Practice Test

Little reason to choose it

Question 1 of 86

2026 Practice Test A Math, question 37

Question

When a student buys 4 notebooks, the total cost is $12. Each notebook costs the same amount. Select each point on the graph that best represents y, the total cost of x notebooks.

Select all the correct answers.

$A graph titled BUYING NOTEBOOKS, with Number of Notebooks on the x-axis from 0 to 6 and Total Cost (dollars) on the y-axis from 0 to 15, with grid lines every 1 unit. Seven points are plotted: (1, 1), (1, 3), (2, 5), (3, 7), (3, 9), (4, 12) and (5, 14).$

On the test you click the points in the figure; here, choose them below. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

A(1, 1)
B(1, 3)
C(2, 5)
D(3, 7)
E(3, 9)
F(4, 12)
G(5, 14)

Show the answer

Correct answer: B, E, and F

Solution:

There are three correct points: (1, 3), (3, 9), and (4, 12).

Find the cost per notebook: $\frac{12}{4} = 3$ .

The equation of the line is $y = 3x$ , where y is the total cost of x notebooks.

Check each point by plugging the x-value into the equation $y = 3x$ . If the y-value from the equation matches the y-value of the point, then the point is on the line $y = 3x$ .

Check all seven points:

(1, 1) $y = 3(1) = 3$ y-values do not match; incorrect point

(1, 3) $y = 3(1) = 3$ y-values match; CORRECT point

(2, 5) $y = 3(2) = 6$ y-values do not match; incorrect point

(3, 7) $y = 3(3) = 9$ y-values do not match; incorrect point

(3, 9) $y = 3(3) = 9$ y-values match; CORRECT point

(4, 12) $y = 3(4) = 12$ y-values match; CORRECT point

(5, 14) $y = 3(5) = 15$ y-values do not match; incorrect point

Question analysis

A graph titled Buying Notebooks shows seven points, and students select every point that fits the total cost of buying notebooks, given the cost of a few notebooks and that each costs the same. It tests finding a unit rate and recognizing points of a proportional relationship. Some points line up in tempting patterns, and the item does not say how many to select.

NYS standards: 7.RP.2b constant of proportionality
7.RP.2c proportional relationship equations
6.RP.3a tables of equivalent ratios
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a unit rate and deciding which points fit a proportional relationship are grade 7 skills, and the arithmetic here is easy. What adds demand is testing all seven points with no count given, since one miss or extra pick loses the item.
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: Every choice must be checked moderate
Reading a figure, table or graph minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Finding the price is one division, 12 ÷ 4 = 3, so the question is really about checking: each chosen dot must fit y = 3x exactly, and the item does not say how many do. The graph punishes judging by eye, since a straight row of dots climbs from (1, 3) at $2 a notebook and two dots sit a single gridline off the true line. Multiplying each x by 3 and reading each y from the axis, not from the pattern, settles all seven.

Every choice must be checked moderate Choosing the answer
Seven dots must each be tested against $3 a notebook, and the item never says how many fit; one missed point or one extra loses the whole item, so no dot can be skipped. (Embretson & Wetzel, 1987)
Reading a figure, table or graph minor Understanding the question
The graph stacks two dots above 1 notebook and two above 3, and each total cost has to be read off a grid ruled at every dollar up to 15. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer minor Choosing the answer
(5, 14) sits one gridline below (5, 15), where 5 notebooks at $3 would be, and (2, 5) sits one below (2, 6), so judging by the line's direction cannot rule either out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the cost of one notebook from the total given in the question, then multiply each point's x-coordinate by that price and keep only the points whose y-coordinate matches exactly.

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 4 notebooks for $12 makes one notebook cost $1, so this lowest dot mostly catches guesses, such as taking the dot nearest the origin as a start.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the straight row (1, 3), (2, 5), (3, 7) as the graph because its dots line up, though it climbs $2 a notebook and would meet the y-axis at $1, not the origin.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Prefers the lower of the two dots above 3 notebooks because it continues the evenly spaced row from (1, 3) and (2, 5), instead of checking that 3 notebooks cost $9.
G Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Finds that 5 notebooks cost 5 × $3 = $15, then reads the dot above 5 as sitting on the top gridline, 15, when it is one line lower, at 14.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit price: $\text{price of one} = \text{total cost} \div \text{number bought}$; The unit price is what one item costs. Divide the total cost by the number of items bought. This works because every notebook costs the same, so the total splits into equal parts.
A proportional relationship: $y = kx$; y is the total cost in dollars, and x is the number of notebooks. k is the price of one notebook, the number you always multiply by. A point fits only if its y is exactly k times its x. The graph is a straight line through (0, 0).

Sort out the question
Four notebooks cost $12, and each one costs the same. On the graph, x is the number of notebooks, and y is the total cost in dollars. The graph shows seven points. You must select every point that fits. The question does not say how many fit, so test all seven.
Find the price of one notebook
Split the $12 equally among the 4 notebooks:
$12 \div 4 = 3$
Each notebook costs $3. So the total cost is 3 times the number of notebooks:
$y = 3x$
Test each point
For each point, multiply its x by 3. Keep the point only if the result equals its y.
Choice Point 3 × x Fits?
A (1, 1) 3 No
B (1, 3) 3 Yes
C (2, 5) 6 No
D (3, 7) 9 No
E (3, 9) 9 Yes
F (4, 12) 12 Yes
G (5, 14) 15 No
Watch for look-alike points
Watch out: the dots (1, 3), (2, 5) and (3, 7) sit in a neat straight row. But that row climbs $2 per notebook, not $3. Lining up is not enough: (2, 5) and (3, 7), choices C and D, do not fit.
Watch out: 5 notebooks cost $5 \times 3 = 15$ dollars. The dot above 5 sits one gridline lower, at 14. Read the y-axis carefully, and you will skip choice G.
Name the points
Three points fit: (1, 3), (3, 9) and (4, 12). On the test, you click each of these dots on the graph. On this site, you check their boxes. So the answers are B, E and F.

Choice	Point	3 × x	Fits?
A	(1, 1)	3	No
B	(1, 3)	3	Yes
C	(2, 5)	6	No
D	(3, 7)	9	No
E	(3, 9)	9	Yes
F	(4, 12)	12	Yes
G	(5, 14)	15	No

Check your answer

All three points fit $y = 3x$ : $3 = 3 \times 1$ , $9 = 3 \times 3$ and $12 = 3 \times 4$ . They also line up with (0, 0), since 0 notebooks cost $0. And (4, 12) is the fact the question gave you.

More questions like this

The original DOE question

Question 37 as shown on the DOE’s digital test, page 108 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test A Math, page 108, question 37.

Question 2 of 86

2026 Practice Test C Math, question 8

Question

Jordan is helping plan snacks for a school dance. He knows that 0.75 of the students voted for pizza, and he wants to record the same amount using a fraction.

Which fractions are equivalent to the decimal 0.75?

Select the four correct answers.

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

A $\frac{3}{4}$
B $\frac{15}{20}$
C $\frac{75}{100}$
D $\frac{2}{5}$
E $\frac{6}{8}$
F $\frac{8}{10}$

Show the answer

Correct answer: A, B, C, and E

Check each fraction to determine whether it is equal to 0.75 when simplified or converted to a decimal.

$\frac{3}{4} \rightarrow 3 \div 4 = 0.75$
This is the most common fraction equal to 0.75.

$\frac{15}{20} \rightarrow 15 \div 20 = 0.75$
Simplifies to 3/4, because 15 and 20 can both be divided by 5.

$\frac{75}{100} \rightarrow 75 \div 100 = 0.75$
This is a direct match to 0.75 when written as a decimal.

$\frac{2}{5} \rightarrow 2 \div 5 = 0.4$ , which is less than 0.75.

$\frac{6}{8} \rightarrow 6 \div 8 = 0.75$
Simplifies to $\frac{3}{4}$ because both 6 and 8 can be divided by 2.

$\frac{8}{10} \rightarrow 8 \div 10 = 0.8$ , which is more than 0.75.

Question analysis

Jordan wants to record 0.75 of the students' pizza votes as a fraction, and students select the four fractions, out of six, that equal 0.75. It tests converting between decimals and fractions and recognizing equivalent fractions. Each choice has to be checked by its value, not by how its numbers look.

NYS standards: 7.NS.2d fractions to decimals by long division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Matching fractions to a decimal such as 0.75 builds on grade 4 to 6 work, and the denominators here divide easily. The added demand is the format: four of the six must be chosen, and one wrong or missing pick loses the item.
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: Every choice must be checked moderate
A very close wrong answer minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Three of the four right fractions, 3/4, 75/100 and 6/8, are familiar forms of 0.75, so the item comes down to the fourth. Students who stop after the familiar ones must decide between 15/20, which needs dividing by 5 to recognize, and 8/10, which sits close to 0.75 on a number line. Dividing each numerator by its denominator, or scaling 3/4 up by 5, settles it.

Every choice must be checked moderate Choosing the answer
Six fractions with six different denominators have to be tested one by one, by dividing or by simplifying, since 15/20 and 8/10 cannot be judged by how they look. (Embretson & Wetzel, 1987)
A very close wrong answer minor Choosing the answer
8/10 is 0.8, only 0.05 above 0.75, and it follows 6/8 by adding 2 to the top and the bottom, so it looks like one more member of the family. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
Four of the six must be chosen and all four must be right for the item to count, so each verdict has to be kept while the remaining fractions are tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Turn each choice into a decimal by dividing its numerator by its denominator, or reduce it to lowest terms, and keep only the ones whose value is exactly the decimal given.

Traps

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

The wrong answers, classified

D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No usual working links 2/5 to 0.75, since 2 ÷ 5 = 0.4, so it mostly catches students filling the fourth slot by guessing.
F Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Builds 8/10 from 6/8 by adding 2 to the top and the bottom, as if equal additions kept a fraction's value, or accepts 0.8 as close enough.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing a fraction to a decimal: $\frac{a}{b} = a \div b$; a is the numerator, the top. b is the denominator, the bottom. A fraction bar means divide. So dividing the top by the bottom gives the fraction’s value as a decimal.
Equivalent fractions: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top and b is the bottom. n is a number that goes evenly into both. Dividing the top and the bottom by the same number gives an equal fraction. Equivalent fractions name the same amount in different ways.

Sort out the question
Six fractions are given, A to F. You need the four that equal the decimal 0.75. Equivalent means equal in value. Test every choice, because some fractions look alike but are not equal.
Write 0.75 as a fraction
0.75 means 75 hundredths:
$0.75 = \frac{75}{100}$
So choice C works. Now simplify it. 25 goes into both 75 and 100:
$\frac{75 \div 25}{100 \div 25} = \frac{3}{4}$
So choice A, 3/4, works too.
Test choices B and E
Simplify each one to see if it becomes 3/4:
- B: 5 goes into 15 and 20. $\frac{15 \div 5}{20 \div 5} = \frac{3}{4}$ . It works.
- E: 2 goes into 6 and 8. $\frac{6 \div 2}{8 \div 2} = \frac{3}{4}$ . It works.
Test choices D and F
Divide the top by the bottom:
- D: $2 \div 5 = 0.4$ . That is less than 0.75.
- F: $8 \div 10 = 0.8$ . That is more than 0.75.
Watch out: F looks like 6/8 with 2 added to the top and the bottom. Adding the same number to both changes the value. Only multiplying or dividing both keeps it.
Name the four answers
A, B, C and E all equal 0.75. On the test, you check the box beside each of them. On this site, you choose all four, then check your answer.
So the answers are A, B, C and E.

Check your answer

Turn each answer into a decimal. $3 \div 4$ , $15 \div 20$ , $75 \div 100$ and $6 \div 8$ all give 0.75. Exactly four choices passed, and the question asks for four. So none were missed.

More questions like this

The original DOE question

Question 8 as shown on the DOE’s digital test, page 81 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 81, question 8.

Question 3 of 86

2026 Practice Test C Math, question 10

Question

$Two parallel lines, n and p, rise from lower left to upper right, and both cross a horizontal line m. Where line n crosses line m, the angle above line m, between the left part of m and the upper part of n, is marked r°. Farther right, a vertical segment runs from line p down to line m, meeting m at a right angle (marked with a small square); the angle at the top, between line p and the vertical segment, is marked 73°.$

In the figure above, $n \parallel p$ . Straight line m intersects both line n and line p. What is the value of r?

A17
B117
C163
D173

Show the answer

Correct answer: C

The triangle formed by line p, line m, and the line segment connecting the two contains a 73° angle, a right angle (90°), and a missing angle measuring x°. The sum of the interior angles of a triangle is 180°, so begin by finding the value of x:

$180^{\circ} = 73^{\circ} + 90^{\circ} + x$

$180^{\circ} = 163^{\circ} + x$

$17^{\circ} = x$

Line m is a straight line, so it has a measure of 180°. The 17° angle from the triangle and the adjacent angle outside the triangle are supplementary, so they have a sum of 180°. Let s represent the measure of the adjacent angle outside the triangle. The value of s can be determined by subtracting:

$s = 180^{\circ} - 17^{\circ}$

$s = 163^{\circ}$

Given that n and p are parallel lines and that they have been cut by a transversal, line m, then corresponding angles are congruent. Angles r and s are corresponding angles, so the value of r is congruent to the value of s, which is 163°.

Question analysis

Two parallel lines, n and p, cross a straight line m, and a segment drawn at a right angle to m forms a small triangle with lines m and p, where a 73° angle is marked. Students find the angle r at line n. It combines a triangle's angle sum, a straight angle and parallel lines, and the angle wanted sits away from the marked ones.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle problems use supplementary, vertical and adjacent angles. This one also needs the angle sum of a triangle and the angles formed by parallel lines and a transversal, grade 8 facts, chained across a busy figure.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Combines ideas moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each fact needs only a subtraction, but the student has to plan a route from the 73° angle at line p over to r at line n. The first number the route gives, the triangle's 17° angle, is listed, and it is the acute partner of r rather than r itself. The figure helps here: r is drawn plainly obtuse, and 163 is also 90 + 73, the triangle's exterior angle, which confirms it.

Reading a figure, table or graph moderate Understanding the question
The 73° sits between line p and the vertical segment, not between p and m, and r is the obtuse angle above m on the left of line n, far from every marked angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
Three facts chain together: the triangle's angles sum to 180°, the angles along straight line m sum to 180°, and corresponding angles at the parallel lines n and p are equal. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
The angle pairs formed by a line crossing two parallel lines are a grade 8 topic, and the question uses them without naming them. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of the right triangle, use the straight line m to get the angle beside it, then carry that angle over to line n using the parallel lines.

Traps

Stops one step early: A choice is a number from partway through the solution.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the triangle's third angle, 180 − 90 − 73 = 17, and stops, though r is the obtuse angle next to the 17° angle, not that angle itself.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No angle fact in the figure gives 117; even 180 − 73 is 107, so this choice mostly catches guesses at an obtuse angle.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds the 17° angle and subtracts it from 180 but borrows without reducing the 8, so 180 − 17 comes out as 173.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles of a triangle: $a + b + c = 180^{\circ}$; a, b and c are the three angles inside any triangle. They always add up to 180°. If you know two of them, subtract their sum from 180° to find the third.
Angles on a straight line: $a + b = 180^{\circ}$; a and b are two angles side by side on a straight line. Together they make a half turn, which is 180°. Use it to find the angle next to one you already know.
Corresponding angles: $n \parallel p \Rightarrow \text{corresponding angles are equal}$; The symbol ∥ means is parallel to. When line m crosses two parallel lines, it makes the same angles at each crossing. Angles in the same position at the two crossings are called corresponding angles. They are equal.

Sort out the question
Lines n and p are parallel, and line m crosses both. Near line p, a straight-up segment meets line m. The small square shows a right angle, 90°. Line p, line m and that segment make a small triangle. Its top angle is 73°.
You need r, the wide angle where line n crosses line m.
Find the triangle’s third angle
The third angle is where line p crosses line m. Call it x. Use the triangle rule:
$73 + 90 + x = 180$
$x = 180 - 163 = 17$
So that angle is 17°.
Watch out: this is not r. Stopping here gives choice A, 17.
Use the straight line m
The 17° angle sits above line m, on the right of line p. Beside it, on the left of line p, is a wide angle. Call it s. The two angles fill the straight line m, so they add to 180°:
$s = 180 - 17 = 163$
So s is 163°.
Watch out: borrow carefully. 180 − 17 is 163, not 173, choice D.
Carry the angle to line n
Angle s is above line m and to the left of line p. Angle r is above line m and to the left of line n. They are in the same position, so they are corresponding angles. Since n and p are parallel:
$r = s = 163$
So the answer is C, 163.

Check your answer

Look at the figure. Angle r is wider than a right angle, so it must be more than 90°. That rules out 17. Also, the narrow angle beside r, on the right of line n, matches the 17° angle at line p. And $163 + 17 = 180$ , as it should on a straight line.

More questions like this

The original DOE question

Question 10 as shown on the DOE’s digital test, page 83 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 83, question 10.

Question 4 of 86

2026 Practice Test C Math, question 16

Question

$\frac{3+(-3)}{3-(-3)} =$

A−1
B0
C $\frac{1}{6}$
D1

Show the answer

Correct answer: B

Simplify the expression.

$3 + (-3)$ is the equivalent of subtracting 3 from 3, and $3 - (-3)$ is the equivalent of adding 3 to 3:

$\frac{3+(-3)}{3-(-3)} = \frac{3-3}{3+3} = \frac{0}{6}$

The zero property of division states that zero divided by any nonzero number equals zero, so $\frac{0}{6} = 0$ .

Question analysis

This fraction has 3 plus negative 3 on top and 3 minus negative 3 on the bottom, and students find its value. It tests adding and subtracting negative numbers and dividing integers. The top and the bottom look almost alike, so each sign has to be read carefully before anything is divided.

NYS standards: 7.NS.1c rational number subtraction
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding and subtracting a negative integer and dividing are grade 7 basics, and here the numbers are small. It is a short routine calculation, easier than typical grade 7 work, though the two signs invite a quick misreading.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 16+, 2 words.
Corrected from the DOE handbook: A grammatical error in the explanation was corrected.
A rule stated too broadly 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 two one-step sign calculations and a division by 6. The danger is reading the top and the bottom as alike: they differ in a single sign, and treating them the same way gives 1 or −1. Students who rewrite 3 − (−3) as 3 + 3 and 3 + (−3) as 3 − 3 see 0 over 6 at once, and zero divided by a nonzero number is zero.

Error-prone arithmetic minor Working it out
The top adds −3 and the bottom subtracts −3, signs that look nearly the same but give 0 and 6, so each has to be rewritten before anything is divided. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Simplify the top and the bottom separately, rewriting the subtraction of a negative as an addition, and only then divide the top by the bottom.

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 misconception Expected pull: moderate
Adds the sizes on top and keeps the minus, taking 3 + (−3) as −6, so the fraction becomes −6/6 = −1.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No sign rule gives a top of 1, so 1/6 mostly draws students who find the bottom, 6, and are unsure what a top of 0 means.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Treats the top like the bottom, adding 3 and 3 in both places, so the fraction becomes 6/6 = 1.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “3 − (−3) is equivalent of adding 3 to 3”, without the article that the same sentence uses in “3 + (−3) is the equivalent of subtracting 3 from 3”. The site now says “is the equivalent of adding 3 to 3”.
Explanation: The DOE’s explanation says “The zero property of division states that zero divided by any number equals zero.” Zero divided by zero is undefined, not zero, so the rule holds only when the divisor is not zero. The site now says “zero divided by any nonzero number equals zero”. Here the divisor is 6, so 0/6 = 0, as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding a negative: $a + (-b) = a - b$; a and b are numbers. Adding a negative is the same as subtracting. On a number line, adding −3 moves you 3 steps left, just like subtracting 3.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding. Taking away a debt of 3 dollars leaves you 3 dollars better off. On a number line, you move right.
Zero divided by a number: $\frac{0}{b} = 0 \text{ when } b \neq 0$; b is any number except 0. Zero split into b equal parts leaves 0 in each part. The rule needs b not to be 0, because dividing by 0 has no answer.

Sort out the question
The top of the fraction is $3 + (-3)$ , and the bottom is $3 - (-3)$ . A fraction bar means divide. Work out the top and the bottom separately first. Then divide the top by the bottom.
Work out the top
Adding −3 is the same as subtracting 3:
$3 + (-3) = 3 - 3 = 0$
A number plus its opposite is always 0.
Watch out: do not add the 3s and keep the minus sign. That makes the top −6 and leads to choice A, −1.
Work out the bottom
Subtracting −3 is the same as adding 3:
$3 - (-3) = 3 + 3 = 6$
Watch out: the top and bottom look alike, but they are not. Making both of them 6 gives 6/6 = 1, choice D.
Divide the top by the bottom
Now the fraction is 0 over 6. Zero divided by any number other than zero is zero:
$\frac{3+(-3)}{3-(-3)} = \frac{0}{6} = 0$
A top of 0 does not make the fraction 1/6, choice C. So the answer is B, 0.

Check your answer

Check by multiplying. A division is right when the answer times the bottom gives the top. Here $0 \times 6 = 0$ , which is the top. So 0/6 really is 0.

More questions like this

The original DOE question

Question 16 as shown on the DOE’s digital test, page 89 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 89, question 16. This is the DOE’s version, before the corrections listed in the analysis.

Question 5 of 86

2026 Practice Test C Math, question 33

Question

A bag contains 40 tiles that are either red, green, or blue. A tile is selected at random, the color is recorded, and the tile is put back in the bag. This process is carried out 120 times, and 18 of those times a red tile is selected. Based on the information, what is the most likely number of red tiles in the bag?

Show the answer

Correct answer: A

First, determine what fraction of the sample were red tiles, given 18 of the 120 tiles drawn were red.

$\frac{18 \div 6}{120 \div 6} = \frac{3}{20}$

To predict how many of the actual 40 tiles in the bag are red, multiply 40 by the fraction of red tiles in the sample.

$\frac{3}{20}(40) = \frac{120}{20} = 6$

Question analysis

A bag holds 40 red, green and blue tiles, and a tile is drawn and put back 120 times, with red coming up 18 times. Students predict the most likely number of red tiles in the bag. It tests using experimental results to estimate what is in the bag, and the draws outnumber the tiles, which can confuse the scaling.

NYS standards: 6.SP.7 long-run relative frequency
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Predicting from repeated trials is grade 6 probability, and grade 7 students make similar estimates from samples. The twist is scaling down, from 120 draws to a bag of 40 tiles, rather than up to a larger group.
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
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 61 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

There are more draws than tiles, which is what makes this unusual: the 120 draws must be scaled down to a bag of 40, not up to a bigger group. Red came up 18 times because each tile went back in the bag, so 18 is not a count of red tiles. Since there are three draws for every tile, dividing the 18 red draws by 3 gives the same estimate as taking 3/20 of 40.

Turning words into math minor Understanding the question
The 18 red draws come from 120 draws with the tile put back each time, so they give a share, 3/20, to take of the 40 tiles, not a count of red tiles. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Students must know that the share of red in repeated draws estimates the share of red tiles in the bag; the stem never says to multiply 40 by that share. (Anderson & Krathwohl, 2001)

Strategy: Write the red draws as a fraction of all 120 draws, simplify it, and take that same fraction of the 40 tiles in the bag.

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

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that red came up in 18 of 120 draws, or 15%, and gives 15 as the count, stopping before taking that share of the 40 tiles.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives 18, the number of times red was drawn, as the number of red tiles, though each tile went back and could be drawn many times.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 18 red draws in 120 gives 30 red tiles, three quarters of the bag, when red came up in fewer than one draw in six.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from repeated draws: $\text{red tiles} \approx \frac{\text{red draws}}{\text{all draws}} \times \text{tiles in the bag}$; The fraction is the share of draws that came up red. That share is your best guess for the share of tiles that are red. Multiply it by the number of tiles in the bag to predict the red tiles. The sign ≈ means about equal.

Sort out the question
The bag has 40 tiles, each red, green or blue. A tile is drawn, its color is written down, and it goes back in the bag. This happens 120 times, and red comes up 18 times.
You need the most likely number of red tiles in the bag.
See why 18 is not the answer
Each tile goes back in the bag after it is drawn. So the same red tile can be drawn again and again. That means 18 counts draws, not tiles.
Watch out: choice C, 18, mixes up the number of red draws with the number of red tiles.
Find the share of red draws
Red came up 18 times out of 120. Simplify by dividing the top and bottom by 6:
$\frac{18}{120} = \frac{3}{20}$
As a decimal, that is 0.15, or 15%.
Watch out: 15 is a percent here, not a number of tiles. Stopping now gives choice B.
Take that share of the bag
Use the same share of the 40 tiles in the bag:
$\frac{3}{20} \times 40 = \frac{120}{20} = 6$
So about 6 of the 40 tiles are red.
Give the answer
The best prediction is 6 red tiles. The other 34 tiles are probably green or blue.
So the answer is A, 6.

Check your answer

There were 120 draws for 40 tiles. That is 3 draws for every tile. So divide the red draws by 3: $18 \div 3 = 6$ . Same answer. Choice D, 30 red tiles, would be three quarters of the bag. Then red would come up far more than 18 times.

More questions like this

The original DOE question

Question 33 as shown on the DOE’s digital test, page 106 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 106, question 33.

Question 6 of 86

2026 Practice Test C Math, question 37

Question

The table shows the probability of randomly picking each flavor of candy from a bowl.

PICKING CANDY FROM
A BOWL

Flavor	Probability
Butterscotch	0.57
Lemon	0.095
Maple	0.125
Strawberry	0.21

Which flavor is most likely to be randomly picked?

Abutterscotch
Blemon
Cmaple
Dstrawberry

Show the answer

Correct answer: A

The flavor most likely to be picked at random is butterscotch, because its probability is the greatest of the probabilities in the list. More than half the candies in the bowl are butterscotch.

Question analysis

A table lists the probability of picking butterscotch, lemon, maple or strawberry candy at random from a bowl, each as a decimal. Students decide which flavor is most likely to be picked. It tests the meaning of probability and comparing decimals, which have different numbers of digits here and must be compared by place value.

NYS standards: 6.SP.6 probability from 0 to 1
6.NS.7 ordering and absolute value
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Error-prone arithmetic minor
Solution: 1 step
Reading level: Question and choices: about grade 9, 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

One idea answers the question: the most likely flavor has the greatest probability. The only obstacle is comparing decimals of different lengths, where 0.125 and 0.095 look bigger than 0.57 to students who count digits. Writing each as thousandths, 570, 95, 125 and 210, or noticing that 0.57 alone is more than one half, makes the order plain.

Error-prone arithmetic minor Working it out
The probabilities have two and three decimal places, 0.57 beside 0.095 and 0.125, so comparing them by length, or by their digits as whole numbers, ranks them wrongly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Line the four decimals up by place value, adding zeros so each has three decimal places, and compare them digit by digit starting from the tenths place.

Traps

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

The wrong answers, classified

B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads lemon’s 0.095 as 0.95, overlooking the 0 in the tenths place, so it looks almost certain and larger than 0.57.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the decimal with more digits as the larger, or compares 125 with 57 as whole numbers, so 0.125 seems bigger than 0.57.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common misreading makes 0.21 the largest, since it is neither the longest decimal nor the one with the biggest digits, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{candies of that flavor}}{\text{all candies in the bowl}}$; P is the probability of picking a flavor. It is a number from 0 to 1. The more candies of a flavor there are, the bigger its probability. So the flavor with the greatest probability is the most likely to be picked.

Sort out the question
The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.
The greater the probability, the more likely the pick. So you need the greatest of the four decimals.
Line up the decimals
The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:
Flavor Probability
Butterscotch 0.570
Lemon 0.095
Maple 0.125
Strawberry 0.210
Compare the tenths digits
Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.
Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.
Give the answer
Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.
So the answer is A, butterscotch.

Flavor	Probability
Butterscotch	0.570
Lemon	0.095
Maple	0.125
Strawberry	0.210

Check your answer

0.57 is more than 0.5, which is one half. So more than half the candies are butterscotch. The other three flavors add up to $0.095 + 0.125 + 0.21 = 0.43$ , less than half. So butterscotch must be the most likely.

More questions like this

The original DOE question

Question 37 as shown on the DOE’s digital test, page 110 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 110, question 37.

Question 7 of 86

2026 Practice Test C Math, question 43

Question

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

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

Show the answer

Correct answer: A

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 8 of 86

2026 Practice Test D Math, question 3

Question

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

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

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

What is the value of $x - y$ ?

A−7
B−5
C5
D7

Show the answer

Correct answer: B

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

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

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

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

$x + 16 = y + 11$

$x = y - 5$

$x - y = -5$

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 9 of 86

2026 Practice Test D Math, question 13

Question

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

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

A43°
B44°
C46°
D47°

Show the answer

Correct answer: A

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

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

$5x - 5 = 90$

$5x = 95$

$x = 19$

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

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

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

The smallest angle is 43°.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 10 of 86

2026 Practice Test D Math, question 14

Question

James must read 84 pages of a book for a class assignment. If he reads $\frac{1}{4}$ of the pages in the morning and $\frac{1}{3}$ of the remaining pages in the afternoon, how many pages will he still need to read?

Show the answer

Correct answer: B

Pages read in the morning: $\left(\frac{1}{4}\right)(84) = 21$

Pages remaining after the morning: $84 - 21 = 63$

Pages read in the afternoon: $\left(\frac{1}{3}\right)(63) = 21$

Pages he still needs to read: $63 - 21 = 42$

Question analysis

James must read 84 pages for a class assignment: 1/4 of them in the morning, then 1/3 of the remaining pages in the afternoon. Students find how many pages he still has left to read. It tests finding a fraction of a quantity in stages, and the bold word remaining decides which amount the second fraction is taken of.

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking fractions of a quantity in stages is typical grade 7 work. The numbers divide evenly and the bold word flags the trap, so the four short steps match routine 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: Many steps moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 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

Each step is simple, but the second fraction has a different whole from the first. Students who take 1/3 of the full 84 pages get a tidy 28 and a listed answer, despite the bold word that warns against it. Here the pages read and the pages left both come to 42, so confusing those two quantities costs nothing, and checking that 21 + 21 + 42 = 84 confirms the work.

Many steps moderate Working it out
Four operations in order, a quarter of 84, a subtraction, a third of the 63 left, and a second subtraction, each feed the next, so a slip early carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The bold “remaining” moves the base of the second fraction from all 84 pages to the 63 left after the morning, and “still need to read” asks for what is left, not what was read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work in order: find the morning's pages, subtract to get what remains, take 1/3 of that remainder, and subtract again; a quick bar sketch of the 84 pages helps.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Takes 1/3 of all 84 pages for the afternoon, 28, instead of 1/3 of the 63 remaining, so 84 − 21 − 28 = 35.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 45; it is not a fraction or a difference of any page count in the problem, so it mostly draws guesses near 42.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Takes the afternoon's third from all 84 pages and subtracts only that, 84 − 28 = 56, overlooking the 21 pages already read in the morning.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of a number: $\frac{a}{b} \text{ of } n = \frac{a}{b} \times n$; n is the whole amount, and a/b is the fraction you take. The word of means multiply. When the top, a, is 1, this is just n divided by b. For example, 1/4 of 84 is 84 ÷ 4.

Sort out the question
James has 84 pages to read. In the morning he reads 1/4 of all the pages. In the afternoon he reads 1/3 of the remaining pages, the ones still left after the morning. You need the number of pages he still has to read after that.
Find the morning pages
Take 1/4 of 84:
$\frac{1}{4} \times 84 = 84 \div 4 = 21$
He reads 21 pages in the morning. That leaves:
$84 - 21 = 63$
63 pages are left after the morning.
Find the afternoon pages
The afternoon fraction is 1/3 of the 63 pages left, not of all 84:
$\frac{1}{3} \times 63 = 63 \div 3 = 21$
He reads 21 pages in the afternoon.
Watch out: taking 1/3 of all 84 pages gives 28. Then 84 − 21 − 28 = 35, choice A. The bold word remaining tells you to use 63.
Find the pages still to read
Take the afternoon pages away from the 63 pages left after the morning:
$63 - 21 = 42$
He still needs to read 42 pages.
Watch out: subtract both readings, morning and afternoon. Taking only 28 from 84 gives 56, choice D.
So the answer is B, 42.

Check your answer

Add up all the parts: 21 pages in the morning, 21 in the afternoon and 42 still to read. 21 + 21 + 42 = 84, the whole assignment. Another way: after he reads 1/3 of the 63 pages, 2/3 of them are left, and 2/3 of 63 is 42.

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 D Math, page 87, question 14.

Question 11 of 86

2026 Practice Test D Math, question 21

Question

Maxie borrowed $7.75 from her mother, $11.00 from her father, and $4.50 from her brother to purchase a video game. Her grandmother gave her $25.00 as a gift. How much money will Maxie have left or still owe if she uses the money her grandmother gave her to pay back the money she borrowed?

AMaxie will still owe $2.25.
BMaxie will still owe $1.75.
CMaxie will have $1.75 left.
DMaxie will have $2.25 left.

Show the answer

Correct answer: C

First, determine the total amount Maxie borrowed, in dollars:

$7.75 + 11.00 + 4.50 = 23.25$

Then use subtraction to determine the difference between the amount her grandmother gave her and the amount she borrowed:

$25.00 - 23.25 = 1.75$

The difference is positive, which means that Maxie will still have $1.75 left from the amount her grandmother gave her.

Question analysis

Maxie borrowed money from her mother, father and brother to buy a video game, then received $25.00 from her grandmother to pay them back. Students decide how much money she will have left over or still owe. It tests adding and subtracting decimals with money, and the choices also test which way the difference goes.

NYS standards: 6.NS.3 multi-digit decimal operations
6.NS.5 positive and negative numbers
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
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: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Few students miss this one, since it needs only one addition and one subtraction with money. The choices come in pairs, so the right amount must also be matched with the right outcome. Lining up the decimal points, and asking whether $25.00 is more or less than the $23.25 borrowed, settles both the amount and whether Maxie owes or keeps money.

Error-prone arithmetic minor Working it out
The loans add to $23.25, and subtracting that from $25.00 means borrowing across two zeros, a classic place for a digit to go wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Each amount appears twice among the choices, once as money left and once as money owed, so the result must also be read for direction by asking which is larger, the gift or the loans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the three amounts borrowed, compare the total with the $25.00 gift, and decide from which is larger whether Maxie ends up with money left or still owes some.

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 Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two errors at once, a wrong difference such as $2.25 and the reversed outcome, so no single line of working reaches it and it mostly draws guesses.
B Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Finds the $1.75 difference correctly but reverses its meaning, as if the $25.00 gift fell short of the $23.25 borrowed instead of covering it.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts $23.25 from $25.00 column by column, taking the smaller digit from the larger instead of borrowing, so 5 − 0 and 2 − 0 give $2.25.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Money left over: $\text{left} = \text{money she has} - \text{money she owes}$; Money she has is the $25.00 gift. Money she owes is the total she borrowed. If the result is positive, she has that much left. If it were negative, she would still owe that much. Comparing the two amounts first tells you which.

Sort out the question
Maxie borrowed from three people: $7.75, $11.00 and $4.50. Her grandmother gave her $25.00. She uses the $25.00 to pay everyone back. You need to know if money is left over or if she still owes some, and how much.
Add up what she borrowed
Line up the decimal points and add:
$7.75 + 11.00 + 4.50 = 23.25$
She owes $23.25 in all. That is the amount she must pay back.
Compare the gift with the debt
The gift, $25.00, is more than the $23.25 she owes. So she can pay everyone back and still have money left. She will not owe anything.
Watch out: once you find $1.75, keep its meaning straight. Saying she still owes $1.75, choice B, gets the direction backwards.
Subtract to find what is left
Take the debt away from the gift:
$25.00 - 23.25 = 1.75$
You need to regroup (borrow) across the zeros. Think of $25.00 as 24 dollars and 100 cents. Then 100 − 25 = 75 cents, and 24 − 23 = 1 dollar.
Watch out: taking the smaller digit from the larger one in each column gives $2.25, choice D.
So the answer is C: Maxie will have $1.75 left.

Check your answer

Add back. The $23.25 she pays plus the $1.75 she keeps is 23.25 + 1.75 = 25.00. That is the whole gift, so the subtraction is right.

More questions like this

The original DOE question

Question 21 as shown on the DOE’s digital test, page 94 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 94, question 21.

Question 12 of 86

2026 Practice Test D Math, question 22

Question

Seth grows strawberries in his garden.

He picked $2\frac{1}{2}$ cups of strawberries on Monday.
He picked 4 cups of strawberries on Tuesday.
He ate $\frac{3}{4}$ cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.

How many cups of strawberries did Seth have left?

A $3\frac{1}{2}$
B $4\frac{1}{4}$
C6
D9

Show the answer

Correct answer: B

When Seth picks strawberries, the value represents a positive number, and when he eats the strawberries, the value represents a negative number. Write an expression and then evaluate the expression, where the negative value is multiplied by 3 because he ate the same amount on 3 different days.

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

$2\frac{1}{2} + 4 - 2\frac{1}{4}$

$6\frac{1}{2} - 2\frac{1}{4}$

$6\frac{2}{4} - 2\frac{1}{4}$

$4\frac{1}{4}$

If the multiplication is confusing, evaluate the expression below which represents the sum for the 5 days with each day listed separately.

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

$4\frac{1}{4}$

Question analysis

Seth picked strawberries on Monday and Tuesday, in amounts given as a mixed number and a whole number, then ate the same fraction of a cup at breakfast on three days. Students find how many cups he has left. It tests adding, multiplying and subtracting fractions in a short word problem, and the eaten amount repeats each day.

NYS standards: 7.NS.3 four operations with rational numbers
7.EE.3 multi-step rational number problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding mixed numbers and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so 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 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The bulleted list lays out the story day by day, so setting up the problem is easy; the points are lost in the fractions. The number of breakfast days, 3, sits right beside the 3/4 cup, and subtracting 3 cups instead of three quarters of a cup each day is the tempting shortcut. Writing every amount in fourths, 26/4 picked and 9/4 eaten, makes the last step a whole-number subtraction.

Error-prone arithmetic moderate Working it out
Three breakfasts of 3/4 cup make 9/4 = 2 1/4 cups, and taking that from 6 1/2 means renaming the half as 2/4, two fraction steps where a slip is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Picking adds and eating subtracts, and “each day on Wednesday, Thursday, and Friday” means the 3/4 cup counts three times, not once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the two amounts picked, multiply the breakfast amount by the number of days, and subtract that from the total; writing every amount in fourths keeps the arithmetic simple.

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Subtracts 3 cups, the number of breakfast days, instead of 3 × 3/4 = 2 1/4 cups, so 6 1/2 − 3 = 3 1/2.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives exactly 6; it is near the 6 1/2 cups picked and mostly catches students who drop the fractions or guess.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the 2 1/4 cups eaten to the 6 1/2 cups picked instead of subtracting, 8 3/4 cups, and takes the nearest choice, 9.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total of a repeated amount: $\text{total} = \text{number of times} \times \text{amount each time}$; Use it when the same amount happens again and again. Here Seth eats 3/4 cup on each of 3 days, so the number of times is 3 and the amount each time is 3/4 cup.
A whole number times a fraction: $n \times \frac{a}{b} = \frac{n \times a}{b}$; n is a whole number, and a/b is a fraction. Multiply the whole number by the top only, and keep the bottom. It works because n groups of a pieces make n × a pieces, and each piece is still size 1/b.

Sort out the question
Seth picked 2 1/2 cups on Monday and 4 cups on Tuesday. Then he ate 3/4 cup each day on Wednesday, Thursday and Friday. That is 3 days of eating. Picking adds strawberries, and eating takes them away. You need the number of cups he has left.
Add what he picked
Add the two days of picking:
$2\frac{1}{2} + 4 = 6\frac{1}{2}$
He picked 6 1/2 cups in all, before he ate any.
Find how much he ate
He ate 3/4 cup on each of 3 days. Multiply:
$3 \times \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4}$
To change 9/4 to a mixed number, note that 4 fits into 9 twice, with 1 left over. He ate 2 1/4 cups.
Watch out: he ate 3/4 cup three times, not 3 cups. Subtracting 3 gives 3 1/2, choice A.
Take away what he ate
Write the half as fourths, so both fractions have the same bottom: 1/2 = 2/4. Then subtract the wholes and the fourths:
$6\frac{1}{2} - 2\frac{1}{4} = 6\frac{2}{4} - 2\frac{1}{4} = 4\frac{1}{4}$
Watch out: eating takes away. Adding the 2 1/4 cups gives 8 3/4, and the nearest choice is 9, choice D.
So the answer is B, $4\frac{1}{4}$ cups.

Check your answer

Use decimals. 2.5 + 4 = 6.5 cups picked. 3 × 0.75 = 2.25 cups eaten. Then 6.5 − 2.25 = 4.25 cups, which is 4 1/4.

More questions like this

The original DOE question

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

Question 13 of 86

2026 Practice Test D Math, question 25

Question

A bus trip takes 9 hours if the mean speed is 50 miles per hour. How many hours would this trip take if the mean speed was 45 miles per hour?

Show the answer

Correct answer: B

First, find the length of the bus trip in miles by multiplying the time, 9 hours, by the average (mean) speed, 50 miles per hour, to get 450 miles.

Then, divide the total distance, 450 miles, by the average (mean) speed, 45 miles per hour, to get the total time:

$450 \div 45 = 10\text{ hours}$

Question analysis

A bus trip takes 9 hours at a mean speed of 50 miles per hour, and students find how long the same trip takes at 45 miles per hour. It tests speed, distance and time. Working out the fixed distance first is the key, since guessing how the time should change leads many students astray.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
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: No obvious method moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 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 short calculations solve it, but only after noticing that speed and time move in opposite directions: a slower bus takes longer. Students who scale the 9 hours by 45/50, as in a direct proportion, get a little over 8 hours, a result that should feel wrong for a slower trip. Finding the distance first turns the question into one division with no direction to decide.

No obvious method moderate Working it out
The question never mentions the trip's length, yet the 450-mile distance is what stays fixed; setting up a proportion between hours and speeds, the usual reflex, points the wrong way. (Schoenfeld, 1985)
Something you must already know minor Working it out
Students must bring distance = speed × time themselves, to turn 9 hours at 50 miles per hour into miles before dividing by the new speed. (Anderson & Krathwohl, 2001)

Strategy: Find the trip's length in miles from the first speed and time, since that length stays the same, then divide it by the new mean speed.

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: strong
Treats time as proportional to speed, scaling 9 hours by 45/50 to 8.1, about 8, though a slower bus needs more time, not less.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12 hours; it needs a slip such as 540 miles for 9 × 50, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the 5-mile-per-hour drop in speed to the 9 hours, 9 + 5 = 14, combining numbers from the question without a model of the trip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, speed and time: $d = r \times t$; d is the distance traveled. r is the rate, or speed, such as miles per hour. t is the time. It works because at 50 miles each hour, 9 hours covers 9 groups of 50 miles.
Finding the time: $t = d \div r$; This is the same rule turned around. d is the distance and r is the speed. Divide the distance by the speed to find t, the number of hours. Use it when you know how far and how fast.

Sort out the question
A bus trip takes 9 hours at a mean speed of 50 miles per hour. The mean speed is the average speed over the whole trip. You need the time for the same trip at 45 miles per hour. The length of the trip does not change, so find it first.
Find the length of the trip
Use distance equals speed times time, with a speed of 50 miles per hour and a time of 9 hours:
$d = 50 \times 9 = 450$
The trip is 450 miles long.
Find the new time
Now the bus goes 45 miles per hour over the same 450 miles. Divide the distance by the speed:
$t = 450 \div 45 = 10$
The trip takes 10 hours.
Check that it makes sense
A slower bus needs more time for the same trip. 10 hours is more than 9 hours, so it fits.
Watch out: time does not shrink with speed. Scaling 9 hours by 45/50 gives about 8 hours, choice A, which would make the slower bus arrive sooner. Adding the drop of 5 miles per hour to 9 hours gives 14, choice D, which has no reason behind it.
So the answer is B, 10 hours.

Check your answer

Check the distance both ways. At 50 miles per hour for 9 hours, the bus goes 50 × 9 = 450 miles. At 45 miles per hour for 10 hours, it goes 45 × 10 = 450 miles. Same trip, same distance.

More questions like this

The original DOE question

Question 25 as shown on the DOE’s digital test, page 98 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 98, question 25.

Question 14 of 86

2026 Practice Test D Math, question 29

Question

Karl has one red spinner and one blue spinner. Each spinner is divided into 4 equal sections, numbered 1 through 4. He spins each spinner once and writes down the number that each lands on. What is the probability that the two numbers, when multiplied together, will have 4 as a product?

A $\frac{1}{16}$
B $\frac{3}{16}$
C $\frac{4}{7}$
D $\frac{7}{16}$

Show the answer

Correct answer: B

There are four possible outcomes for each spin of the spinner:

1, 2, 3, 4

The possible outcomes of two spins are:

$\begin{array}{cccc} 11 & 12 & 13 & \boxed{14} \\ 21 & \boxed{22} & 23 & 24 \\ 31 & 32 & 33 & 34 \\ \boxed{41} & 42 & 43 & 44 \end{array}$

Of the 16 possible outcomes, only 3 pairings have a product of 4: 14, 22, and 41.

Therefore, the probability that the two numbers resulting from two consecutive spins have a product of 4 is $\frac{3}{16}$ .

Question analysis

Karl spins a red spinner and a blue spinner, each with four equal sections numbered 1 to 4, and multiplies the two results. This compound probability question asks for the chance that the product is 4. Students build the sample space of outcomes, and listing every pair rather than the ones that come to mind is the safe route.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a compound probability from a list or table of equally likely outcomes is standard grade 7 work. Two four-section spinners give a small sample space that is quick to write out, and the event needs one careful scan of the products.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 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 question is decided by the count of favorable outcomes, not by any calculation. Students who think of 4 only as 2 × 2 find one outcome, and those who think in factor pairs find two, which is not a listed choice and should send them back to the table. A 4-by-4 grid of products, red down the side and blue across, shows all three cells at once.

Something you must already know moderate Working it out
Nothing in the stem says that red 1 with blue 4 and red 4 with blue 1 are separate outcomes, or that two four-part spinners give 16 equally likely pairs. (Anderson & Krathwohl, 2001)
Much to keep track of minor Working it out
All 16 pairs must be scanned for a product of 4, and the two pairs that use 1 and 4 are easy to miss beside the obvious 2 times 2. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Make a 4-by-4 table with the red spinner's numbers down the side and the blue spinner's across the top, write each product in its cell, and count the cells that match.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds only the double, 2 and 2, and stops, missing the two outcomes in which a 1 on one spinner meets a 4 on the other.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No count of the 16 outcomes gives a denominator of 7, so 4/7 has no working behind it and mostly attracts guesses.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Counts the 7 outcomes in which either spinner shows a 4, reading “4 as a product” as a pair that simply contains a 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of outcomes you want}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. An outcome is one possible result. This works when every outcome is equally likely, like landing on equal sections of a spinner.
Counting outcomes of two actions: $\text{total outcomes} = a \times b$; a is the number of results for the first action, and b is the number for the second. Each result of the first can go with each result of the second, so you multiply.

Sort out the question
Karl spins two spinners, one red and one blue. Each has 4 equal sections, numbered 1 to 4. He multiplies the two numbers. You need the probability that the product, the answer to a multiplication, is 4.
Count all the outcomes
The red spinner has 4 results, and so does the blue one. Each red result can go with each blue result:
$4 \times 4 = 16$
So there are 16 outcomes, all equally likely. Red 1 with blue 4 is not the same as red 4 with blue 1. They come from different spinners.
Make a table of products
Write each product in a table. Red goes down the side and blue goes across the top.
Red × Blue 1 2 3 4
1 1 2 3 4
2 2 4 6 8
3 3 6 9 12
4 4 8 12 16
Three cells show 4: red 1 with blue 4, red 2 with blue 2, and red 4 with blue 1.
Watch out: thinking of 4 only as 2 × 2 finds one outcome. That gives 1/16, choice A.
Find the probability
Use the probability rule. Three outcomes give 4, out of 16 in all:
$P = \frac{3}{16}$
Watch out: 7 outcomes have a 4 on at least one spinner. But that counts pairs that show a 4, not pairs whose product is 4. That mistake gives 7/16, choice D.
So the answer is B, $\frac{3}{16}$ .

Red × Blue	1	2	3	4
1	1	2	3	4
2	2	4	6	8
3	3	6	9	12
4	4	8	12	16

Check your answer

List the ways to multiply two numbers from 1 to 4 and get 4: $1 \times 4$ , $2 \times 2$ and $4 \times 1$ . That is 3 ways. The table has 16 cells, and 13 of them are not 4. So a small chance like 3/16 makes sense.

More questions like this

The original DOE question

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

Question 15 of 86

2026 Practice Test D Math, question 41

Question

The distance an eagle flies is proportional to the time the eagle spends flying. An eagle flies 90 miles in 3 hours. What is the constant of proportionality for this relationship?

Show the answer

Correct answer: B

The constant of proportionality is the ratio of y (90) to x (3). $90 \div 3 = 30$ . So the constant of proportionality is 30.

Question analysis

An eagle flies 90 miles in 3 hours, and the distance it flies is proportional to its flying time. This proportional relationship question asks for the constant of proportionality. Students need to know that the constant is the unit rate, here miles per hour, and find it from the two values given.

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students find constants of proportionality in tables, graphs and equations. This one is a single division of miles by hours given in a sentence, the same unit rate work students do in grade 6.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 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

One division answers the question, so it is among the easiest on the test. The wrong choices are the two numbers in the stem and one unrelated value, so a student who knows what the constant means is not tempted. Checking that the constant times 3 hours gives back the 90 miles confirms it.

Something you must already know minor Working it out
Students must know that the constant of proportionality is the unit rate, distance over time, here the miles flown in a single hour; the stem gives only the term. (Anderson & Krathwohl, 2001)

Strategy: Divide the distance by the time, because the constant of proportionality is the distance flown in one hour, then check by multiplying it by the 3 hours.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the 3 hours printed in the stem as the constant, instead of the miles flown in each hour.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 90 miles and 3 hours leads to 60, which may echo the 60 minutes in an hour, so it mostly attracts guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives the 90 miles, the distance for the whole 3-hour flight, as if the eagle covered it in a single hour.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx \text{, so } k = \frac{y}{x}$; x is one amount, here the time in hours, and y is the other, here the distance in miles. k is the constant of proportionality. It is the same for every pair, and it equals y divided by x. Here it is the miles flown in 1 hour.

Sort out the question
The eagle flies 90 miles in 3 hours. Proportional means the distance is always the same number times the time. That number is the constant of proportionality. You need to find it.
Divide the distance by the time
Use the formula. The distance is y and the time is x:
$k = \frac{90}{3} = 30$
The constant is 30. It means the eagle flies 30 miles in each hour.
Avoid the numbers in the question
Watch out: 3 is the number of hours, choice A. And 90 is the distance for all 3 hours, choice D. Neither is the distance for 1 hour.
So the answer is B, 30.

Check your answer

Multiply back. At 30 miles each hour, the eagle flies $30 \times 3 = 90$ miles in 3 hours. That matches the question. Hour by hour, it goes 30, 60, then 90 miles.

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 D Math, page 114, question 41.

Question 16 of 86

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 17 of 86

2024 Test A Math, question 15

Question

A program on a computer randomly generates a sequence of whole numbers from 1 to 9, inclusive. If the computer generates a sequence of 300 numbers, what is the best prediction of the number of odd numbers in the sequence?

A120
B133
C150
D167

Show the answer

Correct answer: D

There are five odd numbers in the range 1 through 9: 1, 3, 5, 7, and 9.

So the proportion of odd numbers in the range 1 through 9 is $\frac{5}{9}$ .

The expected number of odd values in the sequence from the computer program is

$300 \bullet \frac{5}{9} = 166.\overline{66}.$

So the best prediction for the number of odd numbers in the sequence is 167.

Question analysis

A probability question: a computer randomly generates 300 whole numbers from 1 to 9, and students predict how many will be odd. It asks students to find the probability of an odd number and use it to predict a count. Counting carefully which of the nine numbers are odd, and rounding the prediction sensibly, are the points to watch.

NYS standards: 6.SP.7 long-run relative frequency
6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a probability to predict how often an event happens is grade 6 and 7 work. This takes two short steps plus a rounding decision, about the demand of a typical grade 7 probability task.
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
Something you must already know minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 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 method is short, a probability times 300, so the question is decided by the probability. Odd and even feel like an even split, and half of 300 is a listed choice, but the numbers 1 to 9 hold five odd and four even. Listing the nine numbers before computing removes the trap, and the result, about 167, then needs only rounding.

Turning words into math minor Understanding the question
The phrase "whole numbers from 1 to 9, inclusive" means nine equally likely values, five of them odd, so the chance of odd is 5/9, not the one half that odd and even suggest. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The stem never says that an expected count is the probability times the number of trials, here 5/9 of the 300 numbers generated. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The product 300 × 5/9 is 166.67, not a whole number, so the prediction has to be rounded to the nearest whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the odd numbers from 1 to 9, write the probability as that count over 9, and multiply by 300, rounding to the nearest whole number.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No count of the nine numbers gives 120, which is 40% of 300; reaching it takes two errors at once, so it mostly attracts guesses.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Counts the four even numbers instead of the five odd ones and predicts 4/9 of 300, about 133.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Assumes odd and even numbers are equally likely and takes half of 300, ignoring that 1 to 9 holds one more odd number than even.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of the results that you want.
Expected count: $E = P \times n$; E is the expected count, your best guess of how many times something will happen. P is its probability, and n is the number of tries. If something happens 5 times out of 9, it should happen about 5 ninths of the time in many tries.

Sort out the question
The computer picks whole numbers from 1 to 9. Inclusive means 1 and 9 count too. Randomly means each number has the same chance. It picks 300 numbers. You need the best guess of how many of them are odd.
Count the odd numbers
An odd number does not split evenly by 2. From 1 to 9, the odd numbers are 1, 3, 5, 7 and 9. That is 5 of the 9 numbers. The even ones, 2, 4, 6 and 8, are only 4. Watch out: odd and even are not equally likely here, since there is one more odd number. Taking half of 300 gives 150, choice C.
Find the chance of odd
$P = \frac{\text{wanted results}}{\text{possible results}}$
$P = \frac{5}{9}$
Each number the computer picks has a 5 in 9 chance of being odd. Watch out: do not use the 4 even numbers. That gives about 133, choice B.
Multiply by the 300 numbers
$E = P \times n$
$E = \frac{5}{9} \times 300 = \frac{1500}{9} \approx 166.7$
The sign ≈ means about. You can’t have part of a number, so round to the nearest whole number. 166.7 is closer to 167 than to 166. So the answer is D, 167.

Check your answer

The odd numbers are a bit more than half of the nine. So the answer should be a bit more than half of 300, which is 150. Only 167 is a bit more than 150. Also, 167 odd plus about 133 even makes 300.

More questions like this

The original DOE question

Question 72 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 72. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 18 of 86

2024 Test A Math, question 17

Question

The probability of drawing a red candy at random from a bag of 25 candies is $\frac{2}{5}$ . After 5 red candies are removed from the bag, what is the probability of randomly drawing a red candy from the bag?

A0
B $\frac{1}{10}$
C $\frac{1}{5}$
D $\frac{1}{4}$

Show the answer

Correct answer: D

Start with a proportion to determine the total number of red candies in the bag:

$\frac{2}{5} = \frac{r}{25}$

$\frac{2}{5}(25) = r$

$10 = r$

There are currently 10 red candies out of 25 candies in the bag. After 5 red candies are removed, there would be 5 red candies left in the bag and 20 total candies.

The probability of randomly drawing a red candy would then be $\frac{5}{20} = \frac{1}{4}$ .

Question analysis

A probability question about a bag of 25 candies in which the chance of drawing red is 2/5; then 5 red candies are taken out. Students find the new probability of drawing a red candy. They must work out how many red candies there were, then rebuild the fraction from the counts left in the bag.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability as a fraction of outcomes is grade 6 and 7 content. This adds a step back from the probability to a count and an update after removal, which is typical grade 7 multi-step work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question gives a probability, not a count, so the first job is to rebuild the bag: 10 red candies among 25. Removing candies changes two numbers at once, and the strongest wrong choice comes from updating the red count while leaving the total at 25. A student who writes both new counts before forming the fraction avoids it.

Turning words into math moderate Understanding the question
The probability 2/5 has to be turned into a count, 10 red of 25, and "after 5 red candies are removed" lowers both the red count and the bag's total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three steps follow in order: 2/5 of 25 = 10, then 5 red among 20 candies left, then 5/20 reduced to lowest terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use 2/5 of 25 to find the number of red candies, subtract 5 from both the red count and the total, and write the new fraction in lowest terms.

Traps

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

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Takes 2/5 of 25 as 25 ÷ 5 = 5, forgetting to multiply by 2, so removing 5 red candies seems to leave none.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working gives 1/10, which would need 2 red candies among the 20 left; it mostly draws students who are guessing.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Subtracts the 5 removed candies from the 10 red ones but keeps 25 as the total, finding 5/25 = 1/5, though the bag now holds 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{red candies}}{\text{all candies}}$; P is the probability, the chance of drawing a red candy. Red candies is how many red ones are in the bag. All candies is how many candies are in the bag in all. Each candy is equally likely, so the chance is the fraction of the bag that is red.

Sort out the question
The bag has 25 candies. The chance of drawing red is $\frac{2}{5}$ . Then 5 red candies are taken out. You need the new chance of drawing red. Taking out candies changes both the red count and the total.
Find the number of red candies
A chance of $\frac{2}{5}$ means 2 out of every 5 candies are red. Call the red count r.
$P = \frac{\text{red candies}}{\text{all candies}}$
$\frac{2}{5} = \frac{r}{25}$
$r = \frac{2}{5} \times 25 = 2 \times 5 = 10$
There are 10 red candies. Watch out: do not stop at $25 \div 5 = 5$ . You must also multiply by 2. Thinking there are 5 red candies leads to 0, choice A.
Take out the 5 red candies
$\text{red: } 10 - 5 = 5$
$\text{all: } 25 - 5 = 20$
Now 5 of the 20 candies are red. Watch out: the total goes down too. Keeping 25 gives $\frac{5}{25} = \frac{1}{5}$ , choice C.
Find the new probability
$P = \frac{5}{20} = \frac{1}{4}$
To simplify, divide the top and bottom by 5. So the answer is D, $\frac{1}{4}$ .

Check your answer

At first, $\frac{2}{5}$ is 40% of the bag. Taking out only red candies should lower the chance. $\frac{1}{4}$ is 25%, which is lower but not zero. That makes sense, because 5 red candies are still in the bag.

More questions like this

The original DOE question

Question 74 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 74. 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 19 of 86

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 20 of 86

2024 Test A Math, question 36

Question

CELL PHONE SALES BY COLOR

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

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

A18%
B20%
C22%
D28%

Show the answer

Correct answer: B

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

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

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

$7k + 65 = 100$

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

$7k = 35$

$k = 5$

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

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

$30 - 10 = 20$

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 21 of 86

2024 Test A Math, question 40

Question

The price of a sandwich was raised from $6.25 to $6.75. What was the percent increase in the price?

A5%
B8%
C7%
D50%

Show the answer

Correct answer: B

Use the percent increase expression to find the answer:

$\frac{\text{new price} - \text{original price}}{\text{original price}}$

$\frac{6.75 - 6.25}{6.25} = \frac{0.50}{6.25} = 0.08 = 8\%$

Question analysis

A sandwich's price went up from $6.25 to $6.75, and this percent change question asks for the percent increase in the price. Students divide the 50-cent increase by one of the prices and write the result as a percent. Picking which price is the base, and doing the decimal division accurately, decide whether the answer comes out right.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a percent increase is a core grade 7 skill, and this is the plain version with no context beyond two prices. Its demand lies in measuring the change against the correct price and dividing by $6.25 without a calculator.
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
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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 50-cent rise is easy to find; the percent depends on what it is compared with. Dividing by the new price, $6.75, gives a value that rounds to a listed choice one point below the answer, and reading 50 cents as 50 percent gives another. The choices are printed out of order, 8% before 7%, so students should read each one rather than scan the list.

A very close wrong answer moderate Choosing the answer
Measured against $6.75 the 50-cent rise is about 7.4%, and against $6.25 it is 8%; both are listed side by side, so only the right base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic moderate Working it out
The division 0.50 ÷ 6.25 is easiest as 50/625, which reduces to 2/25 and then to 8/100; a three-digit decimal divisor invites a slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Put the 50-cent change over the original price, 0.50 over 6.25, then scale the fraction to hundredths or divide to read off the percent.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 5%: dividing the 50-cent rise by either price does not come near it, so it mostly draws guesses.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the 50-cent increase by the new price, $6.75, instead of the original, which gives about 7.4%, so 7%.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 50-cent increase as 50 percent, treating cents as a percent of one dollar rather than of the $6.25 price.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; New is the price after the change, and original is the price before it. The top is the amount of the increase. Dividing by the original compares the increase with where you started. Multiplying by 100 turns that decimal into a percent.

Sort out the question
The sandwich cost $6.25 before. That is the original price. Now it costs $6.75, the new price. You need the percent increase. That compares the increase with the original price. Note that the choices are not in order: C, 7%, comes after B, 8%.
Find the amount of the increase
$6.75 - 6.25 = 0.50$
The price went up $0.50, or 50 cents. Watch out: 50 cents is not 50%. It would be 50% only if the price started at $1.00. That mistake gives choice D.
Divide by the original price
$\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$
$\frac{6.75 - 6.25}{6.25} \times 100 = \frac{0.50}{6.25} \times 100$
In cents, this is 50 out of 625. Divide the top and bottom by 25. Then multiply both by 4 to get hundredths.
$\frac{50}{625} = \frac{2}{25} = \frac{8}{100} = 0.08$
Change to a percent
$0.08 \times 100 = 8$
The price went up 8%. Watch out: divide by the original price, $6.25, not the new one. Dividing by $6.75 gives about 7.4%, which is choice C.
So the answer is B, 8%.

Check your answer

Work forward from the answer. 8% of $6.25 is $0.08 \times 6.25 = 0.50$ , or 50 cents. Then $6.25 + 0.50 = 6.75$ . That is the new price, so 8% is right.

More questions like this

The original DOE question

Question 97 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 2025 Admissions, page 86, question 97. Open page 86 of the handbook (PDF)

Question 22 of 86

2024 Test A Math, question 41

Question

Terrell played 5 computer games and earned a mean score of 8 points per game. If his mean score for the first 4 games was 7 points per game, how many points was his score in the fifth game?

Show the answer

Correct answer: C

If the mean number of points for the first four games is 7, the total number of points earned in those games is 28.

$4 \times 7 = 28$

If the mean number of points for all five games is 8, the total number of points earned is 40.

$5 \times 8 = 40$

To find the number of points in the fifth game, find the difference between 40 and 28.

$40 - 28 = 12$

Question analysis

This mean, or average, word problem says Terrell averaged 8 points over 5 computer games and 7 points over the first 4, and asks for his score in the fifth game. Students turn each average into a total and compare the totals. Combining the two averages directly, without going through totals, is the tempting shortcut that fails.

NYS standards: 6.SP.5c range, center, and overall pattern
6.SP.3 measures of center and variation
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Computing a mean is grade 6 content. Working backward from two means to the totals they stand for, then to one missing score, is a common grade 7 test task, and it comes with a tempting shortcut.
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
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards turning both means into totals, 40 points for five games and 28 for four, after which one subtraction finishes it. Averaging 7 with the unknown score is a quick shortcut that gives a listed choice, but it weighs four games as one. Checking that 28 plus the answer, divided by 5, gives 8 catches that error and the near miss next to the answer.

Turning words into math moderate Understanding the question
Each mean has to become a total before the fifth game can be found: "a mean score of 8 points per game" over 5 games means 40 points in all. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Students must know that a total is the mean times the number of games, and that means over different numbers of games cannot simply be averaged. (Anderson & Krathwohl, 2001)

Strategy: Turn each mean into a total by multiplying by its number of games, then subtract the four-game total from the five-game total.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Averages the four-game mean with the fifth score, (7 + x) ÷ 2 = 8, so x = 9, as if the first four games counted once.
B Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Sees the fifth game must lift the mean 1 point above 7 but spreads that over only 4 games, 7 + 4 = 11, instead of all 5.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 14, which is twice the four-game mean and matches no total or difference, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total from a mean: $\text{total} = \text{mean} \times \text{number of games}$; The mean, or average, is what each game would score if the total were shared out equally. So the total is the mean times the number of games. Use it to turn each mean back into a total. Totals can be subtracted, but means over different numbers of games can’t.

Sort out the question
Terrell played 5 games. His mean score for all 5 was 8 points per game. His mean for the first 4 games was 7 points per game. The mean, or average, is the total split equally among the games. You need his score in game 5.
Find the total for all 5 games
Turn the mean for all 5 games into a total. Multiply the mean by the number of games.
$\text{total} = \text{mean} \times \text{number of games}$
$\text{total} = 8 \times 5 = 40$
He scored 40 points in all 5 games together.
Find the total for the first 4 games
Use the same formula with the mean of 7 and 4 games.
$\text{total} = 7 \times 4 = 28$
He scored 28 points in the first 4 games.
Subtract to find game 5
The 40 points include the first 4 games and game 5. Take away the first 4 games to leave game 5.
$40 - 28 = 12$
He scored 12 points in game 5. Watch out: don’t average 7 and the fifth score as if they were two games. That gives 9, choice A. The 7 stands for 4 games, not 1. Watch out: adding 4 to 7 gives 11, choice B. Each of the first 4 games was 1 point below 8. So game 5 needs 8 plus 4 more points.
So the answer is C, 12 points.

Check your answer

Try scores of 7, 7, 7, 7 and 12. They add up to $28 + 12 = 40$ , and $40 \div 5 = 8$ . That is the mean for all 5 games, so 12 is right.

More questions like this

The original DOE question

Question 98 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 98. 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 23 of 86

2024 Test A Math, question 42

Question

Lian bought enough oranges to fill 4 bags. Each bag contains 8 oranges. The total cost was $11.52. At that rate, how much would Lian pay for 42 oranges?

A$17.28
B$15.12
C$15.02
D$12.52

Show the answer

Correct answer: B

First, find the amount, in dollars, that Lian paid for each orange. There are 4 bags with 8 oranges each.

$4 \times 8 = 32$

Divide the total cost by the total number of oranges to get the cost per orange:

$11.52 \div 32 = 0.36$

Multiply to find the cost of 42 oranges:

$0.36 \times 42 = 15.12$

Question analysis

This rate word problem says Lian paid $11.52 for 4 bags of 8 oranges each and asks how much 42 oranges would cost at the same rate. Students find the price of one orange and multiply. Since 42 oranges is not a whole number of bags, working by the bag instead of by the orange invites mistakes.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a unit price and scaling it up is grade 6 rate work. The decimals are multi-digit, but the plan is routine, with less to manage than grade 7 problems built on percents or negative numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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 plan is familiar, total oranges, price per orange, then the new cost, so the work lies in the decimals. Because 42 oranges is five bags and two oranges, students who price by the bag either round up to six bags or must handle the leftover. Estimating, since 42 is about a third more than 32, gives roughly $15 and rules out the far choices but not the near one.

Error-prone arithmetic moderate Working it out
Two decimal operations are done by hand: 11.52 ÷ 32 = 0.36 for one orange, then 0.36 × 42 = 15.12, where a dropped carry is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The choices $15.12 and $15.02 differ by 10 cents, so a rough estimate of about $15 cannot choose between them; only the exact product can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply to find how many oranges were bought, divide the total cost by that number to get the price of one orange, then multiply by 42.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Prices whole bags, rounding 42 oranges up to 6 bags of 8, and pays 6 × $2.88 = $17.28, missing that "at that rate" means per orange.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Uses the right method but slips one digit in multiplying 0.36 by 42, a tens-place error that lands 10 cents short of the total.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives $12.52, exactly one dollar more than $11.52; no price per orange or per bag leads there, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit price: $\text{unit price} = \frac{\text{total cost}}{\text{number of items}}$; The unit price is what one item costs. Divide the total cost by how many items you got. It works because the total is shared equally among the items.
Cost: $\text{cost} = \text{unit price} \times \text{number of items}$; The unit price is the cost of one item, here one orange. Multiply it by how many items you buy to get the total cost. This works when every item costs the same.

Sort out the question
Lian bought 4 bags with 8 oranges in each bag. She paid $11.52 in all. You need the cost of 42 oranges. “At that rate” means each orange costs the same as before. So find the price of one orange first.
Count the oranges
Each of the 4 bags holds 8 oranges, so multiply to find how many she bought.
$4 \times 8 = 32$
Lian bought 32 oranges for $11.52.
Find the price of one orange
$\text{unit price} = \frac{\text{total cost}}{\text{number of items}}$
$\text{unit price} = \frac{11.52}{32} = 0.36$
This is easier in cents: $1{,}152 \div 32 = 36$ . So one orange costs $0.36, or 36 cents.
Find the cost of 42 oranges
$\text{cost} = \text{unit price} \times \text{number of items}$
$\text{cost} = 0.36 \times 42$
Break 42 into 40 and 2 to make it easier.
$0.36 \times 40 = 14.40$
$0.36 \times 2 = 0.72$
$14.40 + 0.72 = 15.12$
The 42 oranges cost $15.12.
Avoid the traps and answer
Watch out: $15.02, choice C, is only 10 cents away. Multiply and add with care. Watch out: Lian pays for each orange, not for full bags. Buying 6 whole bags would cost $17.28, choice A.
So the answer is B, $15.12.

Check your answer

42 oranges is 10 more than 32. Those 10 oranges cost $3.60, since $10 \times 0.36 = 3.60$ . Then $11.52 + 3.60 = 15.12$ . Same answer, so $15.12 is right.

More questions like this

The original DOE question

Question 99 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 99. Open page 87 of the handbook (PDF)

Question 24 of 86

2024 Test A Math, question 44

Question

A tank with a 500-gallon capacity currently contains 75 gallons of water. Additional water is poured into this tank at a rate of 5 gallons per minute. After 45 minutes of adding water, what percentage of the tank’s total capacity will be filled? (Assume that there is no loss of water from the tank.)

A45%
B55%
C60%
D70%

Show the answer

Correct answer: C

If 5 gallons of water is added to the tank per minute for 45 minutes, then $5 \times 45 = 225$ gallons is added to the tank.

The tank started with 75 gallons of water in it. After the 45 minutes, it contains $75 + 225 = 300$ gallons, assuming no loss of water.

The tank has a capacity of 500 gallons, and $\frac{300}{500} = 0.6$ .

Rewrite decimal 0.6 as a percentage:

$0.6 = \frac{6}{10} = \frac{60}{100} = 60\%.$

Question analysis

A 500-gallon tank holds 75 gallons, and water flows in at 5 gallons per minute for 45 minutes. This rate and percent word problem asks what percentage of the tank's capacity will then be filled. Students find the water added, the new amount in the tank and its percent of 500, keeping track of every quantity of water.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve multistep rate and percent problems. This one combines a pouring rate, a starting amount and a percent of capacity in three short steps with friendly numbers, which matches typical grade 7 work.
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: Much to keep track of minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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 friendly, so the question tests whether students track every quantity of water. Forgetting the 75 gallons that were there from the start is the costly slip, because the share of the tank that was poured in is itself a listed choice. A quick check, that 300 gallons is more than half of 500 but well under three quarters, confirms the answer.

Much to keep track of minor Working it out
The 75 gallons already in the tank, given in the first sentence, must be held aside through the rate calculation and added back before dividing. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Many steps minor Working it out
The chain runs from 5 gallons a minute for 45 minutes to 225 gallons, then to 300 gallons in the tank, then to 300 out of 500 as a percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the rate by the time to get the water added, account for the water already in the tank, and divide the total by the 500-gallon capacity.

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
Finds that the 225 gallons poured in fill 45% of the tank and stops before adding the 15% filled by the 75 gallons already there.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Leaves out the starting 75 gallons and gives the share still empty after pouring, (500 − 225) ÷ 500 = 55%, instead of the share filled.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 70%, which would need 350 gallons, as if water flowed for 55 minutes; it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Amount from a rate: $\text{amount} = \text{rate} \times \text{time}$; The rate is how much water goes in each minute, here 5 gallons. The time is how many minutes it runs. Multiply them to get the amount added. It works because the same amount goes in every minute.
Percent of a whole: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you have, and the whole is the full amount. Dividing gives what fraction of the whole you have. Multiplying by 100 turns it into a percent, which means out of 100.

Sort out the question
The tank holds at most 500 gallons. That is its capacity. It already has 75 gallons in it. Water is added at 5 gallons per minute for 45 minutes. You need the percent of the tank that will be full.
Find the water added
The same amount of water goes in every minute, so multiply the rate by the time.
$\text{amount} = \text{rate} \times \text{time}$
$\text{amount} = 5 \times 45 = 225$
So 225 gallons are poured in during the 45 minutes.
Find the total water in the tank
Add the water that was already there.
$75 + 225 = 300$
The tank holds 300 gallons after 45 minutes. Watch out: don’t forget the first 75 gallons. The 225 gallons alone fill 45% of the tank, choice A.
Find the percent full
$\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$
$\text{percent} = \frac{300}{500} \times 100 = \frac{3}{5} \times 100 = 60$
The tank is 60% full. Watch out: find the part that is full, not the part that is empty. Choice B, 55%, is the empty part if you also forget the first 75 gallons.
So the answer is C, 60%.

Check your answer

10% of 500 gallons is 50 gallons. The tank has 300 gallons, which is 6 groups of 50. So it is 60% full. Also, 300 gallons is less than 500, so the tank never overflows.

More questions like this

The original DOE question

Question 101 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 101. Open page 87 of the handbook (PDF)

Question 25 of 86

2024 Test A Math, question 54

Question

The probability of an event occurring is 0.05. What is the chance that the event will occur?

Alikely
Bunlikely
Cimpossible
Dneither likely nor unlikely

Show the answer

Correct answer: B

The probability of the event is 0.05, which equals 5%, which is less than 50%.

Option A:

An event that is likely has a probability over 50%.

Option B:

An event that is unlikely has a probability less than 50% but greater than 0%.

Option C:

An event that is impossible has a 0% probability.

Option D:

An event that is neither likely nor unlikely has a 50% probability.

Question analysis

This probability question gives an event a probability of 0.05 and asks how likely the event is to occur, described in words. Students place the decimal on the scale from 0, impossible, to 1, certain. Reading the place value of the decimal correctly is the only real hurdle.

NYS standards: 6.SP.6 probability from 0 to 1
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Describing in words how likely an event is from its probability between 0 and 1 is a grade 6 standard. This is one quick judgment with no computation, well below grade 7 probability work with compound events.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One judgment answers the question: 0.05 is a small probability, but it is not zero. The two tempting errors are treating a small chance as no chance at all and misreading 0.05 as 0.5, an even chance. Writing the probability as 5% or as 1 in 20 makes its size plain.

Something you must already know minor Working it out
The probability scale, 0 for impossible, 1/2 for an even chance and 1 for certain, is assumed, and 0.05 has to be placed near its low end. (Anderson & Krathwohl, 2001)

Strategy: Picture the probability scale from 0 to 1, with 0 for impossible, 1/2 for an even chance and 1 for certain, and see where 0.05 falls.

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common reading makes 0.05 likely, which would need a probability above one half, so it mostly draws guesses.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats a probability close to 0 as exactly 0, as if an event with a 1 in 20 chance could never happen.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Reads 0.05 as 0.5, a place-value slip, which would mean an even chance of happening or not.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability scale: $0 \leq P \leq 1$; P is the probability of an event, a number that tells how likely it is to happen. It is always from 0 to 1. 0 means impossible, and 1 means certain. One half, or 0.5, means an even chance. The closer P is to 0, the less likely the event.

Sort out the question
Probability is a number that tells how likely an event is. An event is something that might happen. Here the probability is 0.05. You need the word that best describes that chance.
Change 0.05 to a percent
$0.05 = \frac{5}{100}$
So 0.05 is 5 out of 100, or 5%. That is the same as 1 chance in 20. Watch out: 0.05 is not 0.5. A probability of 0.5 would be an even chance, choice D.
Place it on the scale
$0 \leq P \leq 1$
Probability Chance
0 impossible
more than 0, less than 0.5 unlikely
0.5 neither likely nor unlikely
more than 0.5, less than 1 likely
1 certain
0.05 is more than 0 but much less than 0.5. So the event is unlikely.
Rule out the other choices
Watch out: a small chance is still a chance. 0.05 is more than 0, so the event can happen. It is not impossible, choice C. Likely, choice A, would need more than 0.5.
So the answer is B, unlikely.

Probability	Chance
0	impossible
more than 0, less than 0.5	unlikely
0.5	neither likely nor unlikely
more than 0.5, less than 1	likely
1	certain

Check your answer

Picture 20 tries. With a 5% chance, the event would happen about once, since 5% of 20 is 1. It would not happen about 19 times. That is rare but possible, so unlikely fits.

More questions like this

The original DOE question

Question 111 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 2025 Admissions, page 90, question 111. Open page 90 of the handbook (PDF)

Question 26 of 86

2024 Test B Math, question 9

Question

A car dealership is having a sale where all cars have a 12% discount. What is the discounted price of a car that originally cost $15,600?

A$13,728
B$14,300
C$14,400
D$15,588

Show the answer

Correct answer: A

$15{,}600\times 12\%=$

$15{,}600\times 0.12=1{,}872$

$15{,}600-1{,}872=13{,}728$

Question analysis

A percent word problem about a car sale: every car is 12% off, and one car originally cost $15,600. Students find the sale price of that car. The discount has to be worked out as a dollar amount and then taken off, with care over the decimal point in 12%.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Discounts are one of the percent applications in grade 7, and this is a routine two-step example with whole-dollar amounts. It matches typical grade 7 work, with no reverse percent or second change.
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: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 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

This is a standard two-step discount, and a quick estimate does most of the work: 10% of $15,600 is $1,560, so 12% is a little under $1,900, and only one choice sits about that far below the price. The wrong choices take off $1,300, $1,200 or just $12, so the question mainly catches students unsure what 12% of a price means.

Error-prone arithmetic minor Working it out
Finding 12% of $15,600 by hand, 0.12 × 15,600 = 1,872, means placing a decimal point in a product of large numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “discounted price” asks for what the buyer pays, so the $1,872 discount has to be taken off the price rather than reported. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write 12% as 0.12 and multiply by the original price to get the discount, then subtract it; or multiply the price by 0.88 in one step.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Divides the price by 12, as if 12% meant one twelfth, and subtracts the $1,300 that gives instead of $1,872.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives $14,400; the $1,200 it takes off is 12% of $10,000, not of $15,600, so it mostly catches guesses.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Subtracts $12 from the price, treating the 12 in 12% as a number of dollars instead of 12 hundredths of the price.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $x = \frac{p}{100} \times W$; W is the whole amount, here the first price. p is the percent, and x is the part you want. Percent means out of 100, so p percent is the fraction p over 100. Multiplying the whole by that fraction gives the part.
Sale price: $\text{sale price} = \text{first price} - \text{discount}$; The first price is what the car cost before the sale. The discount is the amount taken off. The sale price is what the buyer pays, so you subtract.

Sort out the question
The car first cost $15,600. A discount is an amount taken off the price. Here it is 12% of the price. You need the discounted price, the amount the buyer pays after the discount.
Find 12% of the price
$x = \frac{p}{100} \times W$
$x = \frac{12}{100} \times 15{,}600 = 0.12 \times 15{,}600$
Break it into easy parts. 10% is $15{,}600 \div 10 = 1{,}560$ . 1% is $15{,}600 \div 100 = 156$ , so 2% is $2 \times 156 = 312$ .
$1{,}560 + 312 = 1{,}872$
The discount is $1,872. Watch out: 12% does not mean $12. Taking off $12 gives $15,588, choice D.
Take the discount off
$\text{sale price} = \text{first price} - \text{discount}$
$15{,}600 - 1{,}872 = 13{,}728$
The buyer pays $13,728. Watch out: 12% is not one twelfth. Dividing by 12 takes off only $1,300 and gives $14,300, choice B.
So the answer is A, $13,728.

Check your answer

After a 12% discount, the buyer pays 88% of the price, because $100 - 12 = 88$ . Then $0.88 \times 15{,}600 = 13{,}728$ . That matches. Also, $13,728 is less than $15,600, as a sale price should be.

More questions like this

The original DOE question

Question 66 as printed on page 201 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 66. Open page 201 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 27 of 86

2024 Test B Math, question 11

Question

In a sports league, each team has 36 players and 3 coaches. There are also a number of team assistants. The ratio of team assistants to players is $\frac{1}{6}$ . What is the ratio of coaches to assistants?

A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D $\frac{5}{6}$

Show the answer

Correct answer: B

Let x be the number of assistants.

Since the ratio of assistants to players is $\frac{1}{6}$ , then:

$\frac{1}{6}=\frac{x}{36}$

$\frac{36}{6}=x$

$6=x$

There are a total of 6 assistants.

So the ratio of coaches to assistants is $\frac{3}{6}=\frac{1}{2}$ .

Question analysis

A ratio word problem about a sports team with 36 players, 3 coaches and an unknown number of assistants, where the ratio of assistants to players is 1 to 6. Students find how many assistants there are, then form a new ratio of coaches to assistants. Keeping the order of each ratio straight matters.

NYS standards: 6.RP.3 ratio and rate reasoning
6.RP.1 ratio concepts and language
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a missing amount from a ratio and then writing a second ratio is grade 6 ratio reasoning with whole numbers. Typical grade 7 work adds fractions, rates or percents, which this two-step problem does not.
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
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two short steps, finding 6 assistants and then writing 3 to 6, make this one of the easier ratio questions. The care needed is in keeping each ratio's order and its two groups straight, since the given ratio compares assistants with players and the asked one compares coaches with assistants. None of the wrong choices matches a common slip closely, so students who find the 6 assistants seldom go wrong.

Turning words into math minor Understanding the question
“The ratio of team assistants to players is 1/6” must become 6 assistants for 36 players, and the question then asks for a ratio in a new order, coaches first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Three groups are named, 36 players, 3 coaches and an unknown number of assistants, and each ratio pairs a different two of them. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Use the 1 to 6 ratio to find the number of assistants from the 36 players, then write coaches over assistants, in that order, 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.
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 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Compares the 9 coaches and assistants together with the 36 players, 9/36 = 1/4, a ratio of all the staff to players that the question never asks for.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/3; it equals the 6 assistants over all 9 staff, which puts the wrong group on top and a total underneath.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reads the ratio 1/6 as the assistants' share of everyone and gives the other 5/6, as if that remaining share described the coaches.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Ratio as a fraction: $\text{ratio} = \frac{\text{first group}}{\text{second group}}$; A ratio compares two amounts. The group named first goes on top, and the group named second goes on the bottom. Order matters. Assistants to players is not the same as players to assistants.
Equal ratios: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a and b are the two numbers in a ratio, written as a fraction. n is any number except zero. Multiplying both numbers by the same n gives an equal ratio. This is how you scale a small ratio up to real counts.

Sort out what you know
Each team has 36 players and 3 coaches. It also has some assistants, but you are not told how many. The ratio of assistants to players is 1/6. That means 1 assistant for every 6 players. You need the ratio of coaches to assistants.
Find the number of assistants
Write the ratio as a fraction, assistants on top and players on the bottom. To get from 6 players to 36, multiply by 6. Do the same to the top.
$\frac{a}{b} = \frac{a \times n}{b \times n}$
$\frac{1}{6} = \frac{1 \times 6}{6 \times 6} = \frac{6}{36}$
So there are 6 assistants for the 36 players.
Write the new ratio
Now the order changes. Coaches come first, so coaches go on top.
$\text{ratio} = \frac{\text{first group}}{\text{second group}}$
$\frac{\text{coaches}}{\text{assistants}} = \frac{3}{6} = \frac{1}{2}$
Divide the top and the bottom by 3 to simplify.
Avoid the traps and answer
Watch out: use only coaches and assistants. Putting all 9 staff over the 36 players gives 1/4, choice A. Putting the 6 assistants over all 9 staff gives 2/3, choice C. And 1/6 is not a share of everyone, so 5/6, choice D, has no meaning here.
So the answer is B, $\frac{1}{2}$ .

Check your answer

A ratio of 1/2 means 1 coach for every 2 assistants. With 3 coaches, that makes 6 assistants. Then 6 assistants to 36 players is $\frac{6}{36} = \frac{1}{6}$ , the ratio the question gave.

More questions like this

The original DOE question

Question 68 as printed on page 201 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 68. Open page 201 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 86

2024 Test B Math, question 38

Question

Michael received a check for $213 for working 20 hours at his part-time job. The amount on the check was his total pay after $32 in taxes was deducted. What was Michael’s hourly pay rate before taxes were deducted?

A$9.05
B$10.65
C$12.25
D$21.35

Show the answer

Correct answer: C

Add the taxes to the amount of the check to find Michael’s total pay before taxes were deducted:

$213+32=245$

Divide the total pay by the number of hours to find the hourly pay rate:

$245\div 20=12.25$

Question analysis

Michael's paycheck shows what he earned for 20 hours after taxes were taken out, and the tax amount is given. Students find his hourly pay rate before taxes. It tests unit rates in a money word problem, where students must decide what to do with the tax that was deducted.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Unit rate problems are grade 6 content, and this one needs one addition and one division with a decimal result. It is easier than typical grade 7 work, apart from deciding which amount is the pay before taxes.
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
Solution: 2 steps
Reading level: Question and choices: about grade 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two operations in the right order, adding the tax back and dividing by the hours, are all this takes. The danger is the word deducted, which pulls students toward subtracting the $32 a second time, and the phrase before taxes, which rules out the take-home rate from dividing the check alone. Checking that $12.25 × 20 − $32 gives back $213 confirms the answer.

Turning words into math minor Understanding the question
"The amount on the check was his total pay after $32 in taxes was deducted" makes $213 what remained, so the $32 is added back, not taken away. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Rebuild Michael's pay before taxes from the check and the tax that was taken out, then divide that total by the 20 hours worked.

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.

The wrong answers, classified

A Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Follows the word deducted and subtracts the $32 from the $213 check again, (213 − 32) ÷ 20 = 9.05, instead of adding it back.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $213 check by 20 hours, which gives Michael's hourly pay after taxes rather than before.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives $21.35, which equals 32 − 10.65; dividing $213 by 10 hours instead of 20, $21.30, comes closest, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Hourly pay rate: $\text{rate} = \frac{\text{total pay}}{\text{hours}}$; rate is the pay for one hour. total pay is the money earned for all the hours. hours is the time worked. Dividing shares the total pay equally among the hours. Use it when the pay for each hour is the same.

Sort out the question
Michael’s check was for $213. That is what was left after $32 in taxes was deducted, which means taken away. He worked 20 hours.
You need his hourly pay before taxes. So first find his full pay before the $32 was taken out.
Add the taxes back
The $32 was taken away from his full pay to leave $213. To undo taking away, add it back.
$213 + 32 = 245$
His pay before taxes was $245. Watch out: The word “deducted” tempts you to subtract again. That gives $(213 - 32) \div 20 = 9.05$ , choice A.
Divide by the hours
$\text{rate} = \frac{245}{20} = 12.25$
He earned $12.25 for each hour before taxes. Watch out: Dividing the $213 check by 20 gives $10.65, choice B. That is his pay for each hour after taxes.
So the answer is C, $12.25.

Check your answer

Work forward. $12.25 \times 20 = 245$ , so 20 hours pay $245. Take away $32 in taxes: $245 - 32 = 213$ . That is the amount on the check.

More questions like this

The original DOE question

Question 95 as printed on page 210 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 95. Open page 210 of the handbook (PDF)

Question 29 of 86

2024 Test B Math, question 54

Question

A rectangular kitchen floor is 12 feet long and 7.5 feet wide. It will be covered completely with new square tiles measuring 0.75 foot on each side. How many tiles will cover the floor?

A68
B78
C120
D160

Show the answer

Correct answer: D

First, determine the area of the kitchen floor in square feet by multiplying the length and width:

$12\times 7.5=90\text{ square feet}$

Next, determine the area of each square tile:

$0.75\times 0.75=\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}$

The number of tiles needed is found by dividing the area of the floor by the area of the tiles:

$90\div\frac{9}{16}=90\times\frac{16}{9}=160$

Question analysis

A rectangular kitchen floor measuring 12 feet by 7.5 feet will be covered with square tiles 0.75 foot on a side. Students find how many tiles are needed. It tests area and division with decimals, and working with a decimal side length while keeping lengths and areas apart is what makes it tricky.

NYS standards: 6.G.1 area of polygons
6.NS.3 multi-digit decimal operations
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Area of a rectangle is grade 6 content, but here it combines with dividing by a decimal tile area, or by the tile's side along each wall. Three decimal steps and a tempting shortcut make it 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: Error-prone arithmetic moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The floor's 90 square feet comes quickly; the trap is what to divide it by. A tile is described by its side, and dividing by 0.75 instead of 0.5625 gives a listed answer that looks reasonable. Counting tiles along each wall, 16 along the length and 10 along the width, avoids both the decimal division and the side-versus-area mistake.

Error-prone arithmetic moderate Working it out
The tile's area, 0.75 × 0.75 = 0.5625 square foot, must divide 90, a decimal division that is easier as 90 × 16/9 or as 16 tiles by 10 tiles. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The tiles are "square tiles measuring 0.75 foot on each side", a side length, which has to be squared before it can be compared with the floor's area in square feet. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many tiles fit along the length and how many along the width by dividing each by 0.75, then multiply the two counts.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Multiplies the 90-square-foot area by the tile's 0.75-foot side instead of dividing, getting 67.5 and rounding to 68.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives 78, which is twice the floor's 39-foot perimeter, 4 × (12 + 7.5), so it mostly draws guesses.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Divides the 90 square feet by the tile's side, 0.75 foot, instead of its area, 0.5625 square foot, getting 120 tiles.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle: $A = l \times w$; A is the area, the flat space inside. l is the length and w is the width. Multiplying counts how many unit squares fit, row by row.
Area of a square: $A = s^2$; A is the area. s is the length of one side. s² means s × s. A square is a rectangle whose length and width are both s.

Sort out the question
The floor is a rectangle 12 feet long and 7.5 feet wide. Each tile is a square, 0.75 foot on each side. You need the number of tiles that cover the whole floor.
Find the floor’s area
Area tells how much flat space the tiles must cover. Use the rectangle formula.
$A = 12 \times 7.5 = 90$
The floor is 90 square feet.
Find one tile’s area
The tile’s side is a length. You need its area to compare it with the floor’s area. Use the square formula.
$A = 0.75^2 = 0.75 \times 0.75 = 0.5625$
As a fraction, $0.75 = \frac{3}{4}$ , so the area is $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ square foot.
Divide the floor by one tile
The number of tiles is the floor’s area divided by one tile’s area. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down.
$90 \div \frac{9}{16} = 90 \times \frac{16}{9} = 10 \times 16 = 160$
Watch out: Dividing 90 by 0.75, the side, gives 120, choice C. You must divide by the tile’s area. Multiplying 90 by 0.75 gives about 68, choice A.
So the answer is D, 160.

Check your answer

Count the tiles in rows. Along the 12-foot length, $12 \div 0.75 = 16$ tiles fit. Along the 7.5-foot width, $7.5 \div 0.75 = 10$ tiles fit. That makes $16 \times 10 = 160$ tiles.

More questions like this

The original DOE question

Question 111 as printed on page 214 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 111. Open page 214 of the handbook (PDF)

Question 30 of 86

2023 Test A Math, question 7

Question

In one week, $1\frac{3}{4}$ inches of rain fell on Monday, $2\frac{2}{3}$ inches fell on Tuesday, and $\frac{7}{8}$ inch fell on Wednesday. How many inches of rain fell during those three days?

A $5\frac{7}{24}$
B $5\frac{1}{24}$
C $3\frac{4}{5}$
D $3\frac{1}{2}$

Show the answer

Correct answer: A

To solve, add the three given fractions:

$1\frac{3}{4} + 2\frac{2}{3} + \frac{7}{8}$

The common denominator is 24:

$= 1\frac{18}{24} + 2\frac{16}{24} + \frac{21}{24}$

$= 4\frac{10}{24} + \frac{21}{24}$

$= 5\frac{7}{24}$

Question analysis

A rainfall word problem gives the inches of rain on Monday, Tuesday and Wednesday as two mixed numbers and a fraction, and asks for the total over the three days. Students add fractions with unlike denominators. Finding a common denominator for fourths, thirds and eighths, and regrouping the fractional part into wholes, is where errors come in.

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding fractions and mixed numbers with unlike denominators is taught before grade 7, and this is a direct word problem. It takes more care than a typical exercise only because three different denominators must be rewritten over a common one.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The setup is plain addition, so the difficulty is all in the fractions. Students who know to rename over a common denominator rarely miss it, while those who add the fraction parts straight across, or put the numerators over 24 without scaling them, reach choices with a whole part of 3. Since 1 3/4 + 2 2/3 alone is more than 4, any total below 4 can be ruled out at a glance.

Error-prone arithmetic moderate Working it out
The fractional parts have denominators 4, 3 and 8, so they must be renamed as 18, 16 and 21 twenty-fourths, and their sum, 55/24, regrouped as 2 7/24 before the wholes are added. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Renaming, adding the numerators, regrouping the improper fraction and adding the whole numbers are four separate moves, and losing the 2 carried from 55/24 drops the total by two inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the three fractional parts over a common denominator of 24, add the whole numbers and the fractions separately, and regroup any extra twenty-fourths as a whole.

Traps

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

The wrong answers, classified

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single renaming mistake gives 49 twenty-fourths; this choice sits exactly 1/4 below the right total, so it mostly catches guesses near it.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the whole numbers to get 3, then adds the fractions straight across, (3 + 2 + 7)/(4 + 3 + 8) = 12/15 = 4/5.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Finds the common denominator 24 but adds the numerators without renaming them, (3 + 2 + 7)/24 = 1/2, and adds that to the wholes, 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; a over b is a fraction, and k is any whole number except 0. Multiplying the top and the bottom by the same number gives an equal fraction. You cut each piece into k smaller pieces. So you have k times as many pieces, each k times smaller.
Adding fractions with the same bottom: $\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}$; d is the denominator, the bottom number. It tells the size of the pieces. a, b and c are the numerators, the top numbers. They count the pieces. When the pieces are the same size, add the counts. The bottom stays the same.
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
Rain fell on three days: $1\frac{3}{4}$ inches, $2\frac{2}{3}$ inches and $\frac{7}{8}$ inch. You need the total rain, in inches. So add the three amounts. Add the whole numbers and the fractions separately. The whole numbers are 1 and 2, so they add to 3 inches.
Find a common denominator
The fractions have denominators, or bottom numbers, 4, 3 and 8. You can only add pieces of the same size. So find the smallest number that 4, 3 and 8 all divide into. Count by 8s: 8, 16, 24. 24 works, since 24 = 4 × 6 and 24 = 3 × 8. So use 24ths.
Rename each fraction in 24ths
Multiply the top and the bottom by the same number.
$\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}$
$\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}$
$\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}$
Watch out: change the tops too. Keeping 3, 2 and 7 over 24 gives $\frac{12}{24} = \frac{1}{2}$ and a total of $3\frac{1}{2}$ , choice D.
Add the fractions
Now the pieces are the same size, so add the tops.
$\frac{18}{24} + \frac{16}{24} + \frac{21}{24} = \frac{55}{24}$
That is more than 1 whole. 24 goes into 55 twice, with 7 left over.
$\frac{55}{24} = 2\frac{7}{24}$
Watch out: never add the bottoms. Adding straight across gives $\frac{12}{15} = \frac{4}{5}$ and a total of $3\frac{4}{5}$ , choice C.
Put the parts together
Add the 3 whole inches from before.
$3 + 2\frac{7}{24} = 5\frac{7}{24}$
So the answer is A, $5\frac{7}{24}$ inches.

Check your answer

Use decimals. $1\frac{3}{4}$ is 1.75, $2\frac{2}{3}$ is about 2.67, and $\frac{7}{8}$ is 0.875. They add to about 5.29. And $\frac{7}{24}$ is about 0.29. So $5\frac{7}{24}$ fits, and $5\frac{1}{24}$ is too small.

More questions like this

The original DOE question

Question 64 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 64. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 31 of 86

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 32 of 86

2023 Test A Math, question 13

Question

The product of two positive integers is 65. Which number could be the sum of the two integers?

Show the answer

Correct answer: B

First, find the pairs of positive integers that multiply to 65: $1 \times 65$ and $5 \times 13$ .

The sums of the pairs are 66 and 18. 18 is the option that is provided in the answer choices.

Question analysis

Two positive integers multiply to 65, and this question asks which number in the choices could be their sum. Students list the factor pairs of 65 and add each pair. The words could be signal that more than one pair exists, and only one of the sums appears among the choices.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding factor pairs of a two-digit number is review by grade 7. The question takes two quick steps, listing the pairs and adding each, with nothing to compute beyond small products and sums.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

With only two factor pairs, 1 × 65 and 5 × 13, the work is a short list and two additions. The word could matters: the sum depends on which pair is chosen, and only one of the two sums appears among the choices. The main slip is answering with a factor rather than with the sum the question asks for.

Something you must already know minor Working it out
The pairs come from factoring 65 as 5 × 13, a fact the stem leaves to the student; the pair 1 and 65 also counts, but its sum, 66, is not offered. (Anderson & Krathwohl, 2001)

Strategy: List every pair of positive integers whose product is 65, starting from 1, and add the two numbers in each pair.

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.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the factor pair 5 and 13 and answers with the factor 5, one of the two integers, instead of their sum.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No pair of positive integers with product 65 adds to 24; reaching it needs a wrong pair such as 3 and 21, whose product is 63, so it mostly catches guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No factor pair of 65 adds to 52, which is 65 − 13 or 4 × 13, and no common working combines them that way, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair: $a \times b = 65$; a and b are positive whole numbers that multiply to make 65. Such a pair is called a factor pair of 65. To find them all, try 1, 2, 3 and so on. Stop when the factors start to repeat.
Dividing by 5: $\text{last digit } 0 \text{ or } 5 \rightarrow \text{divides by } 5$; A whole number splits into groups of 5, with none left over, when its last digit is 0 or 5. So 65 divides by 5, but 64 does not.

Sort out the question
Two positive integers multiply to make 65. Positive integers are the counting numbers 1, 2, 3 and so on. The product is the answer to a multiplication. You need a number that could be their sum, the answer when you add them.
List the factor pairs of 65
Try each small number. See if it divides 65 with nothing left over.
- 1 works: $1 \times 65 = 65$ .
- 2 and 4 don’t work, because 65 is odd. 3 doesn’t work either: $3 \times 21 = 63$ , with 2 left over.
- 5 works, since 65 ends in 5: $5 \times 13 = 65$ .
- 6 and 8 are even, so they don’t work. 7 doesn’t either: $7 \times 9 = 63$ .
Since $8 \times 8 = 64$ , any factor bigger than 8 pairs with one smaller than 8. You already tried those. So the pairs are 1 and 65, and 5 and 13.
Add each pair
$1 + 65 = 66$
$5 + 13 = 18$
The possible sums are 66 and 18. Only 18 is a choice. Watch out: the question asks for the sum, not one of the numbers. Picking the factor 5 is choice A. So the answer is B, 18.

Check your answer

Work back from 18. Look for two numbers that add to 18 and multiply to 65. $5 + 13 = 18$ and $5 \times 13 = 65$ . Both match. For 24 or 52, no pair of whole numbers that multiplies to 65 adds up to them.

More questions like this

The original DOE question

Question 70 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 70. 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 33 of 86

2023 Test A Math, question 14

Question

If n is an odd integer that is less than −3.25, what is the greatest possible value of n?

A−1
B−2
C−3
D−5

Show the answer

Correct answer: D

The two greatest integers less than −3.25 are −4 and −5. Since −4 is even, the answer is −5.

Question analysis

This question asks for the greatest odd integer that is less than −3.25. Students compare negative numbers on a number line and apply two conditions at once, odd and less than. With negatives, the greater number is the one closer to zero, which makes the two conditions easy to misapply.

NYS standards: 6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Ordering negative numbers is grade 6 content, and the question adds only the condition that the integer be odd. It is lighter than typical grade 7 work, though the order of negatives still trips up students who compare their sizes.
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
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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 question is short, but it asks students to reason on the negative side of 0, where greater means closer to 0 and the size of the digits misleads. The integers just below −3.25 are −4 and then −5, and the first of those is even. Sketching a number line, or reading less than as to the left of, keeps the direction straight.

Error-prone arithmetic minor Working it out
On the negative side, less means farther from 0, so the integers below −3.25 run −4, −5, −6; −3, though its digit is closer to 3.25, lies to the right. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
Three conditions apply at once, odd, less than −3.25 and the greatest such value, and checking only two of them leads to a listed choice. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Picture −3.25 on a number line, list the integers to its left starting with the closest, and take the first one that is odd.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Picks the greatest odd number offered, −1, without checking the condition that n be less than −3.25; −1 lies to its right on the number line.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
The choice −2 is even, so it breaks the condition that n be odd, and it is also greater than −3.25; it mostly catches guesses.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Looks for the odd integer just past −3.25 but steps toward 0, taking −3, though −3 is greater than −3.25, not less.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers: $\text{if } a > b > 0 \text{, then } -a < -b$; a and b are positive numbers, and a is bigger. On a number line, numbers get smaller as you move left. A negative number farther from 0 is farther left. So it is smaller, even though its digits look bigger.

Sort out the question
n has to pass three tests. It is an integer, a whole number with no decimal part. It is odd, so it is not a multiple of 2. And it is less than −3.25. Of all the numbers that pass, you need the greatest one. That is the one farthest to the right on a number line.
Picture −3.25 on a number line
−3.25 sits between −4 and −3. Less than −3.25 means to the left of it.
$-5 < -4 < -3.25 < -3$
So −4 is less than −3.25, but −3 is greater. Watch out: −3 is not less than −3.25. It is to the right, closer to 0. For negative numbers, the one closer to 0 is greater. Picking −3 is choice C.
Find the odd integers to the left
The integers less than −3.25, starting with the closest, are −4, −5, −6 and so on. The greatest one is −4. But −4 is even, since 4 = 2 × 2. So skip it. The next one, −5, is odd. Watch out: −1 is odd, but it is greater than −3.25, so choice A breaks the rule. So the answer is D, −5.

Check your answer

Check all three tests for −5. It is an integer. It is odd. It is to the left of −3.25, so it is less. The only integer between −5 and −3.25 is −4, and it is even. So no greater number works.

More questions like this

The original DOE question

Question 71 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 2024 Admissions, page 79, question 71. Open page 79 of the handbook (PDF)

Question 34 of 86

2023 Test A Math, question 16

Question

Between which two consecutive integers is the fraction $\frac{29}{7}$ ?

A2 and 3
B3 and 4
C4 and 5
D5 and 6

Show the answer

Correct answer: C

First, convert $\frac{29}{7}$ to a mixed number: $\frac{29}{7} = 4\frac{1}{7}$ . The two consecutive integers on either side of $4\frac{1}{7}$ are 4 and 5.

Question analysis

This question asks between which two consecutive whole numbers the fraction 29/7 lies. Students divide to turn the improper fraction into a mixed number, or compare it with nearby multiples of 7. It is a quick check of number sense with fractions, where a remainder must be read correctly.

NYS standards: 6.NS.6c positioning rational numbers and points
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Changing an improper fraction to a mixed number and placing it between whole numbers is review well before grade 7. This takes one division with a small divisor and nothing more.
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: A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

Dividing 29 by 7 gives 4 with a remainder of 1, so the fraction is 4 1/7 and lies just past 4. The one decision is which side of 4 the remainder puts it on, and a positive remainder always moves the value up. Writing the mixed number before looking at the choices makes the pair obvious.

A very close wrong answer minor Choosing the answer
The pairs 3 and 4 and 4 and 5 both contain 4, so knowing that 29/7 is about 4 is not enough; only the remainder, 29 − 28 = 1, puts it above 4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Find the largest multiple of 7 that is not more than 29, write the fraction as a mixed number, and read off the whole numbers on each side.

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 division of 29 by 7 lands below 3, since 7 × 3 = 21 is far short of 29; only misreading 29/7 as 2.9 gets there, so it mostly catches guesses.
B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the whole-number part, 4, but takes it as the upper end of the pair, as if 29/7 fell just short of 4 instead of just past it.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing puts 29/7 above 5, since 7 × 5 = 35 is more than 29; even rounding 4 1/7 up gives 5 as the upper end, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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
Consecutive integers are whole numbers that come one right after the other, like 4 and 5. You need the two that $\frac{29}{7}$ falls between. So find the whole number just below it and the one just above it.
Divide 29 by 7
A fraction bar means divide. How many 7s fit in 29?
$7 \times 4 = 28$
$7 \times 5 = 35$
28 is the biggest multiple of 7 that is not more than 29. A multiple of 7 is 7 times a whole number. Then $29 - 28 = 1$ . So 29 ÷ 7 is 4 with 1 left over.
Write it as a mixed number
$\frac{29}{7} = 4\frac{1}{7}$
$4\frac{1}{7}$ is a little more than 4 and less than 5. Watch out: it is just past 4, not just short of it. So 4 is the lower number of the pair. Picking 3 and 4 is choice B. So the answer is C, 4 and 5.

Check your answer

Multiply each end by 7. $4 \times 7 = 28$ and $5 \times 7 = 35$ . The top of the fraction, 29, is between 28 and 35. So $\frac{29}{7}$ is between 4 and 5. As a decimal it is about 4.14.

More questions like this

The original DOE question

Question 73 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 73. Open page 79 of the handbook (PDF)

Question 35 of 86

2023 Test A Math, question 17

Question

A customer wants to buy a pair of hiking boots.

The original price of the boots is $85.75.
The store is offering a 15% discount on all boot purchases.
The customer has a coupon for an additional 25% off the sale price.
The tax rate is 8.5%.

What is the final cost of the boots, including the tax, to the nearest cent?

A$55.82
B$59.08
C$59.32
D$63.17

Show the answer

Correct answer: C

First, determine the sale price after the 15% discount is applied.

Since the customer is saving 15%, this means they are paying 85% $(100 - 15)$ , so multiply the original price by 85% (0.85) to find the discounted price:

$\$85.75 \times 0.85 \approx \$72.89$

Next, find the price after the 25% coupon has been applied:

Since the customer is saving 25% off the sale price, this means they are paying 75% $(100 - 25)$ , so multiply the price found in the first step by 75% (0.75) to find the second discounted price:

$\$72.89 \times 0.75 \approx \$54.67$

Last, find the final cost after an 8.5% tax is added to the second discounted price.

Since the customer is paying 8.5% tax, this means they are paying 108.5% $(100 + 8.5)$ , so multiply the second discounted price by 108.5% (1.085) to find the final cost:

$\$54.67 \times 1.085 \approx \$59.32$

Question analysis

A pair of hiking boots costs $85.75, goes on sale at 15% off, gets an extra 25% off with a coupon, and is then taxed at 8.5%. Students find the final cost to the nearest cent. Each percent applies to the price left after the step before, so the discounts cannot simply be added, and the decimals call for careful multiplication.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 A discount followed by sales tax is standard grade 7 percent work. This item stacks two discounts and a tax, with cents to keep at every stage, so it takes about twice the steps and decimal multiplication of a typical problem.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic major
Turning words into math moderate
Slow to do carefully moderate
Solution: 3 steps
Reading level: Question and choices: about grade 4, 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

The question tests whether students treat successive percents as successive multipliers, and then holds them to three long multiplications. Adding the 15% and 25% into one discount is the tempting shortcut, and a tax read as a dollar amount is the other. The key rounds to the cent at each stage, as a register would; carried exactly, the result is $59.31, so the nearest choice, $59.32, still stands.

Error-prone arithmetic major Working it out
Three products of cents and decimals are worked by hand: 85.75 × 0.85 = 72.8875, then 72.89 × 0.75 = 54.6675, then 54.67 × 1.085 ≈ 59.32, each rounded to the cent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
The phrase "an additional 25% off the sale price" means 25% of the already discounted $72.89, not of the $85.75 original, so the two discounts multiply rather than add. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Slow to do carefully moderate Choosing the answer
A reliable method needs three long decimal multiplications without a calculator, well past the minute and a half a question allows, which tempts students to combine the percents. (Lu & Sireci, 2007)

Strategy: Multiply by 0.85 for the sale, by 0.75 for the coupon on the sale price, and by 1.085 for the tax, rounding to the cent at each stage as a register would.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
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 Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes the 25% coupon off the original $85.75 instead of the sale price, so the discounts add to 40%: 85.75 × 0.60 × 1.085 ≈ $55.82.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives $59.08; it sits 24 cents below the right total and matches no discount or tax mistake, so it mostly catches guesses.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Applies both discounts correctly, reaching $54.67, but adds $8.50 as the tax, reading 8.5% as a dollar amount instead of a percent of the price.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Price after a discount: $\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)$; p is the percent off. A discount is money taken off the price. If you save p percent, you pay the rest, 100 − p percent. For 15% off you pay 85%, so multiply by 0.85.
Price with tax: $\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$; The price is the cost before tax. t is the tax rate, a percent. Tax is extra money added on top. You pay the whole price, 100%, plus t percent more. So for 8.5% tax you multiply by 1.085.

Sort out the question
The boots cost $85.75 at first. The store takes 15% off. Then a coupon takes 25% off the sale price. The sale price is the price after the first discount. Then 8.5% tax is added. You need the final cost to the nearest cent. A cent is $0.01, the second place after the decimal point.
Take off the 15% discount
You save 15%, so you pay 85%. That is 0.85 of the price.
$\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)$
$\text{new price} = 85.75 \times \left(1 - \frac{15}{100}\right) = 85.75 \times 0.85 = 72.8875$
A price can’t have part of a cent, so round to two decimal places. The third digit is 7, which is 5 or more, so round up.
$72.8875 \approx 72.89$
The sign ≈ means about. The sale price is $72.89.
Take 25% off the sale price
The coupon takes 25% off $72.89, so you pay 75% of it.
$72.89 \times 0.75 = 54.6675 \approx 54.67$
The price is now $54.67. Watch out: take the coupon off the sale price, not the first price. Adding 15% and 25% to get 40% off $85.75 leads to $55.82, choice A.
Add the 8.5% tax
You pay the price plus 8.5% more, which is 108.5% of it.
$\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$
$\text{total} = 54.67 \times 1.085 = 59.31695 \approx 59.32$
Watch out: 8.5% is a percent of the price, not $8.50. Adding $8.50 gives $63.17, choice D.
See why rounding each time matters
A store rounds to the cent after each step, as you just did. The test’s answer is worked the same way. Keep every digit and round only at the end, and you get this:
$85.75 \times 0.85 \times 0.75 \times 1.085 = 59.3122\ldots \approx 59.31$
That is $59.31, which is not a choice. The first two roundings each went up by a tiny part of a cent. That was just enough to push the total from $59.31 to $59.32. So here, round at each step, as a store would. So the answer is C, $59.32.

Check your answer

Check the tax a second way. 8.5% of $54.67 is $0.085 \times 54.67 = 4.64695$ , about $4.65. Then $54.67 + 4.65 = 59.32$ . Also estimate: two discounts leave a bit less than two-thirds of $86, near $55. Tax adds under $5, so about $59 or $60 makes sense.

More questions like this

The original DOE question

Question 74 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 2024 Admissions, page 79, question 74. Open page 79 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 86

2023 Test A Math, question 18

Question

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

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

Show the answer

Correct answer: A

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

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

First, subtract 90 from both sides:

$0.15 \times 12x = \$382.59$

$1.8x = \$382.59$

Last, divide both sides by 1.8:

$x = \$212.55$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 37 of 86

2023 Test A Math, question 25

Question

$Line h is horizontal, with arrows at both ends. Line j rises from the lower left to the upper right, and line k rises steeply from the bottom, crossing line h to the right of line j; lines j and k meet at point Q above line h. Where j crosses h, the angle above h and to the left of j is marked 135°. Where k crosses h, the angle below h and to the right of k is marked 100°. At Q, the angle between the parts of lines k and j above Q is marked x°.$

Lines j and k intersect at point Q, and h is a straight line. What is the value of x?

Show the answer

Correct answer: C

First, find the measures of the three angles inside the triangle. The lower left angle is supplementary to the one marked 135°, so the measure of that angle is $180 - 135 = 45$ . The angle on top is a vertical angle with the one marked x°, so the measure of that angle is also x°. The angle on the lower right is a vertical angle with the one marked 100°, so the measure of that angle is also 100°. Since the measures of the three interior angles of a triangle sum to 180°, set up an equation and solve for x:

$45 + x + 100 = 180$

$145 + x = 180$

$x = 35$

Question analysis

Lines j and k cross at point Q above a straight line h, forming a triangle, with a 135° angle marked where j meets h and a 100° angle where k meets h. Students find the angle x at Q. Each marked angle lies outside the triangle, so it has to be turned into an inside angle before the triangle's angle sum can be used.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students use supplementary and vertical angles to find unknown angles. This figure needs both, applied at three different vertices, plus the angle sum of a triangle, which New York teaches in grade 8.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Combines ideas moderate
Something you must already know minor
Solution: 4 steps
Reading level: Question and choices: about grade 2, 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

Every angle the solution needs has to be carried into the triangle from outside, and the two marked angles need different treatment. The 135° is supplementary to the triangle's angle, but the 100° is vertical to it, so subtracting both from 180° out of habit builds the wrong triangle. Checking which side of each line a marked angle sits on, not just how near the triangle it is, separates the two cases.

Reading a figure, table or graph major Understanding the question
None of the three marked angles is inside the triangle formed by h, j and k: the 135° is outside on the left, the 100° is below line h, and x is above Q. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
Two different angle facts are needed: the 135° gives a supplementary interior angle of 45°, while the 100° and x carry into the triangle unchanged as vertical angles. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Supplementary angles, vertical angles and the triangle's 180° sum are all assumed, and the question names none of them. (Anderson & Krathwohl, 2001)

Strategy: Mark the three angles of the triangle formed by lines h, j and k: one is supplementary to 135°, one is vertical to 100°, and the one at Q is vertical to x°.

Traps

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

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Takes the triangle's angle at line k as 180 − 100 = 80, the supplement of the marked angle instead of its vertical angle, so x = 180 − 45 − 80 = 55.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the triangle's angle where line j meets h, 180 − 135 = 45, and gives it instead of going on to the angle at Q.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No angle fact in the figure gives 30, and the angle at Q is drawn close to its true size, so this choice mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the figure
Lines h, j and k make a triangle. Its corners are where j crosses h, where k crosses h, and point Q. The figure marks two angles outside the triangle, 135° and 100°. You need x, the angle marked above Q.
Find the angle where j crosses h
The 135° angle is above h and to the left of j. The triangle’s angle is right next to it, on the other side of j. Together they make the straight line h. Call the triangle’s angle a.
$a + b = 180$
$135 + a = 180$
$a = 180 - 135 = 45$
So that corner of the triangle is 45°. Watch out: this is only one corner, not x. Stopping here gives 45, choice B.
Find the angle where k crosses h
The 100° angle is below h and to the right of k. The triangle’s angle is above h and to the left of k. Lines h and k make an X there, and these two angles are straight across from each other. They are vertical angles, so the triangle’s angle is 100°. Watch out: it is not 180 − 100 = 80. That angle is beside the 100°, not across from it. Using 80 gives 55, choice A.
Find the angle at Q
The x° angle is above Q, between the top parts of lines j and k. The triangle’s angle at Q is between the parts of j and k below Q. These two angles are straight across from each other. So they are vertical angles, and the triangle’s angle at Q is also x°.
Use the triangle’s angle sum
$A + B + C = 180$
$45 + 100 + x = 180$
$145 + x = 180$
$x = 180 - 145 = 35$
The three angles of the triangle must add up to 180°. Only x is unknown. So the answer is C, 35.

Check your answer

Add the triangle’s three angles: $45 + 100 + 35 = 180$ . Also, the angle at Q looks narrow in the figure, much less than a right angle of 90°. 35° fits that.

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 2024 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 38 of 86

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 39 of 86

2023 Test A Math, question 27

Question

A basket contains red balls, green balls, and white balls. There are 12 red balls in the basket. The probability of randomly choosing a red ball is 1 in 3. If the probability of randomly choosing a green ball is 1 in 4, how many green balls are in the basket?

Show the answer

Correct answer: C

First, use the information about the red balls to calculate the total number of balls in the basket:

$\frac{1}{3} = \frac{12}{x}$	Cross multiply.
$x = 36$

There are a total of 36 balls in the basket. The probability of selecting a green ball is 1 in 4. Use that information to calculate how many green balls are in the basket:

$\frac{g}{36} = \frac{1}{4}$	Cross multiply.
$4g = 36$	Apply the multiplicative inverse property; divide both sides of the equation by 4.
$g = 9$

There are 9 green balls in the basket.

Question analysis

A basket holds red, green and white balls, including 12 red ones, and the chances of drawing red and green are 1 in 3 and 1 in 4. Students find how many green balls are in the basket. Getting the total from the red count is the step students skip when they apply a probability to the wrong number.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content, but here it runs backward: a count and a probability give the total, which then gives another count. Two linked steps like this are typical of grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 46 words.
Students, last 30 days: 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 basket's total is never stated, and that missing number is the whole problem: it has to be recovered from the red balls before the green probability means anything. Applying the green chance straight to the 12 red balls is the quick wrong move. Checking that 12 red and 9 green fit in 36 balls with 15 white left over confirms the setup.

Turning words into math minor Understanding the question
The sentence "The probability of randomly choosing a red ball is 1 in 3" has to become 12 = one third of the total, which makes the basket hold 36 balls. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two linked steps: the red balls give the total, 12 × 3 = 36, and only then does the green probability give a count, 36 ÷ 4 = 9. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use the red balls to find the total, since 12 is one third of it, then take one fourth of that total for the green balls.

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Takes one fourth of the 12 red balls, 12 ÷ 4 = 3, applying the green probability to the only count the stem gives instead of to the total.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No use of the two probabilities gives 8, which is one third of the 24 balls that are not red, so it mostly catches guesses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Scales the 12 red balls by 4/3 instead of 3/4, taking the 1-in-4 chance to mean more balls than the 1-in-3 chance: 12 × 4/3 = 16.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{balls of one color}}{\text{all the balls}}$; P is the probability, the chance of picking that color. Each ball is equally likely to be picked. So the chance is the fraction of all the balls that have that color. A chance of 1 in 3 means one third of the balls.

Sort out the question
The basket has red, green and white balls. There are 12 red balls. The chance of picking red is 1 in 3. The chance of picking green is 1 in 4. You need the number of green balls. You don’t know the total number of balls yet, so find it first.
Find the total number of balls
1 in 3 means one third of all the balls are red. So 12 is one third of the total. Call the total number of balls b.
$\frac{12}{b} = \frac{1}{3}$
$b = 12 \times 3 = 36$
There are 36 balls in all.
Find the number of green balls
1 in 4 means one fourth of all 36 balls are green. Call the number of green balls g.
$\frac{g}{36} = \frac{1}{4}$
$g = 36 \div 4 = 9$
Watch out: take one fourth of all 36 balls, not of the 12 red ones. $12 \div 4 = 3$ gives choice A. Multiplying 12 by $\frac{4}{3}$ gives 16, choice D. So the answer is C, 9.

Check your answer

A 1 in 4 chance is smaller than a 1 in 3 chance. So there should be fewer green balls than red ones, and 9 is less than 12. Check: $\frac{9}{36} = \frac{1}{4}$ . The rest, $36 - 12 - 9 = 15$ , are white.

More questions like this

The original DOE question

Question 84 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 84. 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 40 of 86

2023 Test A Math, question 34

Question

$Parallelogram PQRS, with P at the bottom left, Q at the top left, R at the top right and S at the bottom right. The angle at P is marked 56°, the angle at Q is marked x°, and the angle at S is marked y°. The angle at R is not marked.$

In the parallelogram above, what is the value of $x + y$ ?

A112
B124
C148
D248

Show the answer

Correct answer: D

In a parallelogram, opposite angles are congruent, so the measure of the unmarked angle is 56°. The sum of the interior angles of a parallelogram is equal to 360°. Use that information to solve the problem:

$x + y + 56 + 56 = 360$

$x + y + 112 = 360$

$x + y = 248$

Question analysis

This geometry question shows parallelogram PQRS with a 56° angle at P and unknown angles x and y at Q and S, and asks for the sum x + y. Students need angle facts about parallelograms: opposite angles match, and the four angles total 360°. Stopping after finding only one angle, or adding the wrong pair, are the common mistakes.

NYS standards: 7.G.5 angle relationships and equations
8.G.5 angle sums, transversals, angle-angle similarity
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 The parallelogram facts needed, equal opposite angles and a 360° angle sum, rest on grade 8 reasoning about parallel lines and triangles. Once they are known, the arithmetic is short, about the size of a typical grade 7 angle 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: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two short angle facts finish the question, but the figure marks three of the four corners, and the unmarked one at R is easy to forget. Students who find one obtuse angle stop with half the total, and those who add the two 56° angles give the part already known. Checking that the four angles, 56, 124, 56 and 124, add to 360 confirms the sum.

Something you must already know moderate Working it out
The figure gives one angle and no rules: students must bring both that opposite angles of a parallelogram are equal and that a quadrilateral's angles total 360°, or that neighboring angles add to 180°. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The labels x and y sit at Q and S, opposite corners, while the corner R across from the 56° angle at P carries no label at all. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Mark the angle at R equal to the angle at P, since opposite angles of a parallelogram are equal, then take both away from the total of a quadrilateral's four angles.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Adds the two 56° angles at P and R, 56 + 56 = 112, giving the part of the 360° already known instead of the two angles that remain.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds one unmarked angle, 180 − 56 = 124, and stops before adding the equal angle at the opposite corner.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No angle fact in the parallelogram gives 148, which is exactly 100 less than the right total, so it mostly catches guesses or a stray subtraction slip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Opposite angles of a parallelogram: $\angle P = \angle R$; A parallelogram is a four-sided shape whose opposite sides are parallel. P and R are opposite corners: they don’t share a side. Opposite angles of a parallelogram are always equal.
Angles of a four-sided shape: $\angle P + \angle Q + \angle R + \angle S = 360$; P, Q, R and S are the four corners. Their angles always add up to 360 degrees. A line from one corner to the opposite corner cuts the shape into two triangles, and each triangle’s angles add up to 180 degrees.
Neighboring angles of a parallelogram: $\angle P + \angle Q = 180$; P and Q are corners next to each other, joined by one side. In a parallelogram, their angles add up to 180 degrees. The same is true of any two neighboring corners.

Sort out the figure
PQRS is a parallelogram. The angle at corner P is 56°. The angle at Q is x°, and the angle at S is y°. The angle at R has no label. You need x + y, the angles at Q and S added together.
Find the angle at R
R is across from P, so they are opposite corners.
$\angle P = \angle R$
$\angle R = 56$
So the angle at R is 56°, the same as at P.
Use the 360° total
$\angle P + \angle Q + \angle R + \angle S = 360$
$56 + x + 56 + y = 360$
$x + y + 112 = 360$
$x + y = 360 - 112 = 248$
Watch out: 112 is the two angles you already know, not x + y. Answering 112 is choice A.
Find x and y one at a time
Q is next to P, so use the neighbor rule.
$\angle P + \angle Q = 180$
$56 + x = 180, \quad x = 124$
S is also next to P, so y = 124 too. Then $124 + 124 = 248$ , the same total. Watch out: stopping at one angle, 124, gives choice B. The question asks for both added together.
So the answer is D, 248.

Check your answer

Add all four angles: $56 + 124 + 56 + 124 = 360$ . That is the right total for a four-sided shape. Also, the angles at Q and S look wider than a right angle in the figure, and 124° is wider than 90°.

More questions like this

The original DOE question

Question 91 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 91. Open page 84 of the handbook (PDF)

Question 41 of 86

2023 Test A Math, question 37

Question

$A circle with a shaded sector between two radii, one pointing straight up and one pointing up and to the right. The central angle of the shaded sector is marked 45°.$

The shaded sector of the circle shown above has an area of $18\pi$ square feet. What is the circumference of the circle?

A $144\pi$ ft
B $24\pi$ ft
C $18\pi$ ft
D $9\pi$ ft

Show the answer

Correct answer: B

The area of the entire circle is $\pi r^{2}$ .

The fraction of the circle that is shaded is $\frac{45}{360} = \frac{1}{8}$ . So, the area of the sector is $\frac{1}{8}\pi r^{2}$ . Use that information to find r:

$\frac{1}{8}\pi r^{2} = 18\pi$

$r^{2} = 144$

$r = 12$

Use r to calculate the circumference of the circle:

$c = 2\pi r = 2\pi(12) = 24\pi$

The circumference is $24\pi$ feet.

Question analysis

This circle question shows a shaded 45° sector with an area of 18π square feet and asks for the circumference of the whole circle. Students work backward: from the sector to the full area, from the area to the radius, and from the radius to the circumference. Mixing up area and circumference, or stopping partway, gives wrong choices.

NYS standards: 7.G.4 circle area and circumference
6.G.5 perfect squares and perfect cubes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 circle problems usually give the radius and ask for area or circumference. This one runs backward from the area of a 45° sector, needing the sector's share of the circle, a square root and a switch to circumference: four linked steps.
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: Combines ideas moderate
Something you must already know moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question moves from a part of the circle's area to the whole circle's length, so area and circumference both appear and are easy to swap. Many students scale the sector up to 144π and stop, giving an area when a length was asked. Solving r² = 144 also needs a square root, after which 2πr is routine. Asking at the end whether the result is a length, measured in feet rather than square feet, catches the main trap.

Combines ideas moderate Working it out
Three ideas chain together: the 45° sector is 1/8 of the circle, so the whole area is 8 × 18π = 144π, then the radius comes from the area, and the circumference from the radius. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know moderate Working it out
Neither πr² nor 2πr is given, nor that a full turn is 360°, so every link of the chain rests on formulas students must already know. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The stem says only “shaded sector”; its 45° angle is marked at the center in the figure, the one place the fraction of the circle comes from. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Work out what fraction of the circle a 45° sector is, scale its area up to the whole circle, solve for the radius, and only then use the circumference formula.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Scales the sector up to the whole circle, 8 × 18π = 144π, and gives that area, though the question asks for the circumference.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Repeats the sector's area from the stem, 18π, as if the shaded part's area were the circle's circumference.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No step with the 45° angle or the circle formulas gives 9π, which is half the sector's area, so it mostly attracts guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a circle in a sector: $\text{fraction of the circle} = \frac{\text{angle}}{360}$; A sector is a slice of a circle, like a slice of pizza. Its angle is the angle at the center, in degrees. A full turn around the center is 360 degrees. So the slice is that fraction of the whole circle.
Area of a circle: $A = \pi r^{2}$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π, called pi, is a number a little more than 3. r² means r times r.
Circumference of a circle: $C = 2\pi r$; C is the circumference, the distance around the circle. r is the radius and π is pi. Two radii make the diameter, the distance across, so this is the same as π times the diameter.

Sort out the question
The shaded slice has a 45° angle at the center. Its area is $18\pi$ square feet. You need the circumference, the distance around the whole circle. For that you need the radius, so work back to it.
Find what fraction is shaded
$\text{fraction of the circle} = \frac{\text{angle}}{360}$
$\frac{45}{360} = \frac{1}{8}$
The shaded slice is one eighth of the circle. Eight slices like it would fill the circle.
Find the area of the whole circle
$8 \times 18\pi = 144\pi$
The whole circle has an area of $144\pi$ square feet. Watch out: this is an area, not the circumference. Stopping here gives choice A.
Find the radius
$A = \pi r^{2}$
$\pi r^{2} = 144\pi$
$r^{2} = 144$
$r = 12$
Divide both sides by π. Then ask what number times itself makes 144. It is 12, so the radius is 12 feet.
Find the circumference
$C = 2\pi r$
$C = 2 \times \pi \times 12 = 24\pi$
Leave π as it is, because the choices are written with π. The circumference is $24\pi$ feet. So the answer is B, $24\pi$ ft.

Check your answer

Work forward. A radius of 12 feet gives an area of $\pi \times 12 \times 12 = 144\pi$ square feet. One eighth of that is $144\pi \div 8 = 18\pi$ square feet, the shaded area in the question.

More questions like this

The original DOE question

Question 94 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 94. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 42 of 86

2023 Test A Math, question 41

Question

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

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

Show the answer

Correct answer: C

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

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

Question analysis

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

NYS standards: 7.EE.3 multi-step rational number problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 percent work often takes a percent of a percent or finds a percent change. Here both percents are of the same whole book, so they combine directly; the rest is a subtraction and simplifying a fraction, lighter than typical.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 33 words.
Corrected from the DOE handbook: Missing percent signs in the explanation were restored.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

Explanation: The handbook writes “So far, Sarah has read 15 + 17 = 32 of the entire book. That means she has 100 − 32 = 68 left to read.” (p. 143, question 98), with no percent signs, so 32 and 68 read as amounts of the book (pages, say) that the question never gives. The question gives 15% and 17%, and the handbook’s next line begins “68% =”, so the site now reads 15% + 17% = 32% and 100% − 32% = 68%. The answer, 17/25, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction: $p\% = \frac{p}{100}$; p is the percent. Percent means “out of 100,” so p percent is p hundredths. For example, 50% is 50 out of 100, which is one half.
Simplifying a fraction: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top number, b is the bottom, and n is a number that divides both evenly. Dividing the top and bottom by the same number keeps the value the same. A fraction is in lowest terms when no number bigger than 1 divides both.

Sort out the question
Sarah read 15% of her book yesterday and 17% today. Both are percents of the whole book. You need the part still left to read, as a fraction in lowest terms. That means a fraction that can’t be simplified any more.
Find how much she has read
$15\% + 17\% = 32\%$
Both percents are of the same whole book, so you can add them. She has read 32% of it.
Find how much is left
The whole book is 100%.
$100\% - 32\% = 68\%$
Watch out: 32% is the part she has read, not the part left. Rounding 32% to 30% gives $\frac{3}{10}$ , choice B.
Write 68% as a fraction
$p\% = \frac{p}{100}$
$68\% = \frac{68}{100}$
This fraction is not in lowest terms yet, because 68 and 100 are both even.
Write it in lowest terms
The number 4 divides both 68 and 100 evenly.
$\frac{a}{b} = \frac{a \div n}{b \div n}$
$\frac{68}{100} = \frac{68 \div 4}{100 \div 4} = \frac{17}{25}$
Watch out: don’t round 68% to 70%. That gives $\frac{7}{10}$ , choice D. So the answer is C, $\frac{17}{25}$ .

Check your answer

Turn the answer back into a percent: $\frac{17}{25} = \frac{68}{100}$ , which is 68%. Then 68% + 32% = 100%, the whole book. Also, 17 and 25 share no factor except 1, so the fraction is in lowest terms.

More questions like this

The original DOE question

Question 98 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 98. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 43 of 86

2023 Test A Math, question 53

Question

At West College, students are randomly assigned to one of 20 dormitories and one of 6 dining rooms. Kharleen likes 6 of the dormitories and 2 of the dining rooms. What is the probability that she is assigned to both a dormitory and a dining room that she likes?

A10%
B12%
C19%
D38%

Show the answer

Correct answer: A

Kharleen likes $\frac{6}{20} = \frac{3}{10}$ of the dormitories. She likes $\frac{2}{6} = \frac{1}{3}$ of the dining rooms. Multiply those fractions together to calculate the probability of Kharleen being assigned to both a dormitory and a dining room that she likes: $\frac{3}{10} \times \frac{1}{3} = \frac{1}{10} = 10\%$ .

Question analysis

This probability word problem says students at West College are randomly assigned one of 20 dormitories and one of 6 dining rooms, and Kharleen likes 6 dormitories and 2 dining rooms. It asks for the chance that she gets both a dormitory and a dining room she likes. Students combine two independent events and write the result as a percent.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the chance that two independent events both happen is grade 7 compound probability. Each event's chance comes straight from the counts given, and joining them and writing the result as a percent is a typical grade 7 task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question turns on one rule, multiplying the chances of two independent events. Students who know it reduce the fractions first and finish in seconds; students who don't tend to combine the counts, and 6 × 2 = 12 reads naturally as 12%. The other two choices match no common working, so the real contest is between the product of the counts and the product of the chances.

Something you must already know moderate Working it out
The rule that the chance of two independent events both happening is the product of their chances, 3/10 × 1/3, is assumed; nothing in the stem says to multiply. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The phrase “both a dormitory and a dining room that she likes” describes two separate random assignments that must both go her way, so the two chances are combined, not compared. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choice 12% sits only 2 points above 10%, so a rough sense that the answer is small cannot separate them; only computing 3/10 × 1/3 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the chance of a liked dormitory and the chance of a liked dining room as fractions, multiply them because both must happen, and convert the product to a percent.

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.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Multiplies the numbers of liked options, 6 × 2 = 12, and reads that product as 12%, without dividing by the 20 × 6 possible assignments.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No combination of 6/20 and 2/6 gives 19%, a value between the right answer and either single chance, so it mostly draws estimates and guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 38%, which is larger than either single chance, 30% or about 33%, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. Each outcome must be equally likely.
Probability of two separate events: $P(\text{A and B}) = P(\text{A}) \times P(\text{B})$; A and B are two events that don’t affect each other, like two separate random picks. P(A) is the chance of A, and P(B) is the chance of B. The chance that both happen is the two chances multiplied.

Sort out the question
Kharleen gets one of 20 dorms, or dormitories, where students live. She also gets one of 6 dining rooms. Each is picked at random. She likes 6 dorms and 2 dining rooms. You need the chance that she gets a dorm she likes and a dining room she likes.
Find the chance of a liked dorm
$P = \frac{\text{outcomes you want}}{\text{all outcomes}}$
$P(\text{dorm}) = \frac{6}{20} = \frac{3}{10}$
She likes 6 of the 20 dorms. Divide the top and bottom by 2 to simplify.
Find the chance of a liked dining room
Use the same formula for the dining rooms.
$P(\text{dining room}) = \frac{2}{6} = \frac{1}{3}$
She likes 2 of the 6 dining rooms. Divide the top and bottom by 2.
Multiply the two chances
The two picks are separate. One doesn’t change the other.
$P(\text{A and B}) = P(\text{A}) \times P(\text{B})$
$\frac{3}{10} \times \frac{1}{3} = \frac{3}{30} = \frac{1}{10}$
Both must happen, so the chance is smaller than either one alone.
Change to a percent
$\frac{1}{10} = \frac{10}{100} = 10\%$
Watch out: multiplying 6 × 2 = 12 and calling it 12% skips dividing by all the possible assignments. That is choice B. So the answer is A, 10%.

Check your answer

Count every possible assignment. There are $20 \times 6 = 120$ dorm and dining room pairs, all equally likely. She likes $6 \times 2 = 12$ of them. Then $\frac{12}{120} = \frac{1}{10}$ , which is 10%, the same answer.

More questions like this

The original DOE question

Question 110 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 110. 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 44 of 86

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 45 of 86

2023 Test A Math, question 57

Question

Chanelle selects a marble from a container and then returns it to the container. When she does this 3 times, the probability of choosing a red marble all 3 times is $\frac{1}{216}$ . Based on this information, what is the probability of Chanelle choosing a red marble the first time she selects a marble?

A $\frac{1}{72}$
B $\frac{1}{36}$
C $\frac{1}{8}$
D $\frac{1}{6}$

Show the answer

Correct answer: D

Let $\frac{1}{x}$ be the probability of Chanelle selecting a red marble the first time. The probability of selecting a red marble three times is $\frac{1}{x} \times \frac{1}{x} \times \frac{1}{x}$ , which is given as $\frac{1}{216}$ . Since $216 = 6 \times 6 \times 6$ , the value of x is 6, and the probability of Chanelle selecting a red marble the first time is $\frac{1}{6}$ .

Question analysis

This probability question says Chanelle draws a marble and puts it back, three times, and the chance of red all three times is 1/216. It asks for the chance of red on her first draw. Students reverse the multiplication rule for repeated independent draws, which means finding a cube root, and dividing by 3 is the tempting shortcut.

NYS standards: 7.SP.8a compound event as fraction of outcomes
8.EE.2 square and cube roots
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Multiplying probabilities for repeated draws with replacement is grade 7 compound probability. Running it backward, from three draws to one, needs a cube root, a grade 8 idea, though the numbers are friendly and the work is short once set up.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 55 words.
Students, last 30 days: 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 three draws are seen as one chance cubed, the question is a cube root, and 216 is a perfect cube many students know. The risk lies in undoing the cube the wrong way: dividing by 3 because there are three draws, or stopping at 36 after one division by 6. A quick check, 1/6 × 1/6 × 1/6 = 1/216, confirms the answer.

Something you must already know moderate Working it out
Students must know that three draws with the marble returned each time multiply, p × p × p, and must recognize 216 as 6 × 6 × 6; the stem states neither. (Anderson & Krathwohl, 2001)
No obvious method minor Working it out
The usual direction runs from one draw to three; here the three-draw chance is given and the bold first asks for one draw, so students work backward and undo a cube. (Schoenfeld, 1985)

Strategy: Call the chance of red on one draw p, so three draws with replacement give p × p × p; then find the number whose cube equals the given probability.

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 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides 216 by 3 because there are three draws, instead of finding the number that multiplied by itself three times makes 216.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Divides 216 by 6 to get 36 and stops one division short, keeping 1/36, which is the chance of two reds in a row, not one.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with 1/216 gives 1/8, the chance of three heads in three coin tosses, so it mostly attracts guesses at a familiar fraction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of repeated picks: $P(\text{red 3 times}) = p \times p \times p$; p is the chance of red on one pick. Chanelle puts the marble back each time, so every pick has the same chance p, and one pick doesn’t change the next. The chance of red on all three picks is the three chances multiplied.

Sort out the question
Chanelle picks a marble and puts it back, 3 times. The chance of red all 3 times is $\frac{1}{216}$ . You need the chance of red on just the first pick. Call that chance p.
Write the three picks as a product
She puts each marble back, so the container is the same every time. Each pick has the same chance, p.
$P(\text{red 3 times}) = p \times p \times p$
$p \times p \times p = \frac{1}{216}$
Find the number that fits
You need a fraction that, multiplied by itself three times, makes $\frac{1}{216}$ . Its top is 1, since 1 × 1 × 1 = 1. For the bottom, find a number that times itself three times makes 216.
$6 \times 6 \times 6 = 36 \times 6 = 216$
So the bottom is 6, and p is $\frac{1}{6}$ .
Avoid the tempting shortcuts
Watch out: dividing 216 by 3 gives 72, choice A. Three picks means multiplying three times, not dividing by 3. Also, $\frac{1}{36}$ is the chance of red twice in a row, choice B.
So the answer is D, $\frac{1}{6}$ .

Check your answer

Multiply it out: $\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}$ . That matches the question. Choice A fails fast: $\frac{1}{72} \times \frac{1}{72}$ is already far smaller than $\frac{1}{216}$ .

More questions like this

The original DOE question

Question 114 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 114. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 46 of 86

2023 Test B Math, question 6

Question

What is the value of $3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)$ ?

A $-3\frac{1}{12}$
B $-2\frac{11}{12}$
C $6\frac{1}{12}$
D $9\frac{5}{12}$

Show the answer

Correct answer: C

The first step is to rewrite the problem replacing all subtraction with adding the opposite:

$3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)$

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

Then, rewrite all the fractions using a common denominator of 12, which is the least common multiple of the fractions’ denominators, 4, 3, and 2:

$3\frac{3}{12} + \left(-1\frac{8}{12}\right) + 4\frac{6}{12}$

Next, combine the positive numbers and, finally, add the negative number to the sum by subtracting the number with the least absolute value from the number with the greatest absolute value and using the sign of the number with the greatest absolute value:

$7\frac{9}{12} + \left(-1\frac{8}{12}\right)$

$6\frac{1}{12}$

Question analysis

What is 3 1/4 − 1 2/3 − (−4 1/2)? This question on rational-number arithmetic combines mixed numbers with denominators of 4, 3 and 2 and ends by subtracting a negative number. Students must turn that last subtraction into an addition and find a common denominator before combining.

NYS standards: 7.NS.1c rational number subtraction
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Subtracting negative mixed numbers is core grade 7 content. This expression is typical in length, but it needs a common denominator of 12 and careful handling of the rule for subtracting a negative.
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 10, 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 is short, but it mixes a subtraction that stays a subtraction with one that becomes an addition, and that difference is what most wrong answers miss. The fractions add a second load, since twelfths are needed before anything combines. Estimating first, 3 − 1.7 + 4.5 is about 6, shows the result is positive, which rules out both negative values.

Error-prone arithmetic moderate Working it out
Three mixed numbers with denominators 4, 3 and 2 must become twelfths, and the last term, −(−4 1/2), changes sign while the middle subtraction does not. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Rewriting the double negative, renaming each fraction, adding 3 1/4 and 4 1/2, then taking away 1 2/3 are four steps, each able to carry a slip forward. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change subtracting −4 1/2 into adding 4 1/2, rename every fraction in twelfths, add the two positive numbers, then take away 1 2/3.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single error gives −3 1/12; it needs subtracting 4 1/2 and also miswriting 3 1/4 as 3 1/12, two errors at once, so it mostly draws guesses.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Treats subtracting −4 1/2 as subtracting 4 1/2, so 3 1/4 − 1 2/3 − 4 1/2 = −2 11/12.
D Negatives and direction (Brown & Burton, 1978)
Reveals a surface strategy Expected pull: weak
Drops both minus signs and adds the three mixed numbers, 3 1/4 + 1 2/3 + 4 1/2 = 9 5/12, as if every operation were addition.

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 stand for any numbers. Taking away a negative number is the same as adding its opposite, the same number with the other sign. Think of it this way: taking away a debt leaves you richer.
Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a is the top and b is the bottom of a fraction. Multiplying both by the same number n does not change its value. Use it to give fractions the same bottom number, so you can add or subtract them.

Sort out the question
You need the value of $3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)$ . Each number is a mixed number, a whole number plus a fraction. There are two subtractions. The second one takes away a negative number.
Change the double negative
Use the rule for subtracting a negative. Taking away $-4\frac{1}{2}$ is the same as adding $4\frac{1}{2}$ .
$3\frac{1}{4} - 1\frac{2}{3} + 4\frac{1}{2}$
The first minus sign stays as it is. Watch out: if you subtract $4\frac{1}{2}$ instead, you get $-2\frac{11}{12}$ , choice B.
Give the fractions one denominator
The denominator is the bottom number of a fraction. Here they are 4, 3 and 2. The smallest number all three divide into is 12. So change each fraction to twelfths.
$\frac{1}{4} = \frac{3}{12} \qquad \frac{2}{3} = \frac{8}{12} \qquad \frac{1}{2} = \frac{6}{12}$
The problem is now:
$3\frac{3}{12} - 1\frac{8}{12} + 4\frac{6}{12}$
Add the two positive numbers
You may add and subtract in any order, as long as each number keeps its sign. Add the two positive numbers first.
$3\frac{3}{12} + 4\frac{6}{12} = 7\frac{9}{12}$
That is 3 + 4 = 7 wholes and 3 + 6 = 9 twelfths.
Take away the last number
$7\frac{9}{12} - 1\frac{8}{12} = 6\frac{1}{12}$
That is 7 − 1 = 6 wholes and 9 − 8 = 1 twelfth. So the answer is C, $6\frac{1}{12}$ .

Check your answer

Estimate with decimals: $3.25 - 1.67 + 4.5 \approx 6.08$ . And $6\frac{1}{12}$ is about 6.08 too. The answer must be positive, so A and B are out. Choice D comes from adding all three numbers.

More questions like this

The original DOE question

Question 63 as printed on page 193 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 63. Open page 193 of the handbook (PDF)

Question 47 of 86

2023 Test B Math, question 8

Question

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

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

Show the answer

Correct answer: A

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

Question analysis

Given the equation 2x + 4 = y, students solve for x in terms of y. This literal-equation question offers four choices that mix multiplying and dividing by 2 with adding and subtracting 4. Undoing the two operations on x in the right order, with the right inverse each time, is what it tests.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve two-step equations such as 2x + 4 = 10. Here the right side is the letter y, so the same two inverse steps leave an expression, a first taste of the grade 8 skill of rearranging.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The two inverse steps are routine, subtract 4 and then divide by 2, but the letters hide whether they were done correctly. The main hazard is moving the 4 across the equals sign while keeping its plus sign, which produces an expression nearly identical to the right one. Substituting x = 1 and y = 6 tests every choice in seconds.

A very close wrong answer moderate Choosing the answer
(y − 4)/2 and (y + 4)/2 differ only in the sign before 4, and that sign depends on undoing the +4 by subtracting it from both sides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
With y on the right instead of a number, the answer is an expression, so there is no value to check unless the student substitutes a pair such as x = 1 and y = 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Treat y as if it were a number and undo the operations on x in reverse order, using the inverse of each, then check by substituting a value.

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it balanced. Use this to undo, one at a time, each step that was done to x.

Sort out the question
The equation is $2x + 4 = y$ . “Solve for x in terms of y” means get x alone on one side. The other side will have y in it. Treat y as if it were a number.
Undo the adding of 4
Right now, x is doubled and then 4 is added. Undo those steps in reverse order. First, subtract 4 from both sides.
$2x + 4 - 4 = y - 4$
$2x = y - 4$
Watch out: the 4 must be subtracted from y. Writing $2x = y + 4$ leads to choice C.
Undo the doubling
Next, divide both sides by 2. Divide the whole right side, not only the y.
$\frac{2x}{2} = \frac{y - 4}{2}$
$x = \frac{y - 4}{2}$
Give the answer
Watch out: choices B and D multiply by 2. But x was multiplied by 2, so you undo that by dividing. So the answer is A, $x = \frac{y - 4}{2}$ .

Check your answer

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

More questions like this

The original DOE question

Question 65 as printed on page 193 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 65. Open page 193 of the handbook (PDF)

Question 48 of 86

2023 Test B Math, question 11

Question

$n = 0.13 + \frac{207}{100}$

What is the value of n expressed as a fraction?

A $\frac{1}{3}$
B $\frac{11}{5}$
C $\frac{9}{4}$
D $\frac{23}{10}$

Show the answer

Correct answer: B

$n = 0.13 + \frac{207}{100} =$

$\frac{13}{100} + \frac{207}{100} =$

$\frac{220}{100} = \frac{11}{5}$

Question analysis

The number n equals 0.13 plus 207/100, and students must write n as a single fraction. This question mixes a decimal with a fraction in hundredths. Converting both to the same form, adding, and reducing to lowest terms are the steps, and the four fractions offered look quite different at first glance.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting decimals to fractions and simplifying are skills taught before grade 7. Both numbers here are in hundredths, so the addition is immediate and only the final reduction takes thought.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Converting 0.13 to hundredths makes this a single addition, 13 + 207 = 220 hundredths, so the question tests little beyond reducing a fraction. Two of the wrong choices sit close to 2.2 but come from no common working, and the third only echoes the digits of 0.13. A student who checks each choice as a decimal finds only one equal to 2.2.

An answer in an unexpected form minor Choosing the answer
The sum comes out as 220/100 or 2.2, but the choices are in lowest terms, so it must be reduced, or each choice converted, before anything matches. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
One term is a decimal and the other a fraction; writing 0.13 as 13/100 puts both in hundredths, and a misplaced digit there changes the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write 0.13 as 13/100 so both terms are in hundredths, add the numerators, then divide the top and bottom by their greatest common factor.

Traps

Answer in another form: The right answer is written differently from the way you worked it out.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Echoes the digits of 0.13, reading it as the fraction 1/3, and ignores the 207/100 altogether.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9/4, which is 2.25; adding 13 and 207 hundredths gives 2.2, so this mostly draws guesses near the sum.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing leads to 23/10 except 13 + 207 coming out as 230, a slip with no reason behind it, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths: $\text{a decimal with two places} = \frac{\text{its digits}}{100}$; The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
Adding fractions with the same bottom: $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$; a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.

Sort out the question
n equals 0.13 plus $\frac{207}{100}$ . One part is a decimal and one is a fraction. You need n as a fraction. The choices are in lowest terms, which means they can’t be simplified any more.
Write the decimal as a fraction
The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.
$0.13 = \frac{13}{100}$
Now both numbers are in hundredths. Watch out: 0.13 is not $\frac{1}{3}$ . That only copies the digits, choice A.
Add the hundredths
Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.
$\frac{13}{100} + \frac{207}{100} = \frac{13 + 207}{100} = \frac{220}{100}$
Simplify the fraction
$\frac{220}{100}$ is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.
$\frac{220 \div 20}{100 \div 20} = \frac{11}{5}$
So the answer is B, $\frac{11}{5}$ .

Check your answer

Change both to decimals. $\frac{11}{5} = 11 \div 5 = 2.2$ . And $0.13 + 2.07 = 2.20$ . They match. Choices C and D are close, but $\frac{9}{4} = 2.25$ and $\frac{23}{10} = 2.3$ .

More questions like this

The original DOE question

Question 68 as printed on page 194 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 68. Open page 194 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 49 of 86

2023 Test B Math, question 12

Question

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

Show the answer

Correct answer: A

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

$25 + 35 + 10x = 100$

$60 + 10x = 100$

$10x = 40$

$x = 4$

Question analysis

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

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving a two-step equation from a short description is standard grade 7 work. The perimeter setting is simple; the only extra care is counting the sides that remain after the two given ones.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The perimeter and the two named sides make the subtraction obvious, so the real decision comes after it: what to divide the leftover 40 centimeters by. Students who stop at the leftover length, or who lose track of how many sides remain, reach a listed choice. Writing 25 + 35 + 10x = 100 keeps the count of 10 in view.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 50 of 86

2023 Test B Math, question 15

Question

On a map, the distance between Deshawn’s house and Xin’s house is 5.4 inches. If 2 inches on the map represents 3 miles, what is the distance between these two houses, in miles?

A2.6
B3.6
C7.5
D8.1

Show the answer

Correct answer: D

Let d represent the distance between the two houses, in miles. Write a proportion to solve for d:

$\frac{d}{5.4} = \frac{3}{2}$

$d = \frac{3(5.4)}{2} = \frac{16.2}{2} = 8.1$

Question analysis

A map shows Deshawn's and Xin's houses 5.4 inches apart, with 2 inches on the map standing for 3 miles. Students find the actual distance between the houses in miles. This scale question uses a decimal map length, so the multiplication must be exact and the inches-to-miles ratio kept the right way up.

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Converting a map distance with a given scale is routine grade 7 work. This one needs a single proportion and a decimal multiplication, with no rounding or second step, so it matches a typical classroom exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Unlike most map questions, this one asks for an exact distance, so the decimal 5.4 has to be carried all the way through. The two tempting wrong answers come from opposite places: an upside-down scale shrinks the distance, and a rounded map length gives a result close to the true one. Checking that the answer is one and a half times 5.4 catches both.

Error-prone arithmetic minor Working it out
The map distance is the decimal 5.4, and 5.4 × 3 = 16.2 then halved is a decimal multiplication where rounding 5.4 to make it easier changes the answer. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
7.5 and 8.1 miles are close enough that a rough estimate, 5 × 1.5, lands on the wrong one; only the exact product separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Set up miles over inches equal to 3 over 2, then multiply 5.4 by 3 and divide by 2, keeping the decimal exact throughout.

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).
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2.6, which sits just below half of 5.4 and matches no step with the scale, so it mostly draws guesses.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Inverts the scale, multiplying 5.4 by 2/3 instead of 3/2, as if each mile took up 1.5 inches on the map.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds 5.4 inches down to 5 before scaling, so 5 × 3 ÷ 2 = 7.5 miles instead of the exact distance.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion: $\frac{\text{real distance}}{\text{map distance}} = \frac{\text{real distance}}{\text{map distance}}$; A map shrinks every distance by the same amount. So real distance over map distance makes the same fraction everywhere on the map. Put miles on top on both sides and inches on the bottom. Then solve for the missing distance.

Sort out the question
On the map, Deshawn’s house and Xin’s house are 5.4 inches apart. The scale says 2 inches on the map stand for 3 miles. You need the real distance between the houses, in miles.
Set up the proportion
Let d be the real distance, in miles. Put miles on top and inches on the bottom, on both sides.
$\frac{d}{5.4} = \frac{3}{2}$
Solve for d
To get d alone, multiply both sides by 5.4.
$d = \frac{3 \times 5.4}{2} = \frac{16.2}{2} = 8.1$
The houses are 8.1 miles apart. Watch out: keep 5.4 exact. Rounding it to 5 gives 7.5 miles, choice C.
Give the answer
Watch out: don’t flip the scale. Multiplying 5.4 by $\frac{2}{3}$ gives 3.6, choice B. Each inch stands for more than a mile, so the miles must be more than 5.4. So the answer is D, 8.1.

Check your answer

Each inch stands for $3 \div 2 = 1.5$ miles. So 5 inches is 7.5 miles, and 0.4 inch is $0.4 \times 1.5 = 0.6$ mile. Together, $7.5 + 0.6 = 8.1$ miles.

More questions like this

The original DOE question

Question 72 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 72. Open page 195 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 86

2023 Test B Math, question 16

Question

$A wheel: a circle with a small solid hub at its center and 15 straight spokes running from the hub to the rim, equally spaced around it.$

This wheel has 15 spokes, equally spaced. What is the measure of the angle between adjacent spokes?

A30°
B24°
C15°
D12°

Show the answer

Correct answer: B

The total number of degrees in a circle is 360. The measure of the angle between adjacent spokes is $\frac{360}{15} = 24^{\circ}$ .

Question analysis

A wheel has 15 spokes, equally spaced around its hub, as the figure shows. Students find the measure of the angle between two neighboring spokes. The question tests knowing how many degrees make a full turn around a point and sharing them equally among the spokes.

NYS standards: 7.G.5 angle relationships and equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding an angle by dividing a full turn into equal parts is early angle measurement. It takes one division, and the only fact needed is the number of degrees in a circle, so it sits below 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: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 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

One fact and one division answer this question, so it is among the easiest on the test. The wrong choices catch students who use a straight angle’s 180° for the full turn, who take the number of spokes as the angle, or who guess a familiar angle. Multiplying the answer by 15 to get back to 360° is a quick check.

Something you must already know minor Working it out
The question never says that a full turn is 360°; knowing it, and that the 15 spokes split it into 15 equal angles, is the whole method. (Anderson & Krathwohl, 2001)

Strategy: The spokes split the full turn around the hub into 15 equal angles, so divide the number of degrees in a circle by 15.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
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 working with 15 spokes gives 30°, which is 360 ÷ 12, the angle between hours on a clock, so it mostly draws guesses at a familiar angle.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the number of spokes, 15, as the angle in degrees, matching a number in the question without any calculation.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 180 by 15, using a straight angle’s 180° as the full turn around the hub instead of 360°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal angles around a point: $\text{each angle} = 360 \div \text{number of equal angles}$; One full turn around a point is 360 degrees. If lines from the point split that turn into equal angles, each angle gets the same share. So divide 360 by how many angles there are.

Sort out the question
The wheel has 15 spokes. They all start at the hub, the center of the wheel, and they are equally spaced. Adjacent spokes are spokes right next to each other. You need the angle between two adjacent spokes.
Count the angles around the hub
Go once around the hub. Between each spoke and the next is one angle. So 15 spokes make 15 angles, all the same size. Together they make one full turn, which is 360 degrees.
Watch out: a full turn is 360 degrees, not 180. Using 180 gives 12 degrees, choice D.
Share the full turn
Use the formula for equal angles.
$\text{each angle} = 360 \div \text{number of equal angles}$
$\text{each angle} = 360 \div 15 = 24$
Each angle is 24 degrees. Watch out: 15 is the number of spokes, not the size of the angle, so choice C is wrong. So the answer is B, 24°.

Check your answer

Multiply back: 15 angles of 24 degrees make $15 \times 24 = 360$ degrees, a full turn. Also, a clock’s 12 hour marks are 30 degrees apart. This wheel has more spokes than that, so each angle must be smaller than 30 degrees.

More questions like this

The original DOE question

Question 73 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 73. Open page 195 of the handbook (PDF)

Question 52 of 86

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 53 of 86

2023 Test B Math, question 20

Question

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

Show the answer

Correct answer: C

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

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

$5n + 10 = 110$

$5n = 100$

$n = 20$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 54 of 86

2023 Test B Math, question 22

Question

A vendor is selling 10 packages of homemade fudge at a bake sale. The vendor weighed and labeled each package. The package weights are given in ounces:

4.6	6.0	3.8	3.5	4.5
5.4	3.9	6.5	4.8	6.2

The vendor created a box plot of the package weights. Which box plot best represents the data?

A $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at about 4.1, the box runs from 5 to 6 with the median line at 5.5, and the right whisker ends at about 6.3.$
B $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 4.5, and the right whisker ends at 6.5.$
C $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from about 3.9, just left of 4, to 6, with the median line halfway between 4.5 and 5, and the right whisker ends at 6.5.$
D $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 5, and the right whisker ends at 6.5.$

Show the answer

Correct answer: C

A set of 10 data points is given. The first step for determining the quartiles and minimum and maximum values of the box plot is to order the data from smallest to greatest value. For this data set the order is:

3.5	3.8	3.9	4.5	4.6
4.8	5.4	6.0	6.2	6.5

The minimum value is the smallest decimal, 3.5, and the maximum value is the greatest decimal, 6.5.

To determine the median in a set with an even number of values, locate the two middle values (5th and 6th in this case): 4.6 and 4.8. Determine the mean, or average, of the two middle numbers by adding them together and dividing the sum by 2:

$\frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7$

To determine the value of the first quartile (Q1), locate the middle value of the 5 terms to the left of the median:

3.5

3.8

3.9

4.5

4.6

To determine the value of the third quartile (Q3), locate the middle value of the 5 terms to the right of the median:

4.8

5.4

6.0

6.2

6.5

Box plot C identifies the correct minimum, maximum, median, and quartiles of the data set.

Question analysis

A vendor weighs 10 packages of homemade fudge, from 3.5 to 6.5 ounces, and makes a box plot of the weights. Students choose the box plot that matches the data. They must sort the decimals, find the median of an even number of values and both quartiles, then compare plots that differ only slightly.

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Building a box plot from data is grade 7 content. This version needs all five summary values from ten unsorted decimals, including a median between two values, then a close comparison of plots, longer and finer than typical.
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
A very close wrong answer moderate
Reading a figure, table or graph moderate
Slow to do carefully minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 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

Most of the work happens before the plots are even read: the weights must be sorted, and the median of ten values is the average of two of them. After that, three plots look almost identical, so a careless median, taken from the unsorted list or placed halfway between the extremes, settles on a wrong one. The careful student checks the median line and the box’s left edge against values that fall between the tick marks.

Many steps moderate Working it out
Ten unsorted decimals must be ordered, then five values found: the median as the mean of the fifth and sixth, 4.6 and 4.8, and each quartile as the middle of its half. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
Three plots share whiskers at 3.5 and 6.5 and a box ending at 6; they differ only in the median line, at 4.5, near 4.75 or at 5, and in where the box starts. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph moderate Understanding the question
The axis is ticked every half ounce, and the correct plot’s median, 4.7, and lower quartile, 3.9, both fall between ticks, so each line’s position must be judged by eye. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Slow to do carefully minor Choosing the answer
Sorting ten decimals, computing a five-number summary and comparing four plots takes well over a minute and a half when done carefully. (Lu & Sireci, 2007)

Strategy: Sort the ten weights, then find the least and greatest, the median as the mean of the fifth and sixth values, and each quartile as the middle of its half.

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.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from these weights gives whiskers near 4.1 and 6.3 or a median of 5.5, since the data run from 3.5 to 6.5, so it mostly draws guesses.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Skips sorting and takes 4.5, the fifth weight as printed, for the median, so the plot with its median line on 4.5 seems to fit.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Finds the median with the midrange formula, (3.5 + 6.5) ÷ 2 = 5, halfway between the lightest and heaviest packages instead of the middle of the sorted weights.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Median of an even number of values: $\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}$; First put the values in order. With an even number of values, two sit in the middle. The median is their mean, or average: add them and divide by 2. The median splits the data in half.
Quartiles: $Q_1 = \text{middle of the lower half} \qquad Q_3 = \text{middle of the upper half}$; Q1 is the first quartile and Q3 is the third quartile. Split the ordered data in half at the median. Q1 is the middle of the lower half, and Q3 is the middle of the upper half. A box plot’s box runs from Q1 to Q3.

Sort out the question
The vendor weighed 10 packages of fudge, in ounces. You need the box plot that matches the weights. A box plot is built from five numbers. They are the minimum (least), the first quartile Q1, the median, the third quartile Q3 and the maximum (greatest). So find those five numbers first.
Put the weights in order
List the 10 weights from least to greatest.
3.5 3.8 3.9 4.5 4.6
4.8 5.4 6.0 6.2 6.5
Watch out: you must sort first. The fifth weight as printed is 4.5. Using it as the median points to choice B.
Find the least and greatest
The minimum is 3.5 ounces and the maximum is 6.5 ounces. So the whiskers must run from 3.5 to 6.5. Plot A’s whiskers run from about 4.1 to about 6.3, so A is out.
Find the median
With 10 weights, the two middle ones are the 5th and 6th: 4.6 and 4.8.
$\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}$
$\text{median} = \frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7$
The median is 4.7 ounces. Watch out: the median is not halfway between the least and greatest. That gives $(3.5 + 6.5) \div 2 = 5$ and points to choice D.
Find the two quartiles
The lower half is the 5 weights below the median: 3.5, 3.8, 3.9, 4.5 and 4.6. Its middle value is $Q_1 = 3.9$ .
The upper half is the 5 weights above the median: 4.8, 5.4, 6.0, 6.2 and 6.5. Its middle value is $Q_3 = 6.0$ .
Match the summary to a plot
The five numbers are 3.5, 3.9, 4.7, 6.0 and 6.5. Plots B, C and D all run from 3.5 to 6.5, and their boxes end at 6. They differ in the median line and where the box starts.
- B: the box starts at 4, and the median line is at 4.5.
- D: the box starts at 4, and the median line is at 5.
- C: the box starts just left of 4, and the median line is between 4.5 and 5.
C is the closest match to a Q1 of 3.9 and a median of 4.7. So the answer is C.

Check your answer

A median splits the data in half. Five weights, 3.5 to 4.6, are below 4.7. Five, 4.8 to 6.5, are above it. A median line at 5, as in D, would have six weights below it. A line at 4.5, as in B, would have only three below it.

More questions like this

The original DOE question

Question 79 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 2024 Admissions, page 198, question 79. Open page 198 of the handbook (PDF)

Question 55 of 86

2023 Test B Math, question 23

Question

FRUIT SOLD
AT STORE XYZ

Fruit	Number Sold
Apples	8
Bananas	6
Cherries	12
Oranges	6
Peaches	?
Total	50

Peaches made up what percentage of the total number of pieces of fruit sold at Store XYZ, according to the table above?

A18%
B28%
C36%
D64%

Show the answer

Correct answer: C

First, find the number of peaches sold:

$50 - (8 + 6 + 12 + 6) =$

$50 - 32 = 18$

Then, find the percentage:

$\frac{18}{50} = 0.36 = 36\%$

Question analysis

A table lists the fruit sold at Store XYZ, apples, bananas, cherries and oranges, with the number of peaches shown as a question mark and a total of 50. Students find what percentage of the fruit sold were peaches. They first work out the missing count, then turn it into a percent of the total.

NYS standards: 6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a missing value and writing a part as a percent of a whole are grade 6 skills. With a total of 50, the percent step is quick, so the task is lighter than typical grade 7 percent problems.
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: Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The total of 50 makes the percent step easy, since each piece of fruit is 2 percent, but the question first hides the count it needs. Students who find the 18 peaches and stop, or who give the share of the four listed fruits, reach listed choices. Checking that the peaches’ percent and the other fruits’ percent add to 100 confirms the answer.

Reading a figure, table or graph minor Understanding the question
The table gives four counts, a question mark for peaches and a total of 50, so the missing count must be found by subtraction before any percent. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Adding 8, 6, 12 and 6, subtracting from 50, and turning 18 out of 50 into a percent are three steps, and the count of 18 is itself a listed percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the four known counts, subtract from 50 to get the peaches, then double that count, since each piece out of 50 is 2 percent.

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
Finds the 18 peaches but gives that count as the percent, forgetting to compare it with the total of 50 pieces.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 28%, which would need 14 peaches; it takes a slip such as 36 for 8 + 6 + 12 + 6, so it mostly draws guesses.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the share of the other four fruits, 32 of 50 or 64 percent, instead of the share of the peaches.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you are asking about, here the peaches. The whole is the total, here all 50 pieces of fruit. Dividing gives the fraction of the whole. Multiplying by 100 turns it into a percent, which means out of 100.

Sort out the question
The table shows how many of each fruit Store XYZ sold. The number of peaches is missing, shown by a question mark. The total is 50 pieces of fruit. You need the percent of the total that were peaches.
Add the fruits you know
$8 + 6 + 12 + 6 = 32$
Apples, bananas, cherries and oranges add up to 32 pieces of fruit. That is every fruit except the peaches.
Find the number of peaches
The peaches are whatever is left of the 50.
$50 - 32 = 18$
So 18 peaches were sold. Watch out: 18 is a count, not a percent. Choosing 18% is choice A.
Turn the count into a percent
$\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$
$\frac{18}{50} \times 100 = 0.36 \times 100 = 36$
Peaches were 36% of the fruit sold. Watch out: 32 out of 50 is 64%, choice D. That is the other four fruits, not the peaches. So the answer is C, 36%.

Check your answer

Out of 50 pieces, each piece is 2% of the total, because $100 \div 50 = 2$ . So 18 peaches are $18 \times 2 = 36$ percent. Also, 36% for peaches plus 64% for the rest makes 100%.

More questions like this

The original DOE question

Question 80 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 2024 Admissions, page 198, question 80. Open page 198 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 56 of 86

2023 Test B Math, question 25

Question

The town of Waterville covers an area of 50 square miles. In 2005, its population was 84,000. In 2006, its population had decreased so that the average population per square mile was equal to 1,500. What was the total decrease in population from 2005 to 2006?

A7,500
B8,400
C9,000
D9,200

Show the answer

Correct answer: C

Let p represent the population of Waterville in 2006. The population per square mile is:

$\frac{p}{50} = 1{,}500$

Solve for p:

$p = 1{,}500(50) = 75{,}000$

The population decrease from 2005 to 2006 is $84{,}000 - 75{,}000 = 9{,}000$ .

Question analysis

The town of Waterville covers 50 square miles and had 84,000 people in 2005. By 2006 the population had dropped so that there were 1,500 people per square mile on average. Students find how much the population decreased, turning the population density into a total population before subtracting.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a rate such as people per square mile is grade 6 content. Here it is one step inside a two-step problem with large round numbers, which matches typical grade 7 multi-step rate work.
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
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 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 question gives a total for 2005 but only a rate for 2006, so the two years cannot be compared until the rate becomes a population. After that, one subtraction finishes it. Students who stop at the 2006 population, or who reach for a number shaped like 84,000, land on listed choices, while the near miss above the answer comes from no sensible working.

Turning words into math moderate Understanding the question
“The average population per square mile was equal to 1,500” is a density, which has to become a total, 1,500 × 50 = 75,000, before it can be compared with 84,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
The 2006 population is only a midpoint; subtracting it from 84,000 is a second step, and the question asks for that decrease, not either population. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply 1,500 people per square mile by 50 square miles to get the 2006 population, then subtract it from the 2005 population.

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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 2006 population, 1,500 × 50 = 75,000, and stops; 7,500, one zero short, is the choice whose digits match it.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes a tenth of the 2005 population, 84,000 ÷ 10, a number built from the question rather than from the density.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9,200; it lies just above the right decrease and needs an arithmetic slip with no reason behind it, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Population per square mile: $\text{people per square mile} = \frac{\text{population}}{\text{area}}$; The population is the number of people who live in a place. The area is its size, here in square miles. Dividing gives the average number of people for each square mile. To get back the population, multiply people per square mile by the area.

Sort out the question
Waterville covers 50 square miles. In 2005, 84,000 people lived there. In 2006, there were 1,500 people per square mile, on average. You need how many fewer people lived there in 2006 than in 2005.
Find the 2006 population
Let p be the 2006 population. Use the formula.
$\text{people per square mile} = \frac{\text{population}}{\text{area}}$
$1{,}500 = \frac{p}{50}$
To undo dividing by 50, multiply both sides by 50.
$p = 1{,}500 \times 50 = 75{,}000$
So 75,000 people lived there in 2006.
Find the decrease
A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.
$84{,}000 - 75{,}000 = 9{,}000$
The population went down by 9,000 people.
Give the answer
Watch out: 75,000 is the 2006 population, not the decrease. Choice A, 7,500, looks like it but has one zero fewer. The question asks how much the population dropped. So the answer is C, 9,000.

Check your answer

Work with people per square mile instead. In 2005, there were $84{,}000 \div 50 = 1{,}680$ people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and $180 \times 50 = 9{,}000$ fewer people.

More questions like this

The original DOE question

Question 82 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 82. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 57 of 86

2023 Test B Math, question 27

Question

$\mathrm{N} = \{6, 7, \ldots, 21, 22\}$

Set L (not shown) is a list of 25 consecutive integers. The median of the integers in set N above is equal to the least integer in set L. What is the median of the integers in set L?

Show the answer

Correct answer: B

First, find the median of set N. There are 17 integers in set N, so the 9th integer is the median. The 9th integer is 14.

The least integer in set L is equal to the median of set N, so the least integer in set L is 14.

There are 25 integers in set L, so the median is the 13th integer. That integer is 26.

Question analysis

Set N lists the consecutive integers from 6 to 22, and set L is a list of 25 consecutive integers whose least member equals the median of N. Students find the median of set L. It chains two median questions together, and counting positions in a run of consecutive integers invites off-by-one slips.

NYS standards: 6.SP.5c range, center, and overall pattern
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Medians of short lists are grade 6 content. This item hides both lists, one behind set notation with an ellipsis and one not shown at all, and links them, so it needs more planning and careful counting 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: A very close wrong answer moderate
No obvious method moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question chains two medians, and both depend on counting members correctly in runs of consecutive integers. The easy slip is between position and distance: the 13th member of L is 12 steps above its first, not 13. With four consecutive choices, every counting error has a home, so writing out the first and last members of L, 14 and 38, and averaging them is the safest check.

A very close wrong answer moderate Choosing the answer
The choices 25, 26, 27 and 28 are consecutive, so a count that is off by one position lands on a neighboring choice with nothing to flag it. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
No obvious method moderate Working it out
Set L is never listed; its median has to be found from its least member, 14, and its size, 25, by working out that the middle is the 13th member. (Schoenfeld, 1985)
Reading a figure, table or graph minor Understanding the question
Set N is written with an ellipsis, {6, 7, …, 21, 22}, so its 17 members must be counted, and 22 − 6 = 16 undercounts by one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find N's median by averaging its least and greatest members. L's median is its middle member, so work out that member's position and count up from L's least member.

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives 25, the number of integers in set L stated in the question, instead of finding which integer sits in the middle.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts 13 places up from 14 instead of 12, mixing up the median’s position, 13th, with its distance from the first integer.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 28, which is twice N’s median of 14 and matches no count along set L, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting the integers in a list: $\text{count} = \text{last} - \text{first} + 1$; This counts the whole numbers in a list that goes up by 1, like 6, 7, 8. Subtracting gives the number of steps from the first number to the last. You add 1 because the first number counts too.
Position of the median: $\text{position} = \frac{\text{count} + 1}{2}$; The count is how many numbers are in a list that is in order. When the count is odd, the median is at this position. It has the same number of values before it and after it.

Sort out the question
Set N is 6, 7, and so on up to 22. The dots stand for every integer in between. Integers are whole numbers. Set L has 25 consecutive integers, one right after another. Its least number equals the median of N. You need the median of L.
Count the numbers in set N
$\text{count} = \text{last} - \text{first} + 1$
$22 - 6 + 1 = 17$
Set N has 17 integers. Watch out: $22 - 6 = 16$ misses one, because the first number counts too.
Find the median of set N
$\text{position} = \frac{\text{count} + 1}{2}$
$\text{position} = \frac{17 + 1}{2} = 9$
The median is the 9th number. 6 is the 1st, so the 9th is 8 steps up: 6 + 8 = 14. The median of N is 14.
Find the middle place in set L
So set L starts at 14 and has 25 integers. Find the position of its median.
$\text{position} = \frac{25 + 1}{2} = 13$
The median of L is its 13th number.
Count up to the 13th number
14 is the 1st number. The 13th is 12 steps after it.
$14 + 12 = 26$
Watch out: the 13th number is 12 steps up, not 13. Adding 13 gives 27, choice C. And 25 is how many numbers L has, not its median. So the answer is B, 26.

Check your answer

Set L runs from 14 to $14 + 24 = 38$ . For consecutive integers, the median is halfway between the ends: $(14 + 38) \div 2 = 26$ . That matches. The same trick gives N’s median: $(6 + 22) \div 2 = 14$ .

More questions like this

The original DOE question

Question 84 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 84. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 58 of 86

2023 Test B Math, question 32

Question

$A number line with arrows at both ends and three points marked on it. From left to right, point P is at 1 1/3, point Q is at 3, and point R is at 5 1/6.$

On the number line above, how many units longer is $\overline{QR}$ than $\overline{PQ}$ ?

A $\frac{1}{6}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D3

Show the answer

Correct answer: B

First, find the lengths of the given segments:

$\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}$

$\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}$

Then, find the difference:

$2\frac{1}{6}-1\frac{2}{3}=$

$\frac{13}{6}-\frac{5}{3}=$

$\frac{13}{6}-\frac{10}{6}=$

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

Question analysis

A number line shows three points labeled with mixed numbers and a whole number, and students find how many units longer one segment is than another. It tests subtracting mixed numbers with unlike denominators to find distances. Regrouping when the fraction being subtracted is the larger one is where most errors happen.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding distances between points on a number line is grade 7 work. Here the points are mixed numbers with different denominators, so finding two distances and their difference makes it a typical but careful grade 7 task.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is three subtractions, so most errors are in the fractions rather than the plan. Taking 1 1/3 from a whole number such as 3 tempts students to subtract only the whole parts and keep the 1/3, which reverses which segment looks longer. Rewriting every coordinate in sixths, 8/6, 18/6 and 31/6, turns the work into whole-number subtraction of numerators.

Error-prone arithmetic moderate Working it out
Both lengths subtract mixed numbers with different denominators, 5 1/6 − 3 and 3 − 1 1/3, and the second needs regrouping before the two results are compared in sixths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph minor Understanding the question
The three coordinates come from the labels under P, Q and R, and the segments QR and PQ must each be measured from Q at 3, the point they share. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find each length by subtracting the left endpoint from the right one, writing the mixed numbers in sixths, then subtract the shorter length from the longer.

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts only the whole numbers in 3 − 1 1/3 and keeps the 1/3, giving PQ as 2 1/3, then compares it with 2 1/6 and gets 1/6.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/3; it is only the fraction part of PQ’s length, 1 2/3, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 3, the coordinate of point Q, a number read off the line rather than a difference between two lengths.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Length on a number line: $\text{length} = \text{right point} - \text{left point}$; The length of a segment is the distance between its two end points. Subtract the number on the left from the number on the right. Subtracting tells how many units you move to get from one point to the other.
Mixed number to improper fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number part. b over c is the fraction part, with b on top and c on the bottom. Each whole holds c parts, so a wholes hold a × c parts. Then add the b extra parts. An improper fraction has a top bigger than its bottom.

Read the number line
Point P is at $1\frac{1}{3}$ . Point Q is at 3. Point R is at $5\frac{1}{6}$ .
The bar over QR means segment QR, the piece of the line from Q to R. You need how much longer QR is than PQ. So find both lengths, then subtract.
Find the length of QR
R is on the right, and Q is on the left. Use the length rule.
$\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}$
QR is $2\frac{1}{6}$ units long.
Find the length of PQ
Q is on the right, and P is on the left.
$\text{PQ} = 3 - 1\frac{1}{3}$
Take it in two parts. First, 3 − 1 = 2. Then take away the extra $\frac{1}{3}$ . That gives $2 - \frac{1}{3} = 1\frac{2}{3}$ . So PQ is $1\frac{2}{3}$ units long.
Watch out: Don’t just do 3 − 1 and keep the $\frac{1}{3}$ . That gives $2\frac{1}{3}$ , which is too long. It leads to choice A.
Write both lengths in sixths
The fractions have different bottoms, 6 and 3. Change both lengths to sixths so you can subtract. Use the mixed number rule.
$2\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}$
$1\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}$
Subtract the lengths
Take the shorter length away from the longer one.
$\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}$
QR is $\frac{1}{2}$ unit longer than PQ.
So the answer is B, $\frac{1}{2}$ .

Check your answer

Write every point in sixths. P is $\frac{8}{6}$ , Q is $\frac{18}{6}$ and R is $\frac{31}{6}$ . QR is 31 − 18 = 13 sixths. PQ is 18 − 8 = 10 sixths. The difference is 3 sixths, which is one half.

More questions like this

The original DOE question

Question 89 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 89. Open page 200 of the handbook (PDF)

Question 59 of 86

2023 Test B Math, question 33

Question

$A line segment with point W at its left end and point Y at its right end. Point X lies on the segment between W and Y, closer to Y.$

The distance from W to X is 70% of the distance from W to Y. If the distance from X to Y is 15 miles, what is the distance, in miles, from W to X?

A10.5
B35
C45
D50

Show the answer

Correct answer: B

Since the distance from W to X is 70% of the distance from W to Y, the distance from X to Y is 30% of the distance from W to Y. Let d represent the distance from W to Y. Then 15 miles is 30% of d.

$0.30d=15$

$d=\frac{15}{0.30}=50$

So, the distance from W to X, in miles, is $50-15=35$ .

Question analysis

A percent question with a line segment: point X lies between W and Y, the distance from W to X is 70% of the whole, and the distance from X to Y is 15 miles. Students find the distance from W to X. Matching the 15 miles to the right percent before finding the whole is the step students most often skip.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 percent problems often give a part and its percent and ask for the whole. This one gives the distance for the part the percent does not describe and asks for the other part, so the whole has to be found on the way.
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: 3 steps
Reading level: Question and choices: about grade 5, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The percent and the given distance describe different pieces of the road, which is the whole difficulty. Students who apply the 70% straight to the 15 miles get a small number fast, and students who solve correctly for the whole of WY must still remember to take XY away. Checking that 35 is 70% of 50 confirms the answer and catches both slips.

Turning words into math moderate Understanding the question
The 70% describes WX, but the 15 miles given is XY, so the sentence has to be recast as “XY is 30% of WY” before the 15 miles can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Finding the complement, 30%, then the whole, 15 ÷ 0.30 = 50, then the part, 50 − 15, gives three results, and the middle one is itself a choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out what percent of WY the 15-mile stretch from X to Y must be, use it to find the whole of WY, then find the part from W to X.

Traps

Stops one step early: A choice is a number from partway through the solution.
Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: moderate
Takes 70% of the 15 miles, 0.7 × 15 = 10.5, applying the percent to XY instead of to the whole distance from W to Y.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 45; it would make XY a quarter of WY instead of 30%, so it mostly draws guesses.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the whole distance from W to Y, 15 ÷ 0.30 = 50 miles, and stops before subtracting the 15 miles from X to Y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the full amount. The part is a piece of it. The percent tells how many hundredths of the whole the part is. For example, 30% means 30 out of every 100, or 0.30. Use it whenever a percent describes a piece of a whole.

Sort out the distances
Point X sits between W and Y. So the trip from W to Y has two pieces: W to X, then X to Y.
W to X is 70% of the whole distance from W to Y. X to Y is 15 miles. You need the distance from W to X, in miles.
Find the percent for X to Y
The whole distance from W to Y is 100%. W to X uses 70% of it. The rest is X to Y.
$100 - 70 = 30$
So the 15 miles from X to Y is 30% of the whole distance.
Find the whole distance
Let d be the distance from W to Y. Use the percent formula. The part is 15, and the percent is 30.
$15 = \frac{30}{100} \times d = 0.30 \times d$
Divide both sides by 0.30 to get d by itself.
$d = 15 \div 0.30 = 50$
The distance from W to Y is 50 miles.
Find the distance from W to X
Take the 15 miles from X to Y away from the whole.
$50 - 15 = 35$
W to X is 35 miles.
Watch out: 50 miles is the whole distance from W to Y, choice D. You still have to subtract. Also, the 70% is of the whole distance, not of the 15 miles. Taking 70% of 15 gives 10.5, choice A.
So the answer is B, 35 miles.

Check your answer

Is 35 really 70% of 50? $0.70 \times 50 = 35$ . Yes. And 35 + 15 = 50, the whole distance. The figure also shows X closer to Y. So W to X should be longer than 15 miles, and it is.

More questions like this

The original DOE question

Question 90 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 90. Open page 201 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 86

2023 Test B Math, question 40

Question

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

A45°
B67.5°
C75°
D135°

Show the answer

Correct answer: B

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

$A+B=90$

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

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

$4B=270$

$B=67.5$

Question analysis

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

NYS standards: 7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation from complementary angles is a grade 7 standard. This is the familiar version without a figure: one equation from the definition, then a division and a multiplication, typical of grade 7 classwork.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 97 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 97. Open page 202 of the handbook (PDF)

Question 61 of 86

2023 Test B Math, question 41

Question

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

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

Show the answer

Correct answer: C

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 98 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 98. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 62 of 86

2023 Test B Math, question 43

Question

A certain square puzzle has an area of 64 square inches, and a cubic block has a volume of 64 cubic inches. What is the ratio of the length of a side of the puzzle to the length of a side of the block?

A $1:3$
B $1:2$
C $2:3$
D $2:1$

Show the answer

Correct answer: D

The length of one side of the puzzle is $\sqrt{64}=8$ inches. The length of one side of the block is $\sqrt[3]{64}=4$ inches.

The ratio of the side of the puzzle to the side of the block is $8:4$ , which reduces to $2:1$ .

Question analysis

A square puzzle has an area of 64 square inches, and a cube-shaped block has a volume of 64 cubic inches. Students find the ratio of a side of the puzzle to a side of the block. Both figures are described by the number 64, but turning an area and a volume back into lengths takes different steps, and the order of the ratio matters.

NYS standards: 8.EE.2 square and cube roots
6.G.5 perfect squares and perfect cubes
6.RP.1 ratio concepts and language
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 area and volume problems go from side lengths to an area or a volume. Going backward needs a square root and a cube root, and cube roots are grade 8 content, although 64 is both a perfect square and a perfect cube.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Because 64 is both a perfect square and a perfect cube, the computation is easy once students know to undo squaring and cubing. The risk is at the end: with sides of 8 and 4 inches in hand, writing the block first turns the ratio upside down. Students who never find the side lengths are left with ideas such as two dimensions against three.

Something you must already know moderate Working it out
Neither formula is given: the puzzle’s side is the number whose square is 64 and the block’s edge the number whose cube is 64, and cube roots are grade 8 work. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
A square and a cube share the number 64 in different units, so two different inverse operations feed one ratio, which must list the puzzle’s side first. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the side of the square from its area and the edge of the cube from its volume by testing small whole numbers, then write the puzzle's side first and simplify.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 1 : 3; it makes the puzzle’s side the shorter one, the reverse of 8 and 4 inches, so it mostly draws guesses.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Finds both lengths correctly, 8 and 4 inches, but writes the block’s side first, turning the ratio upside down to 1 : 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the square to 2 and the cube to 3, the powers behind square inches and cubic inches, and writes 2 : 3 without finding any length.

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, in square units. s is the length of one side. All sides of a square are equal, so the area is the side times itself. To find the side from the area, ask which number times itself gives the area.
Volume of a cube: $V = s \times s \times s$; V is the volume, the space inside, in cubic units. s is the length of one edge. A cube’s length, width and height are all equal, so the volume is the edge used three times. To go backward, ask which number used three times gives the volume.

Sort out the shapes
The puzzle is a square with an area of 64 square inches. The block is a cube with a volume of 64 cubic inches. A cube is a box with every edge the same length. You need the ratio of the puzzle’s side to the block’s side, with the puzzle first.
Find the puzzle’s side
Use the area formula backward. Which number times itself makes 64?
$8 \times 8 = 64$
The puzzle’s side is 8 inches. This number is the square root of 64.
Find the block’s side
Use the volume formula backward. Which number used three times makes 64? Try small numbers. 3 × 3 × 3 = 27 is too small.
$4 \times 4 \times 4 = 64$
The block’s side is 4 inches. This number is the cube root of 64.
Write and simplify the ratio
Put the puzzle first, as the question asks. Then divide both parts by 4.
$8 : 4 = (8 \div 4) : (4 \div 4) = 2 : 1$
Watch out: Writing the block’s side first gives 1 : 2, choice B. The order in a ratio matters.
So the answer is D, 2 : 1.

Check your answer

Check both sides. 8 × 8 = 64 square inches, and 4 × 4 × 4 = 64 cubic inches. The puzzle’s side is longer, so its number in the ratio must be bigger. Only choice D puts the bigger number first.

More questions like this

The original DOE question

Question 100 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 100. 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 63 of 86

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 64 of 86

2023 Test B Math, question 51

Question

A number, n, is divided by 25. If the quotient is 10 with a remainder of 5, what is the quotient expressed as a decimal?

A10.20
B10.25
C10.50
D10.75

Show the answer

Correct answer: A

A quotient of 10 with a remainder of 5 means the quotient is

$10+\frac{5}{25}=10\frac{1}{5}=10.20$

Question analysis

A number is divided by 25, giving a quotient of 10 with a remainder of 5, and students write the quotient as a decimal. It tests what a remainder means and how it becomes part of a decimal quotient. Connecting the remainder to the decimal part of the answer is the key idea.

NYS standards: 6.NS.2 multi-digit division
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
Solution: 2 steps
Reading level: Question and choices: about grade 5, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question checks one idea: a remainder is a fraction of the divisor, not a string of digits to place after the point. Students who write the remainder 5 as tenths, or set the divisor 25 after the point, land on listed choices. Rebuilding n = 25 × 10 + 5 = 255 and dividing by 25 confirms the decimal.

Something you must already know moderate Working it out
A remainder of 5 means 5 out of the 25 needed for another whole, so it becomes 5/25 = 0.2, a link between remainders and decimals the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Rebuild n from the quotient and the remainder, then divide it by 25 to a decimal; or write the remainder as a fraction of the divisor.

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

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Writes the divisor, 25, after the decimal point, setting the stem’s numbers 10 and 25 side by side instead of turning the remainder into a fraction.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Writes the remainder 5 as the tenths digit, 10.5, believing a remainder becomes the digits after the decimal point.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 10.75; it treats the remainder as three quarters of the divisor, which nothing in the stem suggests, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Quotient with a remainder: $\text{quotient} = q + \frac{r}{d}$; d is the divisor, the number you divide by. q is the whole-number part of the quotient, the answer to the division. r is the remainder, what is left over. The leftover still has to be divided by d, so it becomes the fraction r over d.

Sort out the words
n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.
Turn the remainder into a fraction
The 5 left over still has to be divided by 25. So it becomes the fraction $\frac{5}{25}$ . Use the formula.
$\text{quotient} = 10 + \frac{5}{25}$
Change the fraction to a decimal
Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.
$\frac{5}{25} = \frac{1}{5} = \frac{20}{100} = 0.20$
So the quotient is 10 + 0.20 = 10.20.
Give the answer
Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.
So the answer is A, 10.20.

Check your answer

Rebuild n. Ten 25s plus the 5 left over is $25 \times 10 + 5 = 255$ . Now divide: $255 \div 25 = 10.2$ . That matches 10.20.

More questions like this

The original DOE question

Question 108 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 108. Open page 205 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 65 of 86

2023 Test B Math, question 52

Question

$An equilateral triangle drawn with dashed sides. On each side a semicircle bulges outward, with that side as its diameter. The three curved arcs together form the outline of the garden.$

The garden design shown above is created in the shape of three semicircles whose diameters lie on the sides of an equilateral triangle. If each diameter is 2 meters long, what is the perimeter of the garden, in meters?

A6
B12
C $3\pi$
D $6\pi$

Show the answer

Correct answer: C

The perimeter (or circumference) of one of the three semicircles is $\frac{1}{2}(\text{diameter})\pi=\frac{1}{2}(2\pi)=\pi$ . Since there are 3 semicircles that make up the perimeter, the perimeter is $3\pi$ .

Question analysis

A garden is outlined by three semicircles whose diameters, each 2 meters long, are the sides of an equilateral triangle. Students find the perimeter of the garden in meters. It tests the circumference of a circle applied to half circles, and the figure mixes straight dashed sides with curved ones, so deciding what counts as the outline matters.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying the circumference formula is a grade 7 standard. Here it is used for a semicircle and repeated three times, a short extension of routine classwork, with a dashed triangle inside the figure to look past.
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: Reading a figure, table or graph minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The perimeter is just three half-circles, and each is half of π × 2. The figure’s dashed triangle invites measuring straight sides instead, and remembering the whole circle’s circumference without halving it gives a choice twice too large. Tracing the garden’s outer edge with a finger before computing shows that only the curves count.

Reading a figure, table or graph minor Understanding the question
The dashed triangle lies inside the garden, so its three 2-meter sides are not part of the perimeter; only the three curved arcs bound the garden. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The circumference formula is not given, and each arc is half of a circle with a 2-meter diameter, so students must take half of π × 2. (Anderson & Krathwohl, 2001)

Strategy: Find the circumference of a full circle with a 2-meter diameter, take half of it for one semicircle, and multiply by the number of semicircles in the outline.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Adds the three dashed sides of the triangle, 3 × 2 = 6, measuring the triangle inside the garden instead of its curved outline.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12, which is twice the triangle’s perimeter and matches no arc length, so it mostly draws guesses.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses a whole circumference, π × 2, for each of the three arcs, forgetting that each one is only half a circle, which doubles the perimeter.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = \pi \times d$; C is the circumference, the distance around a circle. d is the diameter, the straight line across the circle through its center. π is a number a little more than 3.14. The distance around any circle is π times as long as the distance across it.
Arc of a semicircle: $\text{arc} = \frac{1}{2} \times \pi \times d$; A semicircle is half of a circle. Its arc is the curved part. d is the diameter, the straight side. The arc is half of the full circle’s circumference, so take half of π times d.

Sort out the figure
The garden’s edge is made of three semicircles, or half circles. Each one sits on a side of the dashed triangle. Each side is a diameter, 2 meters long. The perimeter is the distance around the outside. Here that is only the three arcs, the curved parts. The dashed triangle is inside, so it does not count.
Find a full circle’s circumference
Picture the whole circle that one semicircle comes from. Its diameter is 2 meters. Use the circumference formula.
$C = \pi \times 2 = 2\pi$
The full circle would measure $2\pi$ meters around.
Find the length of one arc
Each semicircle’s arc is half of that circle. Use the arc formula.
$\text{arc} = \frac{1}{2} \times \pi \times 2 = \pi$
One arc is $\pi$ meters long.
Watch out: Using the full circle’s $2\pi$ for each arc doubles the answer. That gives $6\pi$ , choice D.
Add the three arcs
The garden’s edge is three equal arcs.
$\pi + \pi + \pi = 3\pi$
The perimeter is $3\pi$ meters.
Watch out: The dashed sides add to 2 + 2 + 2 = 6 meters, choice A. That is the triangle, not the curved edge.
So the answer is C, $3\pi$ meters.

Check your answer

Estimate. $3\pi$ is about 3 × 3.14 = 9.42 meters. A curved path is longer than the straight line across it. So the edge should be longer than the triangle’s 6 meters, and 9.42 is.

More questions like this

The original DOE question

Question 109 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 2024 Admissions, page 206, question 109. Open page 206 of the handbook (PDF)

Question 66 of 86

2023 Test B Math, question 56

Question

$A coordinate grid with the origin O, the x-axis marked 1, 2, and 3, and the y-axis marked 1 through 5. Line t passes through the point (0, 1) and line s passes through the point (1, 0); they intersect at P (3, 5), and dashed lines run from P across to 5 on the y-axis and down to 3 on the x-axis. The shaded region is the four-sided figure with vertices O, (0, 1), P (3, 5), and (1, 0).$

Lines s and t intersect at point P (3, 5). What is the area, in square units, of the shaded region in the graph above?

Show the answer

Correct answer: C

First, find the area of the rectangle in the graph:

$3\times 5=15$

Then, find the areas of the two right triangles that are not shaded:

Upper triangle:

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

Lower triangle:

$\frac{1}{2}\times 2\times 5=5$

The area of the shaded region, in square units, is calculated by subtracting the total area of the two right triangles from the area of the rectangle:

$15-(6+5)=15-11=4$

Question analysis

Two lines meet at the point (3, 5) on a coordinate grid, and the shaded region between them and the axes is a four-sided shape. Students find its area in square units. The shape is not a standard figure, so no single area formula applies to it directly.

NYS standards: 6.G.1 area of polygons
6.G.3 polygons in the coordinate plane
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Areas of triangles and rectangles are grade 6 content, and grade 7 students often find composite areas. This region is an irregular four-sided shape with two slanted sides, so students need a plan, and the lengths come from coordinates rather than labels.
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
Reading a figure, table or graph moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards a plan more than computation: the region has no name, so students must see it inside the dashed rectangle from O to P, with two right triangles cut away. Reading each triangle’s legs from the axes is the second hurdle, since the 1-unit intercepts shorten one leg of each, to 4 instead of 5 and to 2 instead of 3. Only one wrong choice, from using one coordinate as both heights, has a clear working behind it, so most errors here are guesses after a failed plan.

No obvious method moderate Working it out
The shaded shape has four sides, two of them slanted, and no formula of its own; it must be boxed in the 3-by-5 rectangle with two triangles removed, or cut along OP. (Schoenfeld, 1985)
Reading a figure, table or graph moderate Understanding the question
No lengths are labeled; every one comes from the coordinates of P (3, 5), the y-intercept 1 of line s and the x-intercept 1 of line t. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
The rectangle, two triangle areas and a subtraction make four computations, 15, 6 and 5, then 15 − 11, each built from different coordinates. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Draw the rectangle from O to P, find the areas of the two unshaded right triangles inside it from the coordinates, and subtract both from the rectangle's area.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2; no rectangle or triangle built from the coordinates leaves that area, so it mostly draws students who judge the narrow region by eye.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Cuts the region along OP into two triangles with bases of 1 but uses 3, the x-coordinate of P, as the height of both: 1.5 + 1.5 = 3.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 7; half of the 3-by-5 rectangle, 7.5, comes closest, so it mostly draws students who guess the region fills about half.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle: $A = l \times w$; A is the area, in square units. l is the length and w is the width. Multiplying them counts the unit squares that fill the rectangle.
Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, one side. h is the height: the distance from the base straight up to the far corner. A right triangle has a square corner, and the two sides that meet there are the base and height. A triangle is half of a rectangle with the same base and height.

Sort out the shaded region
The shaded region has four corners: O at (0, 0), (0, 1) on the y-axis, P at (3, 5), and (1, 0) on the x-axis. Line t meets the y-axis at (0, 1), and line s meets the x-axis at (1, 0). There is no easy formula for this shape. But the dashed lines make a rectangle around it. So find the rectangle’s area and take away the white parts.
Find the rectangle’s area
The rectangle has corners O, (3, 0), P and (0, 5). It is 3 units wide and 5 units tall.
$A = 3 \times 5 = 15$
The rectangle’s area is 15 square units.
Find the upper triangle’s area
The white triangle at the top left has corners (0, 1), (0, 5) and P. Its square corner, or right angle, is at (0, 5). One side runs up the y-axis from 1 to 5, so it is 4 units. The other runs across from 0 to 3, so it is 3 units.
$A = \frac{1}{2} \times 4 \times 3 = 6$
Its area is 6 square units.
Find the lower triangle’s area
The white triangle at the bottom right has corners (1, 0), (3, 0) and P. Its square corner is at (3, 0). One side runs along the x-axis from 1 to 3, so it is 2 units. The other goes up from 0 to 5, so it is 5 units.
$A = \frac{1}{2} \times 2 \times 5 = 5$
Its area is 5 square units.
Watch out: The lines cross the axes at 1, not at O. So these sides are 4 and 2, not 5 and 3.
Subtract the white triangles
Take both triangles away from the rectangle.
$15 - (6 + 5) = 15 - 11 = 4$
The shaded area is 4 square units.
So the answer is C, 4.

Check your answer

Cut the shaded region along a line from O to P. One piece has a base of 1 on the x-axis and a height of 5, since P is 5 units up. Its area is 2.5. The other has a base of 1 on the y-axis and a height of 3, since P is 3 units across. Its area is 1.5. Then 2.5 + 1.5 = 4. Using 3 for both heights gives 3, choice B.

More questions like this

The original DOE question

Question 113 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 113. Open page 208 of the handbook (PDF)

Question 67 of 86

2018 Test A Math, question 18

Question

$Coordinate grid with the origin O, the x-axis and y-axis each marked 1 and 2, and grid lines at 1 and 2. A straight line starts at O and rises through the grid points (1, 1) and (2, 2). The triangle under that line with corners O, (1, 0) and (1, 1) is shaded.$

What is the area of the shaded region in the graph above?

A0.25 square unit
B0.5 square unit
C1 square unit
D1.5 square units

Show the answer

Correct answer: B

The shaded region is a right triangle. Each leg is 1 unit in length.

So the area is

$A=\frac{1}{2}bh=\frac{1}{2}(1)(1)=\frac{1}{2}$ or 0.5 sq unit

Question analysis

A coordinate grid question: find the area of a shaded right triangle with corners at the origin, (1, 0) and (1, 1), under a diagonal line. Students read the base and height from the grid and apply the triangle area formula. The figure does most of the work; the task is recalling the formula correctly.

NYS standards: 6.G.1 area of polygons
6.G.3 polygons in the coordinate plane
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a triangle's area from a grid is grade 6 work, and this takes one step with whole-unit legs. It is easier than typical grade 7 area problems, which use composite figures or scale drawings.
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
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 2, 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

This is a quick formula question with legs of 1. The main risk is the one-half in the triangle formula: leaving it out gives the area of the whole grid square, which is a listed choice. Seeing that the diagonal cuts the unit square exactly in half gives a check that needs no formula at all.

Reading a figure, table or graph minor Understanding the question
The legs of the shaded triangle, from O to (1, 0) and up to (1, 1), are read off the grid lines; the line running on to (2, 2) is not part of the region. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The triangle area formula, one half times base times height, is needed and not given; the shaded triangle is half of one grid square. (Anderson & Krathwohl, 2001)

Strategy: Read the two legs straight off the grid, then apply the triangle area formula, base times height divided by two.

Traps

A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Judges the shaded triangle by eye as a quarter of the 1-by-1 grid square instead of the half that the diagonal from O to (1, 1) cuts off.
C Wrong formula (King et al., 2004)
Reveals a slip Expected pull: strong
Multiplies base by height, 1 × 1 = 1, and leaves out the one-half, giving the whole unit square instead of the triangle.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 1.5, which is more than the whole grid square around the shaded triangle; it mostly draws guesses.

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, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

Sort out the shaded region
The shaded region is a triangle. Its corners are O, the point (1, 0) and the point (1, 1). O is the origin, the point (0, 0) where the axes meet. The line keeps going up to (2, 2), but only the small triangle is shaded. You need its area in square units.
Read the base and height
The base lies along the x-axis, from O to 1. So the base is 1 unit. The height goes straight up from (1, 0) to (1, 1). So the height is 1 unit too. The base and height meet at a right angle, as the grid lines show.
Use the triangle formula
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} = 0.5$
The area is 0.5 square unit. Watch out: don’t forget the one half. $1 \times 1 = 1$ is the whole grid square, choice C. So the answer is B, 0.5 square unit.

Check your answer

The grid square from 0 to 1 has an area of 1 square unit. The slanted line cuts it corner to corner into two equal triangles. The shaded one is half the square, so 0.5 square unit.

More questions like this

The original DOE question

Question 75 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 75. Open page 84 of the handbook (PDF)

Question 68 of 86

2018 Test A Math, question 21

Question

In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without replacement, what is the probability that both cards are not blue?

A $\frac{2}{15}$
B $\frac{4}{25}$
C $\frac{3}{10}$
D $\frac{1}{3}$

Show the answer

Correct answer: A

The probability that both cards are not blue is the same as the probability that both cards are red. There are 4 red cards out of the 10, so the probability of the first card being red is

$\frac{4}{10}$

Now there are 9 cards left, and 3 of those are red, so the probability of the second card being red is

$\frac{3}{9}$

Multiply the two probabilities to find the probability that both cards are red (not blue):

$\left (\frac{4}{10} \right )\left ( \frac{3}{9} \right )=\frac{12}{90}=\frac{2}{15}$

Question analysis

This probability question draws 2 cards, one at a time and without replacement, from 10 cards: 4 red and 6 blue. Students find the chance that neither card drawn is blue. The second draw has one fewer card, and the word not changes which color to count.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability is usually found with a list or tree of equally likely outcomes. Dependent draws without replacement, plus a negative phrasing to translate, go beyond that typical work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Turning words into math minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two ideas stack here: restating not blue as red, and changing the second draw's numbers after a card is removed. Students who know to multiply but forget the removal reach a listed choice, and so do students who compute both blue. Writing the second fraction explicitly as 3 red out of 9 cards, before multiplying, is what separates right from nearly right.

Something you must already know moderate Working it out
Multiplying the first draw's chance by the second draw's, with the second adjusted for the card already taken, is a rule the test assumes and does not give. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
2/15 and 4/25 are about 0.13 and 0.16, too close to tell apart by estimate; only the second draw's fraction, 3 red of 9 cards rather than 4 of 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
"Both cards are not blue" has to be restated as both cards red, and the bold not also invites computing the chance that both are blue. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The product comes out as 12/90, and it has to be reduced to lowest terms before it matches any of the listed fractions. (Enright, Morley & Sheehan, 2002)

Strategy: Restate not blue as red, then multiply the chance of red on the first draw by the chance of red on the second, with one card fewer.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

B Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Multiplies 4/10 × 4/10 = 16/100 = 4/25, as if the first red card were put back before the second draw.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives 3/10: it is not the chance of red or of blue on one draw or on two, so it mostly catches guesses.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Computes the chance that both cards are blue, 6/10 × 5/9 = 30/90 = 1/3, missing the bold word not.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{favorable}}{\text{total}}$; P is the probability, the chance that something happens. Favorable is the number of ways to get what you want, here a red card. Total is the number of all possible picks, here all the cards you can pick from. It works when every pick is equally likely.
Two events in a row: $P(\text{A and B}) = P(\text{A}) \times P(\text{B after A})$; A is the first event and B is the second. P( ) means the probability of what is inside. The chance that both happen is the first chance times the second. For the second chance, count only what is left after the first event.

Sort out the question
There are 10 cards: 4 red and 6 blue. You pick 2 cards “one at a time without replacement”. That means the first card is not put back. “Both cards are not blue” means both are red, since red is the only other color. You need the probability, or chance, of that.
Find the chance the first card is red
4 of the 10 cards are red. Each card is equally likely to be picked. Use the probability formula.
$P(\text{red first}) = \frac{4}{10}$
Find the chance the second card is red
One red card is gone now, and it is not put back. So 9 cards are left, and 3 of them are red.
$P(\text{red second}) = \frac{3}{9}$
Watch out: using 4/10 again acts as if the card went back. That gives 4/25, choice B.
Multiply the two chances
Both cards must be red. So multiply the first chance by the second. To multiply fractions, multiply the tops, then multiply the bottoms.
$\frac{4}{10} \times \frac{3}{9} = \frac{4 \times 3}{10 \times 9} = \frac{12}{90}$
Simplify the fraction
None of the choices is 12/90, so simplify it. Divide the top and the bottom by 6, the biggest number that goes into both.
$\frac{12 \div 6}{90 \div 6} = \frac{2}{15}$
Watch out: the chance that both are blue is $\frac{6}{10} \times \frac{5}{9} = \frac{1}{3}$ , choice D. The question says not blue. So the answer is A, $\frac{2}{15}$ .

Check your answer

Count the ways. There are $10 \times 9 = 90$ ways to pick two cards in order. Red then red has $4 \times 3 = 12$ ways. That gives 12/90, or 2/15. The answer is small, which makes sense, because there are fewer red cards than blue ones.

More questions like this

The original DOE question

Question 78 as printed on page 85 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 78. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 69 of 86

2018 Test A Math, question 24

Question

$Circle graph titled "How People Get to Work in Center City": Drive Alone 49%, Walk 22%, Car Pool 15%, Bus 10%, Bicycle 4%. Below the graph: Total number of people working in Center City = 15,000.$

How many more people in Center City walk to work than ride their bicycle to work?

A2,500
B2,700
C2,800
D3,000

Show the answer

Correct answer: B

According to the chart, 22% of people walk to work and 4% ride a bicycle. Subtract to find the percentage of how many more people walk than bicycle:

22% − 4% = 18%

To find the exact number of people, multiply 18% (0.18) by the number of people working in Center City (15,000):

$(15000)(0.18)=2700$

Question analysis

This circle graph question shows how the 15,000 people working in Center City get to work, with 22% walking and 4% riding a bicycle. Students find how many more people walk than bike by combining a percent difference with a percent of a total. The choices are close together, so estimating is risky.

NYS standards: 6.RP.3c percent as a rate per 100
6.SP.5 summarizing data sets
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading a circle graph and taking a percent of a total are grade 6 skills. With only two short steps and friendly numbers, this is easier than typical grade 7 multistep percent work.
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: A very close wrong answer moderate
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 2 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 method is short: subtract the percents first, or find both groups and subtract, and both reach the same number. What makes the question less safe than it looks is the tight spread of choices, all within 500 of each other, so rounding 18% or slipping on one product lands on a wrong answer. Subtracting the percents first leaves one multiplication to get right instead of two.

A very close wrong answer moderate Choosing the answer
2,700 sits only 100 from 2,800 and 300 from 3,000, so estimating 18% of 15,000 as about a fifth cannot pick it; only the exact product does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
The Walk 22% and Bicycle 4% labels must be picked out of five sectors, and the 15,000 total sits in a caption below the circle graph. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
The percent gap has to become a count, 0.18 × 15,000, a decimal times a five-digit number worked by hand without a calculator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the two percents first, then take that single percent of 15,000, instead of finding each group and subtracting.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2,500, which is 1/6 of 15,000 and matches no sector or difference of sectors; it mostly catches estimates.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Takes 4% of 15,000 as 500 instead of 600 and subtracts it from the 3,300 walkers, getting 2,800.
D Rounds too early (King et al., 2004)
Reveals a slip Expected pull: strong
Rounds the 18-point gap up to 20% to make the multiplication easy and takes 20% of 15,000, which is 3,000.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $x = \frac{p}{100} \times W$; W is the whole amount. p is the percent, and x is the part you want. Percent means out of 100, so p percent is the fraction p over 100. Multiplying the whole by that fraction gives the part.

Sort out the circle graph
A circle graph shows how a whole splits into parts. This one shows how 15,000 workers get to work. Walk is 22%. Bicycle is 4%. You need how many more people walk than ride a bicycle.
Subtract the percents first
Both percents are parts of the same 15,000 people. So you can subtract the percents first. Then you only need one multiplication.
$22\% - 4\% = 18\%$
The difference is 18% of all the workers.
Find 18% of 15,000
$x = \frac{p}{100} \times W$
$x = \frac{18}{100} \times 15{,}000$
First find 1% by dividing by 100. Then multiply by 18.
$15{,}000 \div 100 = 150$
$150 \times 18 = 2{,}700$
That is 2,700 people.
Avoid rounding and answer
Watch out: don’t round 18% up to 20% to make it easy. 20% of 15,000 is 3,000, choice D. The choices are close together, so you need the exact number. So the answer is B, 2,700.

Check your answer

Find each group, then subtract. Walkers: 22% of 15,000 is 3,300. Bike riders: 4% of 15,000 is 600. $3{,}300 - 600 = 2{,}700$ . You get the same answer.

More questions like this

The original DOE question

Question 81 as printed on page 86 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 81. Open page 86 of the handbook (PDF)

Question 70 of 86

2018 Test A Math, question 25

Question

Which of the following numbers has factors that include the smallest factor (other than 1) of 91?

Show the answer

Correct answer: B

To find the smallest factor of 91, list the factors:

1, 7, 13, and 91

The smallest factor (other than 1) is 7.

Of the options listed (30, 35, 39, and 44), only 35 is a multiple of 7.

Question analysis

A number properties question about factors: students find the smallest factor of 91 other than 1, then pick the number that shares it. Because 91 looks like a prime number, many students skip the factoring, and the choices include numbers built from factors that 91 might seem to have.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Factors and divisibility are grade 4 to 6 skills, and this takes two quick checks. It is easier than typical grade 7 work, though recognizing that 91 is not prime catches many students.
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: Complex sentences moderate
Something you must already know moderate
Solution: 2 steps
Reading level: Question and choices: about grade 8, 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 question tests a grade 6 skill inside a hard-to-parse sentence. Once the stem is unpacked, the work is finding the smallest prime factor of 91 and checking the choices for it. Students lose points in two ways: 91 looks prime or divisible by 3, so they guess at its factors, or they find 13 before 7 and accept the choice that shares it.

Complex sentences moderate Understanding the question
The stem nests three ideas, "has factors that include the smallest factor (other than 1) of 91", so it has to be unpacked before any math; this is reading difficulty, not number difficulty. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Something you must already know moderate Working it out
Students must find that 91 = 7 × 13, and 91 is a number many students wrongly take to be prime or a multiple of 3. (Anderson & Krathwohl, 2001)

Strategy: Test 91 against the small primes in order, starting with 2, until one divides it evenly, then check which choice is a multiple of that prime.

Traps

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

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Wrongly takes 91 as a multiple of 3, as 90 is, and picks 30, which 3 divides; in fact 91 ÷ 3 is not a whole number.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Factors 91 as 7 × 13 but matches 39 = 3 × 13, which shares the larger factor, overlooking the word smallest.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Neither of its prime factors, 2 and 11, divides 91, an odd number, so no common working leads to 44; it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

What a factor is: $b = a \times c$; a, b and c are whole numbers. If b = a × c, then a and c are factors of b. That means a divides b evenly, with no remainder. Every whole number has 1 and itself as factors.

Unpack the long sentence
A factor of a number divides it evenly, with nothing left over. For example, 3 is a factor of 12, because $12 \div 3 = 4$ . The question has two steps. First, find “the smallest factor (other than 1) of 91”. Then find the choice that also has that factor.
Test small numbers on 91
Try dividing 91 by 2, 3, 5 and so on, until one works.
- 2: no. 91 is odd. That also rules out 4 and 6, which are even.
- 3: no. A number is a multiple of 3 only if its digits add to a multiple of 3. Here 9 + 1 = 10, which is not.
- 5: no. 91 does not end in 0 or 5.
- 7: yes!
$91 = 7 \times 13$
So 7 is the smallest factor of 91 other than 1.
Avoid the 3 trap
Watch out: 91 is close to 90, and 3 goes into 90. But $91 \div 3$ is 30 with 1 left over. So 3 is not a factor of 91. That trap leads to 30, choice A.
Check each choice for 7
Now look for a multiple of 7 among the choices.
- 30: no. $7 \times 4 = 28$ and $7 \times 5 = 35$ .
- 35: yes. $35 = 7 \times 5$ .
- 39: no. $39 = 3 \times 13$ .
- 44: no. $7 \times 6 = 42$ .
Watch out: 39 shares 13 with 91, but 13 is the larger factor, not the smallest. That is choice C. So the answer is B, 35.

Check your answer

Multiply to be sure: $7 \times 13 = 91$ and $7 \times 5 = 35$ . So 7 goes into both numbers evenly. No number from 2 to 6 goes into 91, so 7 really is the smallest.

More questions like this

The original DOE question

Question 82 as printed on page 86 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 82. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 71 of 86

2018 Test A Math, question 28

Question

$2\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}$

What is the value of the expression shown above?

A $14\frac{7}{20}$
B $14\frac{2}{5}$
C $15\frac{7}{20}$
D $15\frac{2}{5}$

Show the answer

Correct answer: D

$2\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}$

Convert all the fractions to a common denominator (10):

$2\frac{2}{10}+3\frac{3}{10}+4\frac{4}{10}+5\frac{5}{10}$

$=(2+3+4+5)+\left(\frac{2+3+4+5}{10}\right)$

$=14+1\frac{4}{10}=15\frac{2}{5}$

Question analysis

A fractions question: add four mixed numbers whose fractions have denominators of 5, 10, 5 and 2. Students rewrite the fractions over a common denominator, add the whole numbers and the fractions, regroup any extra whole, and simplify. Regrouping and lowest terms both matter, since the choices differ in exactly those places.

NYS standards: 7.NS.1d properties for adding and subtracting
7.EE.3 multi-step rational number problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding mixed numbers with unlike denominators is grade 5 content. With four terms and a regrouping step it takes care, but it asks less than typical grade 7 work with negative rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The numbers are friendly: every fraction converts to tenths, and the whole parts add to 14. The risk is in the bookkeeping, either forgetting to carry the extra whole from the fractions or adding numerators over a made-up denominator. Converting to decimals, 2.2 + 3.3 + 4.4 + 5.5, is a fast check that avoids both.

Error-prone arithmetic moderate Working it out
Four mixed numbers with denominators 5, 10, 5 and 2 must be rewritten as tenths, 2 + 3 + 4 + 5 tenths, and the 14/10 regrouped into a whole and a fraction. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
15 2/5 and 14 2/5 are written alike and share the fraction 2/5; only the whole carried from 14/10 separates them, so a missed carry is invisible at a glance. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
An answer in an unexpected form minor Choosing the answer
The fractions total 1 4/10, but every choice is in lowest terms, so the 4/10 has to be reduced to 2/5 before it matches. (Enright, Morley & Sheehan, 2002)

Strategy: Convert each mixed number to a decimal, since these fractions all convert exactly, add them, and turn the decimal part back into a fraction in lowest terms.

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators straight across, 1 + 3 + 2 + 1 = 7, and puts them over 20, a common denominator (10 × 2), without rescaling them, so nothing carries.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Adds the fractions to 14/10 but keeps only the 4/10 = 2/5 left over, forgetting to carry the whole 1 into 14.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Pairs a correct carry into 15 with the 7/20 from adding numerators, two steps no single line of working combines; it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a is the top number, called the numerator. b is the bottom number, called the denominator. n is any whole number except zero. Multiplying the top and bottom by the same n gives an equal fraction. Use it to give fractions the same denominator so you can add them.

Sort out the question
You need to add four mixed numbers. A mixed number is a whole number plus a fraction, like $2\frac{1}{5}$ . The fractions have different denominators, or bottom numbers: 5, 10, 5 and 2. To add fractions, they need the same denominator.
Add the whole numbers
Add the whole numbers first. You will add the fractions on their own after that.
$2 + 3 + 4 + 5 = 14$
The whole numbers make 14.
Rewrite each fraction in tenths
Use 10 as the denominator, since 5 and 2 both go into 10.
$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}$
$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$
$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$
$\frac{3}{10}$ is already in tenths. Watch out: when you change a bottom number, change the top too. Adding the old tops, 1 + 3 + 2 + 1 = 7, and putting them over 20 gives 7/20. That is in choices A and C.
Add the fractions
$\frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{14}{10}$
14 tenths is more than 1 whole. Ten tenths make 1, so 14 tenths is 1 whole and 4 tenths.
$\frac{14}{10} = 1\frac{4}{10} = 1\frac{2}{5}$
To simplify 4/10, divide the top and bottom by 2.
Put it all together
Add the 14 wholes to the sum of the fractions.
$14 + 1\frac{2}{5} = 15\frac{2}{5}$
Watch out: carry the extra whole into the 14. Missing it gives $14\frac{2}{5}$ , choice B. So the answer is D, $15\frac{2}{5}$ .

Check your answer

Change each mixed number to a decimal: $2.2 + 3.3 + 4.4 + 5.5 = 15.4$ . The 0.4 is 4 tenths, which is 2/5. So the sum is $15\frac{2}{5}$ .

More questions like this

The original DOE question

Question 85 as printed on page 86 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 85. Open page 86 of the handbook (PDF)

Question 72 of 86

2018 Test A Math, question 49

Question

A car travels at 4,400 feet per minute. The radius of each tire on the car is 1 foot. How many revolutions does one of these tires make in 1 minute? (Use the approximation $\frac{22}{7}$ for $\pi$ .)

A700
B1,925
C13,828
D15,400

Show the answer

Correct answer: A

One revolution is equal to the circumference of the tire:

$C=2\pi r=2(1)(\frac{22}{7})=\frac{44}{7}$ ft

The car travels at 4,400 ft per minute. To calculate the number of revolutions, divide the speed by the circumference:

$4,400\div \frac{44}{7}=4,400\times \frac{7}{44}=700$

700 revolutions.

Question analysis

This circle word problem says a car travels 4,400 feet per minute on tires with a radius of 1 foot and asks how many revolutions each tire makes in a minute, using 22/7 for pi. Students must see that one revolution covers the tire's circumference, then divide. Dividing by a fraction is the step to check.

NYS standards: 7.G.4 circle area and circumference
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Circumference is grade 7 content. This item adds the idea that each revolution rolls the tire one circumference forward and requires dividing by a fraction, which is typical of a solid grade 7 application 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: Turning words into math moderate
Something you must already know moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question tests whether students connect distance traveled to turns of a wheel, a link the stem leaves unstated. Every wrong choice is larger than the answer, and two exceed 4,400, which is impossible when each turn covers more than 6 feet. A sense check on size rules those out before any fraction work, leaving only the division by 44/7 to do carefully.

Turning words into math moderate Understanding the question
The stem never says that one revolution rolls the car forward one circumference; students must supply that link to turn 4,400 feet into a count of turns. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know moderate Working it out
The circumference formula, C = 2πr, is not given; with r = 1 and 22/7 for π it gives 44/7 feet per turn. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
Dividing 4,400 by 44/7 means multiplying by 7/44, which is clean only if students see that 4,400 ÷ 44 = 100 before multiplying by 7. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the distance covered in one turn with the circumference formula, 2πr, then divide the distance traveled in a minute by that circumference.

Traps

A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard path gives 1,925, which equals 4,400 × 7 ÷ 16, so it mostly draws students guessing a mid-sized count.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Multiplies 4,400 by π, 22/7, instead of dividing by the circumference, and truncates 13,828.57 to 13,828.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides 4,400 by 2/7 instead of 44/7, dropping the 22 of π while flipping 2 × 22/7, which gives 4,400 × 7 ÷ 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \times \pi \times r$; C is the circumference, the distance around a circle. r is the radius, from the center to the edge. π (pi) is about 22/7. The distance around is a bit more than 3 times the distance across, which is 2 × r.
Number of turns: $n = \frac{d}{C}$; n is the number of turns. d is the distance the car travels. C is the circumference. Each full turn rolls the car forward one circumference, so you divide the distance by it.

Sort out what you know
The car goes 4,400 feet in 1 minute. Each tire has a radius of 1 foot. A revolution is one full turn of the tire. You need how many turns a tire makes in 1 minute.
Find how far one turn goes
When a tire turns once, the car rolls forward the distance around the tire. That distance is the circumference. Use $\frac{22}{7}$ for π:
$C = 2 \times \frac{22}{7} \times 1 = \frac{44}{7}$
One turn moves the car $\frac{44}{7}$ feet, about 6.3 feet.
Divide the distance by one turn
$n = 4{,}400 \div \frac{44}{7}$
To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down):
$4{,}400 \times \frac{7}{44} = 100 \times 7 = 700$
Notice that 4,400 ÷ 44 = 100. That makes it easy.
Give the answer
The tire makes 700 turns. Watch out: multiplying 4,400 by π instead of dividing by the circumference gives about 13,828, choice C. So the answer is A, 700.

Check your answer

Estimate. One turn is about 6 feet. 4,400 ÷ 6 is about 730, close to 700. There should be fewer turns than feet, since each turn covers several feet. Choices C and D are bigger than 4,400, so they cannot be right.

More questions like this

The original DOE question

Question 106 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 106. 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 73 of 86

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 74 of 86

2018 Test B English, question 26

Passage Excerpt from “Niagara Falls”

Excerpt from “Niagara Falls”

by Rupert Brooke

The real secret of the beauty and terror of the Falls is not their height or width, but the feeling of colossal power and of unintelligible disaster caused by the plunge of that vast body of water. If that were taken away, there would be little visible change, but the heart would be gone.

The American Falls do not inspire this feeling in the same way as the Canadian. It is because they are less in volume, and because the water does not fall so much into one place. By comparison their beauty is almost delicate and fragile. They are extraordinarily level, one long curtain of lacework and woven foam. Seen from opposite, when the sun is on them, they are blindingly white, and the clouds of spray show dark against them. With both Falls the colour of the water is the ever-altering wonder. Greens and blues, purples and whites, melt into one another, fade, and come again, and change with the changing sun. Sometimes they are as richly diaphanous $^1$ as a precious stone, and glow from within with a deep, inexplicable light. Sometimes the white intricacies of dropping foam become opaque and creamy. And always there are the rainbows. If you come suddenly upon the Falls from above, a great double rainbow, very vivid, spanning the extent of spray from top to bottom, is the first thing you see. If you wander along the cliff opposite, a bow springs into being in the American Falls, accompanies you courteously on your walk, dwindles and dies as the mist ends, and awakens again as you reach the Canadian tumult. And the bold traveller who attempts the trip under the American Falls sees, when he dare open his eyes to anything, tiny baby rainbows, some four or five yards in span, leaping from rock to rock among the foam, and gambolling beside him, barely out of hand’s reach, as he goes. One I saw in that place was a complete circle, such as I have never seen before, and so near that I could put my foot on it. It is a terrifying journey, beneath and behind the Falls. The senses are battered and bewildered by the thunder of the water and the assault of wind and spray; or rather, the sound is not of falling water, but merely of falling; a noise of unspecified ruin. So, if you are close behind the endless clamour, the sight cannot recognise liquid in the masses that hurl past. You are dimly and pitifully aware that sheets of light and darkness are falling in great curves in front of you. Dull omnipresent foam washes the face. Farther away, in the roar and hissing, clouds of spray seem literally to slide down some invisible plane of air.

Beyond the foot of the Falls the river is like a slipping floor of marble, green with veins of dirty white, made by the scum that was foam. It slides very quietly and slowly down for a mile or two, sullenly exhausted. Then it turns to a dull sage green, and hurries more swiftly, smooth and ominous. As the walls of the ravine close in, trouble stirs, and the waters boil and eddy. These are the lower rapids, a sight more terrifying than the Falls, because less intelligible. Close in its bands of rock the river surges tumultuously forward, writhing and leaping as if inspired by a demon. It is pressed by the straits into a visibly convex form. Great planes of water slide past. Sometimes it is thrown up into a pinnacle of foam higher than a house, or leaps with incredible speed from the crest of one vast wave to another, along the shining curve between, like the spring of a wild beast. Its motion continually suggests muscular action. The power manifest in these rapids moves one with a different sense of awe and terror from that of the Falls. Here the inhuman life and strength are spontaneous, active, almost resolute. . . . A place of fear.

One is drawn back, strangely, to a contemplation of the Falls, at every hour, and especially by night, when the cloud of spray becomes an immense visible ghost, straining and wavering high above the river, white and pathetic and translucent. The Victorian lies very close below the surface in every man. There one can sit and let great cloudy thoughts of destiny and the passage of empires drift through the mind; for such dreams are at home by Niagara. I could not get out of my mind the thought of a friend, who said that the rainbows over the Falls were like the arts and beauty and goodness, with regard to the stream of life—caused by it, thrown upon its spray, but unable to stay or direct or affect it, and ceasing when it ceased. In all comparisons that rise in the heart, the river, with its multitudinous waves and its single current, likens itself to a life, whether of an individual or of a community. A man’s life is of many flashing moments, and yet one stream; a nation’s flows through all its citizens, and yet is more than they. In such places, one is aware, with an almost insupportable and yet comforting certitude, that both men and nations are hurried onwards to their ruin or ending as inevitably as this dark flood. Some go down to it unreluctant, and meet it, like the river, not without nobility. And as incessant, as inevitable, and as unavailing as the spray that hangs over the Falls, is the white cloud of human crying. . . . With some such thoughts does the platitudinous $^2$ heart win from the confusion and thunder of a Niagara peace that the quietest plains or most stable hills can never give.

From LETTERS FROM AMERICA by Rupert Brooke—Public Domain

$^1$ diaphanous: sheer, translucent

$^2$ platitudinous: clichéd, common

Question

Read this sentence from paragraph 2.

They are extraordinarily level, one long curtain of lacework and woven foam.

What is the effect of comparing the American Falls to a “long curtain of lacework and woven foam”?

AIt demonstrates the timelessness of the American Falls.
BIt conveys the secretive nature of the American Falls.
CIt illustrates the elegant uniformity of the American Falls.
DIt communicates the intense strength of the American Falls.

Show the answer

Correct answer: C

The question asks for the effect of the comparison being made in the sentence in paragraph 2. The word “lacework” in the sentence suggests an elegant piece of craftsmanship, and the “one long curtain” and “extraordinarily level” aspects of the Falls mean that the Falls are uniform or consistent; therefore, Option C is correct.

Option A is incorrect because neither “lacework and woven foam,” nor the quality of being extremely level, are related to timelessness.

Option B is incorrect because, while a curtain may suggest the idea of something being hidden or secret, the description of a delicate “lacework” curtain does not suggest this purpose.

Option D is incorrect because “lacework” is generally fragile and does not suggest great strength.

Question analysis

In this figurative language question, students explain the effect of a comparison in Rupert Brooke’s “Niagara Falls”: the American Falls as “one long curtain of lacework and woven foam.” They need to picture the image and read the sentences around it, since a curtain, lace and foam can each call up quite different qualities.

NYS standards: 7R4 word meaning and choice
7L5 figurative language and word relationships
Grade level: Grade 7 reading and task
Compared with grade 7: At grade 7 Explaining the effect of a metaphor in a quoted sentence is routine grade 7 work. The older prose is harder than most grade 7 texts, but the sentences around this one give direct context clues.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Figurative or literary language moderate
Hard words minor
Evidence: Answered from one sentence or line
Reading level: Passage: about grade 9, 987 words; question and choices: about grade 10, 70 words.
Students, last 30 days: 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 image is clear once pictured, and the sentences around it, about the American Falls being “delicate and fragile” and “extraordinarily level,” confirm what it means. Students go wrong by bringing in what they associate with curtains or with Niagara in general, hiding or power, instead of what this curtain is made of. The right choice also uses an abstract word that has to be connected back to “level.”

Figurative or literary language moderate Understanding the question
Students must translate “one long curtain of lacework and woven foam” into qualities: lace is delicate and finely made, and one long level curtain is even from end to end. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Hard words minor Understanding the question
The right choice uses “uniformity,” which students must link to “extraordinarily level” and “one long curtain,” just as “elegant” must be linked to the lacework. (Perfetti, 2007; Freedle & Kostin, 1993)

Strategy: Picture the image first, then reread the few sentences around it about the American Falls and keep only the choice that both the image and those sentences support.

Traps

Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
Dictionary meaning, not meaning here: A common meaning of the word that does not fit its use in this text.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing in lace, foam or a level edge suggests lasting through time, so this attracts mainly students guessing that a natural wonder must seem timeless.
B Dictionary meaning, not meaning here (Perfetti, 2007)
Reveals a surface strategy Expected pull: moderate
Uses the common sense of a curtain as something that hides, but this curtain is lacework, open and decorative, and nothing behind it is concealed.
D Wrong part of the text (Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: strong
Carries over the “colossal power” of paragraph 1, though paragraph 2 says the American Falls do not inspire that feeling as the Canadian do and calls their beauty fragile.

How the questions are analyzed, and the research behind it

Working it out, step by step

Figure out what the question asks
This asks about the effect of a comparison. The author compares the American Falls to “one long curtain of lacework and woven foam.” What quality does that picture give them?
Read the sentence and its neighbor
The full sentence is “They are extraordinarily level, one long curtain of lacework and woven foam.” The sentence just before it says, “By comparison their beauty is almost delicate and fragile.”
Picture the comparison
Lacework is fancy cloth with fine, delicate patterns. It looks elegant, which means graceful and beautiful. “Extraordinarily level” and “one long curtain” mean the Falls are even all the way across, with no breaks. That evenness is uniformity: being the same all the way through. So the picture shows elegant uniformity.
Check each choice
- A is wrong. Nothing about lace or foam suggests lasting forever.
- B is wrong. A curtain can hide things, but lace is open and see-through. Nothing is hidden.
- C is right. Lace is elegant, and a level curtain is uniform.
- D is wrong. Lace is fragile, not strong.
Watch out: Paragraph 1 talks about “colossal power,” so D can seem right. But paragraph 2 says the American Falls “do not inspire this feeling” the way the Canadian Falls do.
Choose the answer
The lace curtain makes the American Falls look graceful and even. So the answer is C.

Why this answer and not the closest wrong one

D is the closest wrong answer, because the Falls seem powerful in other parts of the essay. But this sentence is about the American Falls, whose beauty is “delicate and fragile.” Lace is delicate, not strong. A smooth, level curtain of lace shows grace and evenness, as C says.

More questions like this

The original DOE question

Question 26 as printed on page 135 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 135, question 26. Open page 135 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).

The passage “Excerpt from “Niagara Falls”” as printed (pages 134–135)

The passage “Excerpt from “Niagara Falls”” as printed on page 134 of the DOE handbook.

The passage “Excerpt from “Niagara Falls”” as printed on page 135 of the DOE handbook.

Question 75 of 86

2018 Test B English, question 44

Passage Snowy Mountains

Snowy Mountains

by John Gould Fletcher

Higher and still more high,
Palaces made for cloud,
Above the dingy city-roofs
Blue-white like angels with broad wings,
Pillars of the sky at rest
The mountains from the great plateau
Uprise.

But the world heeds them not;
They have been here now for too long a time.
The world makes war on them,
Tunnels their granite cliffs,
Splits down their shining sides,
Plasters their cliffs with soap-advertisements,
Destroys the lonely fragments of their peace.

Vaster and still more vast,
Peak after peak, pile after pile,
Wilderness still untamed,
To which the future is as was the past,
Barrier spread by Gods,
Sunning their shining foreheads,
Barrier broken down by those who do not need
The joy of time-resisting storm-worn stone,
The mountains swing along
The south horizon of the sky;
Welcoming with wide floors of blue-green ice
The mists that dance and drive before the sun.

“Snowy Mountains” by John Gould Fletcher—Public Domain

Question

Which detail from the poem reflects the speaker’s view that people often fail to appreciate what is familiar?

A“The mountains from the great plateau” (line 6)
B“They have been here now for too long a time.” (line 9)
C“Splits down their shining sides,” (line 12)
D“To which the future is as was the past,” (line 18)

Show the answer

Correct answer: B

The question asks for the detail from the poem that reflects the speaker’s view that people often fail to appreciate that which is familiar.
Line 9 (Option B) most directly supports this idea by implying that the mountains have been here for so long that people are accustomed to seeing them and no longer appreciate their beauty and value, making Option B the correct response.
Option A is incorrect because line 6 describes the mountains’ location, rather than their longevity or familiarity.
Option C is incorrect because, in line 12, the mention of harm that has been done to the mountains does not connect to a sense of overfamiliarity.
Option D is incorrect because, although line 18 alludes to the longevity of the mountains, it does not connect the longevity with a sense of overfamiliarity or the failure of people to appreciate the mountains’ worth.

Question analysis

Students pick the detail from the poem “Snowy Mountains” that reflects the speaker’s view that people often fail to appreciate what is familiar. The idea is never stated outright, so students must read each quoted line in its context and decide which one implies both familiarity and a loss of appreciation.

NYS standards: 7R1 textual evidence and inferences
7R6 point of view and author position
Grade level: Grade 7 reading and task
Compared with grade 7: At grade 7 Evidence questions like this are typical grade 7 work, but the support here is implied rather than stated, which adds an inference step beyond simple matching of words.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly evaluate
What makes it hard: Reading between the lines moderate
Every choice must be checked minor
Evidence: Answered from one sentence or line
Reading level: Passage: 161 words (a poem; reading level not measured); 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

The claim joins two ideas, long familiarity and failing to appreciate, and the right line shows them only together with the line before it. Students who match the idea of time pick the line about the future and the past, and those who match neglect pick a line of damage. Reading line 9 as the explanation for line 8 is the key move.

Reading between the lines moderate Working it out
The line never says “familiar”; its meaning comes from its place right after “But the world heeds them not,” where being here “too long” reads as the reason for that neglect. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)
Every choice must be checked minor Choosing the answer
Each quoted line has to be read in its stanza, because out of context lines 9 and 18 both sound like statements about time. (Embretson & Wetzel, 1987)

Strategy: Restate the claim as “people stop valuing something because they are used to it,” then check each quoted line within its stanza for a reason people might stop noticing the mountains.

Traps

Supports only part of the claim: In a question about evidence: a line that backs one half of the claim but not the other, or a related point rather than the claim itself.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Line 6 only places the mountains above “the great plateau” and touches neither familiarity nor appreciation, so it mainly draws students who are guessing.
C Supports only part of the claim (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Reveals partial knowledge Expected pull: moderate
Shows people harming the mountains, which fits the speaker’s complaint that they are not valued, but “Splits down their shining sides” gives no sign that familiarity is the cause.
D Supports only part of the claim (King et al., 2004; Thissen, Steinberg & Fitzpatrick, 1989)
Reveals partial knowledge Expected pull: strong
Notices that line 18 stresses how long the mountains endure, but it describes their permanence, not people growing used to them and ceasing to value them.

How the questions are analyzed, and the research behind it

Working it out, step by step

Figure out what the question asks
This asks for a line that shows one of the speaker’s views: people often fail to appreciate what is familiar. Familiar means something you know well and see all the time. To appreciate is to value something. So the line must show people no longer valuing the mountains because they are used to them.
Read each line in its place
Line 9 says, “They have been here now for too long a time.” Read it with line 8 just before it: “But the world heeds them not.” “Heeds” means pays attention to. So the world ignores the mountains. Line 9 gives the reason. They have been around so long that people are used to them.
Explain what line 9 means
The mountains have not done anything wrong by being old. The words “too long” show how people feel about them. Something you see every day starts to seem plain. You stop noticing it. That is exactly the idea in the question: people stop valuing what is familiar.
Check each choice
- A is wrong. Line 6 only tells where the mountains are, above “the great plateau.”
- B fits. It explains that people ignore the mountains because they have seen them for so long.
- C is wrong. Line 12 shows harm, but it does not say familiarity is the cause.
- D is wrong. Line 18 means the mountains do not change over time. It is about how long they last, not about people getting used to them.
Choose the answer
Line 9 explains why the world ignores the mountains: people are too used to them. So the answer is B, “They have been here now for too long a time.”

Why this answer and not the closest wrong one

D is the closest trap, since lines 9 and 18 both talk about time. But line 18 praises the mountains for lasting. Line 9 follows “the world heeds them not” and explains it. Only line 9 links long familiarity to people not caring.

More questions like this

The original DOE question

Question 44 as printed on page 144 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 144, question 44. Open page 144 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).

The passage “Snowy Mountains” as printed (page 142)

The passage “Snowy Mountains” as printed on page 142 of the DOE handbook.

Question 76 of 86

2018 Test B Math, question 7

Question

$Parallelogram PQRS with Q and R along the top and P and S along the bottom. Point T is on the bottom side PS, and segment QT cuts off triangle PQT. Angle PQT, at Q, is marked 50°; angle PTQ, at T, is marked 70°; angle QRS, at R, is marked with an arc and no measure.$

In the figure above, PQRS is a parallelogram. The measure of ∠PQT is 50°, and the measure of ∠PTQ is 70°. What is the measure of ∠QRS?

A60°
B70°
C80°
D120°

Show the answer

Correct answer: A

Find the missing angle, angle QPT, of triangle PQT:

180° − 70° − 50° = 60°

In parallelogram PQRS, angle QPT is congruent to angle QRS, so the measure of angle QRS is also 60°.

Question analysis

This geometry question uses parallelogram PQRS and a triangle PQT inside it, giving two of the triangle's angles and asking for angle QRS. Students need the triangle angle sum and the fact that opposite angles of a parallelogram are equal. The challenge is connecting the angles found in the triangle to the angles of the parallelogram.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle work uses vertical, adjacent and supplementary angles. This also needs the triangle angle sum and the opposite angles of a parallelogram, ideas New York places in grade 8, though the arithmetic is short.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Combines ideas moderate
A very close wrong answer moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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

Few steps are needed, but the figure misleads. It draws triangle PQT as isosceles, so the angle at P looks the same as the marked 70°, and the angle at R looks the same again; a student who trusts the drawing never computes. Those who do compute the angle at P must still choose the opposite angle of the parallelogram, not the angle next to it, which is where 120° comes from.

Reading a figure, table or graph major Understanding the question
The figure carries facts the stem leaves out: T lies on side PS, the 50° is only part of the angle at Q, and the 70° is at T, not at a corner of the parallelogram. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
Two separate facts link up: the angles of triangle PQT sum to 180°, and the angle at P, once known, equals the opposite angle at R of parallelogram PQRS. (Embretson, 1998; Daniel & Embretson, 2010)
A very close wrong answer moderate Choosing the answer
60° and 70° are close enough that judging the angle at R by eye cannot separate them, and the drawing even favors 70°; only computing the triangle's third angle does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Neither the triangle angle sum nor the rule that opposite angles of a parallelogram are equal is given; students must bring both to the figure. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of triangle PQT from the angle sum, work out which angle of the parallelogram it matches, then use the parallelogram's angle facts to reach angle QRS.

Traps

Trusting the picture: The figure suggests something it does not state; work from the given facts.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a surface strategy Expected pull: strong
Trusts the drawing, where triangle PQT looks isosceles, so takes the angle at P, and its opposite angle at R, to be 70° like the angle at T.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard step gives 80°; at most a student treats the triangle as having two 50° angles and computes 180 − 100, which nothing in the figure supports.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the 60° angle at P but answers with the angle at Q, 180 − 60 or 50 + 70 = 120, the consecutive angle instead of the opposite one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles in a triangle: $a + b + c = 180$; a, b and c are the three angles inside any triangle. They always add up to 180 degrees. If you know two of the angles, subtract them from 180 to find the third.
Opposite angles of a parallelogram: $\angle P = \angle R$; A parallelogram is a four-sided shape whose opposite sides are parallel, meaning they never meet. In parallelogram PQRS, the corners at P and R are opposite each other. Opposite angles of a parallelogram are always equal.

Sort out the figure
PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, meaning they never meet.
Point T is on the bottom side PS. Segment QT cuts off triangle PQT on the left. In that triangle, angle PQT is 50° and angle PTQ is 70°.
You need angle QRS, the corner at R.
Find the third angle of the triangle
Triangle PQT has angles at Q, T and P. You know two of them. The angle at P is angle QPT. All three must add to 180°.
$a + b + c = 180$
$\angle QPT + 50 + 70 = 180$
$\angle QPT = 180 - 50 - 70 = 60$
So angle QPT is 60°.
Match angle P to angle R
T lies on side PS. So angle QPT is also the corner of the parallelogram at P. Angle P and angle R are opposite corners. Opposite angles are equal, so angle QRS is also 60°.
Watch out: Angle Q is next to angle P, not opposite it. Angle Q is 120°, choice D, but the question asks about angle R. Also, do not trust the drawing. Angle R may look like 70°, choice B.
So the answer is A, 60°.

Check your answer

The four corners of the parallelogram are 60°, 120°, 60° and 120°. They add to 360°, as the angles of any four-sided shape must. Also, the 50° angle at Q fits inside the 120° corner at Q.

More questions like this

The original DOE question

Question 64 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 64. Open page 154 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 77 of 86

2018 Test B Math, question 15

Question

One day, the Early Bird Restaurant used 15 dozen eggs for 200 breakfast customers. At this rate, approximately how many dozen eggs are needed for 300 breakfast customers?

Show the answer

Correct answer: B

Let $x$ be the number of dozens of eggs for 300 customers. Set up a proportion:

$\frac{x}{300}=\frac{15}{200}$

$200x=4500$

$x=22.5\text{ dozen eggs}$

Round up to 23, because 22 dozen eggs would not be enough for 300 customers.

Question analysis

This proportion word problem says a restaurant used 15 dozen eggs for 200 breakfast customers and asks approximately how many dozen are needed for 300 customers. Students scale a rate up and round sensibly. Scaling by the right factor, rather than doubling or dividing the wrong pair of numbers, is the key step.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Scaling a rate from one quantity to another is grade 6 ratio reasoning. This needs one multiplication and a rounding decision, so it is lighter than the multistep proportion problems typical of grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 25 words.
Corrected from the DOE handbook: The explanation’s reason for rounding was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The proportion is simple, since 300 customers is one and a half times 200, and the only judgment is what to do with 22.5 dozen. The choices are spread wide enough that 23 is clearly nearest once 22.5 is found. Errors come mostly from students who skip the proportion and combine the numbers 300, 200 and 15 in some quicker way.

An answer in an unexpected form minor Choosing the answer
The scaled amount is 22.5 dozen, which is not a choice; the word “approximately” signals that the answer is the nearest whole number of dozens. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Scaling 15 dozen by 300/200 needs the factor 1.5, and 15 × 1.5 = 22.5 brings in a half that students must handle before matching a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Compare 300 customers with 200 to get the scale factor, multiply the 15 dozen eggs by it, and choose the closest answer.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Divides 300 customers by 15 dozen, combining two numbers from the question without a model of eggs per customer.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard route gives 25; at most a student who finds 22.5 rounds it to the nearest five instead of the nearest whole dozen.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds the scale factor 300 ÷ 200 = 1.5 up to 2 before multiplying, doubling 15 dozen to 30.

What was corrected from the DOE handbook, and why

Explanation: The handbook says to round 22.5 up to 23 “because you can't have half an egg.” But 22.5 dozen is 270 eggs, a whole number; the half is half a dozen, or 6 eggs. The site gives the real reason: 22 dozen eggs would not be enough for 300 customers, so the amount rounds up to 23 dozen.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling a rate: $\text{new amount} = \text{old amount} \times \frac{\text{new count}}{\text{old count}}$; The old amount is the eggs used before, 15 dozen. The old count is the customers then, 200. The new count is the new number of customers, 300. The fraction tells how many times bigger the new group is. The eggs must grow by that same number of times.

Write down what you know
15 dozen eggs fed 200 customers. A dozen is 12, but you can keep everything in dozens.
You need the dozens of eggs for 300 customers, at the same rate.
Compare the customer counts
How many times bigger is 300 than 200? Divide to find out. This number is called the scale factor.
$\frac{300}{200} = 1.5$
There are 1.5 times as many customers. So you need 1.5 times as many eggs.
Scale up the eggs
$\text{new amount} = \text{old amount} \times \frac{\text{new count}}{\text{old count}}$
$15 \times 1.5 = 22.5$
You need 22.5 dozen eggs.
Watch out: Do not round 1.5 up to 2 first. That doubles 15 and gives 30, choice D.
Round to a choice
22.5 is not a choice. The question says “approximately,” so pick the closest whole number. 22.5 rounds up to 23.
Rounding up also makes sense, because 22 dozen would be too few.
So the answer is B, 23.

Check your answer

Use another route. 15 dozen for 200 customers is 7.5 dozen for every 100 customers. For 300 customers, that is $3 \times 7.5 = 22.5$ dozen. That also rounds to 23.

More questions like this

The original DOE question

Question 72 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 72. Open page 155 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 78 of 86

2018 Test B Math, question 21

Question

Points $X$ , $Y$ , and $Z$ are on a straight line, and $Y$ is between $X$ and $Z$ . Length $\overline{YZ}=\frac{3}{5}\overline{XY}$ , and length $\overline{XY}=20$ centimeters. What is the length of $\overline{XZ}$ ?

A12 cm
B24 cm
C30 cm
D32 cm

Show the answer

Correct answer: D

Since $\overline{XY}=20$ cm, use that to find $\overline{YZ}$ :

$\overline{YZ}=\frac{3}{5}\overline{XY}=\frac{3}{5}(20)=12\text{ cm}$

$\overline{XZ}=\overline{XY}+\overline{YZ}=20+12=32\text{ cm}$

Question analysis

This segment question places points X, Y and Z on a line with Y between the others, gives XY as 20 centimeters, and says YZ is three fifths of XY. Students find the length of the whole segment XZ. Keeping track of which segment the question asks for matters more than the arithmetic.

NYS standards: 7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a fraction of a length and adding segments takes only two short steps with friendly numbers. Grade 7 rational-number problems are usually longer or involve negatives, so this sits below grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

There is no figure, so students who sketch the three points see at once that XZ is made of the two pieces XY and YZ. The arithmetic, three fifths of 20, is friendly. Most errors come from answering with the piece that was just calculated instead of the whole segment the question asks for.

Turning words into math minor Understanding the question
The condition “Y is between X and Z” has to become XZ = XY + YZ, and the overbar notation for segment lengths may be unfamiliar to some students. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find three fifths of 20 to get YZ, then add the two parts, since Y lies between X and Z.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds YZ, three fifths of 20, and gives it as the answer without adding XY to get the whole segment.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Takes one fifth of 20 but forgets to multiply by 3, so adds 4 cm instead of 12 cm to XY.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard route gives 30; a student would have to take YZ as half of XY rather than three fifths, adding 10 cm.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a number: $\frac{a}{b} \text{ of } n = n \div b \times a$; n is the starting amount. b is the bottom of the fraction: split n into b equal pieces. a is the top: take a of those pieces. Here n is 20, b is 5 and a is 3.
Adding segment lengths: $\overline{XZ} = \overline{XY} + \overline{YZ}$; X, Y and Z are points on one line, with Y between X and Z. XY and YZ are the lengths of the two short pieces. XZ is the whole length. The two pieces join end to end to make the whole.

Picture the line
Points X, Y and Z are on one straight line, with Y between X and Z. So the order is X, then Y, then Z.
A segment is the straight piece between two points. The bar over two letters means the length of that segment.
XY is 20 cm. You need XZ, the whole length.
Find the length of YZ
YZ is three fifths of XY. To find three fifths of 20, first find one fifth. Then take 3 of those.
$\frac{3}{5} \text{ of } 20 = 20 \div 5 \times 3 = 4 \times 3 = 12$
So YZ is 12 cm.
Watch out: 12 cm is only one piece of the line. Stopping here gives choice A.
Add the two pieces
Y is between X and Z, so the two pieces join to make the whole.
$\overline{XZ} = \overline{XY} + \overline{YZ} = 20 + 12 = 32$
XZ is 32 cm.
So the answer is D, 32 cm.

Check your answer

Check the fraction: $\frac{12}{20} = \frac{3}{5}$ , so YZ really is three fifths of XY. Also, XZ must be longer than XY, since it holds XY and more. 32 is longer than 20.

More questions like this

The original DOE question

Question 78 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 78. Open page 157 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 79 of 86

2018 Test B Math, question 22

Question

Bryana bought $1\frac{3}{4}$ yards of cloth at $8.00 per yard. If there was an 8% sales tax, what was the total cost of the cloth?

A$12.96
B$14.08
C$15.12
D$16.08

Show the answer

Correct answer: C

Calculate the cost of the cloth before tax:

$1\frac{3}{4}\times 8=\frac{7}{4}\times 8=\$14$

Now find the tax for $14 worth of cloth:

$14\times 8\%=14\times\frac{8}{100}=\$1.12$

Finally, add the cost of the fabric and the tax:

$14 + $1.12 = $15.12

Question analysis

This percent word problem has Bryana buy 1 3/4 yards of cloth at $8.00 a yard and pay 8% sales tax. Students multiply a mixed number by a price, find the tax and add it to get the total cost. Handling the mixed number and taking the tax as a percent of the price both need care.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Sales tax on a purchase is a standard grade 7 percent problem. This one adds a mixed-number multiplication first, but the numbers are friendly and the steps are routine 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: Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is standard for a sales-tax question, so the work lies in the number handling. Converting 1 3/4 yards correctly and taking 8% of the cost, rather than adding 8 cents, are the two places students slip. The choices punish each slip but not stopping early, since neither $14 nor $1.12 appears, which helps students who check their result against the list.

Error-prone arithmetic moderate Working it out
Two number forms meet: the mixed number 1 3/4 times $8 must become $14, and 8% of $14 must become $1.12, a decimal with two places. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The steps, price before tax, then tax, then total, must be done in order, and stopping after any of them gives a number that is not among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Turn 1 3/4 into 7/4 or 1.75 and multiply by $8, then multiply that cost by 1.08 to include the tax in one step.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Multiplies $8 by 1.5 instead of 1.75, pricing the cloth at $12, then adds the 8% tax correctly to reach $12.96.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Prices the cloth correctly at $14 but adds 8 cents, treating 8% as $0.08 instead of 8 hundredths of the price.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two unrelated errors, rounding 1 3/4 yards up to 2 for $16 and then adding 8 cents instead of 8%, so few students land here.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost: $\text{cost} = \text{price per unit} \times \text{number of units}$; Price per unit is the cost of one unit, here $8 for one yard. Number of units is how many you buy, here yards of cloth. Multiply them to get the total before tax.
Sales tax: $\text{tax} = \frac{\text{tax rate}}{100} \times \text{price}$; The tax rate is the percent, here 8. Percent means out of 100, so 8% is 8/100, or 0.08. The price is the cost before tax. The tax is added on top of the price.

Write down what you know
Bryana bought $1\frac{3}{4}$ yards of cloth. Each yard costs $8.00. There is an 8% sales tax. Sales tax is an extra amount added to the price.
You need the total cost, with the tax.
Find the cost before tax
First change the mixed number to a fraction. A mixed number is a whole number plus a fraction. $1\frac{3}{4}$ is the same as $\frac{7}{4}$ , because 1 whole is 4 quarters.
$\text{cost} = \text{price per unit} \times \text{number of units}$
$\text{cost} = 8 \times \frac{7}{4} = \frac{56}{4} = 14$
The cloth costs $14 before tax.
Watch out: $1\frac{3}{4}$ is 1.75, not 1.5. Using 1.5 gives $12.96, choice A.
Find the tax
$\text{tax} = \frac{\text{tax rate}}{100} \times \text{price}$
$\text{tax} = \frac{8}{100} \times 14 = 0.08 \times 14 = 1.12$
The tax is $1.12.
Watch out: 8% of $14 is not 8 cents. Adding only $0.08 gives $14.08, choice B.
Add the tax to the price
The total is the price before tax plus the tax.
$14 + 1.12 = 15.12$
The total is $15.12.
So the answer is C, $15.12.

Check your answer

Estimate. 10% of $14 is $1.40, so 8% is a little less, about $1.10. Then $14 plus about $1.10 is about $15.10. Only $15.12 is that close.

More questions like this

The original DOE question

Question 79 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 79. Open page 157 of the handbook (PDF)

Question 80 of 86

2018 Test B Math, question 25

Question

Jack scored a mean of 15 points per game in his first 3 basketball games. In his 4th game, he scored 27 points. What is his mean score for the first 4 games?

Show the answer

Correct answer: C

Jack scored a mean of 15 points per game in each of the first 3 games, so he earned a total of 45 points for the first 3 games. Use that information to calculate the mean over the four games:

$\frac{45+27}{4}=\frac{72}{4}=18$

Question analysis

This statistics question gives Jack's mean score for his first three basketball games and his score in the fourth, and asks for his mean over all four games. Students recover a total from a mean, add the new score and divide again. The earlier mean stands for three games, not one, which is easy to forget.

NYS standards: 6.SP.5c range, center, and overall pattern
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a mean is grade 6 content, but working back from a mean to a total and then updating it is a common grade 7 test item, with a tempting shortcut that gives the wrong value.
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
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 1, 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 tempting shortcut, averaging 15 and 27, is quick, feels reasonable and gives a choice, so the question rewards students who slow down. Its error is weighting: the 15 stands for three games and the 27 for one. Rebuilding the total from the mean and dividing by four games takes only a moment longer.

Turning words into math moderate Understanding the question
The phrase “a mean of 15 points per game in his first 3 games” has to be turned into a total, 45 points, before the fourth game's 27 can be added. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Students must know that a mean is a total divided by a count, so a mean cannot be averaged with a single score as if both had equal weight. (Anderson & Krathwohl, 2001)

Strategy: Multiply the first mean by 3 to get the points so far, add the fourth game's points, and divide the new total by 4.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives back the first mean, 15, as if one high-scoring game could not change the average.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard computation gives 17; it lies between the old mean and the right one, so it mostly draws students who estimate.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Averages the old mean and the new score, (15 + 27) ÷ 2, giving one game the same weight as the first three together.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean (average): $\text{mean} = \frac{\text{sum of the values}}{\text{number of values}}$; The mean is the average. Add all the values, here the points in each game. Then divide by how many values there are, here the number of games. Turned around, it also says: sum = mean × number of values.

Write down what you know
Jack’s mean for his first 3 games was 15 points. The mean is the average: the total shared out equally among the games.
In game 4 he scored 27 points. You need his mean for all 4 games.
Find the total for 3 games
A mean of 15 over 3 games means the total is 3 times 15. You need the total, because a new mean is built from the total.
$15 \times 3 = 45$
He scored 45 points in the first 3 games.
Add the fourth game
Now add the 27 points from game 4 to get the total for all 4 games.
$45 + 27 = 72$
He scored 72 points in 4 games.
Divide by 4 games
$\text{mean} = \frac{\text{sum of the values}}{\text{number of values}} = \frac{72}{4} = 18$
Watch out: Do not average 15 and 27. That gives 21, choice D. It treats one game as if it counted as much as three.
So the answer is C, 18.

Check your answer

The mean should go up a little, since 27 is above 15. It went from 15 to 18, so that makes sense. Also, $18 \times 4 = 72$ , which matches the total points.

More questions like this

The original DOE question

Question 82 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 82. Open page 157 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 81 of 86

2018 Test B Math, question 26

Question

A cylindrical oil drum can hold 4,320 liters when it is completely full. Currently, the drum is $\frac{1}{3}$ full of oil. How many kiloliters of oil need to be added in order to fill the drum completely?

A1.44
B2.88
C4.32
D14.10

Show the answer

Correct answer: B

Find the number of liters that need to be added. Since $\frac{1}{3}$ of the oil drum is full, $\frac{2}{3}$ of the drum remains empty:

$\frac{2}{3}\times 4{,}320=2{,}880\text{ liters}$

Use the conversion 1 kL = 1,000 L to find the number of kL:

$\frac{2{,}880}{1{,}000}=2.88\text{ kL}$

Question analysis

This fractions and unit conversion problem describes a cylindrical oil drum that holds 4,320 liters when full and is now one third full, and asks how many kiloliters must be added to fill it. Students find the empty fraction, multiply and convert liters to kiloliters. Answering for the right part of the drum, in the right unit, is the challenge.

NYS standards: 7.NS.3 four operations with rational numbers
6.RP.3d converting measurement units
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking a fraction of a quantity and converting metric units are both familiar from grades 5 and 6. Together they make a three-step problem at about the level of routine grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is short, but the question describes the drum one way and asks about it another. Students who multiply by the fraction they are given find the oil already in the drum, and that result, converted to kiloliters, is a choice. Since no choice is in liters, students who forget to convert are warned before they answer.

Turning words into math moderate Understanding the question
The stem gives how full the drum is, 1/3, but asks how much must be added, so students must switch to the empty 2/3 before multiplying by 4,320. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The bold word “kiloliters” calls for the conversion 1 kiloliter = 1,000 liters, which the question does not give. (Anderson & Krathwohl, 2001)

Strategy: Work out what fraction of the drum is still empty, take that fraction of 4,320 liters, and divide by 1,000 to change liters to kiloliters.

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.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Multiplies 4,320 by the given 1/3, finding the oil already in the drum instead of the amount still needed.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Converts the full capacity of 4,320 liters to kiloliters without taking away the third of the drum that is already full.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No conversion of 1,440, 2,880 or 4,320 liters gives 14.10; it is over three times the full drum's capacity, so only guessers choose it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of an amount: $\frac{a}{b} \text{ of } n = \frac{a}{b} \times n$; n is the whole amount, here the full drum. The fraction a over b tells what share of it you want. Multiply to find that share. For example, 2/3 of 9 is 6.
Liters to kiloliters: $\text{kiloliters} = \text{liters} \div 1{,}000$; Kilo means 1,000, so 1 kiloliter holds 1,000 liters. To change liters into kiloliters, divide by 1,000. That moves the decimal point 3 places to the left.

Sort out the question
The full drum holds 4,320 liters. It is now $\frac{1}{3}$ full. You need how much oil to add to fill it.
The answer must be in kiloliters, not liters.
Find the empty fraction
The drum is $\frac{1}{3}$ full. The rest is empty, and that is the part you must fill.
$1 - \frac{1}{3} = \frac{2}{3}$
Watch out: $\frac{1}{3}$ of 4,320 is 1,440 liters. That is the oil already in the drum. It leads to choice A.
Find the liters to add
Take two thirds of the full drum.
$\frac{2}{3} \text{ of } 4{,}320 = \frac{2}{3} \times 4{,}320 = 2{,}880$
One third of 4,320 is 1,440, and two thirds is double that. You need to add 2,880 liters.
Change liters to kiloliters
$\text{kiloliters} = 2{,}880 \div 1{,}000 = 2.88$
You need 2.88 kiloliters.
Watch out: 4.32 is the whole drum in kiloliters. Choice C forgets that part of the drum is already full.
So the answer is B, 2.88.

Check your answer

Add the oil already in the drum to the oil you add: $1.44 + 2.88 = 4.32$ kiloliters. That is 4,320 liters, the full drum. It checks out.

More questions like this

The original DOE question

Question 83 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 83. Open page 158 of the handbook (PDF)

Question 82 of 86

2018 Test B Math, question 30

Question

Three students stand at the starting line of a running track and begin running laps at the same time. Ann completes 1 lap every 2 minutes, Jack completes 1 lap every 3 minutes, and Lee completes 1 lap every 4 minutes. How many laps does Ann complete before all three runners are once again at the starting line at the same time?

Show the answer

Correct answer: B

Find the least common multiple of 2, 3, and 4 — which is 12. So, it takes 12 minutes before all three are back at the starting line. Ann completes 1 lap every 2 minutes, so in 12 minutes she has completed 6 laps.

Question analysis

Three runners start together and finish laps every 2, 3 and 4 minutes, and the question asks how many laps Ann, the fastest, completes before all three are at the starting line together again. It tests least common multiples in a word problem. The catch is keeping track of which quantity the question wants once the common time is found.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Sixth and seventh graders find least common multiples of two numbers in isolation; this applies the idea to three runners in a story and adds a conversion from minutes to laps.
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
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 56 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two quantities are in play, minutes and laps, and the question asks for the second while the natural first result is the first. Students who find the 12-minute meeting time often stop there, and students who do count laps can count Jack's or Lee's by mistake. The question rewards tracking whose laps are asked for and converting the time into laps at the very end.

Turning words into math moderate Understanding the question
"All three runners are once again at the starting line at the same time" has to be recognized as the first common multiple of 2, 3 and 4 minutes; the wording never says multiple. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding a least common multiple is assumed; multiplying the three lap times instead gives 24 minutes, which is a common time but not the first one. (Anderson & Krathwohl, 2001)

Strategy: Find the first time that is a multiple of all three lap times, then divide that time by Ann's lap time to count her laps.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Divides the 12-minute meeting time by Jack's 3-minute lap, giving his 4 laps instead of Ann's.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the least common multiple, 12 minutes, and gives it as the answer without dividing by Ann's 2-minute lap to count her laps.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No method using the lap times 2, 3 and 4 gives 20; it would mean 40 minutes for Ann, not the first time all three meet.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of laps: $\text{laps} = \text{total time} \div \text{time for one lap}$; Total time is how long the runner runs. Time for one lap is how long one trip around the track takes. Dividing tells you how many of those trips fit into the total time.

Sort out what you know
Ann runs 1 lap every 2 minutes. Jack runs 1 lap every 3 minutes. Lee runs 1 lap every 4 minutes. They all start together.
A runner is back at the start line at the end of every lap. First find when all three are “once again at the starting line at the same time”. Then count Ann’s laps up to that moment.
List when each runner is back
Write the minutes when each runner reaches the line.
- Ann: 2, 4, 6, 8, 10, 12 …
- Jack: 3, 6, 9, 12 …
- Lee: 4, 8, 12 …
The first number in all three lists is 12. This is the least common multiple of 2, 3 and 4. It is the smallest number that all three divide into with nothing left over. So they meet after 12 minutes.
Count Ann’s laps
Use the laps formula. Ann runs for 12 minutes, and each lap takes her 2 minutes.
$\text{laps} = 12 \div 2 = 6$
Ann runs 6 laps.
Watch out: 12 is the number of minutes, not laps. Choosing it gives C. Dividing by Jack’s 3 minutes gives his 4 laps, which is A.
Give the answer
The question asks only about Ann, and she runs 6 laps in 12 minutes. So the answer is B, 6.

Check your answer

Check that everyone is back at 12 minutes. Ann: $6 \times 2 = 12$ . Jack: $4 \times 3 = 12$ . Lee: $3 \times 4 = 12$ . No earlier time works. At 6 minutes, Lee is only halfway through his second lap.

More questions like this

The original DOE question

Question 87 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 87. Open page 158 of the handbook (PDF)

Question 83 of 86

2018 Test B Math, question 31

Question

Simplify this expression:

$4\left ( 7-3x \right )-\left ( 5-x \right )$

A $23-4x$
B $23-11x$
C $28-4x$
D $28-12x$

Show the answer

Correct answer: B

4( 7 − 3x ) −( 5 − x) = 28 − 12x − 5 + x = 23 − 11x

Question analysis

This question asks students to simplify an algebraic expression by distributing a number into one set of parentheses, subtracting another set, and combining like terms. It checks the distributive property with a minus sign in front of parentheses. The usual slip is subtracting only the first term inside the second parentheses, which changes the sign of an x term.

NYS standards: 7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Expanding and combining linear expressions is core grade 7 work. This is a standard example, two distributions and one subtraction, with nothing beyond what a grade 7 lesson on the distributive property would cover.
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
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 16+, 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 subtracting a whole group is where students go wrong: the minus sign has to reach the −x as well as the 5. Each choice pairs a constant with an x coefficient, so getting the constant right only narrows the field to two. The question rewards writing out both distributions in full rather than doing them mentally.

Error-prone arithmetic moderate Working it out
The minus sign in front of (5 − x) must change both terms, so −x becomes +x and then combines with −12x from the first group. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Two distributions, then collecting the constants 28 and −5 and the x terms separately: four small operations, each a place where a sign can slip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
23 − 4x shares the right constant with 23 − 11x, so getting 23 does not settle it; only the x coefficient, −12x + x rather than −3x − x, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite the subtraction as adding −1 times the second group, distribute both multipliers, then collect the constant terms and the x terms separately.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Applies each multiplier to the first term only, the 4 to 7 but not −3x and the minus to 5 but not −x, so 28 − 3x − 5 − x = 23 − 4x.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Pairs 4 × 7 = 28 with −3x − x = −4x and loses the −5 altogether; no single consistent mistake produces this expression.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Distributes the 4 correctly to get 28 − 12x and stops there, never subtracting the second group, 5 − x.

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. b and c are the terms inside. You multiply a by each term inside, not only the first one. This works because a groups of (b − c) is a groups of b take away a groups of c.
Subtracting a group: $-(b - c) = -b + c$; b and c are the terms inside the parentheses. A minus sign in front means take away the whole group. So you change the sign of each term inside. Taking away a minus c is the same as adding c.

Sort out the question
You need to simplify $4(7-3x)-(5-x)$ . Simplify means write it in the shortest way. A term is one piece of the expression, like 7 or $-3x$ . Like terms have the same letter part, so $-12x$ and $x$ are like terms. Plain numbers are like terms too.
Multiply the 4 into the first group
Use the distributive property. The 4 multiplies both terms inside.
$4(7-3x) = 4 \times 7 - 4 \times 3x = 28 - 12x$
Watch out: the 4 must multiply the $3x$ too. Writing only $28-3x$ is the first slip that leads to choice A.
Take away the second group
The minus sign in front of $(5-x)$ means subtract the whole group. So change the sign of each term inside.
$-(5-x) = -5 + x$
Taking away a minus $x$ gives plus $x$ .
Watch out: if you stop before this step, you get $28-12x$ . That is choice D.
Combine the like terms
Now you have $28-12x-5+x$ . Put the plain numbers together. Then put the $x$ terms together.
$28 - 5 = 23$
$-12x + x = -11x$
So the expression simplifies to $23-11x$ .
Give the answer
The plain number is 23, and the $x$ term is $-11x$ . So the answer is B, $23-11x$ .

Check your answer

Put in an easy number, like $x = 1$ . The original gives $4(7-3)-(5-1) = 16-4 = 12$ . Choice B gives $23-11 = 12$ . They match. Choice A would give $23-4 = 19$ , which does not match.

More questions like this

The original DOE question

Question 88 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 88. 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 84 of 86

2018 Test B Math, question 38

Question

SURVEY OF CATS PER FAMILY

Number of Cats	Number of Families
0	15
1	42
2	35
3 or more	8

The table above shows the number of cats per family in 100 households in the Blaine neighborhood. By what percentage is the number of families with 1 cat greater than the number of families with 2 cats?

A7%
B10%
C17%
D20%

Show the answer

Correct answer: D

To find by what percent the number of families with 1 cat is greater than the number of families with 2 cats, calculate the difference between the two numbers and divide by the number of families with 2 cats:

$\frac{42-35}{35}=\frac{7}{35}=\frac{1}{5}=0.20\text{ or }20\%$

Question analysis

A table lists how many cats live in each of 100 households in the Blaine neighborhood, and the question asks by what percent the number of one-cat families exceeds the number of two-cat families. It tests percent difference. The tricky part is choosing the right base for the comparison when the table's total is a convenient 100.

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Percent increase and decrease is grade 7 content, and this is one standard comparison. What makes it harder than a routine exercise is choosing the base and ignoring the tempting 100-family total.
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
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is a single subtraction and division, so the question is really about choosing the base of a percent comparison. The 100-household total makes the count difference look like a percent already, and dividing by the larger count is the other classic slip. Students who say "greater than the 2-cat count" to themselves pick 35 as the base.

Turning words into math moderate Understanding the question
"Greater than the number of families with 2 cats" fixes the base of the percent as 35, the count being compared against, not 42 or the 100 households. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
17% and the right 20% both come from dividing the 7-family gap by a count, 42 or 35; only choosing 35, the count being compared against, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
Only two rows of the four-row table matter, the 42 families with 1 cat and the 35 with 2; the 0-cat and 3-or-more rows go unused. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the difference between the two counts, decide which count the comparison is measured against, divide by it, and convert to a percent.

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.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Subtracts 42 − 35 = 7 and, since there are 100 households, reads that 7-point gap in shares as a 7% increase.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No division of the 7-family gap by a count in the table gives 10%, so this mostly attracts students estimating a round percent.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: moderate
Divides the 7-family difference by 42, the larger count, measuring the change from the wrong amount: 7/42 is about 16.7%, so 17%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent greater than: $\text{percent} = \frac{\text{difference}}{\text{amount you compare to}} \times 100$; The difference is how much bigger one amount is than the other. The amount you compare to is the one after the word than. Dividing shows what share of that amount the difference is. Times 100 turns that share into a percent.

Sort out the question
From the table, 42 families have 1 cat and 35 families have 2 cats. You need the percent by which 42 is “greater than the number of families with 2 cats”. The rows for 0 cats and 3 or more cats do not matter here.
Find the difference
Subtract the two counts.
$42 - 35 = 7$
There are 7 more families with 1 cat.
Watch out: 7 is a number of families, not a percent. There are 100 households, so 7% looks tempting. It is choice A, and it is wrong.
Divide by the number you compare to
The question compares with the families that have 2 cats. So divide by 35, not 42.
$\frac{7}{35} = \frac{1}{5}$
Watch out: dividing by 42 gives about 17%. That is choice C.
Change the fraction to a percent
Use the percent formula. One fifth of 100 is 20.
$\text{percent} = \frac{7}{35} \times 100 = \frac{1}{5} \times 100 = 20$
So the answer is D, 20%.

Check your answer

Find 20% of 35. That is $0.20 \times 35 = 7$ . Add it on: $35 + 7 = 42$ . That is the number of families with 1 cat, so 20% is right.

More questions like this

The original DOE question

Question 95 as printed on page 160 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 95. Open page 160 of the handbook (PDF)

Question 85 of 86

2018 Test B Math, question 45

Question

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

A50
B60
C100
D500

Show the answer

Correct answer: D

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

18 + 0.024x

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

18 + 0.024x = 30

0.024x = 12

x = 500

Phan used his service for 500 minutes.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 86 of 86

2018 Test B Math, question 49

Question

ANIMAL CARDS

Number of Cards	Picture on Card
8	cat
6	dog
5	bird
4	fish
1	horse

The cards in the table above are mixed in a box. Which animal pictured on a card has exactly a 1 in 4 chance of being picked at random from the box?

Acat
Bdog
Cfish
Dhorse

Show the answer

Correct answer: B

The total number of cards in the box is

8 + 6 + 5 + 4 + 1 = 24

Set up a proportion to figure out which card has exactly a 1 in 4 chance of being picked at random.

$\frac{x}{24}=\frac{1}{4}$

$x=6$

The dog card has a 1 in 4 chance of being randomly selected.

Question analysis

A table lists picture cards for five animals, from 8 cat cards down to 1 horse card, all mixed in one box. Students decide which animal has exactly a 1 in 4 chance of being drawn at random. It tests simple probability as a fraction of the total, and the numbers in the table invite matching the wrong count.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Simple probability from a list of counts is grade 6 work. This takes one sum and one division with small whole numbers, a shorter and more routine task than typical grade 7 probability problems.
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: Turning words into math minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 1, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the 24 cards are totaled, one fourth of them is 6 and the answer is a lookup. The wrong choices mostly attract students who match the numbers 1 and 4 in the phrase to the horse and fish rows, or who assume the most common card is the one asked about. Little time is needed, so the main risk is skipping the total.

Turning words into math minor Understanding the question
"Exactly a 1 in 4 chance" has to be turned into a count, one fourth of the 24 cards, rather than matched to a 1 or a 4 in the table. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The counts sit in the table's first column with the animals in the second, and the total of 24 has to be added up from all five rows. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add the cards to find the total in the box, find one-fourth of that total, and look for the animal with exactly that many cards.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Picks the most common card, but cats are 8 of 24 cards, a 1 in 3 chance, and no reading of the table gives them 1 in 4.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Matches the 4 in "1 in 4" to the 4 fish cards, confusing a number of cards with a probability.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the 1 in "1 in 4" to the single horse card instead of comparing each count with the total of 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

Sort out the question
The table shows 8 cat cards, 6 dog, 5 bird, 4 fish and 1 horse. All the cards are mixed in a box. You need the animal with “exactly a 1 in 4 chance” of being picked. A 1 in 4 chance means a probability of $\frac{1}{4}$ .
Find the total number of cards
Add every row of the table. This total is the bottom of the probability fraction.
$8 + 6 + 5 + 4 + 1 = 24$
There are 24 cards in the box.
Find one fourth of the cards
The animal you want must have one fourth of all the cards.
$24 \div 4 = 6$
So you need the animal with exactly 6 cards. The dog has 6 cards.
$P = \frac{6}{24} = \frac{1}{4}$
Rule out the other choices
Watch out: do not match the numbers 1 and 4 to the table. Fish has 4 cards, but its chance is $\frac{4}{24} = \frac{1}{6}$ . So C is wrong. Horse has 1 card, a chance of $\frac{1}{24}$ , so D is wrong. Cat has $\frac{8}{24} = \frac{1}{3}$ , so A is wrong.
Give the answer
Only the dog has exactly one fourth of the cards. So the answer is B, dog.

Check your answer

Four groups of 6 cards make 24, the whole box. So 6 cards is one of four equal shares. That is a 1 in 4 chance. The dog cards are exactly one of those shares.

More questions like this

The original DOE question

Question 106 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 106. 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).