + 4.75 \times 4 = 31.50$ , which is not over 35.00. For 5 notebooks, $12.50 + 4.75 \times 5 = 36.25$ , which is too much. So Elijah can buy up to 4 notebooks. That makes sense for a budget of $35.00.

More questions like this

The original DOE question

Question 41 as shown on the DOE’s digital test, page 114 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 114, question 41.

Question 22 of 69

2026 Practice Test C Math, question 42

Question

Tariq is making a clay block in the shape of a cube with a side length of $20\text{ cm}$ . He hollows out the center by removing clay to form a rectangular prism-shaped hole with a base of $5\text{ cm} \cdot 5\text{ cm}$ and a height of $20\text{ cm}$ .

$A cube labeled 20 cm along its front right edge. A tall rectangular prism-shaped hole runs from the top face of the cube down to the bottom face, the full height of the cube; its square opening on the top face is shaded gray and labeled 5 cm across. Hidden edges of the cube and the hole are dashed.$

What is the volume, in cubic centimeters, of the block after the clay is removed from the center?

Enter your answer in the space.

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

Show the answer

Correct answer: 7,500

First, find the volume of the cube using the formula $V = lwh$ .

The length, width, and height of the cube are each equal to 20 cm.

$V_c = 20\text{ cm} \times 20\text{ cm} \times 20\text{ cm} = 8{,}000\text{ cm}^3$

Next, find the volume of the cut-out square prism, with equal length and width of 5 cm and height of 20 cm.

$V_p = 5\text{ cm} \times 5\text{ cm} \times 20\text{ cm} = 500\text{ cm}^3$

Finally, subtract the volume of the prism from the volume of the cube.

$8{,}000\text{ cm}^3 - 500\text{ cm}^3 = 7{,}500\text{ cm}^3$

Question analysis

Tariq makes a clay cube 20 centimeters on each side and removes a rectangular prism from its center, with a 5 cm by 5 cm base and a height of 20 cm. Students type the volume of the block that is left, in cubic centimeters. It tests the volume of rectangular prisms, and the hole's size must be read carefully from the words and figure.

NYS standards: 7.G.6 area, surface area, and volume
6.G.2 volume with fractional edge lengths
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the volume of a solid made from rectangular prisms, here by taking one away, is named in the grade 7 standards. This is a direct case with whole numbers, two volumes and one subtraction.
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: Something you must already know minor
Reading a figure, table or graph minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 66 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 direct composite-volume problem with whole numbers: one cube, one hole, one subtraction. What it checks is that both volumes are real volumes, the hole as tall as the cube, and that the subtraction is done at the end. Students who stop at the hole’s volume, or who subtract only the square face, give amounts that are easy to type and hard to doubt without choices.

Something you must already know minor Working it out
The volume of a rectangular prism, length times width times height, is not given, and it is needed twice, once for the 20 cm cube and once for the hole. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The figure labels only the 5 cm opening and the cube’s 20 cm edge; the hole’s 20 cm height comes from the words, while the drawing shows it reaching from top to bottom. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
An answer in an unexpected form minor Choosing the answer
The volume is typed with no choices to compare, so the hole’s 500 cubic centimeters or the cube’s 8,000 can be entered without any warning. (Enright, Morley & Sheehan, 2002)

Strategy: Find the volume of the full cube, then the volume of the square hole that runs from the top face to the bottom, and subtract the hole from the cube.

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 figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

500 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the volume of the hole, 5 × 5 × 20 = 500 cubic centimeters, and gives the clay removed instead of the clay that is left.
7,875 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Treats the hole as a 5 cm cube, 5 × 5 × 5 = 125, though it runs the full 20 cm height, so 8,000 − 125 = 7,875.
7,975 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Subtracts only the area of the square opening, 5 × 5 = 25, from the cube’s 8,000, mixing a face’s area with the hole’s volume.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a rectangular prism: $V = l \times w \times h$; V is the volume, the space inside a solid. l is the length, w is the width and h is the height. A cube is a prism with all three the same. Volume is measured in cubic centimeters, tiny cubes 1 cm on each side.

Sort out the question
Tariq’s block starts as a cube 20 cm on each side. He removes a hole shaped like a rectangular prism, a box shape with flat rectangle sides. Its base is 5 cm by 5 cm, and its height is 20 cm. The figure shows it running from the top face to the bottom face.
You need the volume of the clay that is left, in cubic centimeters.
Find the volume of the cube
Use the volume formula. The length, width and height are all 20 cm:
$V = 20 \times 20 \times 20 = 8{,}000$
The full cube holds 8,000 cubic cm of clay.
Find the volume of the hole
The hole is a prism too. Its length and width are 5 cm, and its height is 20 cm:
$V = 5 \times 5 \times 20 = 500$
So 500 cubic cm of clay are removed.
Watch out: the hole is 20 cm tall, not 5 cm. Treating it as a small 5 cm cube gives 125 and the wrong answer 7,875. And 5 × 5 = 25 is only the area of the opening, not a volume. Subtracting it gives 7,975, also wrong.
Subtract the hole from the cube
The clay left is the cube minus the hole:
$8{,}000 - 500 = 7{,}500$
The block has 7,500 cubic cm of clay left.
Watch out: 500 is the clay taken out, not the clay that is left.
Enter the number
Type only the number, with no units. You can type 7500 or 7,500.
So you would enter 7,500.

Check your answer

Think of the cube as 20 flat layers, each 1 cm thick. Each layer is a 20 by 20 square, 400 square cm, with a 5 by 5 hole, 25 square cm. So each layer holds $400 - 25 = 375$ cubic cm of clay. Then $20 \times 375 = 7{,}500$ . It matches.

More questions like this

The original DOE question

Question 42 as shown on the DOE’s digital test, page 115 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 115, question 42.

Question 23 of 69

2026 Practice Test D Math, question 4

Question

If $x = \frac{1}{4}$ , what is the value of $\frac{2}{2 - x}$ ?

A $\frac{3}{4}$
B $\frac{8}{9}$
C $1\frac{1}{7}$
D $3\frac{1}{2}$

Show the answer

Correct answer: C

The first step is to substitute the value for x into the expression to get $\frac{2}{2 - \frac{1}{4}}$ .

Then, evaluate the denominator and finally divide the numerator by the denominator.

$\frac{2}{\left(\frac{7}{4}\right)} = 2 \div \left(\frac{7}{4}\right) = 2\left(\frac{4}{7}\right) = \frac{8}{7} = 1\frac{1}{7}$

Question analysis

This substitution question sets x equal to one fourth and asks for the value of 2 divided by (2 − x). Students substitute, subtract a fraction from a whole number, and divide by the result. The expression is a stacked fraction, so treating the whole denominator as one quantity and dividing rather than multiplying are where students slip.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Substituting a value and dividing by a fraction are grade 6 skills, and here they take three short steps with friendly numbers. The stacked fraction asks for care, but the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
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: 3 steps
Reading level: Question and choices: about grade 10, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Substitution is one step, so the question is about handling a fraction inside a fraction. The bar over 2 − x groups the whole denominator, and the division by 7/4 must flip it; both errors lead to listed choices. A size check helps: 2 − 1/4 is a little less than 2, so the value must be a little more than 1, which rules out every wrong choice.

Error-prone arithmetic moderate Working it out
The denominator 2 − 1/4 has to become 7/4 before anything is divided, and dividing 2 by 7/4 means multiplying by 4/7; multiplying by 7/4 instead is the usual slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The work ends at the improper fraction 8/7, but the choices show it only as a mixed number, listed among 3/4, 8/9 and 3 1/2. (Enright, Morley & Sheehan, 2002)

Strategy: Simplify the denominator to a single fraction first, then divide 2 by that fraction by multiplying by its reciprocal; treating the fraction bar as a grouping symbol avoids the common slip.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Lets the fraction bar cover only the first 2, working 2/2 − 1/4 = 1 − 1/4 = 3/4 instead of dividing 2 by the whole denominator.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Adds 1/4 to 2 instead of subtracting it, so the denominator becomes 9/4, and 2 ÷ 9/4 = 8/9.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds the denominator 7/4 correctly but multiplies 2 by it instead of dividing, 2 × 7/4 = 14/4 = 3 1/2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b/c is a fraction. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing asks how many copies of b/c fit into a.
Changing an improper fraction to a mixed number: $\frac{a}{b} = q\frac{r}{b}$; Use it when the top, a, is bigger than the bottom, b. q is how many whole times b fits into a. r is the remainder, what is left over. For 8/7, the 7 fits once with 1 left, so 8/7 = 1 1/7.

Sort out the question
You know $x = \frac{1}{4}$ . You need the value of 2 divided by (2 − x). A fraction bar means divide. It also groups, so you work out the whole bottom, 2 − x, before you divide.
Work out the bottom
Put in 1/4 for x. To subtract, write 2 as fourths, since 2 = 8/4:
$2 - \frac{1}{4} = \frac{8}{4} - \frac{1}{4} = \frac{7}{4}$
The bottom is 7/4.
Watch out: the sign is minus. Adding 1/4 gives 9/4 and then 8/9, choice B.
Divide 2 by 7/4
Now the problem is 2 ÷ 7/4. Use the rule for dividing by a fraction. Flip 7/4 to get 4/7, then multiply:
$2 \div \frac{7}{4} = 2 \times \frac{4}{7} = \frac{8}{7}$
Watch out: multiplying by 7/4 instead of 4/7 gives 14/4 = 3 1/2, choice D.
Write it as a mixed number
The choices use mixed numbers. 7 fits into 8 once, with 1 left over:
$\frac{8}{7} = 1\frac{1}{7}$
Watch out: the fraction bar covers all of 2 − 1/4. Working 2/2 = 1 first and then taking away 1/4 gives 3/4, choice A.
So the answer is C, $1\frac{1}{7}$ .

Check your answer

Estimate. The bottom, 1 3/4, is a bit less than 2. So 2 divided by it should be a bit more than 1, and 1 1/7 fits. Also check by multiplying: 7/4 × 8/7 = 56/28 = 2. That is the top, so the division is right.

More questions like this

The original DOE question

Question 4 as shown on the DOE’s digital test, page 77 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 77, question 4.

Question 24 of 69

2026 Practice Test D Math, question 5

Question

In a school raffle, 1 out of every 25 students wins a prize. Each prize includes a randomly selected trading card from a set of 100 different cards, labeled C1 through C100, with each card equally likely to be chosen. If one student is selected at random, what is the probability that the student both wins a prize and receives Card C17?

Write an expression that represents the probability that a student wins the prize and receives Card C17.

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

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

Show the answer

Correct answer: 1/25 · 1/100

To calculate the joint probability, first determine the probability of each independent event.

The probability that a randomly selected student wins the prize is $\frac{1}{25}$ .

The probability that a randomly selected student will receive Card C17 is $\frac{1}{100}$ , because Card C17 is 1 card out of 100 labeled cards.

The probability that a student wins the prize and receives Card C17 is the product of those two probabilities, $\frac{1}{25} \cdot \frac{1}{100}$ .

Question analysis

In a school raffle, 1 out of every 25 students wins a prize, and each prize includes one of 100 equally likely trading cards. Students type an expression for the probability that a randomly chosen student both wins and gets one particular card. It tests the probability of two independent events, and combining the two chances correctly is the main hurdle.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.8a uniform probability models
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students find compound probabilities, usually by listing outcomes for coins or number cubes. These two stages have 25 and 100 outcomes, far too many pairs to list, so a rule must be used, and the answer is typed as an expression.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 87 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 chance is stated almost directly, 1 out of 25 and 1 out of 100, so the question tests one rule: both events must happen, and the chances multiply. The word “and” pulls students toward adding, which gives a result larger than either chance, a sign that something is wrong. Others write only the card's chance and forget that the student must first win a prize.

Something you must already know moderate Working it out
The question never says that the chance of two independent events both happening is the product of their chances, so students must bring that rule to the 1/25 and the 1/100 themselves. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The answer is typed as an expression with no choices to compare, so a sum of the two chances, which looks just as finished, gets no warning on screen. (Enright, Morley & Sheehan, 2002)

Strategy: Write each event's chance as a fraction, then apply the rule for two independent events that must both happen, and check that the result is smaller than either chance alone.

Traps

Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Stops one step early: A choice is a number from partway through the solution.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

1/25 + 1/100 Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Reads “both wins a prize and receives Card C17” as a cue to add, so sums the chances, though needing both events must be less likely than either.
1/100 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Writes only the chance of receiving Card C17 and leaves out that the student must first be one of the 1 in 25 who win a prize.
1/125 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Combines “1 out of 25” and “1 out of 100” by adding the totals, 25 + 100, a faulty rule that gives neither the sum nor the product.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance that something happens, from 0 to 1. The top counts the outcomes you want. The bottom counts all the outcomes that could happen. Use it when every outcome is equally likely, like one card picked from 100.
Probability that two independent events both happen: $P(A \text{ and } B) = P(A) \times P(B)$; A and B are two events. Independent means one does not change the chance of the other. P(A) is the chance of A, and P(B) is the chance of B. Multiply them to get the chance that both happen. The result is smaller than either chance, since both things must happen.

Sort out the question
Two things must both happen. First, the student wins a prize. Second, the card in that prize is Card C17. You need an expression for that chance. An expression is a math phrase made of numbers and signs, with no equal sign. You do not have to work it out to one number.
Find the chance of winning
The question says “1 out of every 25 students wins a prize.” One student is the part that wins, and 25 is the whole group. So:
$P(\text{win}) = \frac{1}{25}$
Find the chance of Card C17
Each prize has one card, picked at random from 100 cards. Every card is equally likely. Card C17 is 1 of those 100 cards:
$P(\text{C17}) = \frac{1}{100}$
Multiply the two chances
Winning a prize does not change which card is inside it. So the two events are independent. To find the chance that both happen, multiply:
$P(\text{win and C17}) = \frac{1}{25} \cdot \frac{1}{100}$
The dot · means times. Worked out, this is 1/2500.
Watch out: the word and does not mean add here. 1/25 + 1/100 is bigger than either chance, but needing both things must be less likely. Writing only 1/100 forgets that the student must win a prize first.
Enter the expression
On the test, press the fraction key, a small box over a small box. Type 1 on top and 25 on the bottom. Move the cursor out of the fraction with the → key. Press the dot key for times. Then use the fraction key again for 1 over 100. On this site, type 1/25 * 1/100 or 1/25 · 1/100. The value 1/2500 also counts.
So you would enter 1/25 · 1/100.

Check your answer

Picture 2,500 students. 1 in 25 wins, so about 100 students win a prize. Of those 100, about 1 gets Card C17. That is 1 student out of 2,500, which matches 1/25 · 1/100 = 1/2500. It is smaller than 1/25 and 1/100, as it should be.

More questions like this

The original DOE question

Question 25 of 69

2026 Practice Test D Math, question 7

Question

A middle school library has a shelf with 40 books, each labeled with a different number from 201 to 240. One book is selected at random. What is the probability, as a percentage, of selecting book number 223?

Enter your answer in the space provided.

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

Show the answer

Correct answer: 2.5%

The probability that book number 223 will be selected at random from 40 books is 1 out of 40. To find the probability as a percentage, convert the fraction to a decimal.

The chance of selecting one specific book is $\frac{1}{40} = 0.025$ .

Convert the decimal to a percentage: $0.025 \times 100 = 2.5\%$

Question analysis

A library shelf holds 40 books numbered 201 to 240, and one is picked at random. Students find the probability of picking one particular book and type it as a percentage. It tests simple probability and converting a fraction to a percent; the book numbers play no part, and the decimal point needs care.

NYS standards: 6.SP.8a uniform probability models
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of one equally likely outcome is grade 6 content. The only further step is rewriting the fraction as a percent, which needs care with place value, so it is lighter than grade 7 compound probability.
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: An answer in an unexpected form moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The book numbers, 201 to 240, and the chosen number 223 play no part, since every book is equally likely and the shelf holds 40. The chance is 1 in 40, and the real demand is the form: changing it to a percent means a decimal with a zero in the tenths place, then a shift of two places. Students who know that 1/4 is 25% can check that 1/40 must be a tenth of that.

An answer in an unexpected form moderate Choosing the answer
Only the bold word percentage sets the form, since the empty box shows no % sign, so the probability 1/40 must be turned into 2.5, not entered as the fraction or as 0.025. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Dividing 1 by 40 gives 0.025, with a zero right after the decimal point, and losing that zero turns the percent into 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the books, write the chance of picking one particular book as a fraction, then turn the fraction into a decimal and multiply by 100 to get the percent.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

0.025 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the probability as a decimal, 1 ÷ 40 = 0.025, and enters it without multiplying by 100, though the box asks for a percentage.
25 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Loses the zero after the decimal point when dividing, taking 1 ÷ 40 as 0.25, as 1 ÷ 4 would be, and so enters 25.
1/40 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Enters the probability as the fraction 1/40, the form it was found in, overlooking the bold word percentage in the question.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance of an event. The top counts the outcomes you want. The bottom counts all the equally likely outcomes. Use it when each outcome has the same chance, like picking one book at random.
Changing a fraction to a percent: $\text{percent} = \frac{a}{b} \times 100$; a/b is the fraction. Percent means out of 100. Multiplying by 100 tells you how many out of 100 the fraction is. For example, 1/2 × 100 = 50, so 1/2 is 50%.

Sort out the question
The shelf has 40 books, numbered 201 to 240. One book is picked at random, so every book has the same chance. You need the chance of picking book 223, written as a percentage. Book 223 is on the shelf, since 223 is between 201 and 240.
Write the probability as a fraction
Only 1 book is number 223, and there are 40 books in all:
$P = \frac{1}{40}$
The numbers on the labels do not matter beyond this. They just give each book its own name.
Change the fraction to a decimal
Divide 1 by 40. It helps to start from 1/4 = 0.25. Since 40 is ten times 4, 1/40 is one tenth of 1/4:
$1 \div 40 = 0.025$
Watch out: there is a 0 right after the decimal point. Losing it gives 0.25 and then 25, which is wrong.
Change the decimal to a percent
Multiply by 100 to get a percent. That moves the decimal point two places to the right:
$0.025 \times 100 = 2.5$
So the chance is 2.5%.
Watch out: 0.025 is the probability as a decimal, and 1/40 is the same chance as a fraction. The question asks for a percentage, so neither one is the answer here.
Enter the number
The box shows no % sign. Type 2.5. Adding the sign, as 2.5%, also counts.
So you would enter 2.5.

Check your answer

2.5% means 2.5 out of every 100. Out of 40 books, that is 40 × 0.025 = 1 book, the one numbered 223. Also, one book out of 40 should be a small chance, and 2.5% is small.

More questions like this

The original DOE question

Question 7 as shown on the DOE’s digital test, page 80 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 80, question 7.

Question 26 of 69

2026 Practice Test D Math, question 8

Question

A square metal plate has a side length of 3 inches. A circular hole with a diameter of 2 inches is cut out from the center of the plate.

$A square, shaded gray, with a white circle cut out of its center. A side of the square is labeled 3 in., and the diameter of the circle is labeled 2 in. The shaded region is the part of the square outside the circle.$

What is the area of the shaded region, to the nearest hundredth of a square inch?

Enter your answer in the space.

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

Show the answer

Correct answer: 5.86

To determine the area of a square, use $A = s^{2}$ .

Begin by finding the length of the side of the square, which is 3 inches. The area of the square is $(3\text{ in})^{2} = 9\text{ square inches}$ .

To determine the area of the circle, use $A = \pi r^{2}$ .

The radius of the circle (the hole) is one-half the diameter or $2\text{ in.} \div 2 = 1\text{ in.}$ , so the area of the entire circle is $\pi(1\text{ in.})^{2} \approx 3.14\text{ sq in.}$

To determine the area of the shaded region, subtract the area of the hole in the square from the area of the square:

$9\text{ sq in.} - \pi\text{ sq in.} =$

$9\text{ sq in.} - (3.14)\text{ sq in.} =$

$5.86\text{ sq in.}$

The shaded area to the nearest hundredth of a square inch $= 5.86$ sq in.

Question analysis

A 3-inch square metal plate has a circular hole 2 inches across cut from its center, and students find the area of the metal that is left, to the nearest hundredth of a square inch. It tests the areas of a circle and a square. Using the radius, not the diameter, and subtracting the right areas are the keys.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a circle's area is a grade 7 standard, and subtracting a hole from a shape is common grade 7 practice. The radius must come from the diameter and the result must be rounded, but the steps are routine.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 49 words.
Corrected from the DOE handbook: A wrong symbol 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

Every number is small, so the work is choosing the right pieces: the area of the square, the area of the hole, and the difference. A radius of 1 makes the circle's area simply π, which is easy to confuse with the circumference, 2π, when the formulas are half remembered. The region asked for is the gray metal, not the hole, and the final rounding goes up, since the third decimal place is 8.

Something you must already know moderate Working it out
Neither area formula is given, and the hole is labeled by its 2-inch diameter, so students must recall A = πr² and halve the diameter to get the radius 1 first. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
The shaded metal is a square with a circle removed, so two area formulas are combined by subtraction rather than either one used alone. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
The result is typed to the nearest hundredth with no choices to compare, so the long decimal 5.858… must be rounded up, not cut off after two places. (Enright, Morley & Sheehan, 2002)

Strategy: Halve the diameter to get the radius, find the area of the square and of the circle, subtract the circle from the square, and round only at the end.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

2.72 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses the circumference, 2π × 1 ≈ 6.28, in place of the circle's area and subtracts it from the square's 9 square inches, getting about 2.72.
3.14 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the area of the circular hole, π × 1² ≈ 3.14 square inches, and enters it instead of the area of the shaded metal left around it.
5.85 Rounds too early (King et al., 2004)
Reveals a slip Expected pull: weak
Works the area out correctly as about 5.858 but cuts off the extra digits instead of rounding up to the nearest hundredth.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation gives the area of the hole as “π(1 in.)² ≅ 3.14 sq in.” The symbol ≅ means “is congruent to”, which an area and a number cannot be; the line means “is approximately equal to”, written ≈. The site now says “π(1 in.)² ≈ 3.14 sq in.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s^{2}$; A is the area, the amount of flat space inside. s is the length of one side. s² means s × s. It works because a square with side s holds s rows of s little unit squares.
Area of a circle: $A = \pi r^{2}$; A is the area. r is the radius, the distance from the center to the edge. It is half the diameter, the distance straight across through the center. π (pi) is a number a little more than 3, about 3.14. r² means r × r.

Sort out the question
The plate is a square with 3-inch sides. A circle with a 2-inch diameter is cut out of the middle. The diameter is the distance straight across the circle through its center. The shaded region is the metal that is left. You need its area to the nearest hundredth of a square inch.
Find the area of the square
Use the formula for the area of a square, with s = 3 inches:
$A = s^{2} = 3^{2} = 3 \times 3 = 9$
The whole plate, before the hole is cut, is 9 square inches.
Find the radius of the hole
The circle formula needs the radius, not the diameter. The figure labels the 2 in. distance across the hole. The radius is half of it:
$r = 2 \div 2 = 1$
The radius is 1 inch.
Find the area of the hole
Use the formula for the area of a circle, with r = 1 inch:
$A = \pi r^{2} = \pi \times 1^{2} = \pi \approx 3.14$
The sign ≈ means is about equal to. The hole is about 3.14 square inches.
Watch out: 2πr is the circumference, the distance around the circle, not its area. Using it gives 9 − 6.28 = 2.72. And 3.14 is the hole, not the metal, so do not stop there.
Subtract the hole from the square
The shaded metal is the square with the circle taken away:
$9 - \pi \approx 9 - 3.14 = 5.86$
The shaded region is about 5.86 square inches.
Round and enter the answer
With more digits of π, 3.1416, you get 9 − 3.1416 = 5.8584. The digit after the hundredths place is 8, which is 5 or more, so round up to 5.86. Chopping it to 5.85 is wrong. Type only the number, with no units.
So you would enter 5.86.

Check your answer

The hole is a bit more than a third of the plate, since 3.14 is a bit more than 9 ÷ 3 = 3. So the metal left should be a bit less than two thirds of 9, which is 6. The answer 5.86 is just under 6, so it makes sense.

More questions like this

The original DOE question

Question 8 as shown on the DOE’s digital test, page 81 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 81, question 8. This is the DOE’s version, before the corrections listed in the analysis.

Question 27 of 69

2026 Practice Test D Math, question 15

Question

$\frac{1}{11} + \frac{1}{22} + \frac{1}{33}$

What is the value of the expression shown above?

A $\frac{1}{22}$
B $\frac{1}{11}$
C $\frac{1}{6}$
D $\frac{3}{22}$

Show the answer

Correct answer: C

To add fractions, find a common denominator and rename the fractions using the common denominator. The least common denominator for 11, 22, and 33 is 66. After adding the fractions, the final fraction can be reduced to lowest terms.

$\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}$

$\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}$

$\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}$

$\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66} = \frac{1}{6}$

Question analysis

This fractions question asks for the value of 1/11 + 1/22 + 1/33. Students find a common denominator for three fractions, rename each one, add and simplify. The denominators are all multiples of 11, and adding the tops and bottoms straight across, or forgetting to rename a numerator, leads to tempting wrong choices.

NYS standards: 7.NS.1d properties for adding and subtracting
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding fractions with unlike denominators is taught before grade 7. Finding a common denominator for three related numbers and simplifying is short, routine work, lighter than typical grade 7 problems with signed rational numbers.
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: 3 steps
Reading level: Question and choices: about grade 9, 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 checks whether students rename fractions before adding, and each wrong choice comes from a shortcut that skips that step or stops early. Seeing that 22 and 33 are multiples of 11 points to 66 as the common denominator. A size check helps too: the sum must be more than 1/11, its largest term, which rules out two of the wrong choices at once.

Error-prone arithmetic moderate Working it out
Three unlike denominators must be renamed over 66, not over the product of all three, and the numerators 6, 3 and 2 have to be found one by one before adding. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The sum comes out as 11/66, which must be reduced to match a choice; its simplest form has a denominator, 6, unlike any in the expression. (Enright, Morley & Sheehan, 2002)

Strategy: Notice that 22 and 33 are multiples of 11, rename all three fractions over their least common multiple, add the numerators, and simplify 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.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the numerators and the denominators straight across, 3 over 11 + 22 + 33 = 66, and simplifies 3/66 to 1/22.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Puts the three 1s over the largest denominator, 33, without renaming the other two fractions, so 3/33 simplifies to 1/11.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds the first two fractions correctly, 2/22 + 1/22 = 3/22, and stops before adding the third fraction, 1/33.

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/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction. It works because n/n is 1, and multiplying by 1 changes nothing. Use it to give fractions the same bottom.
Adding fractions with the same denominator: $\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}$; d is the denominator, the bottom number, which tells the size of each piece. a, b and c are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

Sort out the question
You need to add three fractions: 1/11, 1/22 and 1/33. Their denominators (bottom numbers) are all different. You can only add fractions when the pieces are the same size. So first give all three the same denominator.
Find a common denominator
A common denominator is a number that 11, 22 and 33 all divide into. Notice that 22 = 2 × 11 and 33 = 3 × 11. So the smallest number that all three divide into is:
$2 \times 3 \times 11 = 66$
Rename each fraction
Multiply the top and bottom of each fraction by the number that makes the bottom 66:
$\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}$
$\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}$
$\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}$
Add the tops
Now every piece is a sixty-sixth. Add the tops and keep the bottom:
$\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66}$
Watch out: never add the bottoms. Adding straight across gives 3/66 = 1/22, choice A. Putting the three 1s over 33 gives 3/33 = 1/11, choice B.
Simplify the fraction
To simplify, divide the top and bottom by a number that goes into both. Here 11 goes into 11 and into 66:
$\frac{11}{66} = \frac{11 \div 11}{66 \div 11} = \frac{1}{6}$
Watch out: add all three fractions. Stopping after the first two gives 3/22, choice D.
So the answer is C, $\frac{1}{6}$ .

Check your answer

Estimate with decimals. 1/11 is about 0.091, 1/22 is about 0.045 and 1/33 is about 0.030. They add to about 0.166. And 1/6 is 0.1666…, a match. The other choices are all 0.136 or less.

More questions like this

The original DOE question

Question 28 of 69

2026 Practice Test D Math, question 16

Question

The table shows the proportional relationship between x cups of oatmeal and y ounces of raisins in a bread recipe.

BREAD RECIPE

Cups of Oatmeal (x)	Ounces of Raisins (y)
2	1.0
3	1.5
6	3.0

What is the constant of proportionality of the number of ounces of raisins to the number of cups of oatmeal?

Enter your answer in the space.

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

Show the answer

Correct answer: 0.5

To find p, the constant of proportionality of ounces of raisins, y, to cups of oatmeal, x, use the equation $y = px$ .

The table shows that there is 1 ounce of raisins for 2 cups of oatmeal.

$1 = 2p$

$\frac{1}{2} = \frac{2p}{2}$

$0.5 = p$

The constant of proportionality is 0.5.

Question analysis

A table for a bread recipe lists cups of oatmeal and the ounces of raisins that go with them, in a proportional relationship. Students type the constant of proportionality of raisins to oatmeal. It tests reading a unit rate from a table, and the order in which the question names the two quantities matters.

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from a table is a grade 7 standard, and here a single division in one row gives it. The demand is lighter than typical grade 7 work; the order of the ratio in the wording needs the most care.
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 moderate
An answer in an unexpected form minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 56 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 row gives the same ratio, so the division itself is easy; the question tests which quantity goes on top. Because oatmeal comes first in the table and its numbers are the larger ones, dividing cups by ounces feels natural and produces a clean whole number. Writing the relationship as y = kx, raisins equal k times oatmeal, and checking it on a second row such as 3 cups and 1.5 ounces fixes the order.

Turning words into math moderate Understanding the question
The phrase “of the number of ounces of raisins to the number of cups of oatmeal” fixes the order y ÷ x, while the table lists oatmeal first, inviting 2 ÷ 1.0. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed in with no choices, so an upside-down ratio of 2, a whole number, looks as finished as the decimal the table actually gives. (Enright, Morley & Sheehan, 2002)

Strategy: Divide an ounces-of-raisins value by the cups of oatmeal in the same row, in the order the question names them, and check that every row gives the same value.

Traps

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

The wrong answers, classified

2 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides cups of oatmeal by ounces of raisins, 2 ÷ 1.0, turning the ratio upside down: cups per ounce instead of ounces per cup.
1 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads the first raisin amount, 1.0 ounce, as the constant without dividing it by the 2 cups of oatmeal in the same row.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $y = kx \quad \text{so} \quad k = \frac{y}{x}$; x and y are two amounts that grow together at the same rate. k is the constant of proportionality, the number you always multiply x by to get y. Here y is ounces of raisins and x is cups of oatmeal, so k is ounces of raisins per cup of oatmeal.

Sort out the question
The table shows cups of oatmeal, x, and ounces of raisins, y. Proportional means the raisins are always the same multiple of the oatmeal. That multiple is the constant of proportionality. The question asks for raisins to oatmeal. So you want ounces of raisins for each cup of oatmeal: y ÷ x.
Divide y by x in one row
Use the first row of the table: 2 cups of oatmeal go with 1.0 ounce of raisins.
$k = \frac{y}{x} = \frac{1.0}{2} = 0.5$
So there is 0.5 ounce of raisins for each cup of oatmeal.
Check the other rows
A constant must be the same in every row:
$\frac{1.5}{3} = 0.5 \qquad \frac{3.0}{6} = 0.5$
Every row gives 0.5, so the rule is $y = 0.5x$ .
Watch out: the table lists oatmeal first, which tempts you to divide 2 by 1.0. That gives 2, which is cups per ounce, the ratio upside down. And 1.0 alone is just one raisin amount, not the constant.
Enter the answer
Type the constant in the box as a number, with no units. On this site, 0.5, .5 and 1/2 all count.
So you would enter 0.5.

Check your answer

Multiply each oatmeal amount by 0.5: 2 × 0.5 = 1.0, 3 × 0.5 = 1.5 and 6 × 0.5 = 3.0. Each result matches the raisins in its row. Also, every raisin number is smaller than its oatmeal number, so the constant must be less than 1.

More questions like this

The original DOE question

Question 16 as shown on the DOE’s digital test, page 89 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 89, question 16.

Question 29 of 69

2026 Practice Test D Math, question 17

Question

At Midville High School, 64 students are members of either the orchestra or the choir, or both. Of these students, 38 are members of the orchestra, and 46 are members of the choir. What fraction of the total is the number of students who are members of both the orchestra and the choir?

A $\frac{16}{21}$
B $\frac{10}{19}$
C $\frac{10}{23}$
D $\frac{5}{16}$

Show the answer

Correct answer: D

First, add the number of students that are members of the orchestra and the number of students that are members of the choir.

$38 + 46 = 84$

Then, subtract to find the number of students that are members of both orchestra and choir.

$84 - 64 = 20$

Then, find the fraction of the 64 students that are members of both orchestra and choir.

$\frac{20}{64} = \frac{5}{16}$

Question analysis

At Midville High School, 64 students belong to the orchestra, the choir or both; 38 are in the orchestra and 46 in the choir. Students find what fraction of the 64 belong to both groups. It is a Venn diagram counting question with a fraction, and the students in both groups are included in each membership number.

NYS standards: 7.SP.8b sample spaces for compound events
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Overlapping groups are seldom taught directly in grade 7, though the arithmetic is whole-number addition and subtraction. Students must first find the overlap and then choose the right whole for the fraction, two decisions where errors are common.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 54 words.
Corrected from the DOE handbook: A misstatement in the explanation’s first step was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question hides its key number: the students in both groups appear only as the amount by which 38 + 46 overshoots 64. A sketch of two overlapping circles, with 20 in the middle, 18 in orchestra only and 26 in choir only, shows the whole group at once. After that, the remaining risk is the denominator, since the question asks for a fraction of all 64 students, not of one group.

Turning words into math moderate Understanding the question
“Either the orchestra or the choir, or both” means the 64 counts each student once, while 38 + 46 counts the students in both groups twice; the overlap itself is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method minor Working it out
Nothing names the step that finds the overlap: the excess of 38 + 46 = 84 over 64 is the number in both groups, an idea students must bring themselves. (Schoenfeld, 1985)
An answer in an unexpected form minor Choosing the answer
The fraction 20/64 appears only in lowest terms, so it must be simplified by 4 before it can be matched with a choice. (Enright, Morley & Sheehan, 2002)

Strategy: Add the orchestra and choir counts, subtract the 64 students to find how many were counted twice, then put that number over the total and simplify.

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.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

A Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: strong
Takes 38 + 46 = 84 as the number of students, counting those in both groups twice, and puts the stated 64 over it to get 16/21.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the 20 students in both groups but compares them with the 38 orchestra members instead of all 64 students, so 20/38 = 10/19.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the 20 students in both groups but divides by the 46 choir members, using one group as the whole instead of all 64, so 20/46 = 10/23.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation begins “First, add the number of students that are members of both orchestra and choir. 38 + 46 = 84.” But 38 is the number of orchestra members and 46 the number of choir members; the number of students in both is what the next step finds (84 − 64 = 20). The site now says “First, add the number of students that are members of the orchestra and the number of students that are members of the choir.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap: $\text{total} = O + C - \text{both}$; O is the number in the orchestra, and C is the number in the choir. Students in both groups are counted in O and again in C. Subtracting them once makes each student count one time. Here every student is in at least one group, so this gives everyone.
Fraction of a total: $\text{fraction} = \frac{\text{part}}{\text{whole}}$; The part is the group you are asked about. The whole is the full group it comes from. Here the part is the students in both groups, and the whole is all 64 students. Simplify by dividing the top and bottom by the same number.

Sort out the question
There are 64 students in all. Each one is in the orchestra, the choir or both. 38 are in the orchestra, and 46 are in the choir. Some students are in both, but the question does not say how many. You need that number as a fraction of all 64 students.
Add the two groups
Add the number in the orchestra and the number in the choir:
$38 + 46 = 84$
That is more than the 64 students there are. Why? A student in both groups is counted once with the orchestra and once more with the choir.
Find the students in both
The extra amount over 64 is the students who were counted twice:
$84 - 64 = 20$
So 20 students are in both groups. This is the overlap rule turned around: both = O + C − total.
Watch out: 84 is not the number of students. Putting 64 over 84 gives 16/21, choice A.
Write the fraction and simplify
The part is 20 students, and the whole is all 64 students. Both numbers divide by 4:
$\frac{20}{64} = \frac{20 \div 4}{64 \div 4} = \frac{5}{16}$
Watch out: the whole is all 64 students, not one group. Using 38 gives 10/19, choice B. Using 46 gives 10/23, choice C.
So the answer is D, $\frac{5}{16}$ .

Check your answer

Split the 64 students into three groups that do not overlap. Only orchestra: 38 − 20 = 18. Only choir: 46 − 20 = 26. Both: 20. Add them: 18 + 26 + 20 = 64. That matches the total, so 20 is right.

More questions like this

The original DOE question

Question 17 as shown on the DOE’s digital test, page 90 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 90, question 17. This is the DOE’s version, before the corrections listed in the analysis.

Question 30 of 69

2026 Practice Test D Math, question 18

Question

Which of the following decimals is equivalent to $1\frac{2}{3} + 2\frac{4}{9}$ ?

A $3.\overline{1}$
B $3.\overline{6}$
C $4.\overline{1}$
D $4.\overline{9}$

Show the answer

Correct answer: C

First, add the mixed numbers. Then, convert the fraction $\frac{1}{9}$ to a decimal.

$1\frac{2}{3} + 2\frac{4}{9}$

$1\frac{6}{9} + 2\frac{4}{9}$

$3 + \frac{10}{9}$

$3 + 1\frac{1}{9}$

$4\frac{1}{9} = 4.111\ldots = 4.\overline{1}$

Question analysis

This question asks which repeating decimal equals the sum of the mixed numbers 1 2/3 and 2 4/9. Students rename the fractions with a common denominator, add, and convert the fraction part to a decimal with a bar over its repeating digit. The choices differ in both the whole number and the repeating digit, so each step has to be right.

NYS standards: 7.NS.2d fractions to decimals by long division
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding mixed numbers and writing a fraction as a repeating decimal are standard grade 7 skills. This combines them in three short steps; renaming a fraction and reading the bar notation are the only places that need care.
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: 3 steps
Reading level: Question and choices: about grade 13, 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

Two fraction moves and one conversion stand between the sum and the choices. The carry is the main hazard: 10/9 has to become 1 1/9, and students who keep only the 1/9 land on a choice with the wrong whole number. Estimating first, about 1.7 + 2.4 = 4.1, rules out both choices that begin with 3 and the one that equals 5.

Error-prone arithmetic moderate Working it out
Adding 2/3 and 4/9 needs 2/3 renamed as 6/9, and the sum 10/9 is more than 1, so a whole must be carried into the 3 before converting. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The choices are repeating decimals with bars, so a mixed number in ninths has to be recognized as a digit repeating forever; knowing that ninths repeat their numerator makes the match quick. (Enright, Morley & Sheehan, 2002)

Strategy: Rename 2/3 as ninths, add the whole numbers and the fractions separately, regroup any fraction of 9/9 or more as a whole, then divide the leftover numerator by 9.

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.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Adds the fractions to 10/9 but keeps only the 1/9 left over, forgetting to carry the whole 1 into the 3, so the sum becomes 3 1/9.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators 2 and 4 over 9 without renaming 2/3 as 6/9, so gets 3 6/9, which is 3.666…
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Reaches 4 1/9 but repeats the denominator's 9 instead of the numerator's 1, as if one ninth were 0.999… rather than 0.111…

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/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction, because n/n is 1. Use it to give two fractions the same bottom before you add them.
Ninths as repeating decimals: $\frac{1}{9} = 0.111\ldots = 0.\overline{1}$; 1 divided by 9 is 0.111…, with 1s that go on forever. The bar over a digit means that digit repeats forever. Ninths from 1/9 to 8/9 repeat their top digit, so 4/9 is 0.444… and 2/9 is 0.222…

Sort out the question
You need to add two mixed numbers, 1 2/3 and 2 4/9. A mixed number is a whole number plus a fraction. Then write the sum as a decimal. The choices are repeating decimals, where the bar over a digit means it repeats forever.
Rename two thirds as ninths
The fractions 2/3 and 4/9 have different bottoms. Since 3 × 3 = 9, multiply the top and bottom of 2/3 by 3:
$\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}$
So 1 2/3 is the same as 1 6/9.
Watch out: adding the tops 2 and 4 over 9 without renaming gives 3 6/9, choice B.
Add the mixed numbers
Add the whole numbers, then add the fractions:
$1 + 2 = 3 \qquad \frac{6}{9} + \frac{4}{9} = \frac{10}{9}$
10/9 is more than 1. It is 9/9 plus 1/9, or 1 1/9. Carry the 1 into the whole number:
$3 + 1\frac{1}{9} = 4\frac{1}{9}$
Watch out: forgetting to carry the 1 gives 3 1/9, choice A.
Change it to a decimal
One ninth is 0.111…, with the 1 repeating forever. So:
$4\frac{1}{9} = 4.111\ldots = 4.\overline{1}$
Watch out: ninths repeat the top digit, not the 9. Choice D, 4.999…, does not match 4 1/9.
So the answer is C, $4.\overline{1}$ .

Check your answer

Use rough decimals. 1 2/3 is about 1.667, and 2 4/9 is about 2.444. Their sum is about 4.111. That matches 4.111… and rules out both choices that start with 3. And 4.999… is almost 5, which is too big.

More questions like this

The original DOE question

Question 18 as shown on the DOE’s digital test, page 91 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 91, question 18.

Question 31 of 69

2026 Practice Test D Math, question 26

Question

Jordan has a budget of $120 to rent a food truck. The rental company charges $18 per hour to rent the truck, plus a one-time cleaning fee of $30.

Write an inequality that represents this situation, in which x is the number of hours Jordan can rent the truck.

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

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

Show the answer

Correct answer: 30 + 18x ≤ 120

Let x represent the number of hours Jordan can rent the truck.

The hourly rental cost can be represented as $\$18 \cdot x$ , written as $18x$ .

The one-time cleaning fee of $30 is added to the hourly cost to obtain the total cost.

The sum of the cleaning fee and the hourly rental cost is represented as $30 + 18x$ .

As Jordan has $120 to spend, the total cost must be less than or equal to $120.

This is represented as $30 + 18x \leq 120$ or $120 \geq 30 + 18x$ .

Question analysis

Jordan has $120 to rent a food truck that costs $18 an hour plus a one-time $30 cleaning fee, and students type an inequality for x, the number of hours. It tests writing a two-step inequality from a word problem. Matching each dollar amount to its role and choosing the symbol that fits a spending limit are the tricky parts.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing a two-step inequality from a word problem is the grade 7 standard itself. Nothing is offered to choose from: students type the whole inequality, so the numbers, the variable term and the symbol are all theirs to get right.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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: 3 steps
Reading level: Question and choices: about grade 6, 57 words.
Corrected from the DOE handbook: A spelling error in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is small, so the points ride on two decisions the keypad leaves open: which symbol, and whether to stop before solving. A budget means Jordan may spend all $120 but no more, so the total cost must stay at or under it, and a strict < leaves out the five-hour rental that costs exactly $120. Testing x = 5 and x = 6 in the typed inequality confirms both the symbol and the setup.

Turning words into math moderate Understanding the question
The stem never says “at most”: the limit comes only from “a budget of $120”, while “$18 per hour” must multiply x and the “one-time cleaning fee of $30” is added just once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The whole inequality is typed on a keypad that offers <, ≤, > and ≥ side by side, with no choices to compare, and the solved form x ≤ 5 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Work out what 1 hour and 2 hours would cost Jordan in all, see which amount repeats each hour, build the total for x hours, then compare it with the budget and test a few values.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

30 + 18x < 120 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Builds the cost 30 + 18x correctly but uses a strict less-than sign, which shuts out a five-hour rental costing exactly the $120 Jordan has.
30 + 18x ≥ 120 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Builds the cost correctly but points the symbol the wrong way, treating the $120 budget as an amount Jordan must spend at least instead of a limit.
x ≤ 5 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Writes and solves the inequality correctly, then enters its solution, x ≤ 5, though the item asks for the inequality showing the fee, the hourly charge and the budget.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “The onetime cleaning fee of $30 is added to the hourly cost …”. “Onetime” means “former”; a fee charged once is a “one-time” fee, as the question itself spells it. The site now says “one-time”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee and a rate: $\text{total cost} = \text{fee} + \text{rate} \times x$; The fee is a charge paid only once, so you add it once. The rate is the charge for each hour. x is the number of hours. You multiply the rate by x because it is paid again for every hour.
Staying within a budget: $\text{total cost} \leq \text{budget}$; The sign ≤ means is less than or equal to. A budget is the most money you can spend. You may spend all of it, or less, but never more. So the total cost must be less than or equal to the budget.

Sort out the question
You know three facts:
- Jordan can spend at most $120. That is the budget.
- The truck costs $18 for each hour.
- A cleaning fee of $30 is paid one time only.
x is the number of hours. You need an inequality. That is a math sentence that compares two amounts with a sign such as < or ≤. It must show all three facts.
Work out a few totals
Find the total cost for a few hours. Each hour adds one more $18:
Hours Hourly charges ($) Total cost ($)
1 $18 \times 1 = 18$ $30 + 18 = 48$
2 $18 \times 2 = 36$ $30 + 36 = 66$
x $18x$ $30 + 18x$
The $30 shows up once in every total. The $18 shows up once for each hour. So x hours cost 30 + 18x dollars. Here 18x means 18 times x.
Choose the right sign
The total cost must fit inside the budget. Jordan may spend exactly $120, or less, but not more. That is less than or equal to, the sign ≤. So:
$30 + 18x \leq 120$
Watch out: the sign < means less than, but not equal. $30 + 18x < 120$ leaves out spending exactly $120. The sign ≥ is backward too. It says Jordan must spend at least $120. Also keep each number in its place. $30x + 18$ charges $30 an hour.
Stop before you solve
You could solve the inequality. Subtract 30 from both sides to get $18x \leq 90$ . Then divide both sides by 18 to get $x \leq 5$ . So Jordan can rent the truck for at most 5 hours.
Watch out: the question asks for an inequality that models the situation. To model means to show the facts of the story in math. The solved form, x ≤ 5, is true, but it hides the $18, the $30 and the $120. So x ≤ 5 is not accepted. Enter the inequality before you solve it.
Enter it on the keypad
On the test, build the inequality on the keypad. Press 3, 0, +, 1, 8 and the x key. Then press the ≤ key, in the fourth row of keys. Then press 1, 2, 0. You need no dollar signs and no times sign. On this site, type it, with <= for ≤: 30 + 18x <= 120.
So you would enter 30 + 18x ≤ 120. The same inequality written the other way round, 120 ≥ 30 + 18x, is also right. The solved form, x ≤ 5, is not accepted, because it does not model the situation.

Hours	Hourly charges ($)	Total cost ($)
1	$18 \times 1 = 18$	$30 + 18 = 48$
2	$18 \times 2 = 36$	$30 + 36 = 66$
x	$18x$	$30 + 18x$

Check your answer

Test two numbers of hours. For 5 hours, $30 + 18 \times 5 = 120$ . Is 120 ≤ 120? Yes, so 5 hours fit the budget. For 6 hours, $30 + 18 \times 6 = 138$ . Is 138 ≤ 120? No, and $138 is over budget. The inequality gets both right.

More questions like this

The original DOE question

Question 26 as shown on the DOE’s digital test, page 99 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 99, question 26. This is the DOE’s version, before the corrections listed in the analysis.

Question 32 of 69

2026 Practice Test D Math, question 36

Question

Leah has set a goal to read a total of 500 pages of a textbook. The table shows the book page numbers for the reading she did during three days.

DAILY READING LOG

Day	Pages Read
1	pages 45–90, inclusive
2	pages 100–135, inclusive
3	pages 145–190, inclusive

What percentage of her 500-page goal did Leah read on these three days? Round your answer to the nearest tenth.

Enter your answer in the space.

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

Show the answer

Correct answer: 25.6%

First, determine the number of pages read each day. Since the pages per assignment are inclusive, subtract the first page from the last page assigned and then add 1 to determine the pages read for each day.

Day 1: $(90 - 45) + 1 = 46$ pages

Day 2: $(135 - 100) + 1 = 36$ pages

Day 3: $(190 - 145) + 1 = 46$ pages

Leah read a total of 128 pages, since $46 + 36 + 46 = 128$ .

To determine the percent, express 128 pages out of 500 pages as a fraction.

$\frac{128}{500} = \frac{128 \div 4}{500 \div 4} = \frac{32}{125}$

$32 \div 125 = 0.256 = 25.6\%$

Question analysis

Leah's daily reading log lists three ranges of textbook pages, each marked inclusive, and her goal is 500 pages. This percent question asks what percentage of the goal she read over the three days, rounded to the nearest tenth and typed in. Counting the pages in each inclusive range correctly is the tricky step before the percent.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Finding what percent one number is of another is grade 6 and 7 work. This adds three inclusive page ranges to count, which takes more care than subtracting, then a rounding step, all typed with no choices to check against.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Many steps moderate
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 66 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

Counting pages in a range is the whole difficulty; the percent at the end is routine once the total is right. Subtracting first from last feels natural, and here it even rewards the student with a round 25%, so nothing looks wrong. Checking the rule on a tiny range, pages 1 to 3, which is 3 pages and not 2, fixes the count, and the instruction to round to the nearest tenth hints that the answer is not a whole number.

Something you must already know moderate Working it out
Every row says inclusive, but nothing says that pages 45 to 90 hold 90 − 45 + 1 = 46 pages; subtracting alone loses one page on each of the three days. (Anderson & Krathwohl, 2001)
Many steps moderate Working it out
Three page counts, a sum, a division by 500 and a change to a percent come in sequence, and a one-page error in each count carries through to the end. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The answer is typed with no choices, and the miscounted total, 125 pages, gives a tidy 25% that looks more finished than the right value with its tenth. (Enright, Morley & Sheehan, 2002)

Strategy: Count each day's pages carefully, testing your method on a short range such as pages 1–3, add the three days, divide by 500, and write the result as a percent to the nearest tenth.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

25 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Subtracts the first page from the last for each day, 45 + 35 + 45 = 125 pages, forgetting that an inclusive range counts both end pages, so 125 ÷ 500 = 25%.
0.256 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Counts the 128 pages and divides by 500 correctly, then enters the decimal 0.256, one step short of multiplying by 100 to make a percent.
26 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds 25.6% correctly but rounds it to the nearest whole percent, overlooking the instruction to round to the nearest tenth.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range: $\text{pages} = \text{last page} - \text{first page} + 1$; Inclusive means the first and last pages are both read. Subtracting counts the steps from the first page to the last. That misses one page, so add 1. Test it: pages 1 to 3 are 3 pages, and 3 − 1 + 1 = 3.
Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you have, here the pages read. The whole is the amount you compare with, here the 500-page goal. Dividing gives a decimal. Multiplying by 100 turns it into a percent, a number out of 100.

Sort out the question
Leah’s goal is 500 pages. The log shows three days:
- Day 1: pages 45 to 90
- Day 2: pages 100 to 135
- Day 3: pages 145 to 190
Each range is inclusive. You need the percent of 500 pages she read, rounded to the nearest tenth.
Count the pages each day
Use the rule for an inclusive range: last page minus first page, plus 1.
$90 - 45 + 1 = 46$
$135 - 100 + 1 = 36$
$190 - 145 + 1 = 46$
Watch out: subtracting alone gives 45, 35 and 45 pages. That is 125 pages, which is exactly 25%. A round answer like that can hide the mistake.
Add the three days
Add the pages from all three days to get her total:
$46 + 36 + 46 = 128$
Leah read 128 pages in all. This is the part you compare with the goal.
Change to a percent
Divide the pages read by the goal, then multiply by 100:
$\frac{128}{500} \times 100 = 0.256 \times 100 = 25.6$
Leah read 25.6% of her goal. That already ends in tenths, so no rounding is needed.
Watch out: 0.256 is a decimal, not a percent. And 26 rounds to a whole number, not to the nearest tenth.
Enter the answer
The box has no % sign, but the question asks for a percentage, so the number in the box is the percent. Type it with its decimal point. Typing 25.6% also counts on this site. So you would enter 25.6.

Check your answer

Estimate. 10% of 500 pages is 50 pages. 128 pages is a bit more than 2 × 50 = 100, so the answer is between 20% and 30%. Then check exactly: $0.256 \times 500 = 128$ pages. It matches.

More questions like this

The original DOE question

Question 36 as shown on the DOE’s digital test, page 109 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 109, question 36.

Question 33 of 69

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 34 of 69

2024 Test A Math, question 1

Question

Simplify:

$-3.8 + 2.3 - (-1.1)$

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

Show the answer

Correct answer: -0.4

Subtracting a negative number is the same as adding a positive number. Rewrite the expression to simplify:

$-3.8 + 2.3 - (-1.1)$

$-3.8 + 2.3 + 1.1$

$-3.8 + 3.4$

$-0.4$

Question analysis

This grid-in asks students to simplify a short expression that mixes negative and positive decimals and ends by subtracting a negative number. It tests adding and subtracting rational numbers in order. The double negative, and keeping track of the sign of the final result, are where errors creep in.

NYS standards: 7.NS.1c rational number subtraction
7.NS.1b rational number addition
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding and subtracting negative decimals is core grade 7 content, and this is a routine three-term version. It matches a typical classroom exercise; the only extra demand is gridding a signed decimal correctly.
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 16+, 2 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 three-term expression with small decimals, so the method is never in doubt; the points are lost on signs. The double negative at the end is the classic place to slip; the other is the result itself, which sits just below zero and is easy to grid without its minus sign. Grouping the positive terms first, 2.3 + 1.1, and comparing that sum with 3.8 settles both the size and the sign.

Error-prone arithmetic moderate Working it out
The last term, − (−1.1), has to turn into + 1.1, and the total then crosses zero, since 2.3 + 1.1 = 3.4 falls just short of the 3.8 below zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in it shows no choices, so a double negative read as a single one, or a result gridded without its minus sign, gets no warning from the page. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −1.1 as adding 1.1, combine the two positive decimals first, then compare their total with 3.8 to decide the sign and size.

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

-2.6 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Treats subtracting −1.1 as subtracting 1.1, so it works out −3.8 + 2.3 − 1.1 = −1.5 − 1.1 and moves further below zero.
0.4 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Rewrites the double negative correctly and finds the gap 3.8 − 3.4 = 0.4, but drops the minus sign that the larger negative term gives the result.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a is any number, and b is a positive number, so −b is negative. Taking away a negative number is the same as adding the positive number. Think of money: if someone takes away a debt of b dollars, you are b dollars better off.
Adding a negative and a positive: $-a + b = -(a - b)$; a and b are positive numbers, and a is the larger one. A negative and a positive partly cancel out. So find the difference of their sizes, a − b. The result keeps the sign of the number farther from zero, which here is the negative one.

Sort out the expression
You need to simplify $-3.8 + 2.3 - (-1.1)$ . Simplify means work it out to one number. Two of the numbers are negative: −3.8 and −1.1. One is positive: 2.3. The tricky part is at the end. It says minus negative 1.1. The answer can be negative, so keep track of every sign.
Change minus a negative to plus
Taking away a negative is the same as adding a positive. Use the rule with b = 1.1.
$a - (-b) = a + b$
$-3.8 + 2.3 - (-1.1) = -3.8 + 2.3 + 1.1$
Now only one negative number is left. Watch out: do not just subtract 1.1. That gives $-3.8 + 2.3 - 1.1 = -2.6$ , which is wrong.
Add the two positive numbers
Group the positive numbers first. This makes the rest of the work short.
$2.3 + 1.1 = 3.4$
Now the problem is $-3.8 + 3.4$ . One number is negative and one is positive.
Combine the negative and the positive
They partly cancel. Use the rule with a = 3.8 and b = 3.4.
$-a + b = -(a - b)$
$-3.8 + 3.4 = -(3.8 - 3.4) = -0.4$
The negative number, 3.8, is farther from zero than 3.4 is. So the result is negative. Watch out: keep the minus sign. Dropping it gives 0.4, which is wrong. So you would grid -0.4, with the minus sign in its own column.

Check your answer

Use a number line. Start at −3.8. Adding 2.3 moves you right, to −1.5. Subtracting −1.1 also moves you right, by 1.1, to −0.4. You end just to the left of zero. So a small negative answer makes sense.

More questions like this

The original DOE question

Question 58 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 35 of 69

2024 Test A Math, question 2

Question

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

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

Show the answer

Correct answer: 150

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

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

$r + m = 180$

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

$r = 5m$

$5m + m = 180$

$6m = 180$

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

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

Therefore, the measure of angle R is 150°.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 36 of 69

2024 Test A Math, question 3

Question

A juice mixture contains $\frac{3}{16}$ gallon of apple juice and $\frac{3}{40}$ gallon of cranberry juice. How many gallons of apple juice per gallon of cranberry juice does the mixture contain? (Express your answer as a decimal.)

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

Show the answer

Correct answer: 2.5

Set up a proportion to solve:

$\frac{\frac{3}{16}\text{ apple}}{\frac{3}{40}\text{ cranberry}} = \frac{x\text{ apple}}{1\text{ cranberry}}$

$\frac{3}{16} \times \frac{40}{3} = x$

$\frac{40}{16} = x$

$2.5 = x$

Question analysis

This ratio grid-in describes a juice mixture with a fraction of a gallon of apple juice and a smaller fraction of a gallon of cranberry juice. Students find the unit rate, gallons of apple juice per gallon of cranberry juice, by dividing one fraction by the other and writing it as a decimal. Dividing in the wrong order is the main risk.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of two fractions is exactly the grade 7 standard, and this is a textbook example of it. The only added demands are dividing in the right order and answering as a decimal.
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
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 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 method is one division of fractions, but the question tests whether the division runs the right way. The word per tells which juice goes on top, and a student who divides the smaller amount by the larger gets a tidy decimal for the reverse rate. A sense check helps: the mixture holds more apple juice than cranberry, so the rate must be greater than 1.

Turning words into math moderate Understanding the question
The phrase "gallons of apple juice per gallon of cranberry juice" fixes the order: apple's 3/16 goes on top and cranberry's 3/40 below, though the two fractions invite either order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Dividing 3/16 by 3/40 means multiplying by 40/3; the 3s cancel to leave 40/16, which still has to be reduced and written as a decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The answer must be gridded as a decimal, and with no choices shown, the upside-down 0.4 gives no sign that it answers the reverse question. (Enright, Morley & Sheehan, 2002)

Strategy: Put the apple juice over the cranberry juice and divide the fractions by multiplying by the reciprocal; the matching 3s cancel, leaving a simple fraction to turn into a decimal.

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

0.4 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides cranberry by apple, 3/40 ÷ 3/16 = 16/40 = 0.4, giving gallons of cranberry per gallon of apple, the rate turned upside down.
0.014 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Multiplies the two amounts instead of dividing them, 3/16 × 3/40 = 9/640 ≈ 0.014, as if comparing the juices called for a product.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $r = a \div b$; r is the rate: how much of one thing there is for each 1 of another. a is the amount of the first thing. b is the amount of the second thing, the one named after the word per. Dividing shares a out evenly over each 1 of b.
Dividing by a fraction: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a/b and c/d are fractions, and b, c and d are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. It works because flipping a fraction gives the number that undoes it.

Sort out the question
The mix has $\frac{3}{16}$ gallon of apple juice and $\frac{3}{40}$ gallon of cranberry juice. You need the “gallons of apple juice per gallon of cranberry juice”. Per means for each. So the question asks: for each 1 gallon of cranberry juice, how many gallons of apple juice are there? The answer must be a decimal.
Set up the division
To find an amount for each 1 gallon, divide. Apple juice goes first, because it comes before per. Cranberry juice comes second.
$r = a \div b$
$r = \frac{3}{16} \div \frac{3}{40}$
Watch out: do not flip the order. Cranberry divided by apple gives 0.4. That is gallons of cranberry juice for each gallon of apple juice, the reverse question.
Multiply by the reciprocal
To divide by $\frac{3}{40}$ , multiply by its reciprocal, $\frac{40}{3}$ .
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{3}{16} \times \frac{40}{3} = \frac{3 \times 40}{16 \times 3} = \frac{40}{16}$
The 3 on top and the 3 on the bottom cancel, since $3 \div 3 = 1$ . Watch out: do not just multiply the two amounts. That gives $\frac{9}{640}$ , about 0.014, which is far too small.
Change the fraction to a decimal
Make $\frac{40}{16}$ simpler by dividing the top and bottom by 8. Then divide.
$\frac{40}{16} = \frac{5}{2} = 5 \div 2 = 2.5$
There are 2.5 gallons of apple juice for each gallon of cranberry juice. So you would grid 2.5.

Check your answer

Both fractions have 3 on top. The one with the smaller bottom, $\frac{3}{16}$ , is bigger. So there is more apple juice than cranberry juice, and the answer must be more than 1. That rules out 0.4. Also, $3 \div 40 = 0.075$ and $0.075 \times 2.5 = 0.1875$ , which is $3 \div 16$ .

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 2025 Admissions, page 75, question 60. Open page 75 of the handbook (PDF)

Question 37 of 69

2024 Test A Math, question 4

Question

Mr. Chan’s lawn grows $2\frac{1}{8}$ inches every 2 weeks. He mows his lawn every 2 weeks and cuts off the top $1\frac{3}{4}$ inches of lawn. If Mr. Chan’s lawn was 4 inches tall at the beginning of the season, how many inches tall, in decimal form, is Mr. Chan’s lawn after 8 weeks?

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

Show the answer

Correct answer: 5.5

The lawn grows $2\frac{1}{8}$ inches every 2 weeks. Find the total number of inches it grows in 8 weeks.

$8 \div 2 = 4$

$4 \times 2\frac{1}{8}$

$4 \times 2.125 = 8.5$

The lawn was 4 inches tall at the beginning of the season.

$8.5 + 4 = 12.5$

Mr. Chan mows his lawn every 2 weeks and cuts off the top $1\frac{3}{4}$ inches. Find the total number of inches that are cut off in 8 weeks.

$8 \div 2 = 4$

$4 \times 1\frac{3}{4}$

$4 \times 1.75 = 7$

The height after 8 weeks is the total number of inches grown minus the total number of inches cut off.

$12.5 - 7 = 5.5$

Question analysis

This multistep word problem follows Mr. Chan's lawn over 8 weeks: it grows a mixed number of inches every 2 weeks, and he mows a mixed number of inches off on the same schedule. Students combine the growth, the cuts and the starting height and give the final height as a decimal. Leaving out one of those three pieces is the usual error.

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve multi-step problems with mixed numbers and convert between fractions and decimals. This one strings four such steps together, including counting the 2-week periods, which is typical grade 7 work done carefully.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Turning words into math minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 49 words.
Corrected from the DOE handbook: A misspelled name 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

No single step is hard, but three parts must all reach the final height: the starting 4 inches, four rounds of growth and four rounds of mowing. Leaving out any one of them gives a clean decimal that can be gridded without a second thought. Working one 2-week period at a time, in which the lawn gains 3/8 inch net, keeps the bookkeeping short.

Many steps moderate Working it out
Four pieces have to be combined: count the four 2-week periods in 8 weeks, total the growth, total the mowing, and start from the lawn's first 4 inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
The amounts are mixed numbers with unlike denominators, 2 1/8 and 1 3/4, which must be multiplied by 4 or subtracted and then gridded as a decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase "every 2 weeks" governs both the growing and the mowing, so 8 weeks means four rounds of each, not eight. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in it gives no choices, so a height that leaves out the starting 4 inches, or leaves out the mowing, looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Find the net change for one 2-week period by subtracting the cut from the growth, multiply by the number of such periods in 8 weeks, and add the starting height.

Traps

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

The wrong answers, classified

1.5 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the net gain of 3/8 inch per period and multiplies by the 4 periods, 1.5 inches, but stops before adding the lawn's starting height of 4 inches.
12.5 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds four periods of growth, 8.5 inches, to the starting 4 inches and stops there, never subtracting the 1 3/4 inches cut at each of the four mowings.
7 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Applies the net gain of 3/8 inch to each of the 8 weeks instead of each 2-week period, 8 × 3/8 = 3, then adds the starting 4.

What was corrected from the DOE handbook, and why

Explanation: The handbook’s explanation says “Mr. Chen mows his lawn every 2 weeks.” The question calls him Mr. Chan throughout, so the site now reads “Mr. Chan.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to decimal: $a\frac{b}{c} = a + b \div c$; A mixed number has a whole part, a, and a fraction part, b over c. To write it as a decimal, divide b by c to get the fraction part as a decimal. Then add the whole part. For example, one and a half is 1 + 0.5 = 1.5.
Repeated amount: $\text{total} = n \times a$; n is how many times something happens. a is the amount each time. Adding the same amount n times is the same as multiplying it by n.

Sort out the question
The lawn starts at 4 inches tall. Every 2 weeks it grows $2\frac{1}{8}$ inches. Every 2 weeks Mr. Chan also cuts off $1\frac{3}{4}$ inches. You need the height after 8 weeks, as a decimal. Three things matter: the start, the growing and the cutting.
Count the 2-week periods
Both the growing and the cutting happen every 2 weeks. So count how many 2-week periods fit in 8 weeks.
$8 \div 2 = 4$
There are 4 periods. The lawn grows 4 times and is cut 4 times. Watch out: do not count 8 rounds, one for each week. That leads to the wrong answer 7.
Change the mixed numbers to decimals
The answer must be a decimal, so change both amounts now.
$2\frac{1}{8} = 2 + 1 \div 8 = 2.125$
$1\frac{3}{4} = 1 + 3 \div 4 = 1.75$
Each period, the lawn grows 2.125 inches and loses 1.75 inches.
Add up the growing
The lawn grows 2.125 inches, 4 times.
$\text{total} = n \times a$
$\text{total} = 4 \times 2.125 = 8.5$
It grows 8.5 inches in all. Add the starting height of 4 inches.
$4 + 8.5 = 12.5$
Without mowing, the lawn would be 12.5 inches tall. Watch out: do not forget the starting 4 inches. Leaving them out leads to 1.5.
Add up the cutting
Mr. Chan cuts off 1.75 inches, 4 times.
$\text{total} = 4 \times 1.75 = 7$
He cuts off 7 inches in all.
Subtract the cutting
$12.5 - 7 = 5.5$
The lawn is 5.5 inches tall after 8 weeks. Watch out: 12.5 is not the answer. It leaves out the mowing. So you would grid 5.5.

Check your answer

Work one period at a time. Each period the lawn gains $2.125 - 1.75 = 0.375$ inch. In 4 periods that is $4 \times 0.375 = 1.5$ inches. Add the starting 4 inches: $4 + 1.5 = 5.5$ . Same answer.

More questions like this

The original DOE question

Question 61 as printed on page 76 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 76, question 61. Open page 76 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 38 of 69

2024 Test A Math, question 5

Question

RESULTS FROM SURVEY OF 110 FAMILIES

Number of Children in the Family	Number of Families
0	45
1	32
2	19
3	8
4	6

The table above shows the number of children in each of 110 families. What is the median number of children in these families?

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

Show the answer

Correct answer: 1

Since 110 families were surveyed, the median will have 55 values less than or equal to it, and 55 values greater than or equal to it.

If the values in the table were listed in order, they would look like this:

$\overbrace{0,\ldots,0}^{45},\overbrace{1,\ldots,1}^{32},\overbrace{2,\ldots,2}^{19},\overbrace{3,\ldots,3}^{8},\overbrace{4,\ldots,4}^{6}$

The first row of the table indicates that 45 families had 0 children.

The second row of the table indicates that 32 families had 1 child.

So 77 families had either 0 children or 1 child, and 45 families had 0 children.

Since 55 is between 45 and 77, the median must be a value in the second row of the table.

So the median is 1.

Question analysis

This statistics grid-in gives a frequency table from a survey of 110 families, showing how many families have each number of children from none to four. Students find the median number of children per family. The table lists counts rather than the 110 data values themselves, so locating the middle value takes a running total of the families.

NYS standards: 6.SP.5c range, center, and overall pattern
6.SP.5a number of observations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 median problems usually list a dozen or so values. Here 110 values are hidden in a frequency table, so students must reason about positions with a running count, a less familiar and more error-prone task.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
No obvious method moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Students know a median as the middle of a list, but here the list is never written out. Treating the table's columns as the data gives quick, confident answers, 2 from the middle of the children column or 19 from the middle of the family counts, and neither can be checked against choices. The question rewards counting positions: 45 families have no children, so the 55th and 56th families both fall among the 32 with one child.

Reading a figure, table or graph major Understanding the question
The table hides 110 data values in five rows: the 0 to 4 column lists the possible values, and 45, 32, 19, 8 and 6 say how many families have each. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
No obvious method moderate Working it out
The median of 110 values lies between the 55th and 56th; finding them takes a running count of families, 45 and then 77, a method seldom practiced with tables. (Schoenfeld, 1985)
An answer in an unexpected form minor Choosing the answer
As a grid-in there are no choices to compare, so a student who takes the middle row's 2 has nothing on the page to suggest it is wrong. (Enright, Morley & Sheehan, 2002)

Strategy: With 110 values the median sits between the 55th and 56th; add the family counts row by row until the running total passes 55, and read that row's number of children.

Traps

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

The wrong answers, classified

2 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Takes the middle entry of the children column, 0 to 4, ignoring that each number of children belongs to a different number of families.
19 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Finds the median of the Number of Families column, 6, 8, 19, 32 and 45, a middle family count rather than a number of children.
1.07 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: weak
Computes the mean instead, 118 children among 110 families, about 1.07, a different measure of center from the median asked for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Spot of the median: $\text{middle spot} = \frac{n + 1}{2}$; The median is the middle value when all the values are put in order. n is how many values there are. This gives the spot of the middle value, with the same number of values before and after it. If it ends in a half, the median is halfway between the two values on either side.

Sort out the table
The table covers 110 families. The left column is a number of children. The right column says how many families have that many children. For example, 45 families have 0 children. You need the median number of children. The median is the middle value when all the values are in order.
Picture the values in order
Think of a list with one number for each family: how many children it has. The list has 45 zeros, then 32 ones, then 19 twos, then 8 threes, then 6 fours. It is already in order, from least to most.
$45 + 32 + 19 + 8 + 6 = 110$
There are 110 values. Watch out: the median is not the middle row, 2. Each row stands for a different number of families.
Find the middle spot
Use the median rule with n = 110.
$\text{middle spot} = \frac{n + 1}{2}$
$\frac{110 + 1}{2} = 55.5$
The middle is halfway between the 55th and 56th values. So you need those two values in the list.
Count down the list
Keep a running total of families, row by row. The zeros fill spots 1 to 45. Now add the 32 families with 1 child.
$45 + 32 = 77$
So the ones fill spots 46 to 77. Spot 55 and spot 56 are both in that range. So both values are 1.
Give the median
The 55th and 56th values are both 1. Halfway between 1 and 1 is 1. So the median is 1 child. Watch out: the median is a number of children, not of families. Gridding 19, from the families column, is wrong. So is the mean, or average, which is about 1.07. So you would grid 1.

Check your answer

Half of 110 is 55. Families with 0 children: 45, which is less than half. So the median is not 0. Families with 0 or 1 child: 77, which is more than half. So the median is not more than 1. It must be 1.

More questions like this

The original DOE question

Question 62 as printed on page 76 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 39 of 69

2024 Test A Math, question 46

Question

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

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

Show the answer

Correct answer: B

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 40 of 69

2024 Test B Math, question 1

Question

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

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

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

Show the answer

Correct answer: -0.8

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

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

$\frac{6}{5}x+3+\left(-\frac{10}{5}x\right)=-\frac{4}{5}x+3$ .

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

When expressed as a decimal,

$-\frac{4}{5}=-0.8$ .

Question analysis

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

NYS standards: 7.EE.1 linear expressions with rational coefficients
6.EE.2b parts of an expression
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Expanding and simplifying linear expressions with fraction coefficients is a core grade 7 skill, and this is a typical short example. Writing the coefficient as a decimal in the grid adds care but no new idea.
Estimated difficulty: Straightforward: two steps or one careful reading.
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: 3 steps
Reading level: Question and choices: about grade 9, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The distribution itself is routine; points are lost after it. Students who stop at 6/5x forget the −2x outside the parentheses, and those who combine the x terms correctly can still drop the sign of a negative result. Rewriting 2x as 10/5x keeps the subtraction in fifths, and substituting x = 1 into both forms confirms the coefficient before gridding.

Error-prone arithmetic moderate Working it out
Multiplying 2x by 3/5 gives 6/5x, and taking away 2x means rewriting it as 10/5x; the result is a negative fraction that must then become a decimal with its sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
There are no choices to check against, and the coefficient must be gridded as a decimal with its minus sign, so −4/5 has to become −0.8. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

0.8 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Combines 6/5x and −2x to a coefficient of the right size, 4/5, but drops the minus sign, or subtracts 6/5 from 2, and grids 0.8.
1.2 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Distributes 3/5 to get 6/5x + 3 and grids 6/5 = 1.2, stopping before the −2x outside the parentheses is combined with it.
-1.4 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies 3/5 by 2 as 6/10, doubling the denominator along with the numerator, so the coefficient comes out 0.6 − 2 = −1.4.

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.
Multiplying a fraction by a whole number: $\frac{a}{b} \times c = \frac{a \times c}{b}$; a over b is the fraction, and c is the whole number. Multiply only the top number, the numerator, by c. The bottom number, the denominator, stays the same. The size of each piece does not change. You just have more pieces.

Sort out the question
You start with this expression. An expression is a math phrase with numbers, letters and signs, but no equals sign.
$\frac{3}{5}(2x + 5) - 2x$
To simplify it means to write it in its shortest form. The coefficient of x is the number multiplied by x. In 7x, the coefficient is 7. You need the coefficient of x, written as a decimal.
Multiply out the parentheses
The 3/5 multiplies everything inside the parentheses. Use the distributive property.
$a(b + c) = a \times b + a \times c$
$\frac{3}{5}(2x + 5) = \frac{3}{5} \times 2x + \frac{3}{5} \times 5$
Multiply only the top of 3/5 by 2: $\frac{3}{5} \times 2 = \frac{6}{5}$ . Also, $\frac{3}{5} \times 5 = 3$ . Now the whole expression is:
$\frac{6}{5}x + 3 - 2x$
Watch out: do not double the bottom number too. Writing 6/10 leads to −1.4.
Write 2x in fifths
There are now two x terms: $\frac{6}{5}x$ and $-2x$ . To combine them, they need the same denominator (bottom number). One whole is 5/5, so 2 wholes are 10/5.
$2x = \frac{10}{5}x$
Watch out: the −2x is outside the parentheses, but it still counts. Stopping at 6/5 and gridding 1.2 is a common mistake.
Combine the x terms
Six fifths take away ten fifths leaves negative four fifths.
$\frac{6}{5}x - \frac{10}{5}x = -\frac{4}{5}x$
The 3 has no x, so it stays as it is. The simplified expression is $-\frac{4}{5}x + 3$ . So the coefficient of x is $-\frac{4}{5}$ . Watch out: 6 − 10 is negative. Dropping the minus sign gives 0.8.
Change it to a decimal
A fraction means the top divided by the bottom: $4 \div 5 = 0.8$ . Keep the minus sign.
$-\frac{4}{5} = -0.8$
So you would grid -0.8, with the minus sign in its own column.

Check your answer

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

More questions like this

The original DOE question

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

Question 41 of 69

2024 Test B Math, question 2

Question

Two number cubes have sides labeled 1 through 6. Both number cubes are rolled. How many different outcomes will have a sum of 6?

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

Show the answer

Correct answer: 5

There will be 5 different outcomes with a sum of 6:

$1+5=6$

$2+4=6$

$3+3=6$

$4+2=6$

$5+1=6$

Question analysis

A counting grid-in with two number cubes, each numbered 1 to 6, rolled together. Students count how many different outcomes give a sum of 6. The catch is deciding whether a 2 on the first cube with a 4 on the second counts as a different outcome from the reverse.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students list sample spaces for two number cubes when finding compound probabilities. This asks for one count from that list, a short task, though as a grid-in it offers no choices to expose an incomplete list.
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: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is trivial; the question turns on what counts as a different outcome. Treating the cubes as a first and a second settles it: the first cube runs from 1 to 5 and the second is forced each time. Students who think in pairs of numbers rather than in rolls of two cubes undercount, and those who double every pair overcount by one at the double 3 and 3.

Something you must already know moderate Working it out
Nothing in the stem says that a 2 on the first cube with a 4 on the second differs from the reverse; students must already know that two cubes give 36 ordered outcomes. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in it offers no list of counts to compare, so a student who counts unordered pairs gets no sign that ordered outcomes were wanted. (Enright, Morley & Sheehan, 2002)

Strategy: Treat the cubes as a first and a second cube, run the first cube's value from 1 to 6, and for each value check whether the second cube can complete the sum.

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.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

3 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Counts unordered pairs, 1 and 5, 2 and 4, 3 and 3, treating a 1 on the first cube with a 5 on the second as the same outcome as the reverse.
6 Overlapping groups (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Knows order matters and doubles the 3 pairs to 6, but 3 and 3 reads the same both ways, so that outcome is counted twice.
36 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Grids the total number of outcomes for two cubes, 6 × 6 = 36, instead of only the outcomes whose sum is 6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total outcomes} = a \times b$; a is the number of ways the first cube can land, and b is the number of ways the second cube can land. Multiply them to count every pair. It works because each face of the first cube can go with each face of the second.

Sort out the question
You roll two number cubes. Each one shows 1, 2, 3, 4, 5 or 6. An outcome is one way the two cubes can land. You need the number of outcomes whose sum (the two numbers added) is 6.
The question does not say if order matters. The answer key counts the cubes as a first cube and a second cube. So 1 on the first and 5 on the second is a different outcome from 5 on the first and 1 on the second.
See why order counts
Picture a red cube and a blue cube. Red 1 with blue 5 is not the same roll as red 5 with blue 1. Both can happen, so both count.
This is how all the outcomes of two cubes are counted. Use the counting principle.
$\text{total outcomes} = 6 \times 6 = 36$
Watch out: 36 is every outcome, not just the ones that add up to 6.
List the pairs that make 6
Go through the first cube one number at a time. The second cube has to make up the rest of the 6.
First cube Second cube Sum
1 5 6
2 4 6
3 3 6
4 2 6
5 1 6
The first cube cannot show 6. Then the second cube would need 0, and no face shows 0.
Count the outcomes
The table has 5 rows, so there are 5 outcomes. Watch out: counting 1 and 5 as the same as 5 and 1 gives only 3. Watch out: 3 and 3 is one outcome, not two. Doubling 3 pairs to get 6 counts it twice.
So you would grid 5.

First cube	Second cube	Sum
1	5	6
2	4	6
3	3	6
4	2	6
5	1	6

Check your answer

Each number from 1 to 5 on the first cube goes with exactly one number on the second cube. A 6 on the first cube goes with nothing. So there are exactly 5 outcomes. The table shows the same pattern: as the first number goes up by 1, the second goes down by 1.

More questions like this

The original DOE question

Question 59 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 198, question 59. Open page 198 of the handbook (PDF)

Question 42 of 69

2024 Test B Math, question 3

Question

$Scale drawing of a garden shaped like a rectangle with its lower right corner cut out; all six sides are horizontal or vertical. The top side is labeled 13 cm and the left side 6 cm. From the top right corner, the right side, labeled 3 cm, goes down to an unlabeled horizontal side running left, then an unlabeled vertical side runs down to the bottom side, which is labeled 8 cm.$

The figure above shows a scale drawing of a garden, where 1 centimeter represents 2.5 meters. What is the perimeter of the actual garden in meters?

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

Show the answer

Correct answer: 95

First, find the perimeter of the figure.

Two sides do not have labels. The top of the figure measures 13 cm and the bottom shows a length measuring 8 cm, so the horizontal side without a label measures 5 cm $(13-8)$ .

The left side of the figure measures 6 cm and the right side shows a length measuring 3 cm, so the vertical side without a label measures 3 cm $(6-3)$ .

So the total perimeter of the scale drawing is

$13+3+5+3+8+6=38\text{ cm}$

Use the scale to find the perimeter of the actual garden:

$\frac{1\text{ cm}}{2.5\text{ m}}=\frac{38\text{ cm}}{p\text{ m}}$

$p=2.5\times 38=95$

Question analysis

A grid-in on a scale drawing of an L-shaped garden, where each centimeter stands for 2.5 meters and two sides carry no labels. Students find the perimeter of the drawing, then convert it to the real garden's perimeter in meters. The unlabeled sides are easy to leave out.

NYS standards: 7.G.1 scale drawings
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are grade 7 content, and a typical task converts one or two given lengths. Here two missing sides must be worked out and the perimeter assembled before the scale applies, more steps but nothing beyond grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 4 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 figure hides the work: it labels four of the six sides, and the missing two are short pieces of the notch that are easy to overlook. Once all six are found, the scale step is a single multiplication, but it must go the right way, since the real garden is larger than the drawing. Tracing the outline and counting six sides guards against the most common loss.

Reading a figure, table or graph moderate Understanding the question
Two sides of the notch carry no labels, so students must find them from the opposite sides: 13 − 8 = 5 cm across and 6 − 3 = 3 cm down. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas minor Working it out
The perimeter of an L-shaped drawing has to be joined to a scale of 2.5 meters per centimeter, two separate skills in one answer. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
With no choices, nothing flags a perimeter that is missing two sides or still in centimeters; the student must check the six-sided total alone. (Enright, Morley & Sheehan, 2002)

Strategy: Use the opposite sides to work out the two unlabeled lengths, add all six sides of the drawing, then multiply the total in centimeters by 2.5 meters per centimeter.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
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).

The wrong answers, classified

75 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: strong
Adds only the four labeled sides, 13 + 3 + 8 + 6 = 30 cm, and scales that to 75 m, missing the two unlabeled sides of the notch.
38 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the drawing's perimeter, 38 cm, correctly and grids it without applying the scale of 2.5 meters per centimeter.
15.2 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the 38 cm perimeter by 2.5 instead of multiplying, shrinking the garden though each centimeter of the drawing stands for 2.5 meters.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter: $P = \text{sum of all the side lengths}$; P is the perimeter, the distance all the way around a shape. Add the length of every side, once each. It works for any shape with straight sides, even one with a corner cut out.
Using a scale: $\text{actual length} = \text{drawing length} \times s$; s is the scale, the number of meters that 1 centimeter on the drawing stands for. Here s is 2.5. Each centimeter on the drawing is 2.5 meters in real life, so you multiply.

Sort out what you know
The figure is a scale drawing, a small copy of the real garden. Each 1 cm on the drawing stands for 2.5 m in the real garden. You need the perimeter of the real garden in meters. The perimeter is the distance all the way around.
The shape has 6 sides. Four are labeled: 13 cm, 3 cm, 8 cm and 6 cm. The two sides around the cut-out corner have no labels.
Find the missing side across
Look at the top side, 13 cm. Below it, the bottom side (8 cm) and the short unlabeled side across the notch together reach just as far. So the missing side is the difference.
$13 - 8 = 5$
The missing side across is 5 cm.
Find the missing side going down
Now look at the left side, 6 cm. On the right, the 3 cm side and the unlabeled side going down together make the same height.
$6 - 3 = 3$
The missing side going down is 3 cm.
Add all six sides
$P = \text{sum of all the side lengths}$
$P = 13 + 3 + 5 + 3 + 8 + 6 = 38$
The perimeter of the drawing is 38 cm. Watch out: adding only the four labeled sides gives 30 cm. That leads to the wrong answer 75.
Change to real meters
Each centimeter stands for 2.5 meters, so multiply by 2.5.
$\text{actual length} = \text{drawing length} \times s$
$\text{actual length} = 38 \times 2.5 = 76 + 19 = 95$
Here $38 \times 2 = 76$ and $38 \times 0.5 = 19$ . The real perimeter is 95 meters. Watch out: 38 is still in centimeters, so do not grid it. Dividing by 2.5 gives 15.2, which makes the garden smaller than its drawing.
So you would grid 95.

Check your answer

Slide the two notch sides out to the corner. The distance around stays the same, so it equals a 13 cm by 6 cm rectangle: $2 \times (13 + 6) = 38$ . Also, 95 meters is more than 38, which fits, because each centimeter stands for more than 1 meter.

More questions like this

The original DOE question

Question 60 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 198, question 60. Open page 198 of the handbook (PDF)

Question 43 of 69

2024 Test B Math, question 5

Question

The figure below is a triangular prism. The lateral sides are rectangles.

$A triangular prism with a triangle on the top and bottom and three rectangular sides. On the top triangle, a dashed segment labeled 8 in. runs from the far vertex to the opposite side and meets it at a right angle. The bottom edge below that opposite side is labeled 4 in., and the vertical edge at its left end is labeled 2 in.$

What is the volume of the figure, in cubic inches?

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

Show the answer

Correct answer: 32

The volume of a triangular prism is found by multiplying the area of the triangular base by the height of the prism.

The area of the triangular base of the prism is $\frac{1}{2}bh=\frac{1}{2}(4)(8)=16$ square inches.

The height of the triangular prism is 2 inches.

So the volume of the triangular prism, in cubic inches, is $16\bullet 2=32$ .

Question analysis

A geometry grid-in: find the volume, in cubic inches, of a triangular prism drawn with its triangular faces on top and bottom. The drawing gives a 4-inch side of the triangle, a dashed 8-inch height drawn to that side, and a 2-inch edge. Students must recall the volume formula and match each length to its role.

NYS standards: 7.G.6 area, surface area, and volume
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the volume of a right triangular prism is named in the grade 7 standards, and this is a direct two-step case. The only extra demand is reading a slanted drawing to see which length plays which role.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards knowing that a prism's volume is its base area times its height, then matching each length to its job. The drawing lies on its side, with a short 2-inch height and the triangle flat on top, which makes the base harder to spot. Most lost points come from the one half: leaving it out doubles the volume, and stopping at the triangle's area gives half of it.

Something you must already know moderate Working it out
Neither formula is given: the volume is the area of the triangular base times the prism's height, and that area needs the one half in 1/2 × base × height. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The prism lies with its triangles on top and bottom, so the 2-inch vertical edge is its height, while the dashed 8-inch line is the triangle's height to the 4-inch side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
An answer in an unexpected form minor Choosing the answer
There are no choices, so a product that forgets the one half, or one that stops at the area of the triangle, gives no warning. (Enright, Morley & Sheehan, 2002)

Strategy: Find the area of the triangular face as one half times base times height, using the 4-inch side and the dashed 8-inch height, then multiply by the 2-inch edge of the prism.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

64 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 4 × 8 × 2 = 64 as for a rectangular box, leaving out the one half in the area of the triangular base.
16 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the triangle's area, 1/2 × 4 × 8 = 16 square inches, and stops before multiplying by the prism's 2-inch height.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, one side of the triangle. h is the height, the straight distance from that side to the opposite corner, at a right angle to the side. You take half because a triangle is half of a rectangle with the same base and height.
Volume of a prism: $V = B \times H$; V is the volume, the space inside. B is the area of the base, the shape at each end. H is the height of the prism, the distance between the two ends. It works because the prism is like a stack of layers, each the shape of the base.

Sort out the figure
A prism is a solid with two matching ends joined by flat sides. Here the ends are triangles, on the top and bottom. The other sides are rectangles. You need the volume, the space inside, in cubic inches.
The dashed line on the top triangle is 8 in. It meets one side of the triangle at a right angle (the small square). The bottom edge under that side is 4 in. The edge going straight up is 2 in.
Find the triangle’s base and height
The side that the dashed line meets is the top edge of a rectangle face. Opposite sides of a rectangle are equal. So that side is 4 in., the same as the bottom edge under it.
So the triangle has a base of 4 in. Its height is the dashed line, 8 in., because it meets the base at a right angle.
Find the area of the triangle
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 4 \times 8 = 16$
The triangle’s area is 16 square inches. Watch out: 16 is only the area of one end. Stopping here gives the wrong answer 16.
Multiply by the prism’s height
The height of the prism is the distance between the two triangles. That is the 2 in. edge.
$V = B \times H$
$V = 16 \times 2 = 32$
The volume is 32 cubic inches. Watch out: do not forget the one half. $4 \times 8 \times 2 = 64$ is the volume of a box, not of this prism.
So you would grid 32.

Check your answer

Picture a box that is 4 in. by 8 in. by 2 in. Its volume is $4 \times 8 \times 2 = 64$ cubic inches. The triangle has half the area of a 4 by 8 rectangle, so the prism holds half as much as the box. Half of 64 is 32.

More questions like this

The original DOE question

Question 62 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 199, question 62. Open page 199 of the handbook (PDF)

Question 44 of 69

2024 Test B Math, question 10

Question

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

A $\frac{15}{28}$
B $1\frac{13}{15}$
C $2\frac{1}{3}$
D $3\frac{1}{3}$

Show the answer

Correct answer: B

$4\frac{2}{3}\div 2\frac{1}{2}=\frac{14}{3}\div\frac{5}{2}=$

$\frac{14}{3}\times\frac{2}{5}=\frac{28}{15}=1\frac{13}{15}$

Question analysis

A fractions question asks for 4 2/3 divided by 2 1/2, with mixed numbers on both sides of the division sign. Students convert to improper fractions, multiply by the reciprocal and simplify. Mixed numbers invite shortcuts, so every step has to follow the rule for dividing fractions.

NYS standards: 6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing fractions is a grade 6 standard, and this is one division with small mixed numbers. Grade 7 work usually adds negative numbers or a word problem around it, so this sits below grade 7.
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: 3 steps
Reading level: Question and choices: about grade 16+, 5 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Dividing mixed numbers has a fixed procedure, and every wrong choice is a shortcut that skips part of it. Estimating first helps: 4 2/3 is a little less than twice 2 1/2, so the quotient is a little under 2, which already rules out 15/28 and 3 1/3. The remaining near miss, 2 1/3, comes from dividing by the whole number 2 alone.

Error-prone arithmetic moderate Working it out
Both mixed numbers must become improper fractions, 14/3 and 5/2, before the second is flipped, and 14/3 × 2/5 = 28/15 leaves an improper fraction to deal with. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The choices are written as mixed numbers and one proper fraction, so 28/15 has to be converted to a mixed number before it can be matched. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite both mixed numbers as improper fractions, multiply the first by the reciprocal of the second, then simplify and turn the improper fraction back into a mixed number.

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
Flips the first fraction instead of the second, computing 3/14 × 5/2 = 15/28, which is 2 1/2 divided by 4 2/3.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Divides 4 2/3 by the whole number 2 and drops the 1/2 of the divisor, so 14/3 ÷ 2 = 7/3, or 2 1/3.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Divides the whole parts and the fraction parts separately, 4 ÷ 2 = 2 and 2/3 ÷ 1/2 = 4/3, then adds them to get 3 1/3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number, and b over c is the fraction part. Multiply the whole number by c to turn the wholes into pieces. Then add the b extra pieces. The bottom number stays c.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.

Sort out the problem
You need to divide $4\frac{2}{3}$ by $2\frac{1}{2}$ . Both are mixed numbers, a whole number and a fraction together. Mixed numbers are hard to divide as they are. So first change each one into an improper fraction. That is a fraction whose top is bigger than its bottom.
Change the mixed numbers
$a\frac{b}{c} = \frac{a \times c + b}{c}$
$4\frac{2}{3} = \frac{4 \times 3 + 2}{3} = \frac{14}{3}$
$2\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2}$
Watch out: do not divide the whole parts and the fraction parts on their own. Doing $4 \div 2 = 2$ and $\frac{2}{3} \div \frac{1}{2} = \frac{4}{3}$ , then adding, gives $3\frac{1}{3}$ , choice D.
Flip the second fraction and multiply
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{14}{3} \div \frac{5}{2} = \frac{14}{3} \times \frac{2}{5} = \frac{28}{15}$
Multiply the tops, $14 \times 2 = 28$ , and the bottoms, $3 \times 5 = 15$ . Watch out: flip only the second fraction. Flipping the first one gives $\frac{15}{28}$ , choice A.
Change back to a mixed number
The choices are mixed numbers, so change 28/15 back. 15 goes into 28 one time, with 13 left over.
$\frac{28}{15} = 1\frac{13}{15}$
Watch out: dividing by just 2 and dropping the one half gives $2\frac{1}{3}$ , choice C.
So the answer is B, $1\frac{13}{15}$ .

Check your answer

Estimate. $4\frac{2}{3}$ is about 4.7, and $2\frac{1}{2}$ is 2.5. Two groups of 2.5 make 5, so the answer is a bit less than 2. $1\frac{13}{15}$ is about 1.9, which fits. Or multiply back: $\frac{28}{15} \times \frac{5}{2} = \frac{140}{30} = \frac{14}{3}$ .

More questions like this

The original DOE question

Question 67 as printed on page 201 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 67. Open page 201 of the handbook (PDF)

Question 45 of 69

2023 Test A Math, question 1

Question

Ms. Li opened a retirement account with a deposit of $2,500. This account earns 4% simple interest annually. How many years will it take her to earn $500 on her $2,500 deposit?

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

Show the answer

Correct answer: 5

Simple interest (y) is calculated by multiplying the initial deposit (p), the interest rate (r), and time (t):

$prt = y$

$(2{,}500)(0.04)t = 500$

$100t = 500$

$t = 5$

Question analysis

Ms. Li deposits $2,500 in a retirement account that pays 4% simple interest each year, and this grid-in asks how many years it takes to earn $500 in interest. Students find one year's interest and work out how many years are needed. It turns the usual simple interest question around, asking for the time rather than the interest.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Simple interest is a grade 7 percent application, usually asked as how much interest builds up over a given time. This one gives the interest and asks for the time, a short reverse step with friendly numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once one year's interest is known, $100, the rest is counting how many years make $500, so the arithmetic is light. The question turns the usual simple interest problem around, giving the interest and asking for the time, and that reversal is where students stall or grid a percent instead of a number of years. Writing 4% as 0.04, not 0.4, keeps the yearly interest from coming out ten times too big.

Something you must already know minor Working it out
The stem says 4% simple interest annually but never says that each year adds the same 4% of the $2,500 deposit, $100, so the yearly amount has to come from memory. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The question gives the interest, $500, and asks for the time, so the familiar formula runs backward: 2,500 × 0.04 × t = 500 has to be solved for t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in it shows no choices, so 20, the interest as a percent of the deposit, can be gridded with nothing on the page to say it is not a number of years. (Enright, Morley & Sheehan, 2002)

Strategy: Work out one year's interest as 4% of $2,500, then find how many of those yearly amounts add up to the $500 of interest.

Traps

Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

20 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that the $500 of interest is 20% of the $2,500 deposit and grids 20, stopping before dividing by the 4% earned each year.
0.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Writes 4% as 0.4 instead of 0.04, so one year's interest looks like $1,000 and the $500 seems to take only half a year.
125 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the $500 of interest by the 4 in 4%, combining two numbers from the stem without first finding 4% of the deposit.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest: $I = P \times r \times t$; I is the interest, the extra money the account pays you. P is the deposit, the money you put in. r is the yearly rate as a decimal. t is the time in years. Simple means each year’s interest is figured on the deposit only, so every year adds the same amount.
Percent to decimal: $p\% = \frac{p}{100}$; p is the percent. Percent means out of 100. So divide by 100 to write it as a decimal. For example, 4% is 4 ÷ 100, which is 0.04.

Sort out the question
Ms. Li puts $2,500 into an account. This first amount is called the deposit. Each year the account pays 4% simple interest. Interest is extra money the bank pays you for keeping your money there. You need the number of years it takes to earn $500 of interest. The answer is a number of years, not dollars and not a percent.
Find one year’s interest
First write 4% as a decimal. 4% means 4 out of 100.
$4\% = \frac{4}{100} = 0.04$
Now use the formula for 1 year, so t = 1.
$I = P \times r \times t$
$I = 2{,}500 \times 0.04 \times 1 = 100$
The account earns $100 each year. Watch out: 4% is 0.04, not 0.4. Using 0.4 makes one year’s interest $1,000, and you would grid 0.5 years.
Find how many years make $500
Simple interest adds the same $100 every year. So the time t must make the interest $500.
$2{,}500 \times 0.04 \times t = 500$
$100 \times t = 500$
$t = 500 \div 100 = 5$
It takes 5 years. Watch out: $500 is 20% of $2,500. But 20 is a percent, not a number of years. Don’t grid 20. So you would grid 5.

Check your answer

Count up one year at a time: $100, $200, $300, $400, $500. That is 5 years. Or put t = 5 back into the formula: $2{,}500 \times 0.04 \times 5 = 500$ . That matches the $500 in the question.

More questions like this

The original DOE question

Question 46 of 69

2023 Test A Math, question 2

Question

$6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$

What is the value of the expression shown above?

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

Show the answer

Correct answer: -9

$6 - 9 \div \left\|-3\right\| + (-2)^{3} \times 1\frac{1}{2}$	Calculate the exponent and convert the mixed number to an improper fraction.
$6 - 9 \div \left\|-3\right\| + (-8) \times \frac{3}{2}$	Next, calculate the absolute value.
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$	Perform the indicated multiplication and division.
$6 - 3 + (-12)$	Perform the addition and subtraction.
$3 + (-12) = -9$

Question analysis

This grid-in asks for the value of an arithmetic expression that mixes an absolute value, a negative number raised to the third power, a division and a mixed number. Students apply the order of operations with signed numbers. Cubing a negative base, and doing the division and multiplication before any adding, are where most slips happen.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students routinely evaluate expressions with negative numbers in the right order. This one combines an absolute value, a cubed negative and a mixed number in a single line, so it takes more steps than usual, each with a sign to watch.
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 minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 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 expression packs four kinds of notation into one line: a subtraction, a division by an absolute value, a negative base raised to the third power and a mixed number. None of the operations is hard alone, but each has a rule that is easy to break under time pressure, and the slips lead to very different totals. Rewriting the power and the mixed number first, then dividing and multiplying before any addition, avoids them.

Error-prone arithmetic moderate Working it out
(−2)³ must stay negative, −8, and the mixed number 1½ must become 3/2 before multiplying, so the last product is −12; a sign or fraction slip here moves the total far from −9. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Six numbers and six operations, including a power and an absolute value, have to be worked in a planned order, and every addition waits until the division and multiplication are done. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The value is negative and the grid-in offers no choices, so the minus sign has to be gridded and nothing on the page confirms that the total falls below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the absolute value and the power first, change the mixed number to an improper fraction, do the division and multiplication, and only then add and subtract from left to right.

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

15 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes (−2)³ as positive 8, as if every power of a negative were positive, so the last product is +12 and 6 − 3 + 12 = 15.
-13.5 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Works strictly from left to right: 6 − 9 = −3, divided by 3 is −1, plus −8 is −9, and times 1½ gives −13.5.
-6 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies the base by the exponent, taking (−2)³ as −6, so the product with 3/2 is −9 and the total is 6 − 3 − 9 = −6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{grouping} \rightarrow \text{powers} \rightarrow \times \text{ and } \div \rightarrow + \text{ and } -$; This is the order for working out an expression. First work inside grouping symbols, such as parentheses and absolute value bars. Next do powers. Then multiply and divide from left to right. Last, add and subtract from left to right. Using one order means everyone gets the same answer.
A power: $a^{3} = a \times a \times a$; a is the base, the number being multiplied. The small 3 is the exponent. It says how many times the base is used as a factor, a number being multiplied. It does not mean a times 3.
Absolute value: $\left|a\right| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so the absolute value of −3 is 3.
Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number part, and b over c is the fraction part. Each whole is c pieces, so a wholes are a × c pieces. Add the b extra pieces. The bottom number stays c.

Sort out the expression
You need the value of $6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$ . It has many operations, so the order matters. It has absolute value bars, a power of a negative number and a mixed number. The answer may be negative, so keep track of every sign.
Do the absolute value and the power
Grouping symbols and powers come first.
$\left|-3\right| = 3$
$(-2)^{3} = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$
Two negatives multiply to a positive, 4. One more negative makes it negative again. Watch out: (−2)³ is −8, not +8. Using +8 leads to 15. It is also not −2 × 3. That mistake leads to a total of −6.
Change the mixed number to a fraction
A fraction is easier to multiply than a mixed number.
$1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$
Now the expression looks like this:
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$
Multiply and divide next
Multiplying and dividing come before adding and subtracting.
$9 \div 3 = 3$
$(-8) \times \frac{3}{2} = \frac{-24}{2} = -12$
A negative times a positive is negative. Watch out: don’t just work from left to right. Doing 6 − 9 first, then dividing by 3, leads to −13.5.
Add and subtract last
Now only adding and subtracting are left. Work from left to right.
$6 - 3 + (-12) = 3 + (-12) = -9$
Adding −12 is the same as taking away 12. So you would grid −9, with the minus sign in its own column.

Check your answer

Look at the size of each part. The part 6 − 3 is a small positive number, 3. The last part, −8 times 1½, is a bigger negative number, −12. A big negative plus a small positive is negative. So an answer below zero, −9, makes sense.

More questions like this

The original DOE question

Question 47 of 69

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 48 of 69

2023 Test A Math, question 4

Question

$Line l is vertical and line n is horizontal, and a small square marks the right angle where they meet. Line m rises from the lower left to the upper right; it crosses line l below line n and crosses line n to the right of line l. Where m crosses l, the angle between the lower part of l and the lower-left part of m is marked 35°. Where m crosses n, the angle between the right part of n and the upper-right part of m is marked x°.$

In the figure above, line l is perpendicular to line n. What is the value of x?

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

Show the answer

Correct answer: 55

Since vertical angles are congruent, the right triangle in the figure has acute angles with measures of 35° and x°. The interior angles of a triangle sum to 180°, so set up an equation to solve for x:

$90 + 35 + x = 180$

$125 + x = 180$

$x = 55$

Question analysis

Three lines cross in a figure: line l is perpendicular to line n, and a slanted line m meets them at a marked 35° angle and an unknown angle x. This geometry grid-in asks for the value of x. Students must see the right triangle the lines form and carry the marked angles into it using vertical angles.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle problems use vertical, supplementary and complementary angles. This one also needs the angle sum of a triangle, which New York places in grade 8, and the triangle is not drawn as a separate shape, so students must spot it.
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: Reading a figure, table or graph moderate
Something you must already know moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The figure hides a small right triangle between the three lines, and the question is easy once students see it. Its angles are the right angle, the vertical partner of the 35° angle and the vertical partner of x, so x is what the right angle and 35° leave of 180°. Students who never spot the triangle tend to guess a direct relationship between the two marked angles, which no angle fact supports.

Reading a figure, table or graph moderate Understanding the question
Neither marked angle is inside the right triangle that lines l, m and n form: the 35° opens below the point where m crosses l, and x opens above line n, so each must be carried in as a vertical angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know moderate Working it out
The stem states only that l is perpendicular to n; students must supply that vertical angles are equal and that the three angles of a triangle add to 180°. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in it offers no choices, so 125, the right angle plus the 35° angle, can be gridded with nothing to signal the missing subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Find the triangle formed by the three lines, use vertical angles to move the 35° and x° angles inside it, and apply the triangle's angle sum with the right angle.

Traps

Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

125 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Adds the right angle and the 35° angle, 90 + 35 = 125, and grids that sum without subtracting it from the triangle's 180°.
35 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Assumes line m crosses l and n at equal angles, as it would cross two parallel lines, though l and n are perpendicular.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Vertical angles: $\text{angle } a = \text{angle } b$; When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle: $A + B + C = 180$; A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

Sort out the figure
Line l is perpendicular to line n. Perpendicular means they meet at a right angle, 90°. The small square in the figure marks it. Line m crosses both lines. The three lines make a triangle with a right angle. You need x, the angle marked x° where m crosses n.
Move the 35° angle into the triangle
The 35° angle is below the point where m crosses l. So it is outside the triangle. Lines l and m make an X there. The triangle’s angle is straight across from the 35° angle. They are vertical angles, so they are equal. The triangle’s angle at that corner is 35°.
Move the x° angle into the triangle
The x° angle is above line n, so it is outside the triangle too. Lines m and n make another X. The triangle’s angle at that crossing is straight across from x°. So the third angle of the triangle is also x°. Watch out: line m does not cross l and n at the same angle. Lines l and n are not parallel, so x is not 35.
Use the triangle’s angle sum
The triangle’s three angles are 90°, 35° and x°.
$A + B + C = 180$
$90 + 35 + x = 180$
$125 + x = 180$
$x = 180 - 125 = 55$
Watch out: 125 is only the two known angles added together. You still subtract it from 180. Gridding 125 is wrong. So you would grid 55.

Check your answer

In a right triangle, the two smaller angles add up to 90°. That is because the right angle uses 90° of the 180°. Check: $35 + 55 = 90$ . Also, in the figure the x° angle looks a bit bigger than the 35° angle, and 55 is.

More questions like this

The original DOE question

Question 49 of 69

2023 Test A Math, question 5

Question

The mean value of 8 numbers is 17. Three of these numbers (9, 11, and 20) are discarded. What is the mean of the 5 remaining numbers?

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

Show the answer

Correct answer: 19.2

First, find the sum of the original 8 numbers. Let x equal the sum of those numbers and use the mean formula to solve for x:

$\frac{x}{8} = 17$

$x = 136$

Next, subtract the 3 discarded numbers from the sum:

$136 - 9 - 11 - 20 = 96$

Finally, calculate the mean of the remaining 5 numbers by dividing the new sum, 96, by 5:

$\frac{96}{5} = 19.2$

Question analysis

Eight numbers have a mean of 17, and three of them, 9, 11 and 20, are removed. This statistics grid-in asks for the mean of the five numbers that are left. Students have to rebuild the total from the mean before removing anything, and then divide by the new count rather than the old one.

NYS standards: 6.SP.5c range, center, and overall pattern
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a mean is grade 6 content, but working back from a mean to a total, removing values and averaging again is a typical grade 7 multistep task. As a grid-in, it offers no choices to check a wrong total against.
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
Much to keep track of minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 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

A mean cannot be adjusted directly when numbers are removed, so the question rewards switching from the mean to the total and back. The removed numbers, 9, 11 and 20, average less than 17, so the new mean must rise above 17; that quick check rules out an answer such as 12. Points are lost by dividing by the wrong count or by stopping at the new total.

Many steps moderate Working it out
Three operations come in a fixed order: multiply the mean by 8 to get 136, subtract 9, 11 and 20 to leave 96, then divide by the numbers that remain. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
The count changes partway through, from 8 numbers to 5, and the divisor at the end must be the new count, not the 8 from the first sentence. (Sweller, 1988; Carpenter, Just & Shell, 1990)
An answer in an unexpected form minor Choosing the answer
The answer, 19.2, is a decimal that must be gridded with its point, and with no choices a total of 96 or a mean over 8 numbers looks complete. (Enright, Morley & Sheehan, 2002)

Strategy: Turn the mean into a total by multiplying by 8, subtract the three discarded numbers, and divide what is left by the count of numbers that remain.

Traps

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

The wrong answers, classified

12 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the remaining total, 136 − 40 = 96, correctly but divides it by the original 8 numbers instead of the 5 that are left.
96 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Subtracts the three discarded numbers from the total of 136 and grids the new sum, 96, without dividing by the 5 numbers left.
27.2 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Turns the mean into the total, 136, and divides it by 5 without first subtracting the three discarded numbers, 9, 11 and 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean, or average: $\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}$; The mean is the average. The sum is what you get when you add all the numbers. Divide it by how many numbers there are. The mean is the value each number would have if the total were shared out equally.
Sum from a mean: $\text{sum} = \text{mean} \times \text{how many numbers}$; This is the mean formula turned around. If you know the mean and how many numbers there are, multiply them to get the sum. It works even when you don’t know the numbers themselves.

Sort out the question
There are 8 numbers. Their mean, or average, is 17. You don’t know what the 8 numbers are. Three of them, 9, 11 and 20, are discarded, or thrown away. That leaves 5 numbers. You need the mean of those 5 numbers.
Find the sum of all 8 numbers
You can’t add the 8 numbers, because you don’t know them. But the mean tells you their sum.
$\text{sum} = \text{mean} \times \text{how many numbers}$
$\text{sum} = 17 \times 8 = 136$
The 8 numbers add up to 136.
Take away the three discarded numbers
Subtract the three numbers that were thrown away.
$136 - 9 - 11 - 20 = 136 - 40 = 96$
The 5 numbers left add up to 96. Watch out: 96 is a sum, not a mean, so gridding 96 is wrong. Also, dividing 136 by 5 without subtracting first gives 27.2, which is wrong.
Divide by the 5 numbers left
Now use the mean formula for the numbers that are left.
$\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}$
$\text{mean} = \frac{96}{5} = 19.2$
Watch out: divide by 5, not by 8. Only 5 numbers are left. Dividing by 8 gives 12. So you would grid 19.2, with the decimal point in its own column.

Check your answer

The three numbers thrown away add to 40, so their mean is about 13. That is less than 17. Removing small numbers should make the mean go up, and 19.2 is more than 17. Also, $19.2 \times 5 = 96$ , the sum of the 5 numbers left.

More questions like this

The original DOE question

Question 50 of 69

2023 Test A Math, question 11

Question

$\frac{81}{10} = \frac{9}{n}$

What value of n makes the equation above true?

A1
B $1\frac{1}{9}$
C5
D $10\frac{1}{9}$

Show the answer

Correct answer: B

$\frac{81}{10} = \frac{9}{n}$	Cross multiply.
$81n = 9(10)$	Apply the multiplicative inverse property; divide both sides of the equation by 81.
$n = \frac{90}{81} = 1\frac{9}{81} = 1\frac{1}{9}$	Then simplify the fraction.

Question analysis

This question gives the proportion 81/10 = 9/n and asks for the value of n that makes it true. Students solve a proportion by cross-multiplying and dividing, then simplify the result. Reducing the resulting fraction and matching its form to the choices is where students slip.

NYS standards: 6.EE.7 one-step equations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Solving a proportion is familiar from grades 6 and 7, and this one needs only cross-multiplying and one division. Its one extra demand is simplifying the result and writing it in the form the choices use.
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: An answer in an unexpected form moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 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

The proportion solves in one cross-multiplication, but the result is an improper fraction that must be simplified and rewritten to match. Students who notice that 81 is 9 × 9 can shortcut it: the numerator shrinks by a factor of 9 from left to right, so the denominator must shrink too, giving 10 ÷ 9. The mixed-number form is where answers drift, by dropping the leftover ninth or by writing 10 × 1/9 as 10 1/9.

An answer in an unexpected form moderate Choosing the answer
Cross-multiplying gives n = 90/81, which has to be reduced to 10/9 and rewritten as the mixed number 1 1/9 before it matches any choice. (Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The choices 1 and 1 1/9 differ by one ninth, and estimating 90 ÷ 81 as about 1 cannot separate them; only keeping the remainder does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cross-multiply to get 81n = 90, divide both sides by 81, and simplify the fraction before comparing it with the choices.

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.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

A Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Finds n by dividing 10 by 9 but keeps only the whole-number part of the quotient, 1, dropping the remainder that the exact answer needs.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No cross-multiplication or scaling of 81/10 gives 5, which would make 9/n equal 1.8 instead of 8.1, so it mostly catches guesses.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Reaches n = 10 × 9/81 = 10 × 1/9 but writes the product as the mixed number 10 1/9, as if a whole number beside a fraction meant multiplication.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{, so } a \times d = b \times c$; a over b and c over d are two equal fractions, and b and d are not 0. 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.
Simplifying a fraction: $\frac{a \times k}{b \times k} = \frac{a}{b}$; a over b is a fraction, and k is a number that divides both the top and the bottom. Dividing the top and the bottom by the same number gives an equal fraction with smaller numbers.
Fraction to mixed number: $\frac{a}{b} = q\frac{r}{b}$; a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

Sort out the question
The equation says two fractions are equal: $\frac{81}{10}$ and $\frac{9}{n}$ . n is an unknown bottom number. You need the value of n that makes them equal. Some choices are mixed numbers, so n may not be a whole number.
Cross-multiply the two fractions
Multiply each top by the other fraction’s bottom. This gets n out of the bottom of the fraction.
$\frac{81}{10} = \frac{9}{n}$
$81 \times n = 10 \times 9$
$81n = 90$
Divide both sides by 81
81n means 81 times n. To undo multiplying by 81, divide both sides by 81.
$n = \frac{90}{81}$
Now simplify. 90 and 81 are both in the 9 times table, so divide the top and bottom by 9.
$\frac{90}{81} = \frac{90 \div 9}{81 \div 9} = \frac{10}{9}$
Write it as a mixed number
10 ÷ 9 is 1 with 1 left over. The leftover stays over 9.
$\frac{10}{9} = 1\frac{1}{9}$
Watch out: keep the leftover. Dropping it gives 1, choice A. Also, ten ninths is not 10 and one ninth. That mixed number, $10\frac{1}{9}$ , is choice D, which is far too big. So the answer is B, $1\frac{1}{9}$ .

Check your answer

Estimate. $\frac{81}{10}$ is about 8. For $\frac{9}{n}$ to be about 8, n must be a little more than 1. Check exactly: $9 \div \frac{10}{9} = 9 \times \frac{9}{10} = \frac{81}{10}$ . It matches.

More questions like this

The original DOE question

Question 68 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 68. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 69

2023 Test A Math, question 30

Question

Lamel has a jar containing 6 red chips, 10 blue chips, and 4 yellow chips. If he removes one chip at random, what is the probability that it will not be red?

A $\frac{4}{5}$
B $\frac{7}{10}$
C $\frac{3}{10}$
D $\frac{1}{5}$

Show the answer

Correct answer: B

There are currently 20 chips in the jar ( $6 + 10 + 4$ ). Of those chips, 14 are not red. So the probability of choosing a chip that is not red is $\frac{14}{20} = \frac{7}{10}$ .

Question analysis

In this probability word problem, Lamel's jar holds red, blue and yellow chips, and students find the chance that one chip drawn at random is not red. It tests counting the favorable outcomes against the total and writing the probability as a fraction. Missing the bold word not, or counting the wrong colors, leads to tempting wrong fractions.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Probability of a single event with equally likely outcomes is grade 6 content in New York. Students add three counts and form one fraction, and the only extra care is the bold word not, so it sits below grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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 one-event probability with small whole numbers, so the bold not carries the question. A student who skims past it computes the chance of red, a tidy fraction that is waiting among the choices. The other wrong fractions come from removing the wrong color or keeping only one of the two colors that are not red. Adding blue and yellow first, then dividing by all 20 chips, avoids each of them.

Turning words into math minor Understanding the question
The bold not turns the question around: the chips that count are the 10 blue and 4 yellow ones, 14 in all, or equivalently 1 minus the red share of 6/20. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The count of 14 out of 20 matches no choice until it is reduced to 7/10, so a correct fraction left unsimplified can look missing from the list. (Enright, Morley & Sheehan, 2002)

Strategy: Add all the chips for the total, then count the chips that are blue or yellow, or subtract the chance of red from 1, and simplify the fraction.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Leaves out the 4 yellow chips instead of the 6 red ones, finding the chance of not yellow, 16/20 = 4/5.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Overlooks the bold not and gives the chance that the chip is red, 6 out of 20, which reduces to 3/10.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Counts only the 4 yellow chips as not red and forgets the 10 blue ones, so it finds 4/20 = 1/5, the chance of yellow alone.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. It works when each outcome is equally likely, like pulling one chip without looking.

Sort out the question
Lamel’s jar holds 6 red chips, 10 blue chips and 4 yellow chips. He takes one chip without looking. You need the probability that it is not red. That means the chip is blue or yellow.
Count all the chips
$6 + 10 + 4 = 20$
There are 20 chips in all. That is the bottom of the fraction, because any of the 20 could come out.
Count the chips that are not red
The blue and yellow chips are the ones that are not red.
$10 + 4 = 14$
So 14 chips are not red. Watch out: the question asks for not red. Counting the 6 red chips gives $\frac{6}{20} = \frac{3}{10}$ , choice C. Counting only the 4 yellow chips gives $\frac{1}{5}$ , choice D.
Find and simplify the probability
$P = \frac{\text{outcomes you want}}{\text{all outcomes}}$
$P = \frac{14}{20} = \frac{14 \div 2}{20 \div 2} = \frac{7}{10}$
Dividing the top and bottom by 2 puts the fraction in lowest terms. So the answer is B, $\frac{7}{10}$ .

Check your answer

A chip is either red or not red, so the two chances must add up to 1. The chance of red is $\frac{6}{20} = \frac{3}{10}$ . Then $\frac{3}{10} + \frac{7}{10} = 1$ . It works.

More questions like this

The original DOE question

Question 87 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 87. Open page 83 of the handbook (PDF)

Question 52 of 69

2023 Test A Math, question 41

Question

Yesterday Sarah read 15% of her entire book. Today she read another 17% of the entire book. In lowest terms, what fraction of the book is left for her to read?

A $\frac{7}{25}$
B $\frac{3}{10}$
C $\frac{17}{25}$
D $\frac{7}{10}$

Show the answer

Correct answer: C

So far, Sarah has read $15\% + 17\% = 32\%$ of the entire book. That means she has $100\% - 32\% = 68\%$ left to read.

$68\% = \frac{68}{100} = \frac{17}{25}$

Question analysis

This percent word problem says Sarah read 15% of her book yesterday and 17% today, and asks what fraction of the book she still has to read, in lowest terms. Students combine percents of the same whole, find the part remaining, and convert a percent to a simplified fraction. Rounding, or giving the part already read, leads to wrong choices.

NYS standards: 7.EE.3 multi-step rational number problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 percent work often takes a percent of a percent or finds a percent change. Here both percents are of the same whole book, so they combine directly; the rest is a subtraction and simplifying a fraction, lighter than typical.
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: 3 steps
Reading level: Question and choices: about grade 5, 33 words.
Corrected from the DOE handbook: Missing percent signs in the explanation were 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

The percent work is a sum and a subtraction from 100, so the question mostly tests the last step, writing 68% as a fraction in lowest terms. Students who round to 30% read and 70% left to get a friendlier fraction land on a listed choice, as do those who give the part read. Dividing 68 and 100 by their common factor 4 keeps the answer exact.

Turning words into math minor Understanding the question
Both percents are “of the entire book,” so they simply add to 32%, and “left for her to read” asks for the rest, 100% − 32%, not the part already read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The result has to be written as a fraction in lowest terms, so 68/100 must be divided by 4 to reach a form that appears among the choices. (Enright, Morley & Sheehan, 2002)

Strategy: Add the two percents, since both are of the whole book, subtract the total from 100%, then write the result over 100 and simplify it fully.

Traps

Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
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 Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No sum or difference of 15% and 17% gives 28%, so 7/25 mostly draws students who match the right fraction's 25 without checking the numerator.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Gives the part already read rather than the part left, and rounds that 32% to 30% to write it as 3/10.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Finds that 68% is left but rounds it to 70% to make an easier fraction, giving 7/10 instead of simplifying 68/100 exactly.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes “So far, Sarah has read 15 + 17 = 32 of the entire book. That means she has 100 − 32 = 68 left to read.” (p. 143, question 98), with no percent signs, so 32 and 68 read as amounts of the book (pages, say) that the question never gives. The question gives 15% and 17%, and the handbook’s next line begins “68% =”, so the site now reads 15% + 17% = 32% and 100% − 32% = 68%. The answer, 17/25, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction: $p\% = \frac{p}{100}$; p is the percent. Percent means “out of 100,” so p percent is p hundredths. For example, 50% is 50 out of 100, which is one half.
Simplifying a fraction: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top number, b is the bottom, and n is a number that divides both evenly. Dividing the top and bottom by the same number keeps the value the same. A fraction is in lowest terms when no number bigger than 1 divides both.

Sort out the question
Sarah read 15% of her book yesterday and 17% today. Both are percents of the whole book. You need the part still left to read, as a fraction in lowest terms. That means a fraction that can’t be simplified any more.
Find how much she has read
$15\% + 17\% = 32\%$
Both percents are of the same whole book, so you can add them. She has read 32% of it.
Find how much is left
The whole book is 100%.
$100\% - 32\% = 68\%$
Watch out: 32% is the part she has read, not the part left. Rounding 32% to 30% gives $\frac{3}{10}$ , choice B.
Write 68% as a fraction
$p\% = \frac{p}{100}$
$68\% = \frac{68}{100}$
This fraction is not in lowest terms yet, because 68 and 100 are both even.
Write it in lowest terms
The number 4 divides both 68 and 100 evenly.
$\frac{a}{b} = \frac{a \div n}{b \div n}$
$\frac{68}{100} = \frac{68 \div 4}{100 \div 4} = \frac{17}{25}$
Watch out: don’t round 68% to 70%. That gives $\frac{7}{10}$ , choice D. So the answer is C, $\frac{17}{25}$ .

Check your answer

Turn the answer back into a percent: $\frac{17}{25} = \frac{68}{100}$ , which is 68%. Then 68% + 32% = 100%, the whole book. Also, 17 and 25 share no factor except 1, so the fraction is in lowest terms.

More questions like this

The original DOE question

Question 98 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 98. Open page 86 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 53 of 69

2023 Test B Math, question 2

Question

In a scale drawing of a rectangular garden, the length is 15 inches and the width is 9 inches. In the drawing, 2 inches represents 3 yards. What is the width of the actual garden, in yards?

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

Show the answer

Correct answer: 13.5

Let w represent the width of the actual garden. Set up a proportion based on the information given:

$\frac{w}{9} = \frac{3}{2}$

$w = \frac{3(9)}{2} = \frac{27}{2} = 13.5$

Question analysis

In a scale drawing of a rectangular garden, 2 inches stands for 3 yards, and the drawing measures 15 inches by 9 inches. This grid-in asks for the real width of the garden in yards. Students apply the scale to one side only: the length is given but plays no part, and the scale has to be used the right way round.

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are grade 7 content, and this is a typical one: a single length converted with a stated scale. The only extra demand is using the width rather than the length and entering the result with no choices to check.
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, 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 drawing gives two lengths and only the width matters, so the first decision is which number to scale. The second is direction: the real garden must measure more in yards than the drawing’s 9 in inches, which a student can check before entering anything. Errors come from scaling the length, turning the scale over, or treating 2 to 3 as an increase of one.

Turning words into math minor Understanding the question
“2 inches represents 3 yards” must become 1.5 yards for every inch of drawing, so the 9-inch width is multiplied by 3/2, not by 2/3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The answer is a decimal entered with no choices to compare, so a result from the 15-inch length or from an upside-down scale looks just as finished. (Enright, Morley & Sheehan, 2002)

Strategy: Find how many yards one inch of the drawing stands for, then multiply the drawing's width, not its length, by that number.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

22.5 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Applies the scale to the 15-inch length instead of the 9-inch width: 15 × 3 ÷ 2 = 22.5 yards.
6 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Turns the scale upside down, multiplying the 9 inches by 2/3 instead of 3/2, so the real garden comes out smaller than the drawing’s number.
10 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reads 2 inches to 3 yards as adding 1, keeping a constant difference instead of a constant ratio, so 9 inches becomes 9 + 1 = 10 yards.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion: $\frac{\text{real size}}{\text{drawing size}} = \frac{\text{real size}}{\text{drawing size}}$; A scale drawing shrinks everything by the same amount. So real size over drawing size makes the same fraction for every part of the drawing. Keep the real sizes on top on both sides, and the drawing sizes on the bottom.

Sort out the question
The drawing shows a rectangle, a four-sided shape with square corners. On the drawing, the length is 15 inches and the width is 9 inches. The scale says 2 inches on the drawing stand for 3 yards in real life. You need the real width, in yards.
Only the width matters. The 15-inch length is extra.
Find what one inch stands for
If 2 inches stand for 3 yards, then 1 inch stands for half as much.
$3 \div 2 = 1.5$
So each inch on the drawing is 1.5 yards in real life. That means the number of yards will be bigger than the number of inches.
Set up the proportion and solve
Let w be the real width, in yards. Put yards on top and inches on the bottom, on both sides.
$\frac{w}{9} = \frac{3}{2}$
To get w alone, multiply both sides by 9.
$w = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5$
The real width is 13.5 yards.
Give the answer
Watch out: don’t flip the scale. Using $9 \times \frac{2}{3}$ gives 6, which is smaller than the drawing’s number. And don’t use the length: 15 inches would give 22.5 yards. So you would grid 13.5.

Check your answer

Count groups of 2 inches. The 9-inch width holds 4.5 groups of 2 inches. Each group stands for 3 yards, so the real width is $4.5 \times 3 = 13.5$ yards. That matches.

More questions like this

The original DOE question

Question 59 as printed on page 191 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 59. Open page 191 of the handbook (PDF)

Question 54 of 69

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 55 of 69

2023 Test B Math, question 11

Question

$n = 0.13 + \frac{207}{100}$

What is the value of n expressed as a fraction?

A $\frac{1}{3}$
B $\frac{11}{5}$
C $\frac{9}{4}$
D $\frac{23}{10}$

Show the answer

Correct answer: B

$n = 0.13 + \frac{207}{100} =$

$\frac{13}{100} + \frac{207}{100} =$

$\frac{220}{100} = \frac{11}{5}$

Question analysis

The number n equals 0.13 plus 207/100, and students must write n as a single fraction. This question mixes a decimal with a fraction in hundredths. Converting both to the same form, adding, and reducing to lowest terms are the steps, and the four fractions offered look quite different at first glance.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting decimals to fractions and simplifying are skills taught before grade 7. Both numbers here are in hundredths, so the addition is immediate and only the final reduction takes thought.
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: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Converting 0.13 to hundredths makes this a single addition, 13 + 207 = 220 hundredths, so the question tests little beyond reducing a fraction. Two of the wrong choices sit close to 2.2 but come from no common working, and the third only echoes the digits of 0.13. A student who checks each choice as a decimal finds only one equal to 2.2.

An answer in an unexpected form minor Choosing the answer
The sum comes out as 220/100 or 2.2, but the choices are in lowest terms, so it must be reduced, or each choice converted, before anything matches. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
One term is a decimal and the other a fraction; writing 0.13 as 13/100 puts both in hundredths, and a misplaced digit there changes the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write 0.13 as 13/100 so both terms are in hundredths, add the numerators, then divide the top and bottom by their greatest common factor.

Traps

Answer in another form: The right answer is written differently from the way you worked it out.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Echoes the digits of 0.13, reading it as the fraction 1/3, and ignores the 207/100 altogether.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9/4, which is 2.25; adding 13 and 207 hundredths gives 2.2, so this mostly draws guesses near the sum.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing leads to 23/10 except 13 + 207 coming out as 230, a slip with no reason behind it, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths: $\text{a decimal with two places} = \frac{\text{its digits}}{100}$; The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
Adding fractions with the same bottom: $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$; a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.

Sort out the question
n equals 0.13 plus $\frac{207}{100}$ . One part is a decimal and one is a fraction. You need n as a fraction. The choices are in lowest terms, which means they can’t be simplified any more.
Write the decimal as a fraction
The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.
$0.13 = \frac{13}{100}$
Now both numbers are in hundredths. Watch out: 0.13 is not $\frac{1}{3}$ . That only copies the digits, choice A.
Add the hundredths
Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.
$\frac{13}{100} + \frac{207}{100} = \frac{13 + 207}{100} = \frac{220}{100}$
Simplify the fraction
$\frac{220}{100}$ is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.
$\frac{220 \div 20}{100 \div 20} = \frac{11}{5}$
So the answer is B, $\frac{11}{5}$ .

Check your answer

Change both to decimals. $\frac{11}{5} = 11 \div 5 = 2.2$ . And $0.13 + 2.07 = 2.20$ . They match. Choices C and D are close, but $\frac{9}{4} = 2.25$ and $\frac{23}{10} = 2.3$ .

More questions like this

The original DOE question

Question 68 as printed on page 194 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 68. Open page 194 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 56 of 69

2023 Test B Math, question 37

Question

Which expression is equivalent to $\frac{1}{2}x+7-\left(x-\frac{9}{2}\right)$ ?

A $\frac{1}{2}(x-5)$
B $\frac{1}{2}(x+5)$
C $-\frac{1}{2}(x+23)$
D $-\frac{1}{2}(x-23)$

Show the answer

Correct answer: D

First, distribute the negative sign, which is the same as multiplying both terms in the parentheses by −1.

Next, combine like terms:

$\frac{1}{2}x+7-\left(x-\frac{9}{2}\right)$

$\frac{1}{2}x+7-x+\frac{9}{2}$

$-\frac{1}{2}x+\frac{23}{2}$

Then, factor the common factor of $-\frac{1}{2}$ from both terms, paying careful attention to the signs:

$-\frac{1}{2}x+\frac{23}{2}=-\frac{1}{2}(x-23)$

Question analysis

An algebra question asking which factored expression equals one half x plus 7 minus the quantity x minus nine halves. Students distribute the subtraction, combine like terms with fractions, and match a factored form. Signs are the danger at every step, and the choices differ only in the signs and numbers inside the parentheses.

NYS standards: 7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students expand and simplify linear expressions with fractions. Here a subtracted group must be distributed and the result factored by a negative fraction, two places where a sign is easily lost, a longer chain than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
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 moderate
A very close wrong answer moderate
Solution: 3 steps
Reading level: Question and choices: about grade 15, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two separate sign decisions decide this question: distributing the subtraction over x − 9/2, and factoring a negative half out of +23/2. Students who handle the first correctly often stumble on the second and choose the look-alike with +23. Expanding each choice, rather than factoring by hand, is quicker and safer, since it turns the comparison into matching −1/2x + 23/2 term by term.

Error-prone arithmetic moderate Working it out
The minus before (x − 9/2) must change two signs, and the result mixes halves with whole numbers: 1/2x − x and 7 + 9/2 both need care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form moderate Choosing the answer
The simplified result, −1/2x + 23/2, appears nowhere; every choice is factored by ±1/2, so students must factor out −1/2 themselves or expand each choice to compare. (Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
The answer, −1/2(x − 23), and −1/2(x + 23) differ only in the sign inside the parentheses, and the simplified form’s +23/2 makes the plus sign look right. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Change the sign of both terms inside the parentheses, combine the x terms and the constants, then expand each choice to see which one matches rather than factoring yourself.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Gives the minus to x but not to −9/2, so the constants make 7 − 9/2 = 5/2, then drops the leading negative when factoring: 1/2(x − 5).
B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Combines 1/2x − x as +1/2x, subtracting the smaller coefficient from the larger, and also leaves −9/2 unchanged, so the result is 1/2x + 5/2 = 1/2(x + 5).
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Simplifies correctly to −1/2x + 23/2 but keeps +23 when factoring out −1/2, though dividing +23/2 by −1/2 gives −23.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a group: $-(b - c) = -b + c$; b and c are the two terms inside the parentheses. Terms are the parts joined by plus or minus signs. A minus sign in front of parentheses takes away the whole group. So it changes the sign of every term inside, not only the first one.
Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. Multiply a by each term inside. It works because a groups of (b + c) hold a of each. Use it to multiply out the choices and compare them.

Sort out the question
Equivalent means equal for every value of x. The choices are written in a different form from the problem. So simplify the problem first. Then multiply out the choices and compare.
Remove the parentheses
The minus sign in front of the parentheses takes away both terms inside. So x becomes −x, and $-\frac{9}{2}$ becomes $+\frac{9}{2}$ .
$\frac{1}{2}x + 7 - \left(x - \frac{9}{2}\right) = \frac{1}{2}x + 7 - x + \frac{9}{2}$
Watch out: If you change only the sign of x, the numbers become $7 - \frac{9}{2} = \frac{5}{2}$ . That mistake leads to choice A or B.
Combine the x terms
Like terms have the same letter part. The x terms are $\frac{1}{2}x$ and −x. Write x as $\frac{2}{2}x$ so the fractions match.
$\frac{1}{2}x - \frac{2}{2}x = -\frac{1}{2}x$
The result is negative, because you take away more x than you have.
Combine the numbers
Now combine the plain numbers, 7 and $\frac{9}{2}$ . Write 7 as fourteen halves so the bottoms match. Then add.
$7 + \frac{9}{2} = \frac{14}{2} + \frac{9}{2} = \frac{23}{2}$
The expression simplifies to $-\frac{1}{2}x + \frac{23}{2}$ .
Multiply out choices C and D
Only C and D start with $-\frac{1}{2}$ , so test those. Use the distributive property. A negative times a negative is positive.
$-\frac{1}{2}(x + 23) = -\frac{1}{2}x - \frac{23}{2}$
$-\frac{1}{2}(x - 23) = -\frac{1}{2}x + \frac{23}{2}$
The first line is choice C, and the second is choice D. Only D matches.
Watch out: Choice C looks almost the same. But it gives $-\frac{23}{2}$ , not $+\frac{23}{2}$ .
So the answer is D, $-\frac{1}{2}(x - 23)$ .

Check your answer

Try x = 1 in the problem: $\frac{1}{2} + 7 - \left(1 - \frac{9}{2}\right) = 7\frac{1}{2} + 3\frac{1}{2} = 11$ . Now put x = 1 into choice D: $-\frac{1}{2}(1 - 23) = -\frac{1}{2} \times (-22) = 11$ . Both give 11, so they match.

More questions like this

The original DOE question

Question 94 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 94. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 57 of 69

2018 Test A Math, question 1

Question

$Parallelogram PQRS with side PS on a horizontal line that extends past P and S, and side QR at the top. At S, the angle between side SR and the line extended to the right of S is marked 72°. At Q, a vertical line rises above the figure at a right angle to side QR (right-angle mark), and angle x° is marked on the left of Q, between that vertical line and side QP.$

In the figure above, PQRS is a parallelogram. What is the value of $x$ ?

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

Show the answer

Correct answer: 162

First, find the measure of angle PQR. The measure of angle PQR is equal to the measure of angle PSR.

m∠PSR = 180 − 72 = 108.

So, the measure of angle PQR is also 108.

108 + 90 + x = 360

198 + x = 360

x = 162

Question analysis

A grid-in geometry question about angles around a parallelogram PQRS. A 72° angle sits outside corner S on the extended base, and at corner Q a right angle and the unknown angle x complete the turn around the point. Students chain three angle facts, and the angle asked for lies outside the shape, not inside it.

NYS standards: 7.G.5 angle relationships and equations
8.G.5 angle sums, transversals, angle-angle similarity
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle work usually applies one or two facts, such as supplementary or vertical angles, to find an angle. This chains three, and needs the equal opposite angles of a parallelogram, which rest on grade 8 parallel-line facts.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Combines ideas moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is two subtractions; the difficulty is seeing which angles the marks refer to. Students lose points by treating the 72° as the parallelogram's own angle at S, or by stopping at angle PQR, the most natural number to grid. The question rewards tracing a route from S to Q before computing anything, and checking at the end that x is the angle outside the shape that the arc marks.

Reading a figure, table or graph major Understanding the question
The 72° is marked outside the parallelogram, between side SR and the base extended past S, and x sits outside corner Q, between QP and a vertical ray that meets QR at a right-angle mark. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
Three angle facts have to be chained: a straight line gives the angle inside S, opposite angles of a parallelogram carry it to Q, and the angles around Q make a full 360° turn. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
As a grid-in there are no choices to check against, so a student who stops at angle PQR, inside the parallelogram, gets no signal that the job is unfinished. (Enright, Morley & Sheehan, 2002)

Strategy: Use the straight line at S to find the parallelogram's angle there, carry it to the opposite corner Q, then take that angle and the right angle away from 360.

Traps

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

The wrong answers, classified

108 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds angle PSR = 180 − 72 = 108 and carries it to angle PQR, then grids that interior angle instead of the angle x outside it.
198 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads the 72° as the parallelogram's interior angle at S, so takes angle PQR as 72 and grids 360 − 90 − 72 = 198.
252 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Overlooks the right-angle mark at Q and subtracts only the 108° angle PQR from 360, giving 252.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line: $a + b = 180$; a and b are two angles side by side on a straight line. Together they make a half turn, so they always add up to 180 degrees. Use it when an angle you need shares a straight line with an angle you know.
Opposite angles of a parallelogram: $\angle PQR = \angle PSR$; The symbol ∠ means angle. ∠PQR is the angle at corner Q, between sides QP and QR. ∠PSR is the angle at the opposite corner, S. In a parallelogram, opposite corners always have equal angles, because the shape leans the same way at both.
Angles around a point: $a + b + c = 360$; a, b and c are angles that fit around one point with no gaps and no overlaps. Going all the way around is a full turn, which is 360 degrees. So the angles add up to 360.

Sort out what the figure shows
PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, which means they never meet. You need the value of x, in degrees.
Look closely at the marks. The 72° angle is outside the shape, at S. It sits between side SR and the base line, which keeps going past S. At Q, a vertical line meets side QR. The small square means a right angle, 90°. The angle x is outside the shape too. It runs from that vertical line around to side QP.
Find the angle inside corner S
Angle PSR is the shape’s own angle at S. It sits right next to the 72° angle, on the same straight base line. So the two angles add up to 180°.
$a + b = 180$
$\angle PSR + 72 = 180$
$\angle PSR = 180 - 72 = 108$
So angle PSR is 108°. Watch out: 72° is not the inside angle at S. Using 72 as the inside angle leads to the wrong answer 198.
Carry the angle over to Q
Corners Q and S are opposite each other. In a parallelogram, opposite angles are equal. So angle PQR matches angle PSR.
$\angle PQR = \angle PSR = 108$
So angle PQR is 108°. Watch out: this is not x. Angle PQR is inside the shape, and x is outside it. Stopping here and gridding 108 is the most common mistake.
Add the angles around Q
Three angles fill the space all the way around point Q. They are the right angle (90°), angle PQR (108°) and x. Together they make a full turn.
$a + b + c = 360$
$90 + 108 + x = 360$
$198 + x = 360$
Watch out: don’t skip the right angle at Q. Leaving out the 90 gives 252.
Solve for x
Subtract 198 from 360 to find what is left for x.
$x = 360 - 198 = 162$
So x is 162°. So you would grid 162.

Check your answer

Add the three angles at Q: $90 + 108 + 162 = 360$ . That is a full turn. Also, the arc for x is wider than a right angle but not as wide as a straight line. So x should be between 90° and 180°, and 162° is.

More questions like this

The original DOE question

Question 58 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 58. Open page 81 of the handbook (PDF)

Question 58 of 69

2018 Test A Math, question 2

Question

The owner of a tree farm plants pine trees and oak trees in a ratio of $8:3$ . How many oak trees are planted if 264 pine trees are planted?

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

Show the answer

Correct answer: 99

Let $x$ be the number of oak trees when 264 pine trees are planted. Set up a proportion and solve for $x$ :

$\frac{x}{264}=\frac{3}{8}$

$8x=792$

$x=99$

Question analysis

A ratio word problem about a tree farm that plants pine and oak trees in a fixed ratio of 8 to 3. Given how many pine trees are planted, students find the matching number of oak trees and grid it in. The main risk is pairing each number with the wrong kind of tree.

NYS standards: 6.RP.3a tables of equivalent ratios
6.RP.1 ratio concepts and language
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 This is a one-ratio proportion with whole numbers, the kind of problem grade 6 students solve with equivalent ratios. A typical grade 7 task adds rates with fractions or a multistep percent; this has neither.
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
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 28 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

This is a one-ratio scaling problem with friendly numbers: 264 splits into 8 equal parts of 33. The only real risk is matching 264 to the wrong term of 8:3, and a quick sense check catches it, since oak trees are the smaller group and must number fewer than 264. Students who grid the one-part value or the total of both kinds of tree have answered a question that was not asked.

Turning words into math minor Understanding the question
The ratio 8:3 lists pine first, so the 264 pine trees match the 8 and the oak count matches the 3; reading the terms in the other order flips the setup. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in it offers no choices, so an upside-down proportion that gives 704 oak trees, far more than the 264 pines, has to be caught by the student's own check. (Enright, Morley & Sheehan, 2002)

Strategy: Divide the number of pine trees by the pine part of the ratio to get the size of one part, then multiply by the oak part.

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

The wrong answers, classified

704 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: strong
Turns the ratio upside down, treating the 264 pines as 3 parts: 264 ÷ 3 = 88, then 88 × 8 = 704.
33 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Divides 264 by 8 to get the size of one part, 33 trees, and stops before multiplying by the 3 parts that are oak.
363 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the 99 oak trees correctly and then adds the 264 pines, gridding the total number of trees planted.

What was corrected from the DOE handbook, and why

Explanation: The handbook goes from x/264 = 3/8 to “8x = 762”. Cross-multiplying gives 8x = 3 × 264 = 792, so that line now reads 8x = 792. The handbook’s final answer, x = 99, was already right (792 ÷ 8 = 99).

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a and b are the two numbers in a ratio, written as a fraction. n is any number you choose, except zero. Multiplying both numbers by the same n gives an equal ratio. This is how you scale a small ratio up to real counts.

Sort out the ratio
A ratio compares two amounts. Here the ratio of pine trees to oak trees is 8:3. Pine comes first, so pine goes with the 8. Oak goes with the 3. So for every 8 pine trees, there are 3 oak trees.
The farm planted 264 pine trees. You need the number of oak trees.
Find the size of one part
Think of the trees in equal groups, called parts. Pine trees fill 8 parts, and oak trees fill 3 parts. The 264 pine trees are the 8 parts, so divide by 8.
$264 \div 8 = 33$
Each part is 33 trees. Watch out: 33 is one part, not the number of oak trees. Don’t stop here.
Find the number of oak trees
Oak trees fill 3 parts. Multiply the part size by 3.
$3 \times 33 = 99$
You can see the same thing with equal ratios. Multiply both numbers of 8:3 by 33.
$\frac{8}{3} = \frac{8 \times 33}{3 \times 33} = \frac{264}{99}$
So 264 pine trees go with 99 oak trees.
Avoid the traps and answer
Watch out: if you match 264 to the 3 instead of the 8, you flip the ratio. That gives 704 oak trees. Watch out: the question asks only for oak trees. Adding the 264 pines gives 363. So you would grid 99.

Check your answer

The ratio 8:3 says there are fewer oak trees than pine trees. 99 is less than 264, so that fits. Also, divide both counts by 33: $264 \div 33 = 8$ and $99 \div 33 = 3$ . That gives back 8:3.

More questions like this

The original DOE question

Question 59 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 59. Open page 81 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 59 of 69

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 60 of 69

2018 Test A Math, question 4

Question

A survey asked students what pets they have. Based on the results, the following statements are all true.

20 students have cats.
23 students have dogs.
3 students have both dogs and cats.
5 students have no dogs or cats.

How many students were surveyed?

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

Show the answer

Correct answer: 45

Let x = number of students with only cats as pets.

Let y = number of students with only dogs as pets.

Calculate x and y using the given information:

There are 20 students who have cats, and of those 20 students, 3 have both cats and dogs.

Thus,

$x=20-3=17$

There are 23 students who have dogs, and of those 23 students, 3 have both cats and dogs.

Thus,

$y=23-3=20$

To find the total number of students surveyed, add the number of students who only have cats (x), the number of students who only have dogs (y), the number of students who have both (3), and the number of students who have neither (5):

$3+5+x+y=8+17+20=45$

Question analysis

A grid-in counting question about a survey of students' pets: some have cats, some dogs, some both, and some neither. Students find how many were surveyed by sorting everyone into groups without counting anyone twice. The students who own both pets are the trap in this classic Venn diagram problem.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 classes organize outcomes in lists and tables, but overlapping groups like these are rarely taught directly. The arithmetic is simple whole-number addition and subtraction; the demand lies in spotting the double count.
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
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is small; the trap is in the wording. Because each bullet reads like a separate group, students tend to add the numbers as listed and count the students with both pets twice. A Venn diagram with the overlap filled in first solves the problem, and the remaining risk is forgetting that students with no pets were surveyed too.

Turning words into math moderate Understanding the question
The statement "20 students have cats" already includes the 3 who have both pets, and so does "23 students have dogs", which the separate bulleted list hides. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Four facts, two overlapping groups, their overlap and an outside group of 5 with neither pet, all have to be placed before anything is added. (Sweller, 1988; Carpenter, Just & Shell, 1990)
An answer in an unexpected form minor Choosing the answer
As a grid-in it gives no choices, so a total that counts the 3 twice or leaves out the 5 looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Draw two overlapping circles: put the both-pets group in the middle, subtract it from each pet total to fill the outer parts, then add every region and the students with neither.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

48 Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: strong
Adds 20 + 23 + 5 = 48, counting the 3 students with both pets once among the cat owners and again among the dog owners.
40 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Removes the overlap correctly, 20 + 23 − 3 = 40, but counts only the pet owners, leaving out the 5 surveyed students with no dogs or cats.
51 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds all four numbers in the list, 20 + 23 + 3 + 5 = 51, treating the both-pets group as an extra group of students.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap: $T = C + D - B + N$; T is the total number of students. C is the number with cats, and D is the number with dogs. B is the number with both pets, and N is the number with neither. Students with both are counted in C and again in D, so you subtract B once. Then add N.

Sort out the four facts
You need the number of students who were surveyed. Here is what you know:
- 20 students have cats.
- 23 students have dogs.
- 3 students have both dogs and cats.
- 5 students have no dogs or cats.
The tricky part: the 3 students with both pets are already inside the 20 and inside the 23.
Split the pet owners into groups
Picture two circles that overlap, one for cats and one for dogs. The overlap holds the 3 students with both pets. Take them out of each pet total to find who has only one kind of pet.
$20 - 3 = 17$
So 17 students have only cats.
$23 - 3 = 20$
So 20 students have only dogs.
Add every group once
Now every student is in exactly one group.
Group Students
Only cats 17
Only dogs 20
Both pets 3
Neither pet 5
$17 + 20 + 3 + 5 = 45$
The formula gives the same total.
$T = 20 + 23 - 3 + 5 = 45$
Avoid the traps and answer
Watch out: adding $20 + 23 + 5$ counts the 3 students with both pets twice. That gives 48. Watch out: the 5 students with no pets were surveyed too. Leaving them out gives 40. Adding all four numbers in the list gives 51, which is also wrong. So you would grid 45.

Group	Students
Only cats	17
Only dogs	20
Both pets	3
Neither pet	5

Check your answer

Check each fact against your groups. Cat owners: $17 + 3 = 20$ . Dog owners: $20 + 3 = 23$ . Both: 3. Neither: 5. Every fact matches, so 45 students were surveyed.

More questions like this

The original DOE question

Question 61 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 61. Open page 81 of the handbook (PDF)

Question 61 of 69

2018 Test A Math, question 21

Question

In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without replacement, what is the probability that both cards are not blue?

A $\frac{2}{15}$
B $\frac{4}{25}$
C $\frac{3}{10}$
D $\frac{1}{3}$

Show the answer

Correct answer: A

The probability that both cards are not blue is the same as the probability that both cards are red. There are 4 red cards out of the 10, so the probability of the first card being red is

$\frac{4}{10}$

Now there are 9 cards left, and 3 of those are red, so the probability of the second card being red is

$\frac{3}{9}$

Multiply the two probabilities to find the probability that both cards are red (not blue):

$\left (\frac{4}{10} \right )\left ( \frac{3}{9} \right )=\frac{12}{90}=\frac{2}{15}$

Question analysis

This probability question draws 2 cards, one at a time and without replacement, from 10 cards: 4 red and 6 blue. Students find the chance that neither card drawn is blue. The second draw has one fewer card, and the word not changes which color to count.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability is usually found with a list or tree of equally likely outcomes. Dependent draws without replacement, plus a negative phrasing to translate, go beyond that typical 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: Something you must already know moderate
A very close wrong answer moderate
Turning words into math minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 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

Two ideas stack here: restating not blue as red, and changing the second draw's numbers after a card is removed. Students who know to multiply but forget the removal reach a listed choice, and so do students who compute both blue. Writing the second fraction explicitly as 3 red out of 9 cards, before multiplying, is what separates right from nearly right.

Something you must already know moderate Working it out
Multiplying the first draw's chance by the second draw's, with the second adjusted for the card already taken, is a rule the test assumes and does not give. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
2/15 and 4/25 are about 0.13 and 0.16, too close to tell apart by estimate; only the second draw's fraction, 3 red of 9 cards rather than 4 of 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
"Both cards are not blue" has to be restated as both cards red, and the bold not also invites computing the chance that both are blue. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The product comes out as 12/90, and it has to be reduced to lowest terms before it matches any of the listed fractions. (Enright, Morley & Sheehan, 2002)

Strategy: Restate not blue as red, then multiply the chance of red on the first draw by the chance of red on the second, with one card fewer.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

B Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Multiplies 4/10 × 4/10 = 16/100 = 4/25, as if the first red card were put back before the second draw.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives 3/10: it is not the chance of red or of blue on one draw or on two, so it mostly catches guesses.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Computes the chance that both cards are blue, 6/10 × 5/9 = 30/90 = 1/3, missing the bold word not.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{favorable}}{\text{total}}$; P is the probability, the chance that something happens. Favorable is the number of ways to get what you want, here a red card. Total is the number of all possible picks, here all the cards you can pick from. It works when every pick is equally likely.
Two events in a row: $P(\text{A and B}) = P(\text{A}) \times P(\text{B after A})$; A is the first event and B is the second. P( ) means the probability of what is inside. The chance that both happen is the first chance times the second. For the second chance, count only what is left after the first event.

Sort out the question
There are 10 cards: 4 red and 6 blue. You pick 2 cards “one at a time without replacement”. That means the first card is not put back. “Both cards are not blue” means both are red, since red is the only other color. You need the probability, or chance, of that.
Find the chance the first card is red
4 of the 10 cards are red. Each card is equally likely to be picked. Use the probability formula.
$P(\text{red first}) = \frac{4}{10}$
Find the chance the second card is red
One red card is gone now, and it is not put back. So 9 cards are left, and 3 of them are red.
$P(\text{red second}) = \frac{3}{9}$
Watch out: using 4/10 again acts as if the card went back. That gives 4/25, choice B.
Multiply the two chances
Both cards must be red. So multiply the first chance by the second. To multiply fractions, multiply the tops, then multiply the bottoms.
$\frac{4}{10} \times \frac{3}{9} = \frac{4 \times 3}{10 \times 9} = \frac{12}{90}$
Simplify the fraction
None of the choices is 12/90, so simplify it. Divide the top and the bottom by 6, the biggest number that goes into both.
$\frac{12 \div 6}{90 \div 6} = \frac{2}{15}$
Watch out: the chance that both are blue is $\frac{6}{10} \times \frac{5}{9} = \frac{1}{3}$ , choice D. The question says not blue. So the answer is A, $\frac{2}{15}$ .

Check your answer

Count the ways. There are $10 \times 9 = 90$ ways to pick two cards in order. Red then red has $4 \times 3 = 12$ ways. That gives 12/90, or 2/15. The answer is small, which makes sense, because there are fewer red cards than blue ones.

More questions like this

The original DOE question

Question 78 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 78. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 62 of 69

2018 Test A Math, question 28

Question

$2\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}$

What is the value of the expression shown above?

A $14\frac{7}{20}$
B $14\frac{2}{5}$
C $15\frac{7}{20}$
D $15\frac{2}{5}$

Show the answer

Correct answer: D

$2\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}$

Convert all the fractions to a common denominator (10):

$2\frac{2}{10}+3\frac{3}{10}+4\frac{4}{10}+5\frac{5}{10}$

$=(2+3+4+5)+\left(\frac{2+3+4+5}{10}\right)$

$=14+1\frac{4}{10}=15\frac{2}{5}$

Question analysis

A fractions question: add four mixed numbers whose fractions have denominators of 5, 10, 5 and 2. Students rewrite the fractions over a common denominator, add the whole numbers and the fractions, regroup any extra whole, and simplify. Regrouping and lowest terms both matter, since the choices differ in exactly those places.

NYS standards: 7.NS.1d properties for adding and subtracting
7.EE.3 multi-step rational number problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding mixed numbers with unlike denominators is grade 5 content. With four terms and a regrouping step it takes care, but it asks less than typical grade 7 work with negative rational numbers.
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
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 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 numbers are friendly: every fraction converts to tenths, and the whole parts add to 14. The risk is in the bookkeeping, either forgetting to carry the extra whole from the fractions or adding numerators over a made-up denominator. Converting to decimals, 2.2 + 3.3 + 4.4 + 5.5, is a fast check that avoids both.

Error-prone arithmetic moderate Working it out
Four mixed numbers with denominators 5, 10, 5 and 2 must be rewritten as tenths, 2 + 3 + 4 + 5 tenths, and the 14/10 regrouped into a whole and a fraction. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
15 2/5 and 14 2/5 are written alike and share the fraction 2/5; only the whole carried from 14/10 separates them, so a missed carry is invisible at a glance. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
An answer in an unexpected form minor Choosing the answer
The fractions total 1 4/10, but every choice is in lowest terms, so the 4/10 has to be reduced to 2/5 before it matches. (Enright, Morley & Sheehan, 2002)

Strategy: Convert each mixed number to a decimal, since these fractions all convert exactly, add them, and turn the decimal part back into a fraction in lowest terms.

Traps

Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators straight across, 1 + 3 + 2 + 1 = 7, and puts them over 20, a common denominator (10 × 2), without rescaling them, so nothing carries.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Adds the fractions to 14/10 but keeps only the 4/10 = 2/5 left over, forgetting to carry the whole 1 into 14.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Pairs a correct carry into 15 with the 7/20 from adding numerators, two steps no single line of working combines; it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a is the top number, called the numerator. b is the bottom number, called the denominator. n is any whole number except zero. Multiplying the top and bottom by the same n gives an equal fraction. Use it to give fractions the same denominator so you can add them.

Sort out the question
You need to add four mixed numbers. A mixed number is a whole number plus a fraction, like $2\frac{1}{5}$ . The fractions have different denominators, or bottom numbers: 5, 10, 5 and 2. To add fractions, they need the same denominator.
Add the whole numbers
Add the whole numbers first. You will add the fractions on their own after that.
$2 + 3 + 4 + 5 = 14$
The whole numbers make 14.
Rewrite each fraction in tenths
Use 10 as the denominator, since 5 and 2 both go into 10.
$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}$
$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$
$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$
$\frac{3}{10}$ is already in tenths. Watch out: when you change a bottom number, change the top too. Adding the old tops, 1 + 3 + 2 + 1 = 7, and putting them over 20 gives 7/20. That is in choices A and C.
Add the fractions
$\frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{14}{10}$
14 tenths is more than 1 whole. Ten tenths make 1, so 14 tenths is 1 whole and 4 tenths.
$\frac{14}{10} = 1\frac{4}{10} = 1\frac{2}{5}$
To simplify 4/10, divide the top and bottom by 2.
Put it all together
Add the 14 wholes to the sum of the fractions.
$14 + 1\frac{2}{5} = 15\frac{2}{5}$
Watch out: carry the extra whole into the 14. Missing it gives $14\frac{2}{5}$ , choice B. So the answer is D, $15\frac{2}{5}$ .

Check your answer

Change each mixed number to a decimal: $2.2 + 3.3 + 4.4 + 5.5 = 15.4$ . The 0.4 is 4 tenths, which is 2/5. So the sum is $15\frac{2}{5}$ .

More questions like this

The original DOE question

Question 85 as printed on page 86 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 85. Open page 86 of the handbook (PDF)

Question 63 of 69

2018 Test A Math, question 38

Question

Ryan must read 150 pages for school this weekend. It took him 30 minutes to read the first 20 pages. At this rate, how much additional time will it take him to finish the reading?

A $2\frac{1}{6}\text{ hr}$
B $3\frac{1}{4}\text{ hr}$
C $3\frac{3}{4}\text{ hr}$
D $7\frac{1}{2}\text{ hr}$

Show the answer

Correct answer: B

Ryan has 130 pages left to read (150 − 20). He read 20 pages in 30 minutes, which means he read at a rate of 40 pages per 1 hour. To find out how much longer it will take him to finish the assignment, divide the total number of pages remaining (130) by the number of pages he is able to read per hour (40):

$\frac{130}{40}=3\frac{1}{4}$

Question analysis

This reading-rate word problem says Ryan read 20 of his 150 pages in 30 minutes and asks how much additional time he needs to finish at that rate. Students must find the pages left, use the rate, and convert minutes to hours. The word 'additional' is easy to overlook.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Unit rate problems like this are grade 6 content and common in grade 7. It adds a subtraction step and a conversion to hours, and the word 'additional' must be read closely, which keeps it at grade 7 level.
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
A very close wrong answer moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 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 bold word "additional" carries the question, and missing it costs the point even when the rate work is perfect. The rate itself mixes pages, minutes and hours, so students who skip writing units drift into answers that count half-hour blocks or minutes as hours. Subtracting the pages already read before anything else keeps the work on the right quantity.

Many steps moderate Working it out
The solution chains four moves: subtract the 20 pages read, double 20 pages per half hour to 40 per hour, divide 130 by 40, and write the result in hours. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
The choices 3 1/4 and 3 3/4 hours differ by exactly the half hour already spent on the first 20 pages, so only subtracting those pages first separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
An answer in an unexpected form minor Choosing the answer
A student who works in minutes gets 195 minutes and must convert it to the mixed number 3 1/4 hours to find it among the choices. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract the pages already read, work out how many pages he reads per hour, and divide the remaining pages by that hourly rate.

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.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides the 130 remaining pages by 60, treating the rate as one page a minute, which gives 2 1/6 hours.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Times all 150 pages at 40 pages an hour, ignoring the bold "additional", and gets the whole assignment's 3 3/4 hours instead of the time left.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Divides 150 pages by 20 to get 7.5 blocks of reading and reports them as hours, though each 20-page block takes half an hour.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Time from a rate: $t = \frac{p}{r}$; t is the time needed. p is the number of pages to read. r is the rate: the pages read in one hour. Dividing tells you how many hours of reading the pages need.

Sort out what you know
Ryan must read 150 pages in all. He read the first 20 pages in 30 minutes. The word additional is in bold. It means extra time, after the 30 minutes he already spent.
So you need the time for the pages he has not read yet.
Find the pages left
$150 - 20 = 130$
He still has 130 pages to read. Watch out: timing all 150 pages gives $3\frac{3}{4}$ hours, which is choice C. That counts pages he already read.
Find his rate per hour
The choices are in hours, so find how many pages he reads in an hour. 30 minutes is half an hour. In a full hour he reads twice as much:
$20 \times 2 = 40$
His rate is 40 pages per hour.
Divide the pages by the rate
Dividing tells you how many hours the 130 pages take at 40 pages an hour:
$t = \frac{130}{40} = 3\frac{10}{40} = 3\frac{1}{4}$
It will take $3\frac{1}{4}$ more hours. So the answer is B, $3\frac{1}{4}$ hr.

Check your answer

Work in minutes. 130 pages is 6.5 groups of 20 pages. Each group takes 30 minutes, so 6.5 × 30 = 195 minutes. That is 3 hours and 15 minutes. And 15 minutes is a quarter of an hour, so it matches.

More questions like this

The original DOE question

Question 95 as printed on page 88 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 95. Open page 88 of the handbook (PDF)

Question 64 of 69

2018 Test B Math, question 1

Question

How many 5-digit numbers can be created using the digits 2, 3, 5, 7, and 8 without repeating any digits within that 5-digit number?

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

Show the answer

Correct answer: 120

There are 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit, 2 choices for the fourth digit, and 1 choice for the final digit.

The total number of possibilities is

5 × 4 × 3 × 2 × 1 = 120

Question analysis

This grid-in counting question asks how many different 5-digit numbers can be made from five given digits when no digit may repeat. It tests the counting principle for arrangements rather than arithmetic. The catch is the no-repeat rule: each place has one fewer digit to choose from, and ignoring that gives a far larger count.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability work builds sample spaces with lists and tree diagrams for two or three stages. This needs the multiplication principle with a shrinking number of choices across five places, which listing cannot reach.
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: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 11, 19 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 tiny; the difficulty is knowing that arrangements are counted by multiplying, and that the no-repeat rule removes one digit for each later place. Students who have met the counting principle finish in seconds, while those who start listing numbers run out of time or guess a product they half remember. With no choices to check against, the 3,125 that comes from ignoring the rule looks as convincing as the right count.

Something you must already know moderate Working it out
Nothing in the stem names the counting principle: students must already know to multiply the digits left for each place, 5 × 4 × 3 × 2 × 1, because listing 5-digit numbers by hand cannot finish. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in there are no choices to compare with, so a count of 3,125 or 25 cannot be caught by noticing that it is missing from a list. (Enright, Morley & Sheehan, 2002)

Strategy: Fill the five places from left to right, write how many digits are still available for each place, and multiply those five counts together.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.

The wrong answers, classified

3125 Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Uses the counting principle but lets digits repeat, multiplying 5 × 5 × 5 × 5 × 5 instead of using one fewer digit for each later place.
25 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Multiplies the five digits by the five places, 5 × 5, as if each digit simply took each position once.
15 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Sees that the choices shrink from 5 to 1 but adds them, 5 + 4 + 3 + 2 + 1, instead of multiplying.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total ways} = a \times b \times c \times \ldots$; a is the number of choices for the first spot, b for the second, c for the third, and so on. Multiply them to get every possible way. It works because each choice for one spot can go with each choice for the next spot.

Sort out the question
You have five digits: 2, 3, 5, 7 and 8. A digit is one of the symbols 0 to 9. You build 5-digit numbers from them.
“Without repeating” means each digit is used only once in a number. So 23578 is allowed, but 22578 is not.
You need to find how many different numbers you can make.
Count the choices for each spot
Think of the number as five empty spots. Fill them from left to right.
- First spot: any of the 5 digits, so 5 choices.
- Second spot: one digit is used up, so 4 are left.
- Third spot: 3 are left.
- Fourth spot: 2 are left.
- Last spot: only 1 is left.
Multiply the choices together
Use the counting principle. Each choice for one spot works with every choice for the next spot. So you multiply. You do not add.
$\text{total ways} = 5 \times 4 \times 3 \times 2 \times 1 = 120$
There are 120 different numbers.
Watch out: Adding 5 + 4 + 3 + 2 + 1 gives 15, which is far too small. Using 5 choices for every spot gives 3,125. That count lets digits repeat, which the question does not allow.
So you would grid 120.

Check your answer

Try a smaller case you can list. With the digits 2, 3 and 5, the rule gives $3 \times 2 \times 1 = 6$ numbers. The list is 235, 253, 325, 352, 523 and 532. That is 6, so the rule works.

More questions like this

The original DOE question

Question 58 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 58. Open page 153 of the handbook (PDF)

Question 65 of 69

2018 Test B Math, question 3

Question

$\left | \left ( -6 \right )-\left ( -5 \right )+4.2 \right |-\left | 3-9.6 \right |$

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

Show the answer

Correct answer: -3.4

$=3.2-6.6=-3.4$

Question analysis

This grid-in asks students to evaluate an expression that mixes negative integers, decimals and two absolute values. The hard parts are subtracting a negative number inside the first pair of bars and keeping each absolute value separate until the final subtraction, with no answer choices to check against.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add and subtract signed numbers, and grade 6 students meet absolute value. This combines both in one expression with decimals and several sign decisions, a typical grade 7 exercise with no context 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: Error-prone arithmetic moderate
Many steps moderate
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 16+, 1 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 familiar, but the expression packs four of them with negatives and decimals, and a single sign slip anywhere gives a clean-looking wrong number. The subtlest point comes last: absolute value bars make each part positive, yet the difference of two absolute values can still be negative. Students who believe bars always produce a positive answer reverse the final subtraction, and the grid-in format offers no choices to expose it.

Error-prone arithmetic moderate Working it out
Three sign decisions stack up: (−6) − (−5) must become −6 + 5, then −1 + 4.2 crosses zero, and 3 − 9.6 is negative before its bars make it positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps moderate Working it out
Each pair of bars must be simplified inside and then made positive, and only after both does the outer subtraction happen; doing the bars and the subtraction in the wrong order changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The result is a negative decimal to be gridded with its sign, and with no choices on screen nothing warns a student who has dropped the minus or flipped the subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Simplify inside each pair of absolute value bars first, rewriting minus a negative as plus, then take each absolute value, and only then subtract the second from the first.

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

3.4 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes an expression built from absolute values cannot be negative, so reports 6.6 − 3.2 instead of 3.2 − 6.6.
9.8 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats the bars like parentheses, leaving 3 − 9.6 as −6.6, then subtracts that negative: 3.2 + 6.6.
0.2 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Reads (−6) − (−5) as −6 − 5 = −11, so the first bars give |−6.8| = 6.8, and 6.8 − 6.6 = 0.2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{distance from } a \text{ to } 0$; The two bars around a number mean absolute value. a is any number. Its absolute value is how far it is from 0 on the number line. A distance is never negative. So |−6.6| = 6.6 and |3.2| = 3.2.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Taking away a negative number is the same as adding the positive number. For example, 5 − (−2) = 7. Think of it as taking away a debt: you end up richer.

Plan the order of work
The expression has two pairs of bars. The bars stand for absolute value, a number’s distance from 0.
Work inside each pair of bars first. Then take each absolute value. Subtract the two results last.
Work inside the first bars
Start with $(-6) - (-5) + 4.2$ . Subtracting −5 is the same as adding 5.
$(-6) - (-5) + 4.2 = -6 + 5 + 4.2 = -1 + 4.2 = 3.2$
3.2 is already positive, so $|3.2| = 3.2$ .
Watch out: If you read the start as −6 − 5, you get −11. That path ends at the wrong answer 0.2.
Work inside the second bars
Next, $3 - 9.6 = -6.6$ . This result is negative. Its distance from 0 is 6.6, so the absolute value is positive.
$|-6.6| = 6.6$
Subtract the two results
Now take the second result away from the first.
$3.2 - 6.6 = -3.4$
You take a bigger number away from a smaller one, so the result is negative.
Watch out: The bars made each part positive. They do not make the final answer positive, so 3.4 is wrong. Also, bars are not parentheses. Working out 3.2 − (−6.6) = 9.8 is wrong too.
So you would grid -3.4, with the minus sign.

Check your answer

Estimate. The first part is about 3. The second part is about 7. And 3 − 7 is about −4. The answer −3.4 is close to that. It is also negative, as the estimate says it should be.

More questions like this

The original DOE question

Question 60 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 60. Open page 153 of the handbook (PDF)

Question 66 of 69

2018 Test B Math, question 5

Question

$Four straight lines cross at point P: one horizontal, one vertical and two slanted. Above the horizontal line, from right to left: a 60° angle between the horizontal line and the slanted line rising to the right, a 30° angle between that line and the vertical line, an unlabeled angle between the vertical line and the slanted line rising to the left, and the angle y° between that line and the horizontal line. Below the horizontal line, a 25° angle lies between the vertical line and the slanted line falling to the right.$

Four straight lines intersect at point P as shown above. What is the value of $y$ ?

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

Show the answer

Correct answer: 65

Call the missing angle in the top half of the figure $x$ .

The sum of the four angles on the top of the figure is equal to 180°.

$x+y+30+60=180$

Since $x$ is a vertical angle with the 25° angle, then $x$ is also 25°. Use that to solve for $y$ .

$25+y+30+60=180$

$y+115=180$

$y=65$

Question analysis

This grid-in angle question shows four straight lines crossing at a single point P and asks for the size of an unknown angle y. Students combine vertical angles with the fact that angles along a straight line add to 180 degrees. One angle they need is unlabeled and must be found from its vertical partner.

NYS standards: 7.G.5 angle relationships and equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Using vertical angles and straight angles to write an equation for an unknown angle is a core grade 7 geometry task. Four lines create more angles than usual, but the reasoning stays at grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Combines ideas moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 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 arithmetic is one subtraction; the work is in seeing the figure correctly. Students have to notice that a fourth angle sits between the 30° angle and y, recognize it as the vertical partner of the 25° angle, and then use the straight line. The drawing is not to scale: y is drawn close to the 60° angle across the vertical line, which rewards anyone who computes rather than eyeballs.

Reading a figure, table or graph major Understanding the question
Four lines make eight angles around P but only three are labeled; above the horizontal line sit 60°, 30°, an unlabeled angle and y°, and that unlabeled angle is easy to overlook. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
The solution joins two facts: the unlabeled angle is vertical to the 25° angle below P, and the four angles above the horizontal line make a straight angle of 180°. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
As a grid-in, nothing confirms the result, and the drawing shows y at about 60°, so a reading of the picture produces a number that looks right. (Enright, Morley & Sheehan, 2002)

Strategy: Pick a straight line through P and list the angles on one side of it; replace the unlabeled one with the labeled angle vertical to it, then subtract from 180.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Stops one step early: A choice is a number from partway through the solution.
Trusting the picture: The figure suggests something it does not state; work from the given facts.

The wrong answers, classified

90 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: strong
Misses the unlabeled angle between the 30° angle and y, so subtracts only 30 and 60 from 180.
115 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds the three known angles beside y, 25 + 30 + 60, and grids that sum without subtracting it from 180.
60 Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a surface strategy Expected pull: moderate
Judges from the drawing, where y looks like a mirror image of the 60° angle across the vertical line, instead of computing it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line: $a + b + c + d = 180$; a, b, c and d are angles that sit side by side along one side of a straight line, all meeting at one point. Together they make a half turn, which is 180 degrees. This works for any number of angles, not just four.
Vertical angles: $\angle 1 = \angle 2$; When two lines cross, they make an X. Angle 1 and angle 2 are opposite each other across the crossing point. Angles like these are called vertical angles. They are always equal.

Look closely at the figure
Four lines cross at point P. Look above the flat, or horizontal, line. From right to left there are four angles: 60°, 30°, an angle with no label, and y°.
Below the horizontal line there is a 25° angle. You need to find y.
Find the unlabeled angle
The unlabeled angle sits between the up-and-down, or vertical, line and the slanted line rising to the left.
That slanted line keeps going through P and down to the right. The 25° angle sits between that same line and the vertical line, below P.
So the two angles are opposite each other across P. They are vertical angles, so they are equal. The unlabeled angle is 25°.
Add the angles above the line
The four angles above the horizontal line fill one side of a straight line. So they add up to 180°.
$a + b + c + d = 180$
$y + 25 + 30 + 60 = 180$
$y + 115 = 180$
Solve for y
Take 115 away from 180 to get y alone.
$y = 180 - 115 = 65$
Watch out: Do not skip the unlabeled angle. Using only 30 and 60 gives 90. Do not stop at the sum, 115, either. And do not guess from the picture. Angle y looks like 60°, but the drawing is not exact.
So you would grid 65.

Check your answer

Add all four angles above the horizontal line: $65 + 25 + 30 + 60 = 180$ . They fill a straight line exactly, so 65 fits.

More questions like this

The original DOE question

Question 62 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 62. Open page 153 of the handbook (PDF)

Question 67 of 69

2018 Test B Math, question 15

Question

One day, the Early Bird Restaurant used 15 dozen eggs for 200 breakfast customers. At this rate, approximately how many dozen eggs are needed for 300 breakfast customers?

Show the answer

Correct answer: B

Let $x$ be the number of dozens of eggs for 300 customers. Set up a proportion:

$\frac{x}{300}=\frac{15}{200}$

$200x=4500$

$x=22.5\text{ dozen eggs}$

Round up to 23, because 22 dozen eggs would not be enough for 300 customers.

Question analysis

This proportion word problem says a restaurant used 15 dozen eggs for 200 breakfast customers and asks approximately how many dozen are needed for 300 customers. Students scale a rate up and round sensibly. Scaling by the right factor, rather than doubling or dividing the wrong pair of numbers, is the key step.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Scaling a rate from one quantity to another is grade 6 ratio reasoning. This needs one multiplication and a rounding decision, so it is lighter than the multistep proportion problems typical of grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 25 words.
Corrected from the DOE handbook: The explanation’s reason for rounding was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The proportion is simple, since 300 customers is one and a half times 200, and the only judgment is what to do with 22.5 dozen. The choices are spread wide enough that 23 is clearly nearest once 22.5 is found. Errors come mostly from students who skip the proportion and combine the numbers 300, 200 and 15 in some quicker way.

An answer in an unexpected form minor Choosing the answer
The scaled amount is 22.5 dozen, which is not a choice; the word “approximately” signals that the answer is the nearest whole number of dozens. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Scaling 15 dozen by 300/200 needs the factor 1.5, and 15 × 1.5 = 22.5 brings in a half that students must handle before matching a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Compare 300 customers with 200 to get the scale factor, multiply the 15 dozen eggs by it, and choose the closest answer.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Divides 300 customers by 15 dozen, combining two numbers from the question without a model of eggs per customer.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard route gives 25; at most a student who finds 22.5 rounds it to the nearest five instead of the nearest whole dozen.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds the scale factor 300 ÷ 200 = 1.5 up to 2 before multiplying, doubling 15 dozen to 30.

What was corrected from the DOE handbook, and why

Explanation: The handbook says to round 22.5 up to 23 “because you can't have half an egg.” But 22.5 dozen is 270 eggs, a whole number; the half is half a dozen, or 6 eggs. The site gives the real reason: 22 dozen eggs would not be enough for 300 customers, so the amount rounds up to 23 dozen.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling a rate: $\text{new amount} = \text{old amount} \times \frac{\text{new count}}{\text{old count}}$; The old amount is the eggs used before, 15 dozen. The old count is the customers then, 200. The new count is the new number of customers, 300. The fraction tells how many times bigger the new group is. The eggs must grow by that same number of times.

Write down what you know
15 dozen eggs fed 200 customers. A dozen is 12, but you can keep everything in dozens.
You need the dozens of eggs for 300 customers, at the same rate.
Compare the customer counts
How many times bigger is 300 than 200? Divide to find out. This number is called the scale factor.
$\frac{300}{200} = 1.5$
There are 1.5 times as many customers. So you need 1.5 times as many eggs.
Scale up the eggs
$\text{new amount} = \text{old amount} \times \frac{\text{new count}}{\text{old count}}$
$15 \times 1.5 = 22.5$
You need 22.5 dozen eggs.
Watch out: Do not round 1.5 up to 2 first. That doubles 15 and gives 30, choice D.
Round to a choice
22.5 is not a choice. The question says “approximately,” so pick the closest whole number. 22.5 rounds up to 23.
Rounding up also makes sense, because 22 dozen would be too few.
So the answer is B, 23.

Check your answer

Use another route. 15 dozen for 200 customers is 7.5 dozen for every 100 customers. For 300 customers, that is $3 \times 7.5 = 22.5$ dozen. That also rounds to 23.

More questions like this

The original DOE question

Question 72 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 72. Open page 155 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 68 of 69

2018 Test B Math, question 51

Question

$\frac{14}{21}=\frac{p}{7}$

In the equation above, what is the value of $p$ ?

A $\frac{2}{3}$
B $3$
C $\frac{14}{3}$
D $14$

Show the answer

Correct answer: C

$\frac{14}{21}=\frac{p}{7}$

$21p=7(14)$

$21p=98$

$p=\frac{98}{21}=\frac{14}{3}$

Question analysis

Students solve a proportion that sets 14/21 equal to a fraction with an unknown numerator p over 7. It tests solving a proportion, either by cross-multiplying or by dividing numerator and denominator by the same factor. It is quick, and the main risk is losing track of what p stands for.

NYS standards: 6.EE.7 one-step equations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Solving a one-step proportion is grade 6 and 7 ratio work. This one is purely numerical with one or two steps, shorter than the multistep ratio word problems typical of grade 7.
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: An answer in an unexpected form moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is standard, but the result is not a whole number, and that surprises students who expect the numbers in a proportion to scale evenly. Two intermediate values, the ratio 2/3 and the scale factor 3, appear among the choices. Students who finish the computation and trust a fractional answer get it right.

An answer in an unexpected form moderate Choosing the answer
The answer is the improper fraction 14/3, not a whole number, so students who expect a tidy proportion distrust their result or reach for the factor 3. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Cross-multiplying gives 21p = 98, and 98/21 must be reduced by 7; simplifying 14/21 to 2/3 first and multiplying by 7 is quicker. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Cross-multiply to form an equation in p, divide both sides by p's coefficient, and simplify the fraction you get.

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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Simplifies 14/21 to 2/3 and stops, giving the value of the ratio instead of the numerator p over 7.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Notices that 21 is 3 times 7 and gives that scale factor as p instead of dividing 14 by it.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Copies the left numerator, 14, as if equivalent fractions kept the same numerator when the denominator changes.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross multiplication: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a and c are the top numbers. b and d are the bottom numbers. When two fractions are equal, the top of each times the bottom of the other gives the same answer. It works because you multiply both sides by b and by d.

Sort out the question
Two fractions are equal: $\frac{14}{21}$ and $\frac{p}{7}$ . An equation like this, with two equal fractions, is called a proportion. You need the value of $p$ .
Cross multiply
Use cross multiplication. This gets rid of the fractions. Multiply 21 by $p$ , and multiply 14 by 7.
$21 \times p = 14 \times 7 = 98$
Divide to find p
Divide both sides by 21 to get $p$ alone.
$p = \frac{98}{21}$
Both 98 and 21 can be divided by 7.
$\frac{98 \div 7}{21 \div 7} = \frac{14}{3}$
Make sense of the fraction
The answer is $\frac{14}{3}$ , which is $4\frac{2}{3}$ . It is not a whole number, and that is fine.
Watch out: $\frac{2}{3}$ is the value of $\frac{14}{21}$ , not $p$ . That is choice A. The 3 is what 21 was divided by to get 7. That gives choice B.
Give the answer
The top number over 7 must be $\frac{14}{3}$ to keep the two fractions equal. So the answer is C, $\frac{14}{3}$ .

Check your answer

Think about the bottoms. 7 is one third of 21. So $p$ must be one third of 14, which is $\frac{14}{3}$ . Check it: $\frac{14}{3} \div 7 = \frac{14}{21}$ . Both ways agree.

More questions like this

The original DOE question

Question 108 as printed on page 163 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 108. Open page 163 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 69 of 69

2018 Test B Math, question 54

Question

Raoul is at least 3 years older than Vahn. Which of the following inequalities gives the relationship between Raoul’s age ( $r$ ) and Vahn’s age ( $v$ )?

A $r-v\geq 3$
B $r-v\leq 3$
C $3-v\leq r$
D $3-r\leq v$

Show the answer

Correct answer: A

Raoul is at least 3 years older than Vahn, which can be written as

r ≥ v + 3

Rewrite this inequality to match the answer options:

r − v ≥ 3

Question analysis

Raoul is at least 3 years older than Vahn, and students choose the inequality that links their ages, r and v. It tests translating a phrase like "at least" into an inequality. The correct form may not match the one students write first, so they need to rearrange carefully and keep the direction of the inequality.

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing inequalities from words is grade 6 and 7 work. The extra demand is that "at least" must become the right symbol and the choices are rearranged forms, so students must manipulate inequalities rather than just recognize one.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Turning words into math major
An answer in an unexpected form moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 8, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two skills are tested at once: translating "at least … older than" into an inequality, and recognizing that inequality after a rearrangement. The two choices that begin with 3 − v and 3 − r look like legitimate moves but say only that the two ages add up to at least 3. Testing one pair of ages that fits the sentence and one that does not, such as 20 and 12 against 13 and 12, sorts the choices without algebra.

Turning words into math major Understanding the question
"Raoul is at least 3 years older than Vahn" has to become r ≥ v + 3; "at least" calls for ≥, though the word "least" seems to point the other way. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form moderate Choosing the answer
No choice is written as r ≥ v + 3, so the natural translation has to be rearranged to r − v ≥ 3 before it can be matched. (Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
r − v ≤ 3 is the right r − v ≥ 3 with the symbol reversed; only deciding that "at least" means greater than or equal to separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the sentence directly as Raoul's age compared with Vahn's age plus 3, then rearrange each choice into that form to see which one matches.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Reads "at least 3 years older" as at most 3, reversing the inequality, which would even allow Raoul to be younger than Vahn.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Moves v across r ≥ v + 3 without changing its sign, writing r ≥ 3 − v, which really says only that their ages add up to at least 3.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Makes the same sign slip and also swaps the roles of r and v, giving 3 − r ≤ v, again only a statement about the sum of their ages.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greater than or equal to: $a \geq b$; a and b are two amounts. The sign ≥ means a is greater than or equal to b, so a is b or more. The words at least mean the same thing. The sign ≤ means less than or equal to, or at most.

Sort out the question
The letter $r$ is Raoul’s age, and $v$ is Vahn’s age. Raoul is “at least 3 years older than Vahn”. That means the gap between their ages is 3 years or more. An inequality is a math sentence that uses a sign like $\geq$ instead of =. You need the one that says this.
Write the age gap
Raoul is older, so the gap is his age minus Vahn’s age. The gap is $r - v$ . For example, ages 15 and 10 have a gap of 5.
Turn at least into a sign
At least 3 means 3 or more. So use $\geq$ , which means greater than or equal to.
$r - v \geq 3$
This is choice A.
Watch out: the word least does not mean less. Choice B, $r - v \leq 3$ , says the gap is 3 or less. Then Raoul could even be younger.
Check choices C and D
Add $v$ to both sides of choice C. You get $3 \leq r + v$ . That is about the sum of their ages, not the gap. Choice D works the same way. Neither one says Raoul is older.
Give the answer
Only one choice says the gap is 3 years or more. So the answer is A, $r-v \geq 3$ .

Check your answer

Test ages that do not fit: Raoul 11 and Vahn 10. He is only 1 year older. Choice A gives $11 - 10 = 1$ , and 1 is not 3 or more. So A rightly says no. Choice B says yes, so B is wrong.

More questions like this

The original DOE question

Question 111 as printed on page 164 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 111. Open page 164 of the handbook (PDF)