Practice Test

6.EE expressions, equations, and inequalities

Question 1 of 81

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 81

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 3 of 81

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 4 of 81

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 5 of 81

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 6 of 81

2026 Practice Test C Math, question 1

Question

What is the prime factorization of 756?

A $2^{2} \cdot 3^{2} \cdot 21$
B $2^{2} \cdot 3 \cdot 7 \cdot 9$
C $2^{2} \cdot 7 \cdot 27$
D $2^{2} \cdot 3^{3} \cdot 7$

Show the answer

Correct answer: D

Determine the prime factors of 756 using a factor tree:

$A factor tree for 756: 756 splits into 2 and 378; 378 into 2 and 189; 189 into 3 and 63; 63 into 3 and 21; and 21 into 3 and 7.$

The prime factors are 2, 2, 3, 3, 3, and 7, or $2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7$ , which can be expressed using exponents as $2^{2} \cdot 3^{3} \cdot 7$ .

Question analysis

This number question asks for the prime factorization of 756, written with exponents. Students break the number into prime factors, with a factor tree or repeated division, and match the result to one of four products. Every choice multiplies back to 756, so the real test is whether each factor shown is actually prime.

NYS standards: 6.NS.4 common factors and multiples
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Factor trees and exponent notation are grade 6 skills, and 756 splits quickly by small divisors. What adds a little is that every choice multiplies out to 756, so checking the product alone settles nothing.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 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 factoring itself is quick: halving twice and dividing by 3 three times leaves 7. The question is built so that multiplying back proves nothing, since every choice equals 756, and three of them simply stop the factor tree at a composite number. Students who check only the product, or who take an odd factor such as 21 or 27 for a prime, pick a factorization that is unfinished.

A very close wrong answer moderate Choosing the answer
2² · 3² · 21 shares the right answer's 2² and differs only in 3² · 21 against 3³ · 7, and both equal 756, so neither an estimate nor a product check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that a prime factorization may hold only primes, and must spot that 21, 9 and 27 are composite even though none of them is even. (Anderson & Krathwohl, 2001)

Strategy: Divide 756 by 2 while the result is even, then by 3 while it divides evenly, and keep going until every factor left is prime; then compare factor by factor, not by product.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Divides out 2 twice and 3 twice, then stops at 21, which is 3 × 7, so one composite factor remains in a product that still equals 756.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Splits off a 3 and a 7 but keeps 9 as if it were prime, perhaps because it is odd, though 9 = 3 × 3.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Splits 756 into 4 × 189 and 189 into 7 × 27, then stops, not seeing that 27 is 3 × 3 × 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Writing a repeated factor with an exponent: $a^{n} = a \times a \times \cdots \times a$; a is the number being multiplied, called the base. n is the exponent, the small raised number. It tells how many times a appears in the product. So 2² means 2 × 2, and 3³ means 3 × 3 × 3. Exponents are a short way to write a factor that repeats.

Sort out the question
You need the prime factorization of 756. That means writing 756 as a product of prime numbers only. A product is the answer to a multiplication. A prime number has exactly two factors, 1 and itself. Examples are 2, 3, 5 and 7.
All four choices multiply to 756. So multiplying a choice back does not pick the answer. You must check that every factor is prime.
Divide by 2 while you can
A factor tree splits a number into two factors, again and again. Start with the smallest prime, 2. 756 is even, so 2 goes into it:
$756 = 2 \times 378$
378 is even too:
$378 = 2 \times 189$
189 is odd, so 2 does not go into it. You have pulled out two 2s.
Divide by 3 while you can
Add the digits of 189: $1 + 8 + 9 = 18$ . A number can be divided by 3 when its digits add up to a number that 3 goes into. 3 goes into 18, so it goes into 189:
$189 = 3 \times 63$
$63 = 3 \times 21$
$21 = 3 \times 7$
7 is prime, so the tree stops. You have pulled out three 3s and one 7.
Write the factors with exponents
Collect every prime you pulled out:
$756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7$
Use the exponent rule. Two 2s make $2^{2}$ , and three 3s make $3^{3}$ . The choices use a raised dot, ·, for times. So 756 is $2^{2} \cdot 3^{3} \cdot 7$ .
Choose the answer
Watch out: each wrong choice is a factor tree that stopped too soon. A ends in 21, which is 3 × 7. B has 9, which is 3 × 3. C has 27, which is 3 × 3 × 3. Being odd does not make a number prime.
Only D uses primes alone. So the answer is D, $2^{2} \cdot 3^{3} \cdot 7$ .

Check your answer

Multiply your answer back: $2^{2} \times 3^{3} = 4 \times 27 = 108$ , and $108 \times 7 = 756$ . Then look at each factor. 2, 3 and 7 are all prime. D passes both tests.

More questions like this

The original DOE question

Question 1 as shown on the DOE’s digital test, page 74 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 74, question 1.

Question 7 of 81

2026 Practice Test C Math, question 4

Question

A person buys a used car with a down payment of $1,200 and makes monthly payments of $275 for 3 years. What is the total amount the person pays for the car?

A$2,025
B$9,900
C$11,100
D$13,500

Show the answer

Correct answer: C

Determine the total cost of the car by adding the down payment to the total monthly payments over 3 years.

There are 12 months in a year, so multiply (12 months/year × 3 years) to get a total of 36 months. Determine the total monthly payments by multiplying the monthly payment ($275) × the total number of months (36):

$275 \times 36 = 9{,}900$

Determine the total cost of the car by adding the down payment ($1,200) to the total monthly payments ($9,900):

$9{,}900 + 1{,}200 = 11{,}100$

The total cost of the car is $11,100.

Question analysis

A used car is bought with a down payment of $1,200 followed by monthly payments of $275 for 3 years, and students find the total paid. The work is to count the payments, multiply, and add the down payment. Mixing up years and months, or forgetting the money paid up front, leads to the wrong totals among the choices.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 A total made of one payment up front plus equal monthly payments is a short whole-number calculation, simpler than grade 7 word problems with percents or negatives. The only conversion is from years to months.
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: Much to keep track of minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The situation is familiar and the method direct, so this is one of the easier questions on the test. Points are lost in two places: turning 3 years into 36 payments, and remembering the money paid up front after the long multiplication. Each slip, like paying the down payment once a year, leads to a listed total, so checking that the answer holds exactly one $1,200 and 36 payments of $275 is worth the seconds.

Much to keep track of minor Working it out
The $1,200 down payment is paid once while the $275 repeats every month, so it has to be set aside while 36 payments are totaled and then added at the end. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Error-prone arithmetic minor Working it out
The product 275 × 36 = 9,900 is a two-digit multiplication done by hand; splitting it as 275 × 4 = 1,100 and then 1,100 × 9 makes it quick and safe. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn 3 years into a number of months, multiply the monthly payment by it, and add the down payment once at the end.

Traps

Stops one step early: A choice is a number from partway through the solution.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Units or scale (King et al., 2004)
Reveals a slip Expected pull: moderate
Multiplies the $275 by 3, the number of years, instead of by 36 months, then adds the $1,200 down payment: 825 + 1,200 = 2,025.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 36 monthly payments, 36 × $275 = $9,900, and stops without adding the $1,200 down payment made at the start.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Works out one year as $1,200 + 12 × $275 = $4,500 and multiplies by 3, so the down payment is paid three times instead of once.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing years into months: $\text{months} = 12 \times \text{years}$; Each year has 12 months. Multiply the number of years by 12 to get the number of months. You need months here because a payment is made every month.
Total cost with a down payment: $\text{total} = \text{down payment} + \text{monthly payment} \times \text{months}$; The down payment is money paid once, at the start. The monthly payment is paid every month. Multiply it by the number of months to get all the monthly payments. Then add the down payment only once, because it is paid only once.

Sort out the question
The person pays $1,200 once, at the start. This is the down payment. Then the person pays $275 every month for 3 years. You need the total amount paid for the car.
The payments come once a month, but the time is given in years. So change the years into months first.
Change 3 years into months
Use the months formula:
$\text{months} = 12 \times 3 = 36$
So the person makes 36 monthly payments.
Watch out: multiplying $275 by 3 counts only 3 payments. That leads to choice A, $2,025.
Add up the monthly payments
Multiply $275 by 36. A quick way is to split 36 into 4 × 9:
$275 \times 4 = 1{,}100$
$1{,}100 \times 9 = 9{,}900$
So the 36 payments add up to $9,900.
Watch out: stopping here gives choice B, $9,900. The down payment is still missing.
Add the down payment once
Use the total-cost formula:
$\text{total} = 1{,}200 + 9{,}900 = 11{,}100$
The person pays $11,100 in all. Adding $1,200 for every year counts it three times. That gives choice D, $13,500.
So the answer is C, $11,100.

Check your answer

Estimate. $275 is a little less than $300, and $36 \times 300 = 10{,}800$ . Add the $1,200: $10{,}800 + 1{,}200 = 12{,}000$ . The real total should be a little less than $12,000. $11,100 fits.

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 C Math, page 77, question 4.

Question 8 of 81

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 9 of 81

2026 Practice Test C Math, question 14

Question

If $x = -2$ , what is the value of $|x + x^{2} + x^{3}| - |x + 2x + 3x|$ ?

A−18
B−6
C−2
D2

Show the answer

Correct answer: B

Substitute $-2$ for each value of x in the expression:

$\left|(-2) + (-2)^{2} + (-2)^{3}\right|$
minus
$|(-2) + 2(-2) + 3(-2)|$

Evaluate the terms involving exponents:

$|(-2) + 4 + (-8)|$
minus
$|(-2) + 2(-2) + 3(-2)|$

Evaluate the terms involving multiplication:

$|(-2) + 4 + (-8)|$
minus
$|(-2) + (-4) + (-6)|$

Simplify the expression:

$|2 + (-8)| - |(-2) + (-10)|$

$\lvert -6 \rvert - \lvert -12 \rvert$

The absolute value of $(-6)$ is 6, and the absolute value of $(-12)$ is 12, so $\lvert -6 \rvert - \lvert -12 \rvert$ is the equivalent of subtracting 12 from 6:

$6 - 12 = (-6)$

Question analysis

This substitution question sets x = −2 in an expression that subtracts one absolute value from another: the first holds x, x squared and x cubed, the second x, 2x and 3x. Students evaluate the expression. It tests powers of negative numbers, absolute value and integer subtraction, and the signs of the square and the cube are where answers go wrong.

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 Substituting a negative number and taking absolute values is typical grade 7 sign practice. This expression puts a square and a cube of a negative inside one pair of bars and then subtracts two absolute values, so there are several places to slip.
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
Solution: 4 steps
Reading level: Question and choices: about grade 7, 8 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, but the expression stacks sign decisions, and each wrong choice records a different one going wrong. The powers of −2 alternate in sign, which students who treat every power as positive, or the square as negative, miss. The last step catches students who know the rules: both absolute values are positive, yet 6 − 12 is still negative, so the answer keeps its minus sign.

Error-prone arithmetic moderate Working it out
With x = −2 the square is 4 but the cube is −8, so the first bars hold −2 + 4 − 8 = −6, and the sign changes from each term to the next. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Six terms are substituted, two pairs of bars are simplified and made positive, and only then is 12 subtracted from 6, four stages that must come in a fixed order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out each power of −2 separately, keeping its sign, add the terms inside each pair of bars, take both absolute values, and subtract only at the end.

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Simplifies both insides correctly but treats the first bars like parentheses, keeping −6, so −6 − 12 = −18.
C 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 first bars hold −2 + 4 + 8 = 10, and 10 − 12 = −2.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Takes (−2)² as −4, so the first bars hold −2 − 4 − 8 = −14, and 14 − 12 = 2; dropping the minus from x throughout also gives 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; a is any number. The bars mean absolute value: how far a is from 0 on the number line. A distance is never negative, so |−6| = 6 and |6| = 6. Work out what is inside the bars first, then take the absolute value.
Powers of a negative number: $(-2)^{2} = (-2) \times (-2) \quad (-2)^{3} = (-2) \times (-2) \times (-2)$; The small raised number is the exponent. It tells how many times −2 is multiplied by itself. Two negatives multiply to a positive. So an even number of negative factors gives a positive answer, and an odd number gives a negative one.

Sort out the question
You need the value of $|x + x^{2} + x^{3}| - |x + 2x + 3x|$ when x = −2. The bars | | mean absolute value. Work out each pair of bars on its own. Subtract only at the end.
Find the powers of −2
Put −2 in for x. Keep each negative sign:
$(-2)^{2} = (-2) \times (-2) = 4$
$(-2)^{3} = 4 \times (-2) = -8$
The square is positive, but the cube is negative.
Watch out: taking $(-2)^{3}$ as +8 makes the first bars hold 10. That leads to choice C, −2.
Work out the first bars
Add the three terms inside the first pair of bars:
$|(-2) + 4 + (-8)| = |-6| = 6$
−2 + 4 is 2, and 2 + (−8) is −6. The distance from −6 to 0 is 6.
Work out the second bars
Here 2x means 2 times x, so it is −4. And 3x is −6:
$|(-2) + (-4) + (-6)| = |-12| = 12$
All three terms are negative, so they add to −12. Its absolute value is 12.
Subtract the second value
Now subtract the second value from the first:
$6 - 12 = -6$
Watch out: the bars make each part positive, but the difference can still be negative. Keeping −6 inside the first bars gives −6 − 12 = −18, choice A.
So the answer is B, −6.

Check your answer

Check the second bars a faster way. $x + 2x + 3x = 6x$ , and $6 \times (-2) = -12$ . That matches. Also, the second value, 12, is bigger than the first, 6. So the difference must be negative, as −6 is.

More questions like this

The original DOE question

Question 14 as shown on the DOE’s digital test, page 87 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 87, question 14.

Question 10 of 81

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 11 of 81

2026 Practice Test D Math, question 1

Question

What is the prime factorization of 3,575?

A $5 \cdot 11 \cdot 13$
B $5^{2} \cdot 11 \cdot 13$
C $5 \cdot 715$
D $5^{2} \cdot 143$

Show the answer

Correct answer: B

By observation, 3,575 ends in 75; therefore, 3,575 is divisible by 25.

$3{,}575 \div 25 = 143$

Test 143 for factors and eliminate 3, 5, and 7, but 11 is a factor.

$143 \div 11 = 13$ ; both 11 and 13 are prime numbers.

The first factor, 25, is not prime but is the product of primes, $5 \times 5 = 5^{2}$ .

The prime factorization of 3,575 is $5^{2} \times 11 \times 13$ .

Question analysis

This number question asks for the prime factorization of 3,575, chosen from four products of factors. Students divide by primes, using divisibility tests, until only primes remain. The four products look much alike, so the work must be carried all the way through, keeping count of every prime divided out along the way.

NYS standards: 6.NS.4 common factors and multiples
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Breaking a number into primes is grade 6 number work that relies on divisibility tests. The number here is larger than in most class examples, but the work is still a few divisions, lighter than typical grade 7 problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 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

Three of the four choices multiply back to 3,575, so checking a product rules out only the one that equals 715. The question turns on whether 143 is prime: it survives the quick tests for 2, 3 and 5, and the factor 11 is easy to miss. Students who keep dividing until every factor is prime, and who write a prime as often as it divides, finish in under a minute.

A very close wrong answer moderate Choosing the answer
5² · 143 matches the right factorization in its first factor and multiplies back to 3,575 just as well, so a product check cannot separate them; only splitting 143 into 11 × 13 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Nothing on the page defines a prime factorization or gives a test for 11, and 143 passes the familiar quick tests for 2, 3 and 5, so it looks prime. (Anderson & Krathwohl, 2001)

Strategy: Divide by the smallest prime that goes in evenly, again and again, writing down every prime you use until the quotient reaches 1, then compare that list with each choice.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Lists each prime that divides 3,575 only once, 5, 11 and 13, as if a factorization named which primes divide it, though their product is only 715.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Divides out one 5 and stops at 5 · 715, a true factorization but not a prime one, though 715 ends in 5 and splits further.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Divides by 25 and stops at 143, which passes the tests for 2, 3 and 5 and looks prime, though 143 = 11 × 13.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisibility by 25: $\text{last two digits } 00, 25, 50 \text{ or } 75 \rightarrow \text{divisible by } 25$; A number is divisible by 25 when 25 goes into it with no remainder. You only need to look at the last two digits. If they are 00, 25, 50 or 75, the number is divisible by 25. This works because 100 is 4 × 25, so the hundreds never leave a remainder.
Writing a repeated factor with an exponent: $a^{2} = a \times a$; a is any number. The small raised 2 is an exponent. It tells you how many times a is used as a factor, so 5² means 5 × 5 = 25. Use an exponent in a prime factorization when the same prime shows up more than once.

Sort out the question
A prime number is a whole number greater than 1 whose only factors are 1 and itself. Examples are 2, 3, 5, 7, 11 and 13. The prime factorization of 3,575 writes it as a product (the answer to a multiplication) of prime numbers only. So you must break 3,575 down until every factor is prime.
Divide by 25 first
3,575 ends in 75. So it is divisible by 25:
$3{,}575 \div 25 = 143$
That means 3,575 = 25 × 143. But 25 is not prime. It is 5 × 5, which you can write as 5².
Test 143 for prime factors
Now see if 143 splits. Try the primes in order. 143 is odd, so 2 does not work. Its digits add to 1 + 4 + 3 = 8, which is not a multiple of 3, so 3 does not work. It does not end in 0 or 5, so 5 does not work. Since 7 × 20 = 140, 143 ÷ 7 leaves a remainder of 3, so 7 does not work. Now try 11:
$143 \div 11 = 13$
So 143 = 11 × 13, and both 11 and 13 are prime.
Watch out: 143 looks prime, but it is not. Stopping at 5² · 143 leads to choice D.
Put all the primes together
Collect every prime you found. The 5 appears twice, then 11 and 13 once each. The dot · means times, just like ×.
$3{,}575 = 5 \times 5 \times 11 \times 13 = 5^{2} \cdot 11 \cdot 13$
Watch out: write the 5 twice, since it divides 3,575 twice. Writing it once gives 5 · 11 · 13, choice A, which equals only 715. Choice C, 5 · 715, stops too early, because 715 ends in 5 and splits further.
So the answer is B, $5^{2} \cdot 11 \cdot 13$ .

Check your answer

Multiply the primes back together: 25 × 11 = 275, and 275 × 13 = 3,575. That is the number you started with. Also, 5, 11 and 13 are all prime, so nothing can be split further.

More questions like this

The original DOE question

Question 12 of 81

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 13 of 81

2026 Practice Test D Math, question 6

Question

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

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

Show the answer

Correct answer: B

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

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

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

4 pounds is

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

5 pounds is

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

p pounds is

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 81

2026 Practice Test D Math, question 12

Question

$A circle with center P. Points R and T are on the circle at the ends of a horizontal diameter through P, and point S is on the circle above it. Segment SP is labeled 12. A vertical segment from S meets RT at a right angle, dividing RT into a part labeled x, from R to that point, and a part labeled 5, from that point to T.$

P is the center of the circle above, and R, S, and T are points on the circle. What is the value of x?

Show the answer

Correct answer: D

$\overline{\mathrm{SP}}$ represents the radius of the circle and has length 12.

Both $\overline{\mathrm{RP}}$ and $\overline{\mathrm{PT}}$ are radii and also have length 12.

Therefore, the length of $\overline{\mathrm{RT}}$ , which is a diameter, is 24 and is also represented by $x + 5$ .

$x + 5 = 24$

$x = 19$

Question analysis

A circle with center P has points R, S and T on it, a segment SP labeled 12, and a vertical segment from S that splits RT into a piece labeled x and a piece labeled 5. Students find the value of x. It tests basic circle facts, and the challenge is seeing which lengths in the figure must be equal.

NYS standards: 7.G.4 circle area and circumference
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the fact that every radius has the same length, but spotting which segments are radii in a figure with extra markings is less routine than using a formula.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 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 gives one measurement on a slanted radius and two labels on the diameter, and the right angle at the foot of the vertical segment is a decoy. Once RP and PT are seen as radii, RT is 24, and x is that diameter less the 5. Students who never connect the slanted 12 to the horizontal line fall back on the labeled numbers, adding 12 and 5, or take x to end at the center.

Reading a figure, table or graph moderate Understanding the question
The 12 labels the slanted segment SP, not the horizontal one, and the arrow for x runs from R past the center P to the foot of the vertical segment, so x is neither a radius nor a whole diameter. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
No obvious method minor Working it out
The right-angle mark and the triangle under S suggest the Pythagorean theorem, which is no help; the way in is seeing that RP and PT are radii as well. (Schoenfeld, 1985)

Strategy: Mark every segment from P to the circle as 12, notice that R and T are the ends of a diameter, then write an equation for that diameter using x and the 5.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Information you don't need: The question gives numbers or facts that play no part in the answer.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads x as the radius RP, as if its arrow ended at the center P, though it runs past P to the foot of the vertical segment.
B Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a surface strategy Expected pull: weak
Judges x from the drawing, where it looks about three times as long as the piece labeled 5, instead of working from the radius of 12.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Adds the two labeled numbers, 12 + 5 = 17, without placing the 12 on the diameter, as if x were a radius plus the 5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius and diameter: $d = 2r$; r is the radius, the distance from the center to any point on the circle. d is the diameter, the distance straight across the circle through the center. A diameter is two radii laid end to end, one on each side of the center, so it is twice as long.

Sort out the question
P is the center of the circle, and R, S and T are on the circle. Segment SP is 12. Segment RT goes straight across through P. A short line from S meets RT at a right angle. It splits RT into x, from R to that point, and 5, from that point to T. You need x.
See that SP is a radius
SP runs from the center P to the point S on the circle. So SP is a radius, and the radius is 12. Every radius of a circle has the same length. RP and PT also run from the center to the circle, so RP = 12 and PT = 12.
Find the diameter RT
RT passes through the center, so it is a diameter. Use the formula with r = 12:
$\mathrm{RT} = 2r = 2 \times 12 = 24$
RT is 24 long. It is the radius RP plus the radius PT.
Solve for x
RT is also split into x and 5, so:
$x + 5 = 24$
Subtract 5 from both sides:
$x = 24 - 5 = 19$
Watch out: x runs from R past the center P, so it is longer than a radius. Reading it as 12, choice A, stops at P. Adding 12 + 5 = 17, choice C, puts the 12 in the wrong place.
Work from the numbers, not the picture
The figure is not drawn to scale. In it, x looks about three times as long as 5, which suggests 15, choice B. Trust the labels instead. The right triangle under S is not needed at all.
So the answer is D, 19.

Check your answer

The point where the short line meets RT is 19 from R. P is 12 from R, so that point is 19 − 12 = 7 past the center. Then 7 + 5 = 12, which is PT, a radius. Everything fits.

More questions like this

The original DOE question

Question 12 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 D Math, page 85, question 12.

Question 15 of 81

2026 Practice Test D Math, question 20

Question

What is the value of $\frac{-9(-4)^{2} + 36(-4) + 304}{-4}$ ?

A−76
B−4
C4
D148

Show the answer

Correct answer: B

Use the order of operations to evaluate the expression. It is important to recognize that the fraction bar is a grouping symbol and so the entire numerator should be evaluated first. Exponents are evaluated first.

$\frac{-9(-4)^{2} + 36(-4) + 304}{-4}$

$\frac{-9(16) + 36(-4) + 304}{-4}$

Then multiplication, recognizing that the product of a negative factor and a positive factor is negative.

$\frac{-9(16) + 36(-4) + 304}{-4}$

$\frac{-144 - 144 + 304}{-4}$

Then addition and subtraction in order from left to right, recognizing the rules for adding and subtracting signed numbers.

$\frac{-144 - 144 + 304}{-4}$

$\frac{-288 + 304}{-4} = \frac{16}{-4}$

And finally the fraction can be evaluated by division, recognizing that the quotient of a positive and a negative number is negative.

$\frac{16}{-4} = -4$

Question analysis

This arithmetic question asks for the value of a fraction whose numerator, −9(−4)² + 36(−4) + 304, is divided by −4. Students follow the order of operations with negative numbers: a square, two products, a sum and a division. Every step carries a sign, and one dropped negative changes the result completely.

NYS standards: 7.NS.2a rational number multiplication
7.NS.2b rational number division
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Order of operations with negative numbers is standard grade 7 practice. This expression stacks a squared negative, two products and a division by a negative into one fraction, so it takes more steps than usual, each with a sign to track.
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
A very close wrong answer minor
Solution: 4 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

Nothing here is conceptually new; the question measures care with signs under the order of operations. The parentheses in (−4)² decide whether −9 multiplies 16 or −16, and the two products of −144 then nearly cancel the 304, leaving a small numerator whose sign is easy to lose. Writing each intermediate value with its sign, line by line, prevents most errors.

Error-prone arithmetic moderate Working it out
Five sign decisions sit in one expression: squaring −4, two products with a negative factor, a sum that climbs from −288 past zero, and a division by −4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The numerator alone takes an exponent, two multiplications and two additions before the single division, and all of it must be finished before dividing. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The right value and the choice 4 differ only in sign, so a student with the right size must still check the sign of a positive numerator over −4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Treat the fraction bar as grouping: square −4 first, do the two multiplications, combine the numerator from left to right, and only then divide by −4, writing each sign as you go.

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

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−4)² as −16, so −9 × (−16) = 144, the numerator becomes 304, and 304 ÷ (−4) = −76.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Works the numerator out correctly as 16 but divides by −4 as if the signs did not matter, giving a positive quotient.
D Negatives and direction (Brown & Burton, 1978)
Reveals a surface strategy Expected pull: weak
Drops every negative sign, adding 144 + 144 + 304 = 592 and dividing by 4, so 148 keeps none of the signs in the problem.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{grouping} \rightarrow \text{exponents} \rightarrow \times, \div \rightarrow +, -$; This is the order for working out an expression. Grouping means parentheses and the fraction bar. Exponents are small raised numbers, like the 2 in (−4)². Next, multiply and divide from left to right. Last, add and subtract from left to right.
Squaring a negative number: $(-a)^{2} = (-a) \times (-a) = a^{2}$; a is a positive number, so −a is negative. Squaring means multiplying a number by itself. A negative times a negative is positive, so a negative number squared is positive. For example, (−4)² = 16.
Signs when you multiply or divide: $\frac{\text{positive}}{\text{negative}} = \text{negative}$; When two numbers have different signs, their product or quotient is negative. When they have the same sign, it is positive. A quotient is the answer to a division.

Sort out the question
You need the value of one big fraction. The fraction bar is a grouping symbol, so finish the whole top first. Then divide by −4. Follow the order of operations, and track every minus sign.
Do the exponent first
The only exponent is (−4)². Squaring means multiplying a number by itself:
$(-4)^{2} = (-4) \times (-4) = 16$
Watch out: (−4)² is 16, not −16. Using −16 makes the top 304, which gives −76, choice A.
Do the multiplication
Now multiply. A negative times a positive is negative:
$-9 \times 16 = -144$
$36 \times (-4) = -144$
So the top becomes:
$-144 + (-144) + 304$
Add from left to right
Add the two negatives first:
$-144 + (-144) = -288$
Then add 304. Since 304 is more than 288, the result is positive:
$-288 + 304 = 16$
The whole top is 16.
Divide by −4
Now divide the top by the bottom. A positive divided by a negative is negative:
$\frac{16}{-4} = -4$
Watch out: positive 4, choice C, ignores the negative bottom. Dropping every minus sign gives 592 ÷ 4 = 148, choice D.
So the answer is B, −4.

Check your answer

Multiply back. If the answer is −4, then −4 times the bottom, −4, should give the top. It does: (−4) × (−4) = 16. Also, the top is positive and the bottom is negative, so the answer must be negative.

More questions like this

The original DOE question

Question 20 as shown on the DOE’s digital test, page 93 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 93, question 20.

Question 16 of 81

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 17 of 81

2026 Practice Test D Math, question 28

Question

A graph shows the relationship between the number of gallons of water, y, that has been added to a tank and the number of hours, x, that water has been added to the tank at a constant rate. What does the ordered pair (3, 24) represent?

AThree gallons of water was added in 24 hours.
BThree gallons of water was added per hour.
CTwenty-four gallons of water was added in 3 hours.
DTwenty-four gallons of water was added per hour.

Show the answer

Correct answer: C

It is given that x-values on the graph represent the numbers of hours for which water was added to the tank, and the y-values represent the numbers of gallons of water added in x hours. In the ordered pair (3, 24), 3 represents an x-value and 24 represents a y-value.

Therefore, (3, 24) represents a data point for 24 gallons of water being added to the tank in a 3-hour period.

Question analysis

A graph relates the gallons of water added to a tank, y, to the hours it has been filling at a constant rate, x. Students choose what the ordered pair (3, 24) means in the situation. It tests interpreting coordinates in context, and the four statements mix totals with rates and switch the two numbers around.

NYS standards: 7.RP.2d points on proportional graphs
6.EE.9 dependent and independent variables
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 is a direct case with no graph to read, though the stem introduces the two quantities in an order that needs care.
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: Reading a figure, table or graph minor
A very close wrong answer minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 76 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 fact decides the question: an ordered pair lists x first. The choices then test two confusions, which number is the gallons and whether a point is a rate. Reading the pair aloud as “after 3 hours, 24 gallons”, and noting that 24 ÷ 3 gives a rate of 8 gallons an hour that no choice states, rules out all three wrong statements.

Reading a figure, table or graph minor Understanding the question
No graph is drawn, and the stem names y, the gallons, before x, the hours, so the pair (3, 24) has to be matched to its letters, not to the order the quantities were mentioned. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer minor Choosing the answer
The four statements reuse 3 and 24 in two frames, a total added “in” some hours and an amount “per hour”, so the right statement and its swapped twin differ only in which number counts gallons. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Recall that an ordered pair lists x first and y second, write down what x and y measure here, and read the pair as one sentence about hours and gallons.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Pairs the numbers with the quantities in the order the stem names them, gallons first, so reads (3, 24) as 3 gallons in 24 hours.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes the first coordinate for a rate and pins it to gallons, mixing up one point on the graph with the constant rate, which is 8 gallons an hour.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 24 gallons as the amount added each hour, taking a point’s y-value for the rate, though 24 is the total after 3 hours.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

An ordered pair: $(x, y)$; An ordered pair names one point on a graph. The first number is always the x-value. The second number is always the y-value. Here x counts hours and y counts gallons. So a pair tells how many gallons were added after that many hours.
Rate from a point: $\text{rate} = \frac{y}{x}$; When water is added at a constant rate, divide the y-value by the x-value. The result is the rate: gallons added in 1 hour. A point itself shows a total, not a rate.

Sort out the question
The graph links two amounts:
- x is the number of hours water has been added.
- y is the number of gallons added.
You need to say what the point (3, 24) means. No graph is drawn, so you work from the words.
Match each number to its letter
In an ordered pair, x always comes first and y second. So in (3, 24):
- x = 3, which means 3 hours.
- y = 24, which means 24 gallons.
Watch out: the question names gallons first and hours second. Do not match the numbers to the words in that order. That mistake reads 3 as gallons and 24 as hours, choice A.
Read the pair as a sentence
Put the two meanings together: after 3 hours, 24 gallons have been added in all. The 24 is a total for the whole 3 hours.
Watch out: 24 is not the amount for each hour. The rate per hour is $24 \div 3 = 8$ gallons. So choice D, 24 gallons per hour, is wrong. Choice B, 3 gallons per hour, is also not the rate.
So the answer is C, twenty-four gallons of water was added in 3 hours.

Check your answer

Use the rate of 8 gallons each hour. In 3 hours, $3 \times 8 = 24$ gallons go in. That matches choice C. No choice says 8 gallons per hour, so both “per hour” choices must be wrong.

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 D Math, page 101, question 28.

Question 18 of 81

2026 Practice Test D Math, question 40

Question

A team is participating in a contest to collect canned goods for a food drive. The team has 53 points and must have a total of at least 100 points to win a prize. The team earns 5 points for each can of food a member collects. Which inequality can be used to find x, the number of cans of food the team needs to collect to win a prize?

A $5 + 53x \leq 100$
B $5 + 53x \geq 100$
C $53 + 5x \leq 100$
D $53 + 5x \geq 100$

Show the answer

Correct answer: D

Since each can is worth 5 points, the total number of points earned from all the cans, x, is 5x. The team started with 53 points. That is a constant value and doesn’t change. So the total number of points the team will earn is $53 + 5x$ . The team wants to earn at least (≥) 100 points, so they need $53 + 5x \geq 100$ total points.

Question analysis

A team has 53 points in a canned-food drive, earns 5 points for each can, and needs at least 100 points to win a prize. Students choose the inequality for x, the number of cans the team must collect. It tests writing an inequality from words, where the starting points and the points per can must each take the right place.

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 word problem with a starting amount and a rate is grade 7 work. The four choices pair two expressions with two symbols, so students make one decision about the terms and one about the phrase "at least".
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
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 71 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 choices form a two-by-two grid: which number goes with x, and which way the symbol points. Each decision can be made separately, so a student who settles both gets the item, while one mistake lands on a listed choice. Testing two numbers of cans, 5, which leaves the team at 78 points, and 10, which lifts it to 103, rules out every wrong inequality.

Turning words into math moderate Understanding the question
The 53 points the team “has” are a fixed start, the “5 points for each can” go with x, and “at least 100” calls for ≥, though “least” can sound like less. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The four choices are two expressions, 5 + 53x and 53 + 5x, each with ≤ and with ≥, so the right one and the strongest wrong one differ only in the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the team's total points after 2 cans and then after x cans, translate "at least 100" into a symbol, and compare the result with each choice.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Attaches x to the 53 points the team already has and also reverses “at least”, combining the mistakes behind the other two wrong choices.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Keeps the right symbol but multiplies the number of cans by 53 and adds 5, putting the starting points where the points per can belong.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Writes the points correctly as 53 + 5x but reads “at least 100” as at most 100, reversing the inequality.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total with a start and a rate: $\text{total} = \text{start} + \text{rate} \times x$; The start is the amount you have now, here 53 points. The rate is what each item adds, here 5 points for each can. x is the number of items. The start is added once, and the rate is multiplied by x.
At least: $\text{total} \geq \text{goal}$; The sign ≥ means is greater than or equal to. At least 100 means 100 or more. So the total must be greater than or equal to 100.

Sort out the question
You know:
- The team already has 53 points.
- It earns 5 points for each can.
- It needs at least 100 points to win.
x is the number of cans. You need the inequality that shows this.
Write the total points
Try 2 cans first. The team gets 5 points for each, so $53 + 5 \times 2 = 63$ points. For x cans, the total is:
$53 + 5x$
Watch out: the 53 points do not repeat for each can. $5 + 53x$ gives 53 points for each can, as in choices A and B.
Choose the sign
At least 100 means 100 or more, so use ≥:
$53 + 5x \geq 100$
Watch out: “least” sounds like “less,” but at least means the total can be 100 or higher. Using ≤ gives choice C.
So the answer is D, $53 + 5x \geq 100$ .

Check your answer

Test two numbers of cans. With 10 cans, $53 + 5 \times 10 = 103$ , and 103 ≥ 100 is true. The team wins. With 9 cans, $53 + 5 \times 9 = 98$ , and 98 ≥ 100 is false. The team is short. Choice D gets both right.

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 D Math, page 113, question 40.

Question 19 of 81

2026 Practice Test D Math, question 44

Question

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

$p = 2.50 + 1.25c$

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

Show the answer

Correct answer: C

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

$22.50 = 2.50 + 1.25c$

$22.50 - 2.50 = 1.25c$

$20.00 \div 1.25 = c$

$16 = c$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 20 of 81

2026 Practice Test D Math, question 46

Question

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

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

SHIPPING COSTS

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

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

Show the answer

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

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

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

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

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

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

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

$2 = x - 1$

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

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

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

$8 = x - 1$

$x = 9 \text{ pounds}$

To check:

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

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

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

$3 = x - 1$

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 21 of 81

2024 Test A Math, question 9

Question

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

Show the answer

Correct answer: D

Consider the case where $m = 1$ .

$n - m = 66$

$n - 1 = 66$

$n - 1 + 1 = 66 + 1$

$n = 67$

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 66 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 66. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 22 of 81

2024 Test A Math, question 10

Question

Simplify:

$\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}$

A $x$
B $x - 2$
C $13x - 36$
D $x + \frac{36}{13}$

Show the answer

Correct answer: A

$\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}$

Divide the first 39 by 3.

$\frac{13(x - 3) + 39}{13}$

Distribute 13 through $(x - 3)$ .

$\frac{13x - 39 + 39}{13}$

In the numerator, the terms −39 and 39 cancel out.

$\frac{13x}{13}$

Finally, divide the numerator and denominator by 13 to get x.

Question analysis

An algebra question that asks students to simplify a stacked fraction: 39 times (x − 3) over 3, plus 39, all divided by 13. It tests dividing, distributing and combining like terms in the right order. The nested fraction bars make it easy to divide only some of the terms or to distribute to only one.

NYS standards: 7.EE.1 linear expressions with rational coefficients
6.EE.3 generating equivalent expressions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students expand and simplify linear expressions such as 13(x − 3) + 39. This wraps that work inside two fraction bars, which adds reading load and chances to misstep but no new mathematics.
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
Solution: 3 steps
Reading level: Question and choices: about grade 16+, 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

The expression looks heavier than it is: every number is a multiple of 13, and the constants cancel to leave a single term. What makes it risky is scope, knowing which terms each fraction bar and each multiplier covers. Each wrong choice comes from stretching or shrinking that scope by one term, so working outward from the innermost parentheses, one line per step, is the safe way.

Error-prone arithmetic moderate Working it out
Two fraction bars are nested: 39 ÷ 3 applies only to the first term, the 13 must reach both x and −3, and only then does the whole numerator divide by 13. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Four moves come in a fixed order, divide 39 by 3, distribute 13, cancel −39 with +39, divide by 13, and each is a chance to apply a step to the wrong terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work from the inside out: divide 39 by 3, distribute through the parentheses, combine the constants in the numerator, and only then divide everything by 13.

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

B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads the inner fraction bar as covering the added 39 too, so the numerator becomes 13(x − 3) + 13 = 13x − 26, and dividing by 13 gives x − 2.
C Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Divides only the last 39 by 13, leaving 13(x − 3) + 3 = 13x − 36, as if the outer fraction bar covered a single term.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Distributes the 13 to x but not to −3, so the numerator becomes 13x − 3 + 39 = 13x + 36, and dividing by 13 leaves a fraction.

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. Multiply a by each term inside, and keep the minus sign between them. It works because a groups of (b − c) are a groups of b, take away a groups of c.

Sort out the expression
You need to simplify the big fraction, which means write it in its shortest form. It has a small fraction inside its top. The top is $\frac{39(x - 3)}{3} + 39$ , and the bottom is 13. Work from the inside out, one step at a time.
Divide the first 39 by 3
The small fraction bar is only under $39(x - 3)$ . So divide just that 39 by 3.
$\frac{39(x - 3)}{3} = 13(x - 3)$
Now the top is $13(x - 3) + 39$ . Watch out: the small bar does not reach the added 39. Dividing that 39 by 3 as well leads to $x - 2$ , choice B.
Multiply out the parentheses
$a(b - c) = a \times b - a \times c$
$13(x - 3) = 13 \times x - 13 \times 3 = 13x - 39$
Watch out: multiply the 3 by 13 too, not just the x. Leaving −3 as it is leads to choice D.
Combine the plain numbers on top
$13x - 39 + 39 = 13x$
−39 and 39 add up to 0, so they cancel. The top is now just 13x, which means 13 times x.
Divide by the bottom
Now divide the whole top by 13. Dividing by 13 undoes times 13.
$\frac{13x}{13} = x$
Watch out: the long bar is under the whole top. Dividing only the last 39 by 13 gives $13x - 36$ , choice C. So the answer is A, $x$ .

Check your answer

Try x = 5. Then $x - 3 = 2$ and $39 \times 2 = 78$ . Next, $78 \div 3 = 26$ and $26 + 39 = 65$ . Last, $65 \div 13 = 5$ . You get back 5, the same as x. So x is right.

More questions like this

The original DOE question

Question 67 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 67. Open page 78 of the handbook (PDF)

Question 23 of 81

2024 Test A Math, question 12

Question

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

Show the answer

Correct answer: D

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

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

Question analysis

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

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students compare fractions and decimals routinely, but not with an unknown in the denominators. The comparison is simple once 0.01 is written as a fraction; the unfamiliar setup is the main demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The setup looks unfamiliar, with n in two denominators, but one rewrite makes it routine: 0.01 is 1/100, so the two denominators must sit on either side of 100. The three small choices fail a quick check, since every fraction from 1/1 to 1/4 is far above 0.01, and 0 cannot be a denominator at all. Students who test the choices instead of rewriting the decimal spend a little longer but reach the same place.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 24 of 81

2024 Test A Math, question 18

Question

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

A96
B111
C112
D127

Show the answer

Correct answer: C

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

$2x + 1 = y$

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

$2(63) + 1 = y$

$126 + 1 = y$

$127 = y$

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

$2x + 1 = 63$

Subtract one from both sides of the equation.

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

$2x = 62$

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

$x = 31$

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

$2x + 1 = 31$

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

$2x = 30$

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

$x = 15$

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

$127 - 15 = 112$

The difference is 112.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 25 of 81

2024 Test A Math, question 20

Question

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

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

Show the answer

Correct answer: A

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

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 81

2024 Test A Math, question 22

Question

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

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

A10.0° C
B15.0° C
C20.0° C
D32.4° C

Show the answer

Correct answer: A

The highest temperature was 86° F.

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

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

Subtract 32 from both sides of the equation:

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

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

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

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

$30 = C$

The lowest temperature was 68° F.

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

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

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

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

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

$20 = C$

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

$30 - 20 = 10$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 27 of 81

2024 Test A Math, question 23

Question

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

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

Show the answer

Correct answer: B

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 28 of 81

2024 Test A Math, question 25

Question

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

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

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

Show the answer

Correct answer: A

These facts are given:

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 29 of 81

2024 Test A Math, question 29

Question

Which graph represents the solution to the inequality $x + 4 \geq 3$ ?

A $Number line from −4 to 7 with a tick at every integer. A closed dot at −1, with the line shaded from −1 to the right, ending in an arrow.$
B $Number line from −4 to 7 with a tick at every integer. A closed dot at 1, with the line shaded from 1 to the right, ending in an arrow.$
C $Number line from −4 to 7 with a tick at every integer. A closed dot at 3, with the line shaded from 3 to the right, ending in an arrow.$
D $Number line from −4 to 7 with a tick at every integer. A closed dot at 4, with the line shaded from 4 to the right, ending in an arrow.$

Show the answer

Correct answer: A

Use the properties of equations to solve the inequality for x.

$x + 4 \geq 3$

Subtract 4 from both sides of the inequality.

$x + 4 - 4 \geq 3 - 4$

$x \geq -1$

The graph of $x \geq -1$ will have a point at −1 and will include all values that are greater than, or to the right of, −1. The greater than or equal sign is used, which indicates that the solution includes −1. So the point at −1 will be solid. Option A is the correct answer.

Question analysis

This inequality question asks which number line graph shows the solution set of x + 4 ≥ 3. Students solve a one-step inequality by undoing the addition, then match the result to a graph. All four graphs use a solid dot and an arrow pointing right, so the whole question rests on where the shaded ray begins.

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: Below grade 7 Grade 7 inequality work is usually two-step, such as 3x − 5 ≥ 10, and asks students to choose an open or closed dot and a direction. This one needs a single subtraction, and all four graphs share a closed dot and a rightward arrow.
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: Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 9 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 a single subtraction, and the graphs differ only in where the solid dot sits, so the question comes down to 3 − 4. Students who lose the negative sign land on the graph that starts at 1, while those who never isolate x pick a graph that starts at a number printed in the inequality. Testing 0, which satisfies 0 + 4 ≥ 3, shows that the shading must include 0.

Reading a figure, table or graph minor Understanding the question
All four graphs share a closed dot and a rightward arrow, so the only thing to read is where the dot sits, against tick labels printed with raised minus signs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Undoing the + 4 means computing 3 − 4, the one step that crosses zero; a dropped minus sign moves the dot from −1 to 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 4 from both sides to get x alone, then find the graph whose solid dot sits at that number; test one shaded value in the original inequality to confirm.

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

The wrong answers, classified

B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Computes 4 − 3 instead of 3 − 4, or drops the minus sign of −1, and starts the shaded ray at positive 1.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Starts the ray at 3, the number on the right side of the inequality, as if x ≥ 3, without subtracting the 4.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Starts the ray at 4, the number added to x, reading it off the inequality instead of solving for x.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an inequality balanced: $\text{if } a \geq b \text{, then } a - c \geq b - c$; a and b are the two sides of an inequality, and c is any number. The sign ≥ stands for greater than or equal to. Taking the same number away from both sides keeps the comparison true, like taking the same weight off both pans of a scale. Use it to get x by itself.

Sort out the question
An inequality is a math sentence that compares two amounts. The sign $\geq$ means is greater than or equal to. So $x + 4 \geq 3$ says that x plus 4 is 3 or more.
You need the graph that shows every number x that makes this true. All four graphs have a solid dot and an arrow to the right. Only the place where the dot sits is different.
Get x by itself
Right now 4 is added to x. To undo adding 4, subtract 4 from both sides.
$\text{if } a \geq b \text{, then } a - c \geq b - c$
$x + 4 - 4 \geq 3 - 4$
$x \geq -1$
So x is −1 or any number greater than −1. Watch out: 3 − 4 is −1, not 1. Dropping the minus sign puts the dot at 1, choice B. Putting the dot at 3 or 4 means you read a number off the inequality without solving. That leads to C or D.
Match the solution to a graph
On a number line, greater numbers are to the right. So the shading starts at −1 and goes right. The arrow shows that it keeps going.
The dot is solid, or filled in, because of the or equal to part. A solid dot means −1 itself is a solution, a number that makes the inequality true.
Graph A has a solid dot at −1 and is shaded to the right. So the answer is A, the graph with a solid dot at −1.

Check your answer

Test a number from the shaded part, such as 0. $0 + 4 = 4$ , and 4 is more than 3. Now test the dot: $-1 + 4 = 3$ , and 3 is equal to 3, so −1 works. A number left of the dot, −2, gives $-2 + 4 = 2$ . That is less than 3, so −2 should not be shaded, and it isn’t.

More questions like this

The original DOE question

Question 86 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 86. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 81

2024 Test A Math, question 31

Question

Nura made a square poster with a side length of 13 inches. Latrice made a square poster with a side length of 15 inches. What is the difference, in square inches, between the area of Latrice’s poster and the area of Nura’s poster?

Show the answer

Correct answer: A

Nura’s poster is square, so all the side lengths are 13 inches. The area, in square inches, is

$A = s^{2} = 13^{2} = 169$

Latrice’s poster is square, so all the side lengths are 15 inches.

$A = 15^{2} = 225$

Subtract to determine the difference between the areas.

$225 - 169 = 56$

The difference, in square inches, is 56.

Question analysis

This area word problem compares two square posters, one 13 inches on a side and one 15 inches on a side, and asks for the difference between their areas in square inches. Students square each side length and subtract. Working with the difference in the sides, or with perimeters, instead of the areas is the easy mistake.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Area of a square is elementary content, and squaring 13 and 15 is grade 6 exponent work. Typical grade 7 area problems involve circles, composite shapes or scale drawings; this one has two familiar areas and a subtraction.
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
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 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 method is short, but the question invites a shortcut: the sides differ by 2, so it feels as if the areas should differ by 2 squared. Two choices reward working with the side lengths and one rewards perimeters. Squaring each side before subtracting avoids all three, and a sketch of the 13-inch square inside the 15-inch one shows the gap is an L-shaped strip, far more than 4 square inches.

Something you must already know minor Working it out
The area of a square, side times side, is not given; students must square 13 and 15 themselves rather than work with the side lengths. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
Squaring 13 and 15 by hand gives 169 and 225, and the subtraction 225 − 169 needs a borrow, where an arithmetic slip creeps in. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find each poster's area by squaring its side length, then subtract the smaller area from the larger; work with the areas, not the side lengths.

Traps

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

The wrong answers, classified

B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Compares perimeters instead of areas, 4 × 15 − 4 × 13 = 60 − 52 = 8, a difference in inches, not square inches.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Subtracts the sides first and squares the difference, (15 − 13)² = 4, as if 15² − 13² equaled (15 − 13)².
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the difference in side lengths, 15 − 13 = 2, a length rather than a difference in area.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s^{2} = s \times s$; A is the area, the flat space inside the shape. s is the length of one side. All four sides of a square are equal, so the area is side times side. The small 2 means squared, or times itself. Area is measured in square units, here square inches.

Write down what you know
Nura’s poster is a square with sides of 13 inches. Latrice’s poster is a square with sides of 15 inches. You need the difference between their areas, in square inches. The difference is how much bigger one area is than the other. So find each area, then subtract.
Find the area of Nura’s poster
Use the area formula with the side of 13 inches.
$A = s^{2}$
$A = 13^{2} = 13 \times 13 = 169$
Nura’s poster has an area of 169 square inches.
Find the area of Latrice’s poster
Use the same formula with the side of 15 inches.
$A = 15^{2} = 15 \times 15 = 225$
Latrice’s poster has an area of 225 square inches.
Subtract the two areas
Take the smaller area away from the larger one.
$225 - 169 = 56$
The difference is 56 square inches.
Watch out: don’t subtract the sides first and then square. $(15 - 13)^{2} = 2^{2} = 4$ , choice C, is not the same as $15^{2} - 13^{2}$ . The side difference, 2 inches, is choice D, a length. Comparing perimeters, the distance around, gives 8, choice B.
So the answer is A, 56 square inches.

Check your answer

Picture Nura’s poster laid in a corner of Latrice’s. The extra part is an L-shaped strip 2 inches wide. One arm is 15 by 2 and the other is 13 by 2. $30 + 26 = 56$ , so the strip is 56 square inches. That matches.

More questions like this

The original DOE question

Question 88 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 84, question 88. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 31 of 81

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 32 of 81

2024 Test B Math, question 8

Question

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

Show the answer

Correct answer: C

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

$9\bullet 8=72$

$2x\bullet 3y=6xy$

$72=6xy$

$12=xy$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 33 of 81

2024 Test B Math, question 13

Question

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

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

Show the answer

Correct answer: B

First substitute 2 for r in the equation.

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

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

Then solve for p.

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

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

$p+4=2p+3$

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

$p+1=2p$

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

$1=p$

The value of p is 1.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Adds the outer 1 inside the parentheses before multiplying, writing 2(p + 2) = 2p + 4, so p + 4 = 2p + 4 gives 0.
C Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Multiplies only the p by 2, writing 2p + 1 + 1 = 2p + 2, so p + 4 = 2p + 2 and p comes out as 2.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Drops the 1 inside the parentheses, reading 2(p + 1) as 2p, so the right side is 2p + 1 and p + 4 = 2p + 1 gives 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c$; a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. Adding the same thing, or dividing both sides by the same number that is not zero, works too.

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

Check your answer

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

More questions like this

The original DOE question

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

Question 34 of 81

2024 Test B Math, question 17

Question

A bowl contains 12 green candies, 4 yellow candies, and some red candies. The probability of choosing a green candy at random is twice as great as the probability of choosing a red candy at random. What is the probability of choosing a yellow candy at random?

A $\frac{2}{11}$
B $\frac{2}{9}$
C $\frac{1}{4}$
D $\frac{3}{11}$

Show the answer

Correct answer: A

The probability of choosing a green candy is twice as great as the probability of choosing a red candy. There are 12 green candies. Let r = the number of red candies.

$12=2r$

$r=6$

So there are 6 red candies. 12 green candies + 4 yellow candies + 6 red candies = 22 total candies.

There are 4 yellow candies out of a total of 22 candies. So the probability of choosing a yellow candy is

$\frac{4}{22}=\frac{2}{11}$ .

Question analysis

A probability question about a bowl of 12 green candies, 4 yellow candies and an unknown number of red ones. A green candy is twice as likely to be picked as a red one, and students find the chance of picking a yellow candy. The missing red count has to be worked out before any probability can be found.

NYS standards: 6.SP.8a uniform probability models
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Simple probability is grade 6 content in New York, and grade 7 problems usually give every count. Working backward from a comparison of two probabilities to a missing count, then forward to a third probability, is more than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Combines ideas moderate
Solution: 3 steps
Reading level: Question and choices: about grade 7, 49 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 withholds the red count, and everything depends on recovering it from the comparison of probabilities. Students who cannot make that step tend to leave red out altogether, while those who find 6 red candies can still slip at the end by using the wrong total or the wrong color. Writing all three counts and the total of 22 before forming any fraction prevents both endings.

Turning words into math moderate Understanding the question
“Twice as great as the probability of choosing a red candy” must become 12 = 2 × the red count, which works only because both probabilities share one total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Combines ideas moderate Working it out
An unknown count has to be found from a ratio before any probability can be written, joining proportional reasoning with probability as a part over the total. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Turn twice as likely into a statement about numbers of candies to find the red count, add all three colors for the total, then put the yellow count over that 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.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Finds the 6 red candies but compares the 4 yellow with the 18 green and red ones, 4/18, a part-to-part ratio instead of part to whole.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Leaves out the red candies, whose number is not given, and divides the 4 yellow by the 16 green and yellow candies.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds the 6 red candies and the total of 22 correctly but gives the chance of red, 6/22 = 3/11, instead of yellow.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of the kind you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top is how many candies are the kind you want. The bottom is how many candies there are in all. Each candy is just as likely to be picked, so this fraction gives the chance.

Sort out what you know
The bowl has 12 green candies, 4 yellow candies and some red candies. You are not told how many are red. Let r be the number of red candies.
The chance of picking green is twice the chance of picking red. You need the probability (the chance) of picking yellow.
Compare green and red
Write each chance with the probability formula.
$P = \frac{\text{number of the kind you want}}{\text{total number}}$
Green and red have the same bottom number, the total number of candies. So if the chance of green is twice the chance of red, there are twice as many green candies as red ones.
$12 = 2 \times r$
Find the red candies and the total
$r = 12 \div 2 = 6$
There are 6 red candies. Now add up all the candies.
$12 + 4 + 6 = 22$
There are 22 candies in all. Watch out: do not leave out the red candies just because their number was not given. Using 16 as the total gives 1/4, choice C.
Find the chance of yellow
There are 4 yellow candies out of 22.
$P = \frac{4}{22} = \frac{2}{11}$
Divide the top and the bottom by 2 to simplify. Watch out: the question asks for yellow, not red. 6/22 = 3/11 is the chance of red, choice D. The bottom must be all 22 candies. Using 18 gives 2/9, choice B.
So the answer is A, $\frac{2}{11}$ .

Check your answer

Check the clue. The chance of green is $\frac{12}{22}$ , and the chance of red is $\frac{6}{22}$ . Twelve is twice 6, so green is twice as likely. Also, the three chances add up to 1: $\frac{12}{22} + \frac{4}{22} + \frac{6}{22} = \frac{22}{22} = 1$ .

More questions like this

The original DOE question

Question 74 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 74. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 35 of 81

2024 Test B Math, question 25

Question

Which number line represents the values of x that satisfy the inequality $-5\leq 1-3x\leq 4$ ?

A $Number line from −3 to 3 with a tick at every integer. Open circles at −1 and 2, with the segment between them shaded.$
B $Number line from −3 to 3 with a tick at every integer. Closed circles at −1 and 2, with the segment between them shaded.$
C $Number line from −3 to 3 with a tick at every integer. Open circles at −1 and 2, with the line shaded from −1 to the left and from 2 to the right.$
D $Number line from −3 to 3 with a tick at every integer. A closed circle at −1 and an open circle at 2, with the line shaded from −1 to the left and from 2 to the right.$

Show the answer

Correct answer: B

Split the compound inequality into two inequalities.

$-5\leq 1-3x$ and $1-3x\leq 4$

For each inequality, subtract 1 from each side.

$-6\leq -3x$ and $-3x\leq 3$

For each inequality, divide both sides by −3. Note that dividing each side of an inequality by a negative number changes the direction of the inequality.

$2\geq x$ and $x\geq -1$

Combining the pieces gives the compound inequality $-1\leq x\leq 2$ .

The solution is that x is greater than or equal to −1 and less than or equal to 2, so choose the number line with filled circles at −1 and 2, with the filled segment between the circles.

Question analysis

An inequality question: choose the number line that shows the solutions of the compound inequality −5 ≤ 1 − 3x ≤ 4. Students solve an inequality whose x term has a negative coefficient, then match the endpoints, open or closed circles, and shading to one of four graphs.

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: Above grade 7 Grade 7 students solve and graph two-step inequalities one at a time. This compound inequality needs two solved together and a division by a negative number, which reverses the signs, a step many grade 7 classes only touch on.
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
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 12, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question stacks two classic traps: reversing the inequality signs when dividing by −3, and matching filled or open circles to ≤. All four graphs use the endpoints −1 and 2, so finding those numbers proves little; the choice turns on the circles and the direction of the shading. Testing x = 0, x = 3 and the endpoint x = 2 in the original inequality checks both decisions in under a minute.

Error-prone arithmetic moderate Working it out
Dividing all three parts of −6 ≤ −3x ≤ 3 by −3 reverses both signs and the order, giving 2 ≥ x ≥ −1, which must be rewritten as −1 ≤ x ≤ 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
Two graphs share the endpoints −1 and 2 and the shading between them; only the circles, filled for ≤ and open for <, separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
Each graph must be read for three things at once: the endpoints, the fill of each circle, and whether the shading runs between them or outward. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Subtract 1 from all three parts, divide all three by −3 while reversing both inequality signs, then rewrite the result from smallest to largest before matching circles and shading.

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).

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Solves for the right endpoints and shading but draws open circles, as if ≤ excluded −1 and 2 instead of including them.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Divides by −3 without reversing the signs, so the solutions seem to lie outside −1 and 2; with open circles, this graph shows exactly the values that fail.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Shades outward as if the signs had not reversed, but mixes the circles, filled at −1 and open at 2, which no single reading of ≤ gives.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting from every part: $\text{if } a \leq b \text{, then } a - c \leq b - c$; a and b are the two sides of an inequality, and c is any number. Subtracting the same number from every part keeps the inequality true. The sign does not change.
Dividing by a negative number: $\text{if } a \leq b \text{, then } a \div (-c) \geq b \div (-c)$; a and b are numbers, and c is a positive number, so −c is negative. When you divide both sides by a negative number, the inequality sign flips. For example, 2 is less than 3, but −2 is greater than −3.

Sort out the question
The inequality $-5 \leq 1 - 3x \leq 4$ says that $1 - 3x$ is between −5 and 4. The sign ≤ means less than or equal to. So −5 and 4 themselves are allowed. You need the number line that shows every x that works. To find them, get x alone in the middle.
Subtract 1 from every part
First take away the 1 in the middle. Subtract 1 from all three parts.
$\text{if } a \leq b \text{, then } a - c \leq b - c$
$-5 - 1 \leq 1 - 3x - 1 \leq 4 - 1$
$-6 \leq -3x \leq 3$
Divide by −3 and flip
Now x is multiplied by −3. Divide all three parts by −3. Dividing by a negative number flips each sign.
$\text{if } a \leq b \text{, then } a \div (-c) \geq b \div (-c)$
$\frac{-6}{-3} \geq \frac{-3x}{-3} \geq \frac{3}{-3}$
$2 \geq x \geq -1$
Watch out: if you forget to flip, x seems to lie outside −1 and 2. That leads to choices C and D, which are shaded outward.
Read the result
Turn it around so the smaller number is on the left. It means the same thing.
$-1 \leq x \leq 2$
So x can be any number from −1 to 2. The ends, −1 and 2, are included, because the signs are ≤.
Pick the number line
On a number line, a closed (filled-in) circle means the end number is included. An open circle means it is not. So you need closed circles at −1 and 2, with the part between them shaded. Watch out: choice A has the right shading but open circles, which leave out −1 and 2.
So the answer is B.

Check your answer

Test a few numbers in $1 - 3x$ . For x = 0 you get 1, which is between −5 and 4, and 0 is shaded. For x = 2 you get −5, which is allowed, so the circle at 2 is filled in. For x = 3 you get −8, which is too small, so 3 is not shaded.

More questions like this

The original DOE question

Question 82 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 82. Open page 206 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 81

2024 Test B Math, question 31

Question

If $y=4x$ , what is the value of $3y+2(3y+5)-x$ in terms of x?

A $8x+10$
B $32x+10$
C $35x+10$
D $36x+10$

Show the answer

Correct answer: C

First, substitute $4x$ for each y in the expression:

$3y+2(3y+5)-x=$

$3(4x)+2[3(4x)+5]-x$

Then simplify the expression:

$3(4x)+2[3(4x)+5]-x$

$12x+2(12x+5)-x$

$12x+24x+10-x$

$12x+24x-x+10$

$35x+10$

The simplified expression is $35x+10$ .

Question analysis

Given that y equals 4x, students rewrite the expression 3y + 2(3y + 5) − x using only x. It tests substitution, the distributive property and combining like terms. Errors creep in when a term is dropped or when x terms and y terms are merged before y has been replaced.

NYS standards: 7.EE.1 linear expressions with rational coefficients
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Substituting an expression for a variable and simplifying with the distributive property is standard grade 7 work. Putting the substitution inside parentheses calls for care, but the steps are routine for students who combine like terms fluently.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Substitution and distribution are routine here, so points are lost to bookkeeping rather than method. The expression has three x pieces, 12x, 24x and a lone − x, and the last is easy to overlook after the larger terms. The other two wrong choices come from collecting 3y, 6y and x as if they were like terms, a reason to replace y with 4x before simplifying anything.

A very close wrong answer moderate Choosing the answer
Every choice ends in + 10, and 36x + 10 differs from 35x + 10 by a single x, so only remembering the last − x in the expression separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
The substitution sits inside nested grouping, 2[3(4x) + 5], so the 2 must reach both the 12x and the 5 before the x terms are collected. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Replace every y with 4x in parentheses before doing anything else, then distribute the 2 and collect the x terms and the constant separately.

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: weak
Collects 3y + 6y − x as 8 of a single variable, treating x and y as like terms, and writes the result as 8x without substituting 4x.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Combines 3y + 6y − x into 8y, as if x and y were like terms, and only then substitutes 4x for y, giving 32x + 10.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Substitutes and distributes correctly to reach 12x + 24x + 10 = 36x + 10, then stops before subtracting the final x.

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. Terms are the parts joined by plus or minus signs. Multiply a by each term inside. It works because 2 groups of (12x + 5) hold 2 of the 12x and 2 of the 5.

Sort out the question
$y = 4x$ means y is always 4 times x. You need to rewrite the expression with no y in it. That is what “in terms of x” means. The answer will use only x and numbers.
Replace each y with 4x
Put 4x in parentheses everywhere you see y. The parentheses keep the multiplying clear.
$3y + 2(3y + 5) - x = 3(4x) + 2(3(4x) + 5) - x$
3(4x) means 3 times 4x, which is 12x. It shows up twice.
$= 12x + 2(12x + 5) - x$
Multiply out the parentheses
Use the distributive property. The 2 must multiply both the 12x and the 5.
$2(12x + 5) = 2 \times 12x + 2 \times 5 = 24x + 10$
The whole expression is now:
$12x + 24x + 10 - x$
Combine the like terms
Like terms have the same letter part. Here the x terms are 12x, 24x and −x. The −x means take away one x.
$12x + 24x - x = 35x$
The 10 has no x, so it stays as it is. The result is $35x + 10$ .
Watch out: Don’t forget the −x at the end. Stopping at 12x + 24x gives 36x + 10, choice D. Also, swap in 4x before you combine. Adding 3y, 6y and −x as if they were all y’s mixes unlike terms. It leads to choice B.
So the answer is C, $35x + 10$ .

Check your answer

Test a number. Let x = 1, so y = 4. The original expression gives $3(4) + 2(3 \times 4 + 5) - 1 = 12 + 34 - 1 = 45$ . Your answer gives $35(1) + 10 = 45$ . They match. Choice D would give 46.

More questions like this

The original DOE question

Question 88 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 88. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 37 of 81

2024 Test B Math, question 35

Question

The high temperature on Monday was x° F. On Tuesday, the high temperature was 84° F, which was a 5% increase from Monday’s high temperature. On Wednesday, the high temperature was 10% lower than it was on Monday. How much lower was the high temperature on Wednesday than the high temperature on Monday?

A4.0° F
B8.0° F
C8.4° F
D8.8° F

Show the answer

Correct answer: B

The temperature on Tuesday was 84° F, which was 5% higher than the temperature on Monday.

This can be represented by the equation $84=1.05x$ .

Dividing both sides of this equation by 1.05 gives $x=80$ .

Therefore, the temperature on Monday was 80° F.

Since the temperature on Wednesday was 10% lower than the temperature on Monday, the difference between the two temperatures would equal $(10\%)(80)=8.0^{\circ}\text{ F}$ .

Question analysis

Tuesday's high temperature is given along with its percent increase over Monday's, and Wednesday's high is a percent below Monday's. Students find how many degrees cooler Wednesday was than Monday. It tests percent increase and decrease, and the difficulty is working back to Monday, the base for both percents, rather than starting from Tuesday's number.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually apply a percent to a known amount. Here Monday's temperature must first be recovered from a percent increase and then a second percent applied to it, a reverse step most grade 7 students find unfamiliar.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 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

The question turns on noticing that both percents are measured from Monday, a temperature the stem never states. Students who take 10% of the only temperature given, 84, land on a listed choice just 0.4 degree off. Working back from 84 is forgiving: the exact 84 ÷ 1.05 = 80 and even the rougher 84 × 0.95 = 79.8 lead to the same choice, so the real test is not skipping Monday.

Turning words into math moderate Understanding the question
"84° F, which was a 5% increase from Monday’s high temperature" makes 84 the result of the increase, so Monday comes from solving 1.05x = 84, not from 5% of 84. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
The choices 8.0, 8.4 and 8.8 lie within a degree of each other, so only the base of the 10%, Monday's 80 rather than Tuesday's 84, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
Three days and two percents have to be kept apart: the 5% compares Tuesday with Monday, and the 10% compares Wednesday with Monday. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let Monday's temperature be x, write Tuesday's 84 degrees as 1.05 times x and solve, then take 10 percent of Monday's value.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds Monday's 80° F and then gives the 4-degree rise from Monday to Tuesday, 84 − 80, instead of Wednesday's 10% drop.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes 10% of Tuesday's 84° F, getting 8.4, instead of first working back to Monday's temperature, the base for Wednesday's drop.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Undoes Tuesday's increase in the wrong direction, adding 5% to 84 to get 88.2° F for Monday, then takes 10% of it, about 8.8.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{old} \times \left(1 + \frac{p}{100}\right)$; old is the starting amount. p is the percent it goes up. new is the amount after. The 1 keeps all of the old amount, and p over 100 adds the extra part. So a 5% rise means multiply by 1.05.
Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get, here a number of degrees.

Sort out the question
Monday’s high is unknown. Call it x degrees. Tuesday’s high was 84°F, which is 5% more than Monday. Wednesday’s high was 10% less than Monday.
You need how many degrees lower Wednesday was than Monday. Both percents compare with Monday. So find Monday first.
Write an equation for Tuesday
Tuesday is Monday plus 5% of Monday. That is 105% of Monday. Use the percent increase formula with p = 5.
$84 = x \times \left(1 + \frac{5}{100}\right) = 1.05x$
Solve for Monday
Divide both sides by 1.05 to undo the multiplying.
$x = 84 \div 1.05 = 80$
Monday’s high was 80°F. Watch out: 84 is the result of the increase, not the start. Adding 5% to 84 gives about 88.2°F. Then 10% of that is about 8.8, choice D.
Find Wednesday’s drop
Wednesday was 10% lower than Monday. So the drop is 10% of Monday’s 80°F.
$\text{part} = \frac{10}{100} \times 80 = 8$
Wednesday was 8 degrees lower. Watch out: Taking 10% of Tuesday’s 84 gives 8.4, choice C. The 10% is of Monday’s high. Also, 84 − 80 = 4, choice A, is Tuesday’s rise, not Wednesday’s drop.
So the answer is B, 8.0°F.

Check your answer

Work forward from 80°F. 5% of 80 is 4, and 80 + 4 = 84. That matches Tuesday. Then 80 − 8 = 72, so Wednesday’s high was 72°F. That is 8 degrees below Monday.

More questions like this

The original DOE question

Question 92 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 92. Open page 209 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 38 of 81

2024 Test B Math, question 49

Question

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

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

Show the answer

Correct answer: C

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

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

y can be a positive or a negative number.

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

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

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

Question analysis

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

NYS standards: 7.NS.2a rational number multiplication
6.EE.1 whole-number exponent expressions
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students learn sign rules for products of signed numbers. Here they must reason about four unknown integers under an even power, an odd power and an absolute value, and decide what is always true, which is more abstract than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 58 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One insight decides the question: x² and |w| are positive whatever x and w are, so only the signs of y and z³ matter. Students who count negative letters instead of negative factors pick conditions that look balanced but leave y as the only negative factor. Students who test a single set of values can be fooled by a choice that leaves y and z unspecified.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 106 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 106. Open page 213 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 39 of 81

2024 Test B Math, question 55

Question

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

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

Show the answer

Correct answer: A

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

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

$30{,}000=5x$

$6{,}000=x$

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

$6{,}000=3y$

$2{,}000=y$

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

Question analysis

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

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving a one-variable equation from a word problem is grade 7 work. Two linked multiplicative comparisons, with Saturday measured against Sunday rather than a weekday, add a layer of translation, but the equation itself stays simple.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 12, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 40 of 81

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 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 2024 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)

Question 41 of 81

2023 Test A Math, question 6

Question

$3^{4} + 7^{4} =$

A40
B370
C2,482
D10,000

Show the answer

Correct answer: C

$3^{4} + 7^{4} = 81 + 2{,}401 = 2{,}482$

Question analysis

This exponents question asks for the sum of 3 to the fourth power and 7 to the fourth power. Students evaluate each power separately and then add. The traps are treating an exponent as a multiplier and adding the bases before raising them to a power, and the larger power takes careful multiplication.

NYS standards: 6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Evaluating whole-number powers is grade 6 work, and this question asks only for two of them and a sum. The one demand beyond routine practice is computing a fourth power of 7 accurately without a calculator.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 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

The sum is short, but it tests exactly what an exponent means and when to apply it. The wrong choices come from three misreadings of the notation: treating the power as a multiplier, using one factor too few, and adding the bases before raising them. A size check settles it: 7⁴ alone is 49 × 49, a little under 2,500, so 40 and 370 are far too small and 10,000 far too large.

Error-prone arithmetic moderate Working it out
7⁴ means 7 × 7 × 7 × 7 = 2,401, a four-digit product built by hand, most easily as 49 × 49; 3⁴ = 81 is quick by comparison. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
An exponent counts repeated factors rather than acting as a multiplier, so 3⁴ is 3 × 3 × 3 × 3 = 81, not 3 × 4 = 12; the item assumes that meaning. (Anderson & Krathwohl, 2001)

Strategy: Compute each power on its own, squaring and then squaring again, and add the two results only at the end.

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies each base by its exponent, 3 × 4 + 7 × 4 = 12 + 28 = 40, instead of using each base as a factor four times.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Uses each base as a factor three times instead of four, 27 + 343 = 370, so both powers stop one factor short.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Adds the bases before applying the exponent, treating 3⁴ + 7⁴ as (3 + 7)⁴ = 10,000, though powers come before addition.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A power: $a^{4} = a \times a \times a \times a$; a is the base, the number being multiplied. The small 4 is the exponent. It tells how many times the base is used as a factor, a number being multiplied. So the exponent counts factors. It does not mean multiply by 4.
Powers come before adding: $a^{4} + b^{4} \neq (a + b)^{4}$; a and b are any two positive numbers. The sign ≠ means is not equal to. Powers come before adding. So raise each number to its power first, then add. Adding the bases first gives a much bigger number.

Sort out the question
You need the value of $3^{4} + 7^{4}$ . Each number has a small 4 above it. This 4 is an exponent. Work out each power on its own. Then add the two results.
Find 3 to the fourth power
3⁴ means four 3s multiplied together. Pair them up to make it easier.
$3^{4} = 3 \times 3 \times 3 \times 3 = 9 \times 9 = 81$
Watch out: 3⁴ is not 3 × 4 = 12. Doing that to both powers gives 12 + 28 = 40, choice A.
Find 7 to the fourth power
7⁴ means four 7s multiplied together. Pair them up the same way.
$7^{4} = 7 \times 7 \times 7 \times 7 = 49 \times 49 = 2{,}401$
To find 49 × 49, do 49 × 50 = 2,450. Then take away one 49. That gives 2,401. Watch out: count four 7s, not three. Three 7s make 343. Using three factors in both powers gives 370, choice B.
Add the two powers
$81 + 2{,}401 = 2{,}482$
Watch out: don’t add 3 and 7 first. That gives $(3 + 7)^{4} = 10^{4} = 10{,}000$ , choice D. Powers come before adding. So the answer is C, 2,482.

Check your answer

Estimate. 7⁴ is 49 × 49, a little less than 50 × 50 = 2,500. Adding 81 keeps it near 2,500. Only 2,482 is close. 40 and 370 are far too small, and 10,000 is far too big.

More questions like this

The original DOE question

Question 63 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 63. Open page 77 of the handbook (PDF)

Question 42 of 81

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 43 of 81

2023 Test A Math, question 19

Question

$x:35 = 20:28$

For what value of x is the proportion shown above true?

Show the answer

Correct answer: B

Write the ratios in fraction form and solve for x:

$\frac{x}{35} = \frac{20}{28}$	Cross multiply.
$28x = 700$	Apply the multiplicative inverse property; divide both sides by 28.
$x = \frac{700}{28} = 25$

Question analysis

This question gives the proportion x:35 = 20:28 and asks for the value of x that makes it true. Students use equivalent ratios or cross-multiplication. The trap is reasoning by adding or subtracting the difference between terms rather than multiplying, which gives numbers that look reasonable but break the ratio.

NYS standards: 6.RP.3 ratio and rate reasoning
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a missing term in a proportion is grade 6 ratio reasoning, and here the given ratio simplifies to small numbers. It is easier than typical grade 7 proportional reasoning, which usually involves fractions or a real context.
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
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a single proportion with whole numbers, and the only real trap is reasoning by differences instead of by a common factor. Since 35 is 7 more than 28, adding or subtracting 7 to 20 is tempting, but equal ratios scale; they do not shift. Reducing 20:28 to 5:7 shows at once that the scale factor to 35 is 5.

Error-prone arithmetic minor Working it out
Cross-multiplying gives 28x = 700, and 700 ÷ 28 is not a table fact; simplifying 20:28 to 5:7 first turns it into 5 × 5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
27 is only 2 more than 25, so a rough sense that x is a little more than 20 fits both; only the exact scale factor, 35/28 = 5/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Simplify 20:28 to lowest terms first, then scale it up so its second term becomes 35, or cross-multiply and divide by 28.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reasons with differences: since 35 is 7 more than 28, adds 7 to 20, keeping the gap between the terms instead of their ratio.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Scales 20 by 28/35 instead of 35/28, so 20 × 28 ÷ 35 = 16, which turns the ratio of the second terms upside down.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reasons with differences and also goes the wrong way, subtracting 7 from 20, though x must exceed 20 because 35 exceeds 28.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Ratio as a fraction: $a:b = \frac{a}{b}$; a:b is a ratio, read a to b. It compares two amounts by division. You can write it as the fraction a over b. Two equal ratios make a proportion.
Equivalent fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; a over b is a fraction, and k is any number except 0. Multiplying or dividing the top and the bottom by the same number gives an equal fraction.

Sort out the question
$x:35 = 20:28$ is a proportion. A proportion says two ratios are equal. A ratio compares two numbers, like 20 to 28. You need the value of x that makes the two ratios equal.
Write the ratios as fractions
Write each ratio as a fraction. The first number of the ratio goes on top. The second goes on the bottom.
$\frac{x}{35} = \frac{20}{28}$
Simplify the known fraction
20 and 28 both divide by 4. Smaller numbers are easier to work with.
$\frac{20}{28} = \frac{20 \div 4}{28 \div 4} = \frac{5}{7}$
So now $\frac{x}{35} = \frac{5}{7}$ .
Scale up so the bottom is 35
7 times 5 is 35. To keep the fraction equal, multiply the top by 5 too.
$\frac{5}{7} = \frac{5 \times 5}{7 \times 5} = \frac{25}{35}$
So x = 25. Watch out: a ratio is about multiplying, not adding. 35 is 7 more than 28, but adding 7 to 20 gives 27, choice A. Also, x must be more than 20, because 35 is more than 28. So 16 and 13 are too small. So the answer is B, 25.

Check your answer

Cross-multiply, as the test’s own explanation does. $28 \times 25 = 700$ and $35 \times 20 = 700$ . The two products are equal, so the ratios 25:35 and 20:28 are equal.

More questions like this

The original DOE question

Question 76 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 76. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 81

2023 Test A Math, question 20

Question

$\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =$

A−60
B $-\frac{17}{3}$
C $\frac{77}{9}$
D68

Show the answer

Correct answer: C

$\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =$

$\frac{9 + 64 + 4}{(-3)^{2}} = \frac{77}{9}$

Question analysis

This arithmetic question asks for the value of a fraction whose numerator is the sum of the squares of 3, −8 and 2, and whose denominator is the square of 3 − 8 + 2. Students apply the order of operations with negative numbers. Squaring a negative number and adding inside parentheses before squaring are the points to watch.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Squaring negative numbers and simplifying with the order of operations is standard grade 7 practice. Here the fraction bar groups a numerator and a denominator that each involve a negative sign, but the work stays short.
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
Solution: 3 steps
Reading level: Question and choices: about grade 16+, 3 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 is built so that signs decide everything: the top's 3², (−8)² and 2² are all positive, and so is the square of the bottom's −3. Students who keep a negative after squaring, or who cancel the 9 on top against the 9 below, reach the other choices. The result, 77/9, does not reduce, which can make careful students doubt a correct answer.

Error-prone arithmetic moderate Working it out
Two negatives sit inside squares: (−8)² must be +64, and the denominator's 3 − 8 + 2 = −3 must be squared to +9, so every term ends up positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The numerator needs three squares and a sum, the denominator a sum and then a square, and the fraction bar groups each side, so neither part can be split or cancelled. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Simplify the top and the bottom separately: square each number in the numerator, add inside the parentheses before squaring in the denominator, then write the fraction in lowest terms.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
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 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Squares as if the minus sat outside the parentheses, (−8)² as −64, and also cancels the 9 on top against the 9 below, leaving −64 + 4 = −60.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−8)² as −64, so the top is 9 − 64 + 4 = −51 and −51/9 = −17/3.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Cancels the 9 in the numerator, 9 + 64 + 4, against the denominator's 9, leaving 64 + 4 = 68, though only a factor of the whole top can cancel.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Square of a negative: $(-a)^{2} = (-a) \times (-a) = a^{2}$; a is any number. To square means to multiply a number by itself. A negative times a negative is positive. So the square of a negative number is always positive.
The fraction bar groups: $\frac{\text{top}}{\text{bottom}} = (\text{top}) \div (\text{bottom})$; A fraction bar means divide. It also groups: work out the whole top and the whole bottom first, then divide. You can only cancel a number that multiplies the whole top, not one part of a sum.

Sort out the expression
The top, the numerator, is three squares added together. The bottom, the denominator, is a sum in parentheses that is then squared. To square means to multiply a number by itself. Work out the top and the bottom on their own. Then divide.
Work out the top
Square each number first. Then add.
$3^{2} = 9$
$(-8)^{2} = (-8) \times (-8) = 64$
$2^{2} = 4$
$9 + 64 + 4 = 77$
Watch out: (−8)² is +64, not −64. The minus sign is inside the parentheses, so it is squared too. Using −64 makes the top −51 and gives $-\frac{17}{3}$ , choice B.
Work out the bottom
Work inside the parentheses first. Add and subtract before you square.
$3 - 8 + 2 = -5 + 2 = -3$
$(-3)^{2} = (-3) \times (-3) = 9$
The bottom is 9.
Divide and simplify
$\frac{77}{9}$
77 is 7 × 11 and 9 is 3 × 3. They share no factor, so the fraction is in lowest terms. Watch out: don’t cancel the 9 on top with the 9 below. That 9 is only one part of a sum. Canceling it leaves 64 + 4 = 68, choice D. So the answer is C, $\frac{77}{9}$ .

Check your answer

Every square is positive or zero. So the top and the bottom are both positive, and the answer must be positive. That rules out −60 and $-\frac{17}{3}$ . Also, $\frac{77}{9}$ is about 8.6, which is far from 68.

More questions like this

The original DOE question

Question 77 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 77. Open page 80 of the handbook (PDF)

Question 45 of 81

2023 Test A Math, question 22

Question

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

Show the answer

Correct answer: C

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

Option A:

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

Option B:

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

Option C:

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

Option D:

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

Question 46 of 81

2023 Test A Math, question 26

Question

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

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

Show the answer

Correct answer: A

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

Option A:

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

Option B:

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

Option C:

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

Option D:

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

Option A is the only true inequality.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 47 of 81

2023 Test A Math, question 33

Question

If $m = 5$ and $t = -1$ , what is the value of $\frac{6 - 8(2 - t)}{2m + 4(3 - m)}$ ?

A−9
B−1
C1
D9

Show the answer

Correct answer: A

$\frac{6 - 8(2 - t)}{2m + 4(3 - m)} =$

$\frac{6 - 8[2 - (-1)]}{2(5) + 4(3 - 5)} =$

$\frac{6 - 8(3)}{10 + 4(-2)} =$

$\frac{6 - 24}{10 + (-8)} =$

$\frac{-18}{2} = -9$

Question analysis

This algebra question gives m = 5 and t = −1 and asks for the value of a fraction whose numerator and denominator both contain parentheses. Students substitute the values and follow the order of operations. Subtracting a negative number and multiplying by a negative difference make sign mistakes likely, and the four choices differ only in sign and size.

NYS standards: 6.EE.2c evaluating expressions and order of operations
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Substituting values into expressions with negative numbers is everyday grade 7 work. This fraction has a numerator and a denominator to evaluate separately, with a minus sign before a negative and a negative difference, so it is a longer version of a standard exercise.
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
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 8, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a plain evaluation with no story, and every point is lost on signs. Subtracting the negative t, then handling the two negative intermediate results, each invite a dropped minus, and the four choices are built so that each such slip lands on a listed value. Writing −1 in parentheses when substituting, and finishing the top and the bottom before dividing, keeps the signs in view.

Error-prone arithmetic moderate Working it out
The value t = −1 sits inside 2 − t, so the numerator needs 2 − (−1) = 3, and the work then passes through two negatives, 6 − 24 = −18 and 3 − 5 = −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The numerator and the denominator each take a substitution, a parenthesis, a product and a sum, worked separately before one final division, so a slip in either half changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The choices pair each size with each sign, −9 with 9 and −1 with 1, so getting 18 and 2 right still leaves the sign to decide between two choices. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Substitute with parentheses around −1, simplify the numerator and the denominator separately, innermost parentheses first, and divide only at the end, checking each sign as you go.

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

B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Treats 2 − (−1) as 1 instead of 3, so the numerator becomes 6 − 8 = −2 and the fraction −2 over 2 equals −1.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Takes 6 − 24 as 18 and 3 − 5 as 2, subtracting the smaller number from the larger both times, so the fraction is 18 over 10 + 8 = 18, which is 1.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Gets every step right except 6 − 24, which it writes as 18 instead of −18, so the quotient 18 ÷ 2 comes out positive.

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 and b are numbers. Taking away a negative number is the same as adding the positive number. Think of a debt of 1 dollar being taken away: you are 1 dollar better off.
Dividing a negative by a positive: $\frac{-a}{b} = -\frac{a}{b}$; a and b are positive numbers. A negative number divided by a positive number gives a negative answer. The answer is positive only when both signs are the same.

Sort out the question
You know m = 5 and t = −1. You need the value of the big fraction. Work out the top, called the numerator, and the bottom, called the denominator, one at a time. Then divide the top by the bottom.
Put in the numbers
$\frac{6 - 8(2 - t)}{2m + 4(3 - m)} = \frac{6 - 8(2 - (-1))}{2(5) + 4(3 - 5)}$
Put each number in parentheses, so the minus signs stay clear. Work inside the parentheses first, then multiply, then add or subtract.
Work out the top
$a - (-b) = a + b$
$2 - (-1) = 2 + 1 = 3$
$6 - 8(3) = 6 - 24 = -18$
The top is −18. It is negative because 24 is more than 6. Watch out: 2 − (−1) is 3, not 1. Using 1 makes the top −2 and the answer −1, choice B.
Work out the bottom
$2(5) + 4(3 - 5) = 10 + 4(-2) = 10 - 8 = 2$
Here 3 − 5 = −2, because 5 is more than 3. Then 4 times −2 is −8. The bottom is 2.
Divide the top by the bottom
$\frac{-a}{b} = -\frac{a}{b}$
$\frac{-18}{2} = -9$
Watch out: if you lost the minus sign in 6 − 24, you would get 9, choice D. So the answer is A, −9.

Check your answer

Check the sign first. The top, 6 − 24, is clearly negative. The bottom, 10 − 8, is positive. Negative over positive is negative, so only −1 or −9 can work. Without the signs, the top is 18 and the bottom is 2, and 18 ÷ 2 = 9. So −9 fits.

More questions like this

The original DOE question

Question 90 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 90. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 48 of 81

2023 Test A Math, question 38

Question

Which graph represents the solution to $-3x - 7 > -4$ ?

A $Number line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the left, ending in an arrow.$
B $Number line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the left, ending in an arrow.$
C $Number line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the right, ending in an arrow.$
D $Number line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the right, ending in an arrow.$

Show the answer

Correct answer: A

$-3x - 7 > -4$	Apply the additive inverse property; add 7 to both sides of the inequality.
$-3x > 3$	Apply the multiplicative inverse property; divide both sides of the inequality by −3. Note: when dividing an inequality by a negative number, the inequality sign is reversed.
$x < -1$

The solution is that x is less than −1, so choose the graph that starts at −1 and continues to the left.

Question analysis

This inequality question asks which number line graph shows the solution of −3x − 7 > −4. Students solve a two-step inequality whose x term has a negative coefficient, then match the boundary point and the direction of the shading. Handling the inequality symbol at the division step is where most mistakes happen, and each slip has a matching graph.

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 and graphing a two-step inequality is a core grade 7 standard. This one has a negative coefficient on x, the classic grade 7 inequality trap, and four similar graphs test whether the sign and the direction were both handled.
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: Something you must already know moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 7 words.
Corrected from the DOE handbook: A wrong term 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 solving takes two moves, but the second carries the rule the question tests: dividing an inequality by a negative number flips its direction. The four graphs cross two features, boundary at −1 or 1 and shading left or right, so forgetting the flip, dropping a sign, or both each land on a graph. Testing a shaded value such as x = −2 in the original inequality, where 6 − 7 = −1 is greater than −4, confirms the choice.

Something you must already know moderate Working it out
The step that decides the question, reversing the symbol when dividing both sides by −3, is a rule students must bring themselves; −3x > 3 becomes x < −1, not x > −1. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
The graph shaded to the right from an open circle at −1 matches the right graph's boundary exactly and differs only in direction, the result of skipping the reversal. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
Each graph must be read for two things, where its open circle sits, at −1 or 1, and which way the arrow runs, so a boundary alone does not identify the answer. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add 7 to both sides, divide by −3 and reverse the inequality symbol, then match both the boundary point and the shading direction; test one shaded value in the original.

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

B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Reverses the symbol correctly but drops the negative in 3 ÷ (−3), so the boundary lands at 1 with the shading to its left.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Divides both sides by −3 without reversing the symbol, solving the inequality as if it were an equation, so x > −1 and the shading runs right.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Treats −3x as 3x, so it solves 3x > 3 to get x > 1 with no reversal, which puts both the boundary and the direction wrong.

What was corrected from the DOE handbook, and why

Explanation: The handbook’s steps say “add 7 to both sides of the equation” and “divide both sides of the equation by −3” (p. 142, question 95). But −3x − 7 > −4 is an inequality, not an equation, and the step’s own note is the point of the question: dividing an inequality by a negative number reverses the sign. The site now says “both sides of the inequality” in both steps, as the DOE’s later guides word this step (“Subtract 4 from both sides of the inequality” in the 2024–2025 guide). The working and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding to both sides of an inequality: $\text{if } a > b\text{, then } a + c > b + c$; a and b are the two sides, and c is any number. Adding the same amount to both sides keeps the comparison true, like adding the same weight to both pans of a scale.
Dividing an inequality by a negative number: $\text{if } a > b \text{ and } c < 0\text{, then } \frac{a}{c} < \frac{b}{c}$; a and b are the two sides, and c is a negative number. When you divide both sides by a negative number, the greater-than sign flips to less-than. That is because negatives reverse order: 2 is more than 1, but −2 is less than −1.

Sort out the question
An inequality compares two amounts. The sign > means is greater than. You need the graph of every x that makes $-3x - 7 > -4$ true. The graphs differ in two ways: where the open circle sits, at −1 or 1, and which way the shading goes.
Add 7 to both sides
$\text{if } a > b\text{, then } a + c > b + c$
$-3x - 7 + 7 > -4 + 7$
$-3x > 3$
Adding 7 undoes the −7, so the x term is alone on the left.
Divide by −3 and flip the sign
$\text{if } a > b \text{ and } c < 0\text{, then } \frac{a}{c} < \frac{b}{c}$
$\frac{-3x}{-3} < \frac{3}{-3}$
$x < -1$
So x is any number less than −1. Watch out: if you forget to flip the sign, you get x > −1, and that graph is choice C. Also, 3 ÷ (−3) is −1, not 1. Dropping that minus sign leads to B or D.
Match the solution to a graph
Less than −1 means to the left of −1 on the number line. The circle at −1 is open, or empty, because −1 itself does not work. The sign means greater than, not greater than or equal to.
Graph A has an open circle at −1 and is shaded to the left. So the answer is A, the graph shaded left from an open circle at −1.

Check your answer

Test a number on each side of −1. For x = −2: $-3(-2) - 7 = 6 - 7 = -1$ , and −1 is greater than −4, so −2 works. For x = 0: $-3(0) - 7 = -7$ , and −7 is less than −4, so 0 does not work. The shading must go left.

More questions like this

The original DOE question

Question 95 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 95. Open page 85 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 49 of 81

2023 Test A Math, question 40

Question

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

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

Show the answer

Correct answer: A

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 50 of 81

2023 Test A Math, question 55

Question

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

Show the answer

Correct answer: D

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

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

$4n + 6 = 58$

$4n = 52$

$n = 13$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 112 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 112. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 81

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 52 of 81

2023 Test B Math, question 14

Question

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

Show the answer

Correct answer: D

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

$n = 3 + 2x$

$93 = 3 + 2x$

$90 = 2x$

$45 = x$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 53 of 81

2023 Test B Math, question 17

Question

$m \bullet t = 72$

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

A12
B72
C132
D135

Show the answer

Correct answer: C

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

For $t = 1$ :

$1m = 72$

$m = 72$

For $t = 2$ :

$2m = 72$

$m = 36$

For $t = 3$ :

$3m = 72$

$m = 24$

Then, add those three values together:

$72 + 36 + 24 = 132$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 54 of 81

2023 Test B Math, question 31

Question

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

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

Show the answer

Correct answer: B

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

$xy=30$

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

Then:

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 55 of 81

2023 Test B Math, question 36

Question

The side length of square S is shorter than the side length of square T by 1 centimeter. The area of square S is 9 square centimeters. What is the difference, in square centimeters, between the area of square T and the area of square S?

Show the answer

Correct answer: B

Let x represent the side length of square T, in centimeters. Then the side length of square S is $(x-1)$ .

The area of square S is:

$(x-1)(x-1)=9$

$(x-1)^{2}=9$

$x-1=3$

$x=4$

So the side length of square T is 4 cm, and the area of square T is 16 sq cm.

The difference between the areas, in square centimeters, is $16-9=7$ .

Question analysis

Two squares differ in side length by 1 centimeter, and the smaller one has an area of 9 square centimeters. Students find how much greater the larger square's area is. It tests the link between a square's side and its area, and the extra area is easy to miscount when picturing the border that the longer sides add.

NYS standards: 6.G.5 perfect squares and perfect cubes
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The area of a square and recognizing 9 as a perfect square are grade 6 ideas. This takes three short steps with small whole numbers, lighter than grade 7 work on composite figures, though the first step runs backward from area to side.
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: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every number is small, so the work is keeping track of which quantity the question wants: a difference of areas, not a side or a single area. The added area forms an L-shaped strip around square S, and students who picture it as two strips 4 centimeters long count the corner square twice. Finding both areas, 9 and 16, and subtracting avoids reasoning about the strip at all.

Many steps minor Working it out
The work runs from area to side and back: the square root of 9 gives S’s side, adding 1 gives T’s, squaring gives 16, and subtracting gives the difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The phrase “shorter than the side length of square T by 1 centimeter” describes S from T’s point of view, so T’s side is S’s side plus 1, not minus 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the side of square S from its area, add 1 centimeter to get the side of square T, find that square's area, and subtract the smaller area from the larger.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the side of square S, 3 centimeters, from its area of 9 and gives that length instead of a difference between two areas.
C Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Pictures the added area as two 4-by-1 strips along two sides of square S, 8 in all, counting the 1-by-1 corner where they overlap twice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the stem’s two numbers, 9 + 1 = 10, as if a side 1 centimeter longer made the area 1 square centimeter larger, and gives that area.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s \times s$; A is the area, the space inside the square, in square units. s is the length of one side. All four sides of a square are equal, so the area is the side times itself. To go backward from an area, find the number that times itself gives the area.

Sort out the two squares
Square S has an area of 9 square centimeters. Its side is 1 centimeter shorter than the side of square T. You need the area of square T minus the area of square S, in square centimeters.
Find the side of square S
Use the area formula backward. Which number times itself makes 9?
$3 \times 3 = 9$
So each side of square S is 3 centimeters. This number is called the square root of 9.
Find the side of square T
Square T’s side is 1 centimeter longer than square S’s side.
$3 + 1 = 4$
Each side of square T is 4 centimeters.
Find the area of square T
Use the area formula with a side of 4 centimeters.
$A = 4 \times 4 = 16$
The area of square T is 16 square centimeters.
Subtract the areas
Take the smaller area away from the larger one.
$16 - 9 = 7$
The difference is 7 square centimeters.
Watch out: Picturing the new area as two strips, each 4 by 1, gives 8, choice C. But those strips share a 1-by-1 corner, so it gets counted twice. Also, 3 is a side length, not a difference of areas. That is choice A.
So the answer is B, 7 square centimeters.

Check your answer

Draw square T and mark square S in one corner. The extra space is an L shape. It is a 3-by-1 strip, another 3-by-1 strip and a 1-by-1 corner. That is 3 + 3 + 1 = 7 square centimeters. It matches.

More questions like this

The original DOE question

Question 93 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 2024 Admissions, page 201, question 93. Open page 201 of the handbook (PDF)

Question 56 of 81

2023 Test B Math, question 42

Question

1 gallon = 16 cups

1 cup = 8 fluid ounces

If 1 gallon = $2^{x}$ fluid ounces, what is the value of x?

Show the answer

Correct answer: B

If 1 gallon is equal to 16 cups, and 1 cup is equal to 8 fluid ounces, then

$1\text{ gallon}=$

$(16)(8\text{ fluid ounces})=$

$128\text{ fluid ounces}$

$128=2^{7}$

$1\text{ gallon}=2^{7}\text{ fluid ounces}$

So, $x=7$ .

Question analysis

This question gives two facts, 16 cups in a gallon and 8 fluid ounces in a cup, and says a gallon equals 2 to the power x fluid ounces. Students find the exponent x. It combines a unit conversion with writing a number as a power of 2, and the exponent is easy to confuse with a multiplier.

NYS standards: 6.RP.3d converting measurement units
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with given facts is grade 6 work, and this conversion is a single multiplication. The twist is writing the result as a power of 2, which calls for fluency with powers that grade 7 work rarely asks for.
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: Combines ideas minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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 conversion is one multiplication, so the question really tests whether students can name the exponent for 128. Stopping at the cups, where 16 = 2⁴, or treating 2 to the power x as 2 times x gives a listed choice. Writing 16 as 2⁴ and 8 as 2³ and adding the exponents is the quickest route, and doubling from 1 seven times is a sure check.

Combines ideas minor Working it out
A unit conversion, 16 cups × 8 fluid ounces = 128 fluid ounces, has to be joined to exponents by recognizing 128 as a power of 2. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Students must know or build the powers of 2 up to 128, and see that the exponent counts factors of 2, not the number of terms in a list starting at 1. (Anderson & Krathwohl, 2001)

Strategy: Convert 1 gallon to fluid ounces by multiplying the two facts, then keep halving the result, or keep doubling from 2, counting how many factors of 2 it takes.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Writes the 16 cups in a gallon as 2⁴ and stops before multiplying by the 8 fluid ounces in each cup, which adds 3 to the exponent.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Lists the powers of 2 starting at 1, so counts 1, 2, 4, …, 128 as eight numbers and takes 8 as the exponent, though 1 is 2⁰.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats 2 to the power x as 2 times x, so 2x = 128 gives 64, confusing an exponent with a multiplier.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Meaning of an exponent: $2^x = 2 \times 2 \times \cdots \times 2$; The 2 is the base, the number being multiplied. x is the exponent, the small raised number. It tells how many 2s are multiplied together. For example, 2 to the third power means 2 × 2 × 2, which is 8.
Multiplying powers of the same base: $2^a \times 2^b = 2^{a + b}$; a and b are exponents. The first power has a twos multiplied together. The second has b more twos. Multiplying puts all the twos together, so you add the exponents.

Sort out the question
A gallon is 16 cups. A cup is 8 fluid ounces. A gallon is also $2^x$ fluid ounces. You need x, the exponent. It tells how many 2s are multiplied together.
Change a gallon to fluid ounces
Each of the 16 cups holds 8 fluid ounces. So multiply.
$16 \times 8 = 128$
One gallon is 128 fluid ounces.
Write 128 as a power of 2
Now you need $2^x = 128$ . Keep multiplying by 2, and count the 2s.
Number of 2s 1 2 3 4 5 6 7
Product 2 4 8 16 32 64 128
It takes seven 2s to make 128.
$2^7 = 128$
So x = 7.
Give the answer
Watch out: A gallon is 16 cups, and $16 = 2^4$ . Stopping there gives 4, choice A. And $2^x$ does not mean 2 times x. Solving 2 × x = 128 gives 64, choice D.
So the answer is B, 7.

Check your answer

Use the exponent rule. 16 is $2^4$ and 8 is $2^3$ . So $16 \times 8 = 2^4 \times 2^3 = 2^{4 + 3} = 2^7$ . The exponent is 7 again.

More questions like this

The original DOE question

Question 99 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 99. Open page 203 of the handbook (PDF)

Question 57 of 81

2023 Test B Math, question 44

Question

In a certain class, Rosa will take a total of 4 tests. She has already taken 2 of the tests and earned scores of 81 and 83. What is the least possible score Rosa can earn on the third test and still be able to finish the class with an average score of 85 on all 4 tests? (Assume that test scores can range from 0 to 100.)

Show the answer

Correct answer: A

Let x represent Rosa’s score on the third test, and let y represent her score on the fourth test.

$\frac{81+83+x+y}{4}=85$

$164+x+y=340$

$x+y=176$

If Rosa scores 100 on the fourth test, then the least score she can earn on the third test is 76.

Question analysis

Rosa has scores of 81 and 83 on the first two of four tests and wants a mean of 85 on all four. Students find the lowest score she can earn on the third test and still reach that average, with scores capped at 100. Two tests are still unknown, so the problem needs a plan, not a single equation.

NYS standards: 6.SP.5c range, center, and overall pattern
6.EE.8 writing simple inequalities
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 mean problems usually find one missing score. Here two tests remain, and the least possible score depends on how high the other unknown score is allowed to go, a planning step beyond typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 60 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Most students can find that Rosa needs 176 more points, but the question is about the least third score, which only comes from pushing the fourth score to its limit. Splitting the 176 evenly gives a natural-looking 88 that ignores the word least. The parenthesis about the 0 to 100 range is the key information, so it pays to ask what each stated limit is for.

No obvious method moderate Working it out
The least third score depends on how high the fourth can go; only the parenthesis “test scores can range from 0 to 100” tells students to set the fourth test at 100. (Schoenfeld, 1985)
Many steps minor Working it out
The 340 points needed, the 164 already earned, the 176 still needed and the 100 allowed on the last test make four steps, each a chance to lose track. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the target mean by 4 for the total points needed, subtract the two scores so far, then decide how the remaining points can be split so the third score is as low as possible.

Traps

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

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the mean of the two scores Rosa already has, (81 + 83) ÷ 2 = 82, a number that answers nothing the question asks.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the 176 points still needed and splits them evenly, 88 on each test, missing that the word least lets the fourth test carry more.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 90; leaving out the fourth test gives (81 + 83 + x) ÷ 3 = 85 and x = 91, the nearest route, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average: $\text{average} = \frac{\text{sum of the scores}}{\text{number of scores}}$; The sum is all the scores added together. The number of scores is how many there are. The average, or mean, shares the total equally among them. Turned around, the sum equals the average times the number of scores.

Sort out the question
Rosa takes 4 tests. She scored 81 and 83 on the first two. She wants an average of 85 on all 4. You need the lowest score she can get on test 3 and still reach that average. Each score is from 0 to 100.
Find the total points needed
Turn the average formula around. The sum is the average times the number of tests.
$\text{sum of the scores} = 85 \times 4 = 340$
Rosa needs 340 points in all.
Find the points still needed
Subtract the points she already has.
$81 + 83 = 164$
$340 - 164 = 176$
Tests 3 and 4 must add up to 176 points.
Make test 4 as high as possible
To make test 3 as low as it can be, test 4 must carry as many points as it can. The highest score is 100.
$176 - 100 = 76$
So the lowest score on test 3 is 76.
Watch out: Splitting 176 evenly gives 88 on each test, choice C. That works, but it is not the least score. And 82 is just the average of her first two tests, choice B.
So the answer is A, 76.

Check your answer

Add all four scores: 81 + 83 + 76 + 100 = 340. Then 340 ÷ 4 = 85, the average she wants. A score of 75 on test 3 would need 101 on test 4, which is not possible.

More questions like this

The original DOE question

Question 101 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 101. Open page 203 of the handbook (PDF)

Question 58 of 81

2022 Test A Math, question 2

Question

A homeowner has a budget of $135 for the installation of a light fixture. An electrician charges $60 per hour plus a one-time trip charge of $45. What is the greatest amount of time, in hours, the electrician can work on the installation while staying within the homeowner’s budget? Write your answer as a decimal.

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

Show the answer

Correct answer: 1.5

Let x represent the number of hours the electrician can work.

The cost for the number of hours worked is $60 times the number of hours worked, x. The sum of the one-time trip charge ($45) and the cost of the number of hours worked ( $60x$ ) must be less than or equal to $135. Express this relationship as an inequality:

$45 + 60x \leq 135$

Solve the inequality for x:

$45 + 60x \leq 135$

$60x \leq 90$

$x \leq \frac{90}{60} = \frac{9}{6} = \frac{3}{2}$

Written as a decimal, $\frac{3}{2} = 1.5$ .

Question analysis

A homeowner has $135 to spend, and an electrician charges $60 an hour plus a one-time $45 trip charge. This grid-in asks for the greatest number of hours the electrician can work within the budget, written as a decimal. Students write and solve a two-step inequality, and turning a length of time into decimal hours is the last trap.

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 word problem that leads to an inequality of the form px + q ≤ r is the grade 7 standard itself. The numbers are friendly; the only twists are treating the budget as a limit and writing the time in decimal hours.
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: 3 steps
Reading level: Question and choices: about grade 8, 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

The arithmetic is light, $135 − $45 = $90 and then $90 ÷ $60, so the question is decided by the setup and by the form of the entry. The trip charge is paid once, not every hour, and has to come off the budget before dividing. The time works out to an hour and a half, and students who think in hours and minutes may write it the way a clock shows it rather than as a decimal.

Turning words into math moderate Understanding the question
The stem never says “at most”; the limit comes from “a budget of $135”, and the one-time $45 trip charge is added once while the $60 multiplies the hours, giving 45 + 60x ≤ 135. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The grid-in demands a decimal, so an hour and a half must be entered as 1.5, not as 1:30 or 90 minutes, with no choices to confirm the form. (Enright, Morley & Sheehan, 2002)

Strategy: Take the one-time trip charge off the budget first, divide what is left by the hourly charge, and give the time in hours as a decimal rather than in hours and minutes.

Traps

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

2.25 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole $135 budget by $60 an hour, 135 ÷ 60 = 2.25, leaving out the one-time $45 trip charge that must come off first.
1.3 Units or scale (King et al., 2004)
Reveals a misconception Expected pull: moderate
Finds that the $90 left pays for 1 hour 30 minutes and writes that as 1.3, as if minutes were tenths of an hour rather than sixtieths.
90 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Subtracts the trip charge and enters the $90 left for labor, stopping before dividing by the $60 hourly charge to turn dollars into hours.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a one-time charge: $C = f + r \times h$; C is the total cost. f is the one-time charge, paid once no matter how long the job takes. r is the rate, the cost for each hour. h is the number of hours. The hourly part grows with time, but the one-time charge does not.
Keeping an inequality balanced: $\text{if } a \leq b \text{, then } a - c \leq b - c \text{ and } a \div c \leq b \div c$; The sign ≤ says the left side is less than or equal to the right side. a and b are the two sides. c is a positive number. Subtracting c from both sides, or dividing both sides by c, keeps the statement true.

Sort out the question
The homeowner can spend $135 at most. That is the budget. The electrician has a one-time trip charge of $45. One-time means it is paid once, not every hour. On top of that, the electrician charges $60 for each hour. You need the greatest number of hours that keeps the total at $135 or less. The question asks for a decimal.
Write an inequality
Let h be the number of hours. Use the total cost formula with f = 45 and r = 60. The total must be less than or equal to $135. An inequality is like an equation, but it uses a sign such as ≤ instead of =.
$C = f + r \times h$
$45 + 60 \times h \leq 135$
Take off the trip charge
Subtract 45 from both sides. This keeps the inequality balanced.
$45 + 60 \times h - 45 \leq 135 - 45$
$60 \times h \leq 90$
So $90 is left to pay for the hours. Watch out: don’t divide the whole $135 by $60. That leaves out the trip charge, and you would grid 2.25.
Divide by the hourly rate
60 × h means 60 times h. Divide both sides by 60 to get h by itself.
$h \leq \frac{90}{60} = \frac{3}{2} = 1.5$
The electrician can work at most 1.5 hours. Watch out: 1.5 hours is 1 hour 30 minutes, but don’t grid 1.3. Thirty minutes is half an hour, 0.5, not 0.3. Also don’t grid 90. That is dollars, not hours. So you would grid 1.5.

Check your answer

Put 1.5 hours back in: $45 + 60 \times 1.5 = 45 + 90 = 135$ . That is exactly the budget. Now try 2 hours: $45 + 60 \times 2 = 165$ . That is over $135. So 1.5 hours is the most.

More questions like this

The original DOE question

Question 59 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 59. Open page 79 of the handbook (PDF)

Question 59 of 81

2022 Test A Math, question 7

Question

$\frac{x^{2}y^{6}}{xy^{2}}$

What is the value of the expression above if $x = -2$ and $y = 2$ ?

A−64
B−32
C16
D32

Show the answer

Correct answer: B

First, substitute −2 for x and 2 for y.

$\frac{(-2)^{2}(2)^{6}}{(-2)(2)^{2}}$

Then, simplify.

$\frac{(-2)(-2)(2)^{2}(2)^{4}}{(-2)(2)^{2}}$

$\frac{(-2)(2)^{4}}{1}$

$\frac{-2 \times 16}{1}$

$-32$

Question analysis

This question asks for the value of a fraction built from powers of x and y when x is −2 and y is 2. Students evaluate an expression with exponents and a negative number, or simplify it first by cancelling common factors. Keeping track of the negative sign through the squaring and the division is the tricky part.

NYS standards: 6.EE.2c evaluating expressions and order of operations
7.NS.2a rational number multiplication
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students evaluate expressions with negative values and whole-number exponents. This one is a single fraction of powers, routine in method, but raising a negative base to powers and then dividing leaves several chances to lose the sign.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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 routes work: substitute and compute 256 ÷ (−8), or simplify to xy⁴ first and evaluate (−2)(16). Simplifying first is safer, because it leaves a single negative factor to track. Substituting first brings in (−2)², where treating the minus sign as outside the parentheses flips the final sign, and each listed wrong value comes from an exponent or a sign handled wrongly.

Error-prone arithmetic moderate Working it out
Substituting first gives (−2)²(2)⁶ over (−2)(2)², so the work runs through a squared negative, 2⁶ = 64 and a negative divisor, and one factor of −2 decides the sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The quotient rule for powers is not stated; students who know that x²/x = x and y⁶/y² = y⁴ can reduce the expression to xy⁴ before any number goes in. (Anderson & Krathwohl, 2001)

Strategy: Cancel the factors of x and y that the top and bottom share before substituting, so there is less to compute, and track the sign of every factor of −2 through to the end.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
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 slip Expected pull: weak
Drops the square on y in the denominator, dividing 4 × 64 = 256 by (−2)(2) = −4 instead of by (−2)(4) = −8, giving −64.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Cancels x² against x as if x²/x were 1 rather than x, so only y⁶/y² = y⁴ = 16 is left, with no factor of −2.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−2)² as −4, so −4 × 64 = −256 and −256 ÷ (−8) = 32.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A power: $a^{4} = a \times a \times a \times a$; a is the base, the number being multiplied. The small number is the exponent. It says how many times the base is used as a factor, a number being multiplied. So y⁶ means six y’s multiplied together. It does not mean y times 6.
Dividing powers with the same base: $\frac{a^{m}}{a^{n}} = a^{m - n}$; a is the base. m and n are exponents. The top has m factors of a and the bottom has n. Each matching pair cancels, because a ÷ a = 1. So m − n factors are left. A letter with no exponent, like x, has exponent 1.

Sort out the question
You need the value of $\frac{x^{2}y^{6}}{xy^{2}}$ when x = −2 and y = 2. The top means $x^{2} \times y^{6}$ . The bottom means $x \times y^{2}$ . You could put the numbers in right away. But simplifying first is safer. It leaves only one negative number to track.
Simplify the x part
Use the rule for dividing powers. The x on the bottom has exponent 1.
$\frac{x^{2}}{x} = x^{2 - 1} = x^{1} = x$
You can see it by writing it out: $\frac{x \times x}{x}$ leaves one x. Watch out: x² ÷ x is x, not 1. If you cancel every x, you lose the −2 and get 16, choice C.
Simplify the y part
$\frac{y^{6}}{y^{2}} = y^{6 - 2} = y^{4}$
So the whole fraction becomes a simpler expression:
$\frac{x^{2}y^{6}}{xy^{2}} = x \times y^{4}$
Watch out: the bottom has y², not y. Using y leaves y⁵ and leads to −64, choice A.
Put in the numbers
Now substitute x = −2 and y = 2.
$x \times y^{4} = (-2) \times 2^{4}$
$2^{4} = 2 \times 2 \times 2 \times 2 = 16$
$(-2) \times 16 = -32$
A negative times a positive is negative. So the answer is B, −32.

Check your answer

Substitute first, the long way. Top: $(-2)^{2} \times 2^{6} = 4 \times 64 = 256$ . Bottom: $(-2) \times 2^{2} = (-2) \times 4 = -8$ . Then $256 \div (-8) = -32$ . Watch out: (−2)² is +4, since two negatives multiply to a positive. Writing −4 leads to 32, choice D.

More questions like this

The original DOE question

Question 64 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 81, question 64. Open page 81 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 60 of 81

2022 Test A Math, question 13

Question

Jay sold n tickets. Pilar sold three times as many tickets as Jay did. Together, Jay and Pilar sold ten more tickets than Amie sold. Which expression shows how many tickets Amie sold?

A $10 - 4n$
B $3n^{2} + 10$
C $4n + 10$
D $4n - 10$

Show the answer

Correct answer: D

Since Jay sold n tickets and Pilar sold 3 times the number of tickets Jay sold, Pilar sold $3n$ tickets.

Then, since together Jay and Pilar sold 10 more tickets than Amie, subtract 10 from $(n + 3n)$ to get the expression that shows how many tickets Amie sold.

$(n + 3n) - 10$

Then combine like terms.

$4n - 10$

Question analysis

Jay sells n tickets, Pilar sells three times as many, and together they sell ten more tickets than Amie. Students choose the expression in n for the number of tickets Amie sold. It tests writing and combining algebraic expressions from words, and the last sentence needs a careful reading, because it describes Amie only through the other two.

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 one links three people in a chain, and the last sentence compares the pair's total with Amie instead of describing Amie directly, which grade 7 students often misread.
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
Solution: 3 steps
Reading level: Question and choices: about grade 5, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Building the pair's total, n + 3n = 4n, is routine; the question is decided by which way the last comparison runs. Because the pair sold more than Amie, her expression must come out smaller than 4n, yet the most tempting choice adds the 10. Trying n = 5 settles it: the pair sold 20 tickets, so Amie sold 10, and only one expression gives 10.

Turning words into math moderate Understanding the question
“Together, Jay and Pilar sold ten more tickets than Amie sold” compares the pair's 4n with Amie, so her count is 10 less, though the word more suggests adding. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
Every count is written in terms of n, so no ticket total can be checked unless the student substitutes a value such as n = 5 into each expression. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Write Pilar's tickets and the pair's total in terms of n, then test each choice with a small value such as n = 5 against the sentence comparing the pair with Amie.

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.
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).

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Sees that 10 and the pair's 4n must be subtracted but writes them in the wrong order, 10 − 4n, which is negative once n is 3 or more.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Multiplies Jay's n by Pilar's 3n instead of adding them, getting 3n², and then adds the 10 as the word more suggests.
C Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Follows the word more and adds 10 to the pair's 4n, though the pair sold 10 more than Amie, so Amie sold fewer.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms: $a \times n + b \times n = (a + b) \times n$; n is the unknown number. a and b are the numbers that multiply it. Like terms are terms with the same letter. Add the numbers in front: one n plus three n’s is four n’s, just as 1 apple plus 3 apples is 4 apples.
Ten more than: $\text{if } A = B + 10 \text{, then } B = A - 10$; A and B are two amounts. If A is 10 more than B, A is the bigger one. To get the smaller amount, B, take 10 away from A. The words more than tell you which amount is bigger. They don’t always mean add.

Sort out the question
Jay sold n tickets. The letter n stands for a number you don’t know. Pilar sold three times as many as Jay. Jay and Pilar together sold ten more than Amie. You need an expression for Amie’s tickets. An expression is a math phrase with numbers and letters, like 2n + 1.
Write Pilar’s and the pair’s tickets
Three times as many as n is 3n. So Pilar sold 3n tickets. Together, Jay and Pilar sold n + 3n. Combine like terms:
$n + 3n = 1 \times n + 3 \times n = 4n$
So the pair sold 4n tickets.
Turn the last sentence into math
The pair sold ten more than Amie. So the pair has the bigger number. Amie’s tickets plus 10 make the pair’s 4n.
$\text{Amie} + 10 = 4n$
To find Amie’s tickets, take 10 away from 4n.
$\text{Amie} = 4n - 10$
Watch out: the word more tempts you to add 10. That gives $4n + 10$ , choice C. But Amie sold fewer tickets than the pair, not more.
Rule out the other choices
Choice A, $10 - 4n$ , subtracts in the wrong order. It is negative once n is 3 or more. Choice B, $3n^{2} + 10$ , multiplies n by 3n instead of adding them. So the answer is D, $4n - 10$ .

Check your answer

Try n = 5. Jay sold 5 and Pilar sold 15, so together they sold 20. Amie sold 10 fewer, which is 10. Choice D gives $4 \times 5 - 10 = 10$ . That matches. Choice C gives 30, more than the pair sold.

More questions like this

The original DOE question

Question 70 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 82, question 70. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 61 of 81

2022 Test A Math, question 28

Question

The table shows the income and expenses in a bicycle shop on one day.

BICYCLE SHOP TRANSACTIONS

Description	Income	Expense
Paid employees		$\$120 \times 4$
Sold 2 bicycles	$\$400 \times 2$
Sold clothing	$95
Bought new door		$500
Sold bike parts	p

The sum of all transactions was $70. What is the value of p?

A−$275
B−$85
C$70
D$155

Show the answer

Correct answer: D

The sum of all transactions ($70) includes the income data (positive values) and the expense data (negative values). Write an equation to determine the value of p by adding all the income data and subtracting all the expense data:

$400 \bullet 2 + 95 + p - 120 \bullet 4 - 500 = 70$

Simplify the expression to determine the value of p:

$800 + 95 + p - 480 - 500 = 70$

$895 + p - 980 = 70$

$p - 85 = 70$

$p = 155$

Question analysis

A table lists one day of income and expenses at a bicycle shop, including two products to work out and an unknown amount, p, from selling bike parts. Given that all the transactions add up to $70, students find p. They must treat income as positive and expenses as negative, then solve an equation.

NYS standards: 7.EE.3 multi-step rational number problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve multi-step money problems with positive and negative amounts. This table adds an unknown income entry, so students treat income as positive and expenses as negative before solving a one-step equation.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The table turns a one-variable equation into bookkeeping: each entry must be multiplied out and given a sign by its column before anything is added. The known entries come to a loss of $85, so the bike parts must cover that loss and still leave $70, which is why p is larger than the day's total. Students who stop at the running total, or who copy the $70, never use the equation at all.

Turning words into math moderate Understanding the question
“The sum of all transactions was $70” has to be read with income positive and expenses negative, giving 800 + 95 + p − 480 − 500 = 70. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
Two cells hold products, $120 × 4 and $400 × 2, that must be worked out, and p sits in the Income column beside the sales, so it is added. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
The known entries total 895 − 980 = −85, a negative running sum, so p must exceed 70 by 85 rather than fall short of it. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the two products in the table, add the income entries and subtract the expenses, set that total plus p equal to 70, and solve for p.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single mistake gives −275; it takes counting the $95 clothing sale as an expense and then stopping at the known total, 800 − 95 − 480 − 500.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Totals the known entries correctly, 895 − 980 = −85, and gives that, stopping before solving for the p that brings the sum to $70.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Picks $70, the stated sum of all the transactions, as the value of the bike parts, without setting up an equation.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of all transactions: $\text{sum} = \text{income} - \text{expenses}$; A transaction is money coming in or going out. Income is money that comes in, so it counts as plus. Expenses are money that goes out, so they count as minus. The sum is what is left, like money added to and taken from a wallet.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
The table lists money that came into the shop, called income. It also lists money that went out, called expenses. The bike parts brought in p dollars, an amount you don’t know. All the income minus all the expenses is $70. You need p.
Work out the products in the table
Two cells in the table show multiplication.
$120 \times 4 = 480$
$400 \times 2 = 800$
So paying the employees cost $480. The 2 bicycles brought in $800.
Write the equation
The income is $800, $95 and p. The expenses are $480 and $500. p sits in the Income column, so it is added.
$\text{income} - \text{expenses} = 70$
$800 + 95 + p - 480 - 500 = 70$
Combine the known numbers
$800 + 95 = 895 \qquad 480 + 500 = 980$
$895 - 980 = -85$
The known entries come to −85 dollars. That is a loss of $85. Now the equation is shorter:
$p - 85 = 70$
Watch out: −85 is choice B, but it is not p. You still have to solve for p.
Solve for p
Add 85 to both sides. This keeps the equation balanced.
$p - 85 + 85 = 70 + 85$
$p = 155$
The bike parts brought in $155. Watch out: $70 is the total for the day, not the bike parts. Don’t just copy it, choice C. So the answer is D, $155.

Check your answer

Put 155 back in. Income: $800 + 95 + 155 = 1{,}050$ dollars. Expenses: $480 + 500 = 980$ dollars. $1{,}050 - 980 = 70$ . That matches. And p should be more than 70, because the bike parts had to cover the $85 loss first.

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 NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 86, question 85. Open page 86 of the handbook (PDF)

Question 62 of 81

2022 Test B Math, question 12

Question

Julie is at a clothing store. She wants to buy a bag for $18.75 and some T-shirts for $9.25 each. She can spend no more than $50.00. Which inequality can be used to find the number of T-shirts, x, Julie can purchase?

A $9.25 + 18.75x \leq 50.00$
B $9.25 + 18.75x \geq 50.00$
C $18.75 + 9.25x \leq 50.00$
D $18.75 + 9.25x \geq 50.00$

Show the answer

Correct answer: C

The question uses the phrase “no more than,” which translates to “less than or equal to” (≤).

Since T-shirts cost $9.25 each and Julie bought x T-shirts, the total cost just for the T-shirts is found by multiplying $9.25 by x, or $9.25x$ .

Combining the cost for one bag ($18.75) plus the cost for x T-shirts, the inequality is $18.75 + 9.25x \leq 50.00$ .

Question analysis

Julie wants to buy one bag and some T-shirts at a clothing store, with a price for each, and can spend no more than $50.00. Students choose the inequality for the number of T-shirts, x, she can buy. It tests writing an inequality from a word problem: the choices swap the two prices and the direction of the sign.

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 one fixed cost and a cost per item is standard grade 7 work. With choices given, the task is to match each price to its role and the phrase to its symbol.
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
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The choices form a grid of two decisions: which price goes with x, and which way the symbol points. The prices appear in the order they are used, bag first, so the expression is the easier half and most lost points come from the symbol. Pricing the bag with 2 T-shirts, $37.25, and with 3, $46.50, shows that 9.25 is the amount that repeats with each shirt.

Turning words into math moderate Understanding the question
The $18.75 bag is bought once and the $9.25 is paid for each of x T-shirts, and “no more than $50.00” has to become ≤, which allows spending exactly $50.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choices 18.75 + 9.25x ≤ 50.00 and 18.75 + 9.25x ≥ 50.00 match term for term and differ only in the direction of the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Price one bag and 2 T-shirts by hand to see which price is multiplied by x, then translate “no more than” into an inequality symbol.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Keeps the right symbol but multiplies the number of T-shirts by $18.75 and adds $9.25 once, putting the bag’s price where the price per shirt belongs.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Puts x on the $18.75 bag and also turns “no more than $50.00” into ≥, so both the expression and the symbol are wrong.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Builds the total, 18.75 + 9.25x, correctly but reads “no more than $50.00” as at least, so ≥ would make Julie spend $50.00 or more.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost: $\text{total} = \text{one-time cost} + \text{price each} \times \text{number bought}$; The one-time cost is paid once, like one bag. The price each is paid again for every item, so it is multiplied by the number bought. Add them to get the total. Use it when one cost happens once and another repeats.
Spending limit: $\text{total} \leq \text{limit}$; The total is what Julie spends, and the limit is $50.00. The sign ≤ means less than or equal to. So the total can be under the limit or exactly equal to it, but never over. That is what the words “no more than” mean.

Sort out the question
Julie buys one bag for $18.75. She also buys some T-shirts for $9.25 each. x is the number of T-shirts. She can spend no more than $50.00. You need the right inequality. An inequality is a math sentence that uses a sign like ≤ instead of =.
Write the cost of the T-shirts
Each T-shirt costs $9.25. For x T-shirts, she pays $9.25 x times.
$9.25 \times x = 9.25x$
The bag is bought only once. So its $18.75 is not multiplied by x.
Watch out: putting x on 18.75 prices x bags and one T-shirt. That gives choices A and B.
Add to get the total
Use the total cost formula. The bag is the one-time cost.
$\text{total} = 18.75 + 9.25x$
Test it with 2 T-shirts: $18.75 + 9.25 \times 2 = 18.75 + 18.50 = 37.25$ dollars. That is one bag and two shirts, as it should be.
Choose the sign
“She can spend no more than $50.00” means she can spend less than $50.00, or exactly $50.00. That is less than or equal to, the sign ≤.
$18.75 + 9.25x \leq 50.00$
Watch out: ≥ means greater than or equal to. Choice D says she must spend $50.00 or more, the opposite. So the answer is C, $18.75 + 9.25x \leq 50.00$ .

Check your answer

Test whole numbers of T-shirts. With 3, the total is 18.75 + 27.75 = 46.50 dollars. That is not over $50.00. With 4, it is 18.75 + 37.00 = 55.75 dollars, too much. Choice C allows 3 T-shirts and rules out 4, which fits.

More questions like this

The original DOE question

Question 69 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 200, question 69. Open page 200 of the handbook (PDF)

Question 63 of 81

2022 Test B Math, question 26

Question

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

A $-\frac{2}{115}$
B $-\frac{1}{45}$
C $\frac{2}{9}$
D $\frac{1}{5}$

Show the answer

Correct answer: D

First, simplify the numerator and the denominator separately.

To simplify the numerator, multiply the last two fractions and then subtract from the first fraction.

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

$\frac{2}{3} - \frac{4}{15}$

$\frac{10}{15} - \frac{4}{15}$

$\frac{6}{15} = \frac{2}{5}$

To simplify the denominator, divide the last two fractions and then add to the first fraction.

$\frac{5}{3} + \left(\frac{1}{4} \div \frac{3}{4}\right)$

$\frac{5}{3} + \left(\frac{1}{4} \times \frac{4}{3}\right)$

$\frac{5}{3} + \frac{1}{3}$

$\frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$

Finally, divide the numerator by the denominator.

$\frac{\frac{2}{5}}{2} = \frac{2}{5} \div 2 = \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}$

Question analysis

This fraction arithmetic question gives a complex fraction: a numerator with a subtraction and a multiplication over a denominator with an addition and a division, all with fractions. Students choose its value. It tests order of operations and the four operations with fractions, and working left to right instead of following the order of operations leads to a wrong result.

NYS standards: 7.NS.3 four operations with rational numbers
6.NS.1 dividing fractions by fractions
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students compute with fractions, usually two or three operations at a time. This expression needs five, in an order set by the rules rather than read left to right, so it is longer and more trap-prone than typical work.
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: Many steps moderate
Error-prone arithmetic moderate
Something you must already know minor
Solution: 5 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

Nothing here is beyond grade 7, but the expression stacks five fraction operations, and the top and the bottom each hide an order-of-operations decision. Working left to right in the numerator alone gives a negative value, and doing so in both parts gives an even stranger fraction; both are listed. The sign is a quick check: 4/15 is less than 2/3, so the numerator is positive, and so is the value.

Many steps moderate Working it out
The numerator and the denominator are each simplified on their own, to 2/5 and 2, before the final division, so five operations stand between the expression and its value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Every step is fraction work: 2/3 − 4/15 needs fifteenths, 1/4 ÷ 3/4 needs the divisor flipped, and the last step, 2/5 ÷ 2, is easily turned into 2/5 × 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
No parentheses inside the numerator or the denominator show the order, so students must know to multiply before subtracting and to divide before adding. (Anderson & Krathwohl, 2001)

Strategy: Simplify the numerator and the denominator separately, doing the multiplication and the division before the subtraction and the addition, then divide the two results.

Traps

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

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Works left to right in both parts, (2/3 − 4/5) × 1/3 = −2/45 over (5/3 + 1/4) ÷ 3/4 = 23/9, and −2/45 ÷ 23/9 = −2/115.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Subtracts before multiplying in the numerator, (2/3 − 4/5) × 1/3 = −2/45, but handles the denominator correctly, so −2/45 ÷ 2 = −1/45.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/9: it is positive like the answer, but no single order or fraction slip reaches it, so it mostly draws guesses between the two positive choices.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\times \text{ and } \div \text{ before } + \text{ and } -$; When no parentheses show the order, multiply and divide first. Then add and subtract. Everyone follows the same order, so an expression has only one value.
Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the top numbers, and b and d are the bottom numbers. Multiply top times top and bottom times bottom.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. The reciprocal of c/d is d/c. It works because c/d times d/c is 1. So multiplying by d/c undoes c/d, just as dividing by c/d does.

Sort out the question
The big fraction bar means divide: the top part divided by the bottom part. Work out the top and the bottom on their own. Then divide.
Inside each part, no parentheses show the order. So follow the order of operations.
Multiply first in the top
The top is $\frac{2}{3} - \frac{4}{5} \times \frac{1}{3}$ . Multiply before you subtract.
$\frac{4}{5} \times \frac{1}{3} = \frac{4 \times 1}{5 \times 3} = \frac{4}{15}$
Watch out: subtracting first, 2/3 − 4/5, makes the top negative. That leads to −1/45, choice B. Working left to right in the bottom as well gives −2/115, choice A.
Subtract to finish the top
To subtract, the bottoms must match. Change 2/3 to fifteenths by multiplying its top and bottom by 5.
$\frac{2}{3} - \frac{4}{15} = \frac{10}{15} - \frac{4}{15} = \frac{6}{15} = \frac{2}{5}$
To simplify 6/15, divide its top and bottom by 3. The top part is 2/5.
Divide first in the bottom
The bottom is $\frac{5}{3} + \frac{1}{4} \div \frac{3}{4}$ . Divide before you add. Flip 3/4 and multiply.
$\frac{1}{4} \div \frac{3}{4} = \frac{1}{4} \times \frac{4}{3} = \frac{4}{12} = \frac{1}{3}$
To simplify 4/12, divide its top and bottom by 4.
Add to finish the bottom
The bottoms already match, so add the tops and keep the bottom.
$\frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$
Six thirds make 2 wholes. The bottom part is 2.
Divide the top by the bottom
Now divide 2/5 by 2. Flip 2, which is 2/1, to get 1/2.
$\frac{2}{5} \div 2 = \frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = \frac{1}{5}$
Dividing by 2 is the same as taking half, and half of 2/5 is 1/5. So the answer is D, $\frac{1}{5}$ .

Check your answer

Look at the signs. 4/15 is less than 2/3, so the top is positive. The bottom is positive too. So the answer must be positive, which rules out A and B. Then multiply back: $\frac{1}{5} \times 2 = \frac{2}{5}$ , the top. It matches.

More questions like this

The original DOE question

Question 83 as printed on page 204 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 204, question 83. Open page 204 of the handbook (PDF)

Question 64 of 81

2018 Test A Math, question 6

Question

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

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

Show the answer

Correct answer: B

$2k=m+3$

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

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

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 65 of 81

2018 Test A Math, question 14

Question

Which number line below shows the solution to the inequality $-4< \frac{x}{2}< 2$ ?

A $Number line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −2 and 1, with the segment between them shaded.$
B $Number line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −4 and 2, with the segment between them shaded.$
C $Number line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the line shaded to the left of −8 and to the right of 4.$
D $Number line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the segment between them shaded.$

Show the answer

Correct answer: D

Multiply each term by 2 to eliminate the fraction, and isolate $x$ :

$-4(2)< \left(\frac{x}{2}\right)(2)< 2(2)$

$-8< x< 4$

Therefore, $x$ must be between −8 and 4.

Question analysis

An inequality question: choose the number line that shows the solution of a compound inequality in which x divided by 2 lies between −4 and 2. Students undo the division in every part and match the graph's endpoints and shading. The four number lines differ in their endpoints and in which part of the line is shaded.

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 Grade 7 students solve two-step inequalities and graph them on a number line. This compound inequality needs only one operation applied to all three parts at once, so it matches typical grade 7 work.
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: Reading a figure, table or graph moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 11, 11 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Solving takes one step, multiplying through by 2, so most of the work is matching the result to a picture. The choices pair each common slip with a graph: the bounds left as written, the bounds halved, and the right bounds shaded on the wrong side. Reading the open circles against the labeled ticks at −5 and 5 settles it.

Reading a figure, table or graph moderate Understanding the question
The four number lines are labeled only at −10, −5, 0, 5 and 10, so open circles at −8 and 4, or at −4 and 2, have to be located by counting ticks. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
All three parts of −4 < x/2 < 2 must be multiplied by 2, and the negative bound −4 becomes −8; halving instead gives −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply all three parts of the inequality by 2 to isolate x, then pick the graph whose open circles and shading match the new bounds.

Traps

Stops one step early: A choice is a number from partway through the solution.
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 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides each part of the inequality by 2 instead of multiplying, getting −2 < x < 1, so the open circles land at −2 and 1.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Graphs −4 and 2 as the bounds of x, stopping before the step of multiplying through by 2 to clear the fraction x/2.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Finds −8 and 4 correctly but shades the two rays outside them, as if x had to be less than −8 or greater than 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying an inequality by a positive number: $\text{if } a < b \text{, then } a \times 2 < b \times 2$; a and b are numbers, and the less-than sign means the left side is smaller. If you multiply both sides by the same positive number, like 2, the smaller side stays smaller. So the sign keeps pointing the same way. A negative number would flip it, but 2 is positive.

Sort out the inequality
An inequality compares two amounts that may not be equal. The sign $<$ means less than. Here $\frac{x}{2}$ means x divided by 2. The inequality says x divided by 2 is between −4 and 2. You need the values of x itself. Then you pick the number line that shows them.
Multiply all three parts by 2
To undo dividing by 2, multiply by 2. Multiply all three parts, so the whole statement stays true.
$-4 \times 2 < \frac{x}{2} \times 2 < 2 \times 2$
$-8 < x < 4$
Watch out: multiply, don’t divide. Dividing gives −2 and 1, choice A. Leaving the ends at −4 and 2 is choice B.
Read what the answer means
x is greater than −8 and less than 4. So x is any number between −8 and 4. x cannot equal −8 or 4 exactly. On a number line, that means open circles at −8 and 4, with the part between them shaded.
Find the matching number line
The number lines are labeled only at −10, −5, 0, 5 and 10. Count the tick marks. −8 is 3 ticks left of −5, and 4 is 1 tick left of 5. Watch out: choice C has circles at −8 and 4 but shades the parts outside them. So the answer is D.

Check your answer

Test a number from the shaded part, like 0. $0 \div 2 = 0$ , and 0 is between −4 and 2. Now test 6, which is outside. $6 \div 2 = 3$ , and 3 is not less than 2. So only the part between −8 and 4 works.

More questions like this

The original DOE question

Question 71 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 71. Open page 84 of the handbook (PDF)

Question 66 of 81

2018 Test A Math, question 33

Question

Johan leased a car for three years. He paid a one-time fee of $1,000, and an additional $300 per month for the full three years. At the end of the three years, what is the total amount Johan paid for leasing this car?

A$1,900
B$4,600
C$10,800
D$11,800

Show the answer

Correct answer: D

Three years is 36 months (12 × 3).

Set up an expression to find the total amount Johan paid:

$1{,}000+300(36)=\$11{,}800$

Question analysis

This word problem about leasing a car combines a one-time fee with a monthly payment over three years and asks for the total amount paid. Students must convert years to months, multiply by the monthly payment and add the fee. Leaving out one piece of the cost, or mixing years with months, is the usual slip.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 This is a short total-cost calculation with whole numbers, easier than typical grade 7 word problems, which bring in percents, negative numbers or fractions. The only care needed is turning years into months.
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: Much to keep track of minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is the easiest kind of multi-step word problem: whole numbers, a familiar money situation and a direct method. The only places to slip are changing three years into 36 months and remembering the one-time fee after the long multiplication. Each slip has its own choice, so a quick check that both costs are in the total is worth the seconds.

Much to keep track of minor Working it out
The $1,000 fee is paid once while the $300 repeats monthly, so it must be held aside while 36 × 300 is computed. (Sweller, 1988; Carpenter, Just & Shell, 1990)
A very close wrong answer minor Choosing the answer
The choices $11,800 and $10,800 differ by only $1,000 out of about $11,000, so a rough total cannot separate them; only including the one-time fee does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Count the months in the lease first, multiply the monthly payment by that number, and then add the one-time fee.

Traps

Stops one step early: A choice is a number from partway through the solution.
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 slip Expected pull: moderate
Multiplies $300 by 3 years instead of 36 months, getting $900, then adds the $1,000 fee to reach $1,900.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Charges the $300 for only 12 months, one year rather than the full three, so 3,600 + 1,000 gives $4,600.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 36 monthly payments, 36 × $300 = $10,800, and stops before adding the one-time $1,000 fee.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee: $T = f + m \times n$; T is the total paid. f is the one-time fee, paid only once. m is the monthly payment. n is the number of months. You multiply first because the monthly payment repeats every month. Then you add the fee once.

Sort out what you know
Johan paid $1,000 one time. He also paid $300 every month for 3 years. You need the total he paid.
The payments are counted by the month, but the time is given in years. So first change the years to months.
Count the months
Each year has 12 months:
$3 \times 12 = 36$
So Johan made 36 monthly payments. Watch out: multiplying $300 by 3 years instead of 36 months gives $900. Adding the fee then gives $1,900, which is choice A.
Add up the monthly payments
Multiply the monthly payment by the number of months:
$300 \times 36 = 10{,}800$
The monthly payments add up to $10,800.
Add the one-time fee
Use the formula. The fee goes in once:
$T = 1{,}000 + 300 \times 36 = 1{,}000 + 10{,}800 = 11{,}800$
Watch out: $10,800 is choice C. It leaves out the $1,000 fee. So the answer is D, $11,800.

Check your answer

Round to check. $300 a month for 36 months is a bit more than $10,000. Add the $1,000 fee and you get a bit more than $11,000. $11,800 fits. Choices A and B are far too small for three years of payments.

More questions like this

The original DOE question

Question 90 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 90. Open page 87 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 67 of 81

2018 Test A Math, question 43

Question

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

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

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

Show the answer

Correct answer: D

Solve the equation for $z$ :

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

$wz=xy$

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 68 of 81

2018 Test A Math, question 47

Question

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

Show the answer

Correct answer: B

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

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

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

Question analysis

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

NYS standards: 6.EE.2b parts of an expression
6.EE.6 variables representing unknown numbers
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students work with expressions like 3n + 5 but rarely reason about whether they are even or odd. The idea is simple once the integers are written out, so the demand is about typical for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 12, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 69 of 81

2018 Test A Math, question 50

Question

$100\left ( 2+0.1 \right )^{2}-100=$

A101
B200
C301
D341

Show the answer

Correct answer: D

$100(2+0.1)^2-100=$

$100(2.1)^2-100=100(4.41)-100=441-100=341$

Question analysis

This order of operations question asks students to evaluate 100(2 + 0.1)² − 100, an expression with a decimal, a square and a subtraction. Students must add inside the parentheses first, square the result, multiply by 100, and then subtract. Squaring the two parts of the sum separately is the trap.

NYS standards: 6.EE.1 whole-number exponent expressions
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Evaluating an expression with an exponent and decimals in the correct order is grade 6 content. This is a short routine calculation, so it asks less than typical grade 7 work with negative numbers and multi-step equations.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 3 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

This is a direct order-of-operations calculation with no word problem around it, so the only real risk is the square of a sum. Students who apply the exponent to each part, or only to the 0.1, find a listed answer waiting. Adding inside the parentheses before squaring, and then squaring 2.1 exactly, is the whole task.

A very close wrong answer moderate Choosing the answer
The choices 301 and 341 differ by 40, and squaring 2.1 roughly as 4 lands nearer 301; only the exact square, 4.41, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Squaring 2.1 by hand to get 4.41 puts two decimal places in the product, an easy place to misplace the point. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add inside the parentheses to get a single decimal, square it carefully, then multiply by 100 and subtract 100, following the order of operations.

Traps

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

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Squares only the 0.1, as if the parentheses were not there, so 100 × 2.01 − 100 gives 101.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single slip gives 200; it requires (2 + 0.1)² to come out as 3, so it mostly catches guesses.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Squares each term of the sum separately, 4 + 0.01 = 4.01, instead of squaring 2.1, so 100 × 4.01 − 100 gives 301.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{parentheses} \rightarrow \text{exponents} \rightarrow \times, \div \rightarrow +, -$; This is the agreed order for doing math steps. Work inside parentheses first, then exponents, then multiply and divide, then add and subtract. Everyone follows it, so a problem has only one answer.
Squaring a number: $a^2 = a \times a$; a is any number. The small 2, called an exponent, means multiply the number by itself. So 2.1 squared means 2.1 times 2.1.

Sort out the problem
You need to work out $100(2 + 0.1)^2 - 100$ . The 100 right next to the parentheses means 100 times. Follow the order of operations, one step at a time.
Add inside the parentheses
Parentheses come first in the order of operations. So add the numbers inside them:
$2 + 0.1 = 2.1$
Now the problem is:
$100 \times 2.1^2 - 100$
Square 2.1
$2.1^2 = 2.1 \times 2.1 = 4.41$
Work out 21 × 21 = 441, then put in two decimal places, one for each 2.1.
Watch out: do not square the 2 and the 0.1 one at a time. That gives 4 + 0.01 = 4.01, which is wrong. It leads to 301, choice C.
Multiply, then subtract
The exponent is done, so multiply next. Multiplying by 100 moves the decimal point two places to the right. Subtracting comes last:
$100 \times 4.41 = 441$
$441 - 100 = 341$
So the answer is D, 341.

Check your answer

Estimate. 2.1 is a bit more than 2, so its square is a bit more than 4. 100 times that is a bit more than 400. Take away 100 and you get a bit more than 300. 341 fits. Choices A and B are too small.

More questions like this

The original DOE question

Question 107 as printed on page 90 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 107. Open page 90 of the handbook (PDF)

Question 70 of 81

2018 Test A Math, question 52

Question

Each week, Leon has fixed expenses of $1,250 at his furniture shop. It costs him $150 to make a chair in his shop, and he sells each chair for $275. What is Leon’s profit if he makes and sells 25 chairs in 1 week?

A$1,875
B$2,500
C$3,125
D$4,375

Show the answer

Correct answer: A

Each chair costs Leon $150 to make, and he sells the chair for $275.

His profit is found by subtracting the cost from the price:

$275 − $150 = $125 per chair

If Leon makes and sells 25 chairs in a week, his initial profit is

25 × $125 = $3,125

However, Leon has additional fixed expenses of $1,250 per week, so this cost must also be subtracted to arrive at the profit. His final profit is

$3,125 − $1,250 = $1,875

Question analysis

This profit word problem gives Leon's weekly fixed expenses at his furniture shop, the cost to make each chair and its selling price, then asks his profit for 25 chairs in one week. Students must find the profit on each chair, multiply, and account for the fixed expenses. Forgetting the fixed expenses, or adding them, are the main slips.

NYS standards: 7.EE.3 multi-step rational number problems
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multi-step money problems are standard grade 7 work. This one has three whole-number steps and a familiar profit idea, so it matches typical grade 7 demand, with the fixed cost as the detail to remember.
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: Much to keep track of moderate
Something you must already know 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 arithmetic is whole-number and friendly, so the question is about bookkeeping: which amounts are income, which are costs, and which happen once a week. The fixed expenses are named first in the stem and are easy to forget by the time the per-chair profit is multiplied. A final check that revenue minus all costs, 6,875 − 3,750 − 1,250, gives the same result confirms it.

Much to keep track of moderate Working it out
Three dollar amounts play different roles, a weekly fixed expense, a cost per chair and a price per chair, and each must enter the profit with the right sign. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The bold "profit" is not defined; students must know it means money taken in minus every cost, including costs that do not depend on the number of chairs. (Anderson & Krathwohl, 2001)

Strategy: Subtract the cost from the price to get the profit per chair, multiply by the number of chairs, then subtract the week's fixed expenses.

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).

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the $150 cost by 25 chairs and subtracts the $1,250, treating the cost of making a chair as money earned.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the profit on 25 chairs at $125 each, $3,125, and stops before subtracting the $1,250 fixed expenses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds the $1,250 fixed expenses to the $3,125 chair profit instead of subtracting them, treating an expense as income.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Profit: $P = I - C$; P is the profit, the money Leon keeps. I is the income, the money he takes in from sales. C is the total cost: every dollar he spends, including fixed expenses. Profit is what is left after all costs are paid.

Sort out what you know
Leon pays $1,250 each week no matter what. These are fixed expenses, like rent. Each chair costs $150 to make, and he sells each one for $275.
He makes and sells 25 chairs. Profit is the money left after paying all the costs.
Find the profit on one chair
Each chair sells for $275 but costs $150 to make. Subtract to see what he keeps:
$275 - 150 = 125$
He keeps $125 on each chair he sells.
Find the profit on 25 chairs
$25 \times 125 = 3{,}125$
The chairs bring in $3,125 more than they cost to make. Watch out: $3,125 is choice C, but it is not the final profit. The fixed expenses still have to be paid.
Subtract the fixed expenses
$P = 3{,}125 - 1{,}250 = 1{,}875$
Watch out: the $1,250 is money spent, so subtract it. Adding it gives $4,375, choice D. So the answer is A, $1,875.

Check your answer

Work it the long way. Sales: 25 × $275 = $6,875. Costs: 25 × $150 = $3,750, plus $1,250, which makes $5,000. Profit: $6,875 − $5,000 = $1,875. Same answer.

More questions like this

The original DOE question

Question 109 as printed on page 91 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 109. Open page 91 of the handbook (PDF)

Question 71 of 81

2018 Test B Math, question 6

Question

If $x=9$ and $y=-7$ , what is the value of $x(x-2y)$ ?

A18
B45
C144
D207

Show the answer

Correct answer: D

$x(x-2y)=9[9-2(-7)]=9(9+14)=9(23)=207$

Question analysis

This algebra question asks students to evaluate the expression x(x − 2y) when x is 9 and y is negative 7. It practices substitution and the order of operations with integers. The risk is the sign, since subtracting twice a negative number turns into addition, and dropping that sign or the factor of 2 is easy.

NYS standards: 6.EE.2c evaluating expressions and order of operations
7.NS.2a rational number multiplication
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Substituting values, including a negative one, into an expression is everyday grade 7 work with integers. This has three short steps and one sign decision, matching a typical classroom exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 8, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Substitution itself is routine, so the question tests one thing: subtracting a negative inside the parentheses. A student who turns 9 − 2(−7) into 9 − 14 gets −45 and may then choose 45, since it is the only choice that looks close. Working the parentheses first and writing −(−14) as +14 before multiplying by 9 avoids every wrong choice.

Error-prone arithmetic moderate Working it out
The middle step is 9 − 2(−7): the product 2(−7) is −14, and subtracting it means adding 14, so two negatives must be handled in one short move. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work inside the parentheses first: find 2y, subtract it from x by rewriting minus a negative as plus, and multiply that result by x last.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Drops the factor 2 and adds y instead of subtracting it, working out 9(9 + (−7)) = 9 × 2 = 18.
B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Writes 9 − 2(−7) as 9 − 14 = −5, gets 9 × (−5) = −45, and picks 45 as the nearest choice.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Handles the negative correctly but drops the coefficient 2, computing 9 − (−7) = 16 and then 9 × 16 = 144.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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, 9 − (−14) = 9 + 14. Think of taking away a debt: you end up richer.
Multiplying by a negative: $\text{positive} \times \text{negative} = \text{negative}$; When one number is positive and the other is negative, their product is negative. The product is the answer to a multiplication. So 2 × (−7) = −14.

Write down what you know
You know x = 9 and y = −7. You need the value of $x(x - 2y)$ .
A number written right next to parentheses means multiply. So this is x times the amount inside the parentheses.
Put in the numbers
Replace each letter with its value. Put −7 in parentheses so you do not lose its minus sign.
$x(x - 2y) = 9 \times (9 - 2 \times (-7))$
Work inside the parentheses
Always work inside parentheses first. Multiply before you subtract. A positive times a negative is negative, so $2 \times (-7) = -14$ .
$9 - (-14) = 9 + 14 = 23$
Taking away a negative is the same as adding.
Watch out: If you write 9 − 14 = −5, you end with −45. Then choice B, 45, looks close, but it is wrong.
Multiply by x last
Now multiply 9 by the 23 you found inside the parentheses.
$9 \times 23 = 207$
Watch out: Do not drop the 2. Using 9 − (−7) = 16 gives 144, choice C.
So the answer is D, 207.

Check your answer

Check the inside another way. $9 - 2 \times (-7)$ means start at 9 and move 14 to the right on the number line. You land on 23. Then $9 \times 23 = 180 + 27 = 207$ .

More questions like this

The original DOE question

Question 63 as printed on page 154 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 63. Open page 154 of the handbook (PDF)

Question 72 of 81

2018 Test B Math, question 8

Question

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

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

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

Show the answer

Correct answer: D

Break the equations apart to each equal $M$ :

$M=3N$

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

$M=Q+5$

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

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

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

$M=3N$

$2=3N$

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

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

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

$8=P$

$M=Q+5$

$2=Q+5$

$-3=Q$

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

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

$14=R$

Variable $R$ has the greatest value.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 73 of 81

2018 Test B Math, question 19

Question

How many positive odd numbers satisfy the inequality $3x+8\leq 92$ ?

Show the answer

Correct answer: B

Solve the inequality for x.

3x + 8 ≤ 92

3x ≤ 84

x ≤ 28

The positive odd numbers less than or equal to 28 are 1, 3, 5, …, 25, and 27. There are 14 of them.

Question analysis

This question asks how many positive odd numbers satisfy the two-step inequality 3x + 8 ≤ 92. Students solve the inequality and then count the odd whole numbers in the solution. The counting after the algebra is where errors creep in, especially at the end of the list.

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: At grade 7 Solving a two-step inequality is grade 7 work. Asking for a count of the odd solutions adds a number-property step and an endpoint check, but the total demand stays close to a typical grade 7 problem.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 16, 9 words.
Corrected from the DOE handbook: The explanation’s description of the solutions was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Solving the inequality is routine, and that is the trap: x ≤ 28 looks like an answer and 28 is among the choices. The real task is counting odd numbers from 1 to 27, where the usual shortcut, (27 − 1) ÷ 2, misses the first term. Pairing each odd number with the even number after it, up to 28, gives the count directly and avoids the fencepost slip.

A very close wrong answer moderate Choosing the answer
13 and 14 differ by one, and a rough count of the odd numbers up to 27 cannot separate them; only counting both ends, 1 and 27, gives the exact 14. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Combines ideas minor Working it out
Solving 3x + 8 ≤ 92 is only half the work; counting the positive odd whole numbers up to 28 is a separate number-sense task. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Solve the inequality as you would an equation, keeping the direction of the sign, then count odd numbers from 1 up to the boundary by pairing or halving.

Traps

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

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts the odd numbers from 1 to 27 as (27 − 1) ÷ 2 = 13, forgetting to add one for the first term.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a slip Expected pull: weak
Moves the 8 across without changing its sign, solving 3x ≤ 100, and counts the odd numbers up to 33.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Solves the inequality to x ≤ 28 and gives the boundary 28 instead of counting the positive odd numbers up to it.

What was corrected from the DOE handbook, and why

Explanation: The handbook solves the inequality to x ≤ 28 and then lists “the positive odd numbers less than 28.” The solutions include 28 itself, so the site says “less than or equal to 28.” The count does not change, because 28 is even: the odd numbers are still 1, 3, 5, …, 27, which is 14 numbers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality: $\text{if } a \leq b \text{, then } a - c \leq b - c$; a, b and c are numbers. The sign ≤ means less than or equal to. You may subtract the same number from both sides. You may also divide both sides by the same positive number. The sign stays the same either way.
Counting odd numbers: $\text{odd numbers from 1 to } n = \frac{n}{2}$; n is an even number. Split the numbers from 1 to n into pairs: 1 and 2, 3 and 4, and so on. There are n ÷ 2 pairs, and each pair holds exactly one odd number.

Sort out the question
You need to count the positive odd numbers that make $3x + 8 \leq 92$ true.
The sign ≤ means is less than or equal to. Positive means greater than 0. Odd numbers are 1, 3, 5, 7, and so on.
Solve the inequality
Solve it like an equation. Subtract 8 from both sides. Then divide both sides by 3. Dividing by a positive number keeps the sign the same.
$3x + 8 \leq 92$
$3x \leq 84$
$x \leq 28$
Watch out: 28 is the largest x allowed, not the count. That trap is choice D.
List the odd numbers
The positive odd numbers up to 28 are 1, 3, 5, and so on, up to 25 and 27. The number 28 is even, so it does not count.
Count them by pairing
Split 1 to 28 into pairs: 1 and 2, 3 and 4, and so on, up to 27 and 28. Each pair has one odd number.
$\text{odd numbers from 1 to 28} = \frac{28}{2} = 14$
Watch out: Working out $(27 - 1) \div 2 = 13$ counts the jumps of 2 from 1 to 27, not the numbers. You must add 1 for the first number. That mistake gives choice A.
So the answer is B, 14.

Check your answer

Test the edges. For x = 27: $3 \times 27 + 8 = 89$ , which is at most 92, so it works. For x = 29: $3 \times 29 + 8 = 95$ , which is too big. So 27 is the last odd number that works.

More questions like this

The original DOE question

Question 76 as printed on page 156 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 76. Open page 156 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 74 of 81

2018 Test B Math, question 20

Question

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

Show the answer

Correct answer: A

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

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

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

$12=4x$

$3=x$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 75 of 81

2018 Test B Math, question 27

Question

Nicole’s age now is three times Carmen’s age. If Carmen will be 17 in two years, how old was Nicole 5 years ago?

A38 yr
B40 yr
C45 yr
D50 yr

Show the answer

Correct answer: B

First, find out how old Nicole and Carmen are now.

Let $N$ = Nicole’s age now. Let $C$ = Carmen’s age now.

$C+2=17$

$C=15$ (Carmen’s age now)

$N=3C$

$N=3(15)=45$ (Nicole’s age now)

$N-5=45-5=40$ (Nicole’s age 5 years ago)

Question analysis

This age word problem says Nicole is three times as old as Carmen now, and that Carmen will be 17 in two years. Students work out both present ages and then go back five years for Nicole. Keeping the times straight is the tricky part: two years ahead for Carmen, five years back for Nicole.

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students write simple equations from word problems. Age problems like this one take three short steps, and the shifts in time are the only real hurdle, so the demand is 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: Turning words into math moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 1, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is short, but the question moves through three points in time, and only one of them carries the three-times relationship. Students who find Nicole's present age often stop there, since it is a choice. Labeling each age with its time, now or then, keeps the two-year and five-year shifts with the right person and in the right direction.

Turning words into math moderate Understanding the question
Three points in time appear, now, in two years and 5 years ago, and “three times Carmen's age” holds only for now, so each shift must be applied to the right person. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Students must hold Carmen's future age 17, her present age 15, Nicole's present age 45 and the 5-year shift while deciding which to report. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find Carmen's age now by going back two years from 17, triple it for Nicole's current age, then subtract five.

Traps

Stops one step early: A choice is a number from partway through the solution.
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 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Applies the two-year shift to Nicole as well, subtracting both 2 and 5 from her present age of 45.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds Nicole's present age, three times 15, and stops before going back 5 years.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Adds 5 years to Nicole's present age instead of subtracting, giving her age 5 years from now.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Age at another time: $\text{age in } n \text{ years} = \text{age now} + n$; n is a number of years. To find an age in the future, add n to the age now. To find an age in the past, subtract n instead. Everyone gets older at the same speed, one year per year.

Sort out the times
The question talks about three times: now, two years from now, and 5 years ago. Nicole is 3 times as old as Carmen now.
Carmen will be 17 in two years. You need Nicole’s age 5 years ago.
Find Carmen's age now
In two years Carmen will be 17. So right now she is 2 years younger.
$17 - 2 = 15$
Carmen is 15 now.
Find Nicole's age now
Nicole’s age now is three times Carmen’s age now.
$3 \times 15 = 45$
Nicole is 45 now.
Watch out: 45 is her age now, choice C. The question asks about 5 years ago.
Go back 5 years
Going back in time means subtracting.
$45 - 5 = 40$
Nicole was 40 five years ago.
Watch out: Adding 5 gives 50, choice D. Also, the 2 years only apply to Carmen. Taking 2 more years off Nicole gives 38, choice A.
So the answer is B, 40 yr.

Check your answer

Work forward from your answer. Nicole was 40 five years ago, so she is 45 now. Carmen is a third of that, 15. In two years Carmen will be 17. That matches the question.

More questions like this

The original DOE question

Question 84 as printed on page 158 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 84. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 76 of 81

2018 Test B Math, question 34

Question

$5x^{3}+3x+9+\frac{1}{x^{2}}$

If $x=10$ , what is the value of the expression above?

A2,539.01
B5,039.01
C5,039.1
D5,139

Show the answer

Correct answer: B

$5x^3+3x+9+\frac{1}{x^2}$

$=5(10^3)+3(10)+9+\frac{1}{10^2}$

$=5{,}000+30+9+\frac{1}{100}=5{,}039.01$

Question analysis

Students substitute 10 for x in an expression with a cube, a linear term, a constant and a fraction with x squared in its denominator, then add the results. It tests evaluating expressions with exponents and decimals. The fraction term is where most slips happen, because its place value is easy to get wrong.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Evaluating expressions with whole-number exponents is grade 6 work. The only extra care here is a reciprocal term that turns into a decimal, a single routine substitution that typical grade 7 students handle comfortably.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Solution: 3 steps
Reading level: Question and choices: about grade 7, 11 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Substituting 10 is easy; the risk lies in the two unlike terms at the ends. The cube must be applied to x before multiplying by 5, and the reciprocal of 100 must become a hundredth, not a tenth. With choices that differ by a single decimal place, the question rewards writing each term as a number before adding.

Error-prone arithmetic moderate Working it out
The last term, 1/x² at x = 10, is 1/100, which must be written as 0.01 and lined up beside 5,039; one place too far gives 0.1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
5,039.1 sits beside the right 5,039.01 and differs only in the hundredths place; only writing 1/100 as 0.01, not 0.1, tells them apart, and estimating cannot. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work term by term: compute 10 cubed, then the reciprocal of 10 squared as a decimal, and line up the place values carefully before adding.

Traps

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

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Gets 2,500 for the first term, most likely from (5 × 10)², applying the power to the product 5x and squaring instead of cubing.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Writes 1/100 as 0.1 instead of 0.01, a place-value slip on the last term that puts the total off by 0.09.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Evaluates 1/x² as x² = 100 instead of its reciprocal, so it adds 100 rather than 0.01 to 5,039.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Exponents: $x^3 = x \times x \times x$; x is the base number. The small raised 3 is the exponent. It tells how many times to use x in a multiplication. In the same way, x squared, written with a raised 2, means x times x.

Sort out the question
You need the value of $5x^3+3x+9+\frac{1}{x^2}$ when $x = 10$ . Put 10 in place of every $x$ . Then work out each term, one at a time. A term is a piece joined to the others by a plus sign.
Work out the first term
Do the exponent first. Then multiply by 5.
$10^3 = 10 \times 10 \times 10 = 1{,}000$
$5 \times 1{,}000 = 5{,}000$
Watch out: the exponent belongs only to the $x$ , not to the 5. Squaring $5 \times 10$ gives 2,500, which leads to choice A.
Work out the middle terms
The next term means 3 times $x$ .
$3 \times 10 = 30$
The 9 has no $x$ in it, so it stays 9.
Work out the last term
First find the bottom. Then divide.
$10^2 = 10 \times 10 = 100$
$\frac{1}{100} = 0.01$
One hundredth is 0.01, with the 1 two places after the decimal point.
Watch out: 0.1 is one tenth, not one hundredth. That slip gives choice C. Adding 100 instead of $\frac{1}{100}$ gives choice D.
Add the four terms
Line up the decimal points and add. The 0.01 goes in the hundredths place.
$5{,}000 + 30 + 9 + 0.01 = 5{,}039.01$
So the answer is B, 5,039.01.

Check your answer

The first three terms are whole numbers. They add up to 5,039. The last term is a small part of 1. So the answer is just over 5,039. Only B and C fit. The last term is hundredths, so B is right.

More questions like this

The original DOE question

Question 91 as printed on page 159 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 91. Open page 159 of the handbook (PDF)

Question 77 of 81

2018 Test B Math, question 48

Question

$Coordinate axes with origin O. A shaded triangle stands on the x-axis: its base runs from the point marked m to the point marked n, both to the right of O, and its top vertex touches a dashed horizontal line drawn at 4 on the y-axis.$

What is the area of the shaded triangle shown above?

A $m+n$
B $n-m$
C $2\left ( n-m \right )$
D $4\left ( n-m \right )$

Show the answer

Correct answer: C

The height of the triangle is 4 units. The length of the base is $n-m$ . So the area is

$A=\frac{1}{2}(n-m)(4)=2(n-m)$

Question analysis

Students find the area of a shaded triangle drawn on coordinate axes, with its base between the points m and n on the x-axis and its top vertex at a height of 4. It tests the triangle area formula with an algebraic side length. Writing the base as an expression in m and n is the step that needs care.

NYS standards: 6.G.1 area of polygons
6.EE.2a writing algebraic expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Triangle area is grade 6 content, but a base written with two variables is less familiar. With only two steps, the demand matches typical grade 7 work that mixes geometry with algebraic expressions.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Reading a figure, table or graph moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 7, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is trivial, but students must treat a length as a difference of coordinates and keep a formula's one half when the answer is a letter expression. Those who forget the half, or who stop at the base length, find their result among the choices. Trying m = 1 and n = 4 gives a triangle of area 6 that checks the expressions quickly.

Variables instead of numbers moderate Understanding the question
The base runs from m to n on the x-axis, so its length is the expression n − m, and the area stays in letters with no number to check. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph moderate Understanding the question
The height, 4, is read from the dashed line through the apex rather than labeled on the triangle, and the base endpoints appear only as tick labels. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer moderate Choosing the answer
4(n − m) is the right 2(n − m) doubled, written the same way except for the coefficient; only remembering the one half in a triangle's area separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the base as the difference between its endpoint values, read the height from the figure, and apply one-half times base times height.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two labels, m + n, as if the base were the sum of the endpoints, and goes no further with the height.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the base length n − m correctly and gives it as the area, never multiplying by the height or taking one half.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies the base n − m by the height 4 but forgets the one half in the triangle area formula.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.
Length along the x-axis: $\text{length} = \text{right end} - \text{left end}$; The right end and the left end are the numbers where a segment starts and stops on the x-axis. Subtracting the smaller from the larger tells how long the segment is.

Read the figure
The shaded triangle sits on the x-axis, the flat number line. Its base runs from $m$ to $n$ . Its top corner touches the dashed line at height 4. You need the area, written with the letters $m$ and $n$ .
Find the length of the base
The base starts at $m$ and ends at $n$ . Use the length rule: right end minus left end.
$\text{base} = n - m$
Watch out: adding the ends, $m+n$ , does not give a length. That is choice A.
Find the height
The height is the straight-up distance from the base to the top corner. The top corner is on the dashed line, which is at 4 on the y-axis. So the height is 4.
Find the area
Use the triangle area formula.
$A = \frac{1}{2} \times (n - m) \times 4 = 2(n - m)$
Half of 4 is 2, so the area is $2(n-m)$ .
Watch out: forgetting the one half gives $4(n-m)$ , choice D. Stopping at the base gives $n-m$ , choice B.
Give the answer
The area is 2 times the length of the base. So the answer is C, $2(n-m)$ .

Check your answer

Try numbers. Say $m = 1$ and $n = 4$ . The base is 3 and the height is 4, so the area is $\frac{1}{2} \times 3 \times 4 = 6$ . Choice C gives $2(4-1) = 6$ . It matches. Choice D would give 12.

More questions like this

The original DOE question

Question 105 as printed on page 162 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 105. Open page 162 of the handbook (PDF)

Question 78 of 81

2018 Test B Math, question 50

Question

Which number line below shows the solution set for $2x-2\leq y\leq 4x+10$ when $y=1$ ?

A $A number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to −1 1/2; both ends are closed dots.$
B $A number line from −3 to 3 marked in quarter units, with a solid segment from −1 1/2 to 2 1/4; both ends are closed dots.$
C $A number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to 1 1/2; both ends are closed dots.$
D $A number line from −3 to 3 marked in quarter units, with a closed dot at 1 1/2 and a solid ray from it running left past −2 1/2 to an arrowhead.$

Show the answer

Correct answer: C

Separate the compound inequality into two pieces:

$2x-2\leq y$ and $y\leq 4x+10$

Substitute $y=1$ into each inequality and solve for $x$ :

$2x-2\leq1$

$2x\leq 3$

$x\leq\frac{3}{2}$

$1\leq 4x+10$

$-9\leq4x$

$-\frac{9}{4}\leq x$

The solution is the number line that shows that $x$ is greater than or equal to $-2\frac{1}{4}$ and less than or equal to $1\frac{1}{2}$ .

Question analysis

Students substitute y = 1 into the compound inequality 2x − 2 ≤ y ≤ 4x + 10 and pick the number line that shows the values of x that work. It tests solving two-step inequalities and graphing the solution. The four graphs differ in their endpoints and in whether the shading is closed off on both sides, so each endpoint has to be found exactly.

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: Above grade 7 Grade 7 students solve and graph one two-step inequality. This compound inequality in two variables requires substituting, solving two inequalities and combining them into one interval, a step toward algebra 1.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Reading a figure, table or graph moderate
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 9, 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

Substituting y = 1 is easy; the work is solving two inequalities and matching a fractional interval to a graph. Students who solve only the left inequality find a matching ray, and sign slips on either endpoint produce intervals that shift or flip across zero. The graphs differ by quarter-unit ticks, so reading them carefully matters as much as the algebra.

Many steps moderate Working it out
The compound inequality must be split into 2x − 2 ≤ 1 and 1 ≤ 4x + 10, each solved for x, and the two results joined into one interval. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Both endpoints are fractions, 3/2 and −9/4, and the second comes from dividing −9 by 4, where a lost sign moves the endpoint across zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph moderate Understanding the question
The graphs are marked in quarter units, so −2 1/4 and 1 1/2 have to be found by counting small ticks between the labeled integers. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer minor Choosing the answer
The ray ending at 1 1/2 looks almost like the right segment, sharing its right end; only the left end, an arrow instead of a closed dot at −2 1/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Split the compound inequality into two separate inequalities, solve each for x after putting 1 in for y, then combine the two results into a single interval and match it to a graph.

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

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Gets the left endpoint −2 1/4 right but gives the right endpoint the wrong sign, ending the segment at −1 1/2 instead of 1 1/2.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Gives both endpoints the opposite sign, so the segment runs from −1 1/2 to 2 1/4, the true interval reflected across 0.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Solves only 2x − 2 ≤ 1 and graphs every x up to 1 1/2, leaving out the lower bound that 1 ≤ 4x + 10 sets.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality: $\text{if } a \leq b, \text{ then } a + c \leq b + c$; a, b and c are numbers. The sign ≤ means is less than or equal to. Adding or subtracting the same number c on both sides keeps it true. Dividing both sides by the same positive number also keeps it true.

Sort out the question
The sign $\leq$ means is less than or equal to. An inequality compares two amounts that may not be equal. The statement $2x-2 \leq y \leq 4x+10$ is two inequalities joined together. You put 1 in place of $y$ . Then you need every $x$ that makes both parts true.
Split it into two parts
With $y = 1$ , the statement becomes two inequalities.
$2x - 2 \leq 1$
$1 \leq 4x + 10$
The answer must make both of them true at the same time.
Solve the first part
Add 2 to both sides. Then divide both sides by 2.
$2x \leq 3$
$x \leq \frac{3}{2}$
So $x$ is at most $1\frac{1}{2}$ . This is the right end of the answer.
Watch out: the right end is a positive number. Choice A ends its bar at $-1\frac{1}{2}$ instead.
Solve the second part
Subtract 10 from both sides. Then divide both sides by 4.
$-9 \leq 4x$
$-\frac{9}{4} \leq x$
So $x$ is at least $-2\frac{1}{4}$ . This is the left end.
Watch out: keep the minus sign. Choice B uses the opposite signs and ends at $2\frac{1}{4}$ .
Put the two parts together
The answer is every number from $-2\frac{1}{4}$ to $1\frac{1}{2}$ , with both ends included. On the graph, that is a solid bar with a dot at each end. The small ticks are quarters. So count 1 tick left of −2 and 2 ticks right of 1.
Watch out: choice D uses only the first part. Its arrow goes left forever.
Give the answer
Choice C shows a bar from $-2\frac{1}{4}$ to $1\frac{1}{2}$ , with dots at both ends. So the answer is C.

Check your answer

Test a number on the bar, like $x = 0$ . Then $2(0)-2 = -2$ and $4(0)+10 = 10$ , and 1 is between them. Now test $x = -3$ , which is off the bar. Then $4(-3)+10 = -2$ , and 1 is not less than −2. So the bar must stop on the left, which rules out D.

More questions like this

The original DOE question

Question 107 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 107. Open page 163 of the handbook (PDF)

Question 79 of 81

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 80 of 81

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)

Question 81 of 81

2018 Test B Math, question 56

Question

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

Show the answer

Correct answer: D

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

x − 4 = 18

x = 22

Question analysis

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

NYS standards: 6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 6 students solve equations like x − p = q, and this one needs only that and halving 36. It is a quick, routine word problem, easier than the two-step equations typical of grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Number of 2s	1	2	3	4	5	6	7
Product	2	4	8	16	32	64	128