Practice Test

2024 Handbook

2024 Test B Math

Question 1 of 57

2024 Test B Math, question 1

Question

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

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

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

Show the answer

Correct answer: -0.8

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

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

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

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

When expressed as a decimal,

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

Question analysis

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

NYS standards: 7.EE.1 linear expressions with rational coefficients
6.EE.2b parts of an expression
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Expanding and simplifying linear expressions with fraction coefficients is a core grade 7 skill, and this is a typical short example. Writing the coefficient as a decimal in the grid adds care but no new idea.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Multiplying a fraction by a whole number: $\frac{a}{b} \times c = \frac{a \times c}{b}$; a over b is the fraction, and c is the whole number. Multiply only the top number, the numerator, by c. The bottom number, the denominator, stays the same. The size of each piece does not change. You just have more pieces.

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

Check your answer

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

More questions like this

The original DOE question

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

Question 2 of 57

2024 Test B Math, question 2

Question

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

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

Show the answer

Correct answer: 5

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

$1+5=6$

$2+4=6$

$3+3=6$

$4+2=6$

$5+1=6$

Question analysis

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

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students list sample spaces for two number cubes when finding compound probabilities. This asks for one count from that list, a short task, though as a grid-in it offers no choices to expose an incomplete list.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 3 of 57

2024 Test B Math, question 3

Question

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

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

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

Show the answer

Correct answer: 95

First, find the perimeter of the figure.

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

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

So the total perimeter of the scale drawing is

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

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

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

$p=2.5\times 38=95$

Question analysis

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

NYS standards: 7.G.1 scale drawings
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are grade 7 content, and a typical task converts one or two given lengths. Here two missing sides must be worked out and the perimeter assembled before the scale applies, more steps but nothing beyond grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 9, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The figure hides the work: it labels four of the six sides, and the missing two are short pieces of the notch that are easy to overlook. Once all six are found, the scale step is a single multiplication, but it must go the right way, since the real garden is larger than the drawing. Tracing the outline and counting six sides guards against the most common loss.

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 4 of 57

2024 Test B Math, question 4

Question

At one location, the temperature increased from −7° F to 4° F between 5:00 a.m. and 9:00 a.m. By 11:00 a.m., the temperature was 3.5 times the 9:00 a.m. temperature. What was the total increase in temperature between 5:00 a.m. and 11:00 a.m., in degrees Fahrenheit?

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

Show the answer

Correct answer: 21

The 11:00 a.m. temperature is 3.5 times the 9:00 a.m. temperature. The 9:00 a.m. temperature was 4° F.

$4\times 3.5=14$

The temperature increased from −7° F at 5:00 a.m. to 14° F at 11:00 a.m., which is an increase of 21° F.

Question analysis

A temperature word problem with negative numbers: the temperature rises from −7° F to 4° F by 9:00 a.m., and by 11:00 a.m. it is 3.5 times the 9:00 a.m. reading. Students find the total rise from 5:00 to 11:00 a.m. Subtracting a negative starting temperature is where most slips happen.

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Using signed numbers for temperature change is standard grade 7 work. This adds a decimal multiplier and asks for a change rather than a temperature, two short steps, which keeps it at a typical grade 7 level.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Finding the 11:00 a.m. temperature is a single multiplication; the points are lost in the subtraction that follows. A rise that starts below zero has two parts, 7 degrees up to zero and 14 more above it, and students who treat −7 as 7 lose the first part. The times in the stem also invite measuring from 9:00 a.m. instead of 5:00 a.m., which covers only the last two hours.

Error-prone arithmetic moderate Working it out
The last step is 14 − (−7): subtracting the negative starting temperature means adding 7, and the part of the rise below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
“3.5 times the 9:00 a.m. temperature” multiplies the reading of 4°, not the 11-degree rise, and “total increase” runs from the 5:00 a.m. reading, not from 9:00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Three times of day carry three temperatures, −7°, 4° and 14°, and the answer compares the first with the last, not either of them with the middle one. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find the 11:00 a.m. temperature from the 9:00 a.m. one, then subtract the 5:00 a.m. temperature from it, remembering that subtracting a negative number adds its size.

Traps

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

7 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds 14° at 11:00 a.m. but subtracts 7 instead of −7, so 14 − 7 = 7, leaving out the part of the rise below zero.
14 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 11:00 a.m. temperature, 4 × 3.5 = 14, and grids it without subtracting the 5:00 a.m. temperature to get the increase.
10 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Measures the rise from 9:00 a.m. instead of 5:00 a.m., 14 − 4 = 10, which is only the last two hours of the total increase.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Change in temperature: $\text{change} = \text{end} - \text{start}$; End is the last temperature, and start is the first one. Subtract start from end to find how much it went up. It works with negative numbers too. A positive change means the temperature rose.
Subtracting a negative number: $a - (-b) = a + b$; a is any number, and −b is a negative number. Taking away a negative is the same as adding its size. On a thermometer, going from −7 up to 0 is 7 degrees, so those 7 degrees get added.

Sort out what you know
There are three times and three temperatures, in degrees Fahrenheit (°F).
- 5:00 a.m.: −7°F, which is 7 degrees below zero
- 9:00 a.m.: 4°F
- 11:00 a.m.: 3.5 times the 9:00 a.m. temperature
You need the total increase from 5:00 a.m. to 11:00 a.m. That is how much warmer it got from the first time to the last.
Find the 11:00 temperature
“3.5 times the 9:00 a.m. temperature” means 3.5 times 4°F.
$3.5 \times 4 = 14$
At 11:00 a.m. it was 14°F. Watch out: 14 is a temperature, not the increase. Gridding 14 stops too early.
Subtract the start from the end
The increase runs from 5:00 a.m. to 11:00 a.m. So it starts at −7°F and ends at 14°F.
$\text{change} = \text{end} - \text{start}$
$\text{change} = 14 - (-7)$
Watch out: start at 5:00 a.m., not at 9:00 a.m. Starting from 4°F gives only 10.
Subtract the negative number
Taking away a negative number is the same as adding its size.
$a - (-b) = a + b$
$14 - (-7) = 14 + 7 = 21$
The temperature rose 21°F. Watch out: writing 14 − 7 = 7 loses the minus sign. It skips the 7 degrees the air warmed just to reach zero.
So you would grid 21.

Check your answer

Count up on a thermometer. From −7°F up to 0°F is 7 degrees. From 0°F up to 14°F is 14 more. $7 + 14 = 21$ . Or add the two parts of the morning: it rose 11 degrees by 9:00 and 10 more by 11:00, and $11 + 10 = 21$ .

More questions like this

The original DOE question

Question 61 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 199, question 61. Open page 199 of the handbook (PDF)

Question 5 of 57

2024 Test B Math, question 5

Question

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

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

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

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

Show the answer

Correct answer: 32

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

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

The height of the triangular prism is 2 inches.

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 6 of 57

2024 Test B Math, question 6

Question

This graph shows y, the weight of a pole of length x.

$Graph titled WEIGHT OF POLE. The x-axis, Length (feet), is marked 1, 2, 3 and 4, and the y-axis, Weight (ounces), is marked 4, 8, 12 and 16, with grid lines at each mark; the origin is labeled O. A straight line starts at O and rises through the grid corners (1, 4), (2, 8), (3, 12) and (4, 16), ending in an arrow, and the point (1, 4) is marked with a dot and labeled (1, 4).$

What is represented by the point with coordinates (1, 4)?

AThe unit rate is 4 ounces per foot.
BThe y-intercept is 4.
CA pole 4 feet long weighs 1 ounce.
DThe length increases 4 feet for each 1 ounce of weight.

Show the answer

Correct answer: A

The x-axis represents the length of the pole, in feet, and the y-axis represents the weight of the pole, in ounces.

The point (1, 4) means that a 1-foot pole would weigh 4 ounces.

Since the graph is linear, the pole weighs 4 ounces for every additional 1 foot of length.

So the unit rate is 4 ounces per foot.

Question analysis

A graph of a pole's weight in ounces against its length in feet is a straight line through the origin, with the point (1, 4) marked. Students decide what that point says about poles, choosing among statements about rates, intercepts and particular poles, so the meaning of each coordinate has to be pinned down.

NYS standards: 7.RP.2d points on proportional graphs
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on a proportional graph means in the situation is a named grade 7 skill. This is a direct, one-step example with a clear graph, 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 understand
What makes it hard: Reading a figure, table or graph moderate
A very close wrong answer minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 2, 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

Nothing needs computing; the question tests whether a point on a graph can be put into words. Each wrong statement misuses the same two numbers, as an intercept, with the coordinates swapped, or as a rate turned upside down. Saying the point aloud with its units, a 1-foot pole weighs 4 ounces, and comparing that sentence with each statement settles it quickly.

Reading a figure, table or graph moderate Understanding the question
The axes carry the meaning: length in feet runs along x and weight in ounces up y, so (1, 4) describes a 1-foot pole, and the line through O makes that a rate. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer minor Choosing the answer
Both rate statements pair 4 with 1, one as ounces per foot and the other as feet per ounce; only the axis labels say which way the rate runs. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
The choices use the terms unit rate and y-intercept, which the question does not define, and two statements hang on knowing what each one means. (Anderson & Krathwohl, 2001)

Strategy: Note which quantity each axis measures, say in words what the x-coordinate and the y-coordinate of the marked point stand for, then compare that sentence with each choice.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
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
Takes the 4 in the point's label for the y-intercept, though the line starts at the origin, where a pole of no length weighs 0 ounces.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads the coordinates the wrong way round, taking 1 as the weight in ounces and 4 as the length in feet.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Turns the rate upside down, describing feet of length per ounce of weight instead of ounces of weight per foot of length.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{y}{x}$; For a point on the line, y is the weight in ounces and x is the length in feet. The unit rate is the weight of 1 foot of pole. This works for a straight line that starts at the origin, the point (0, 0).

Read the axes
The graph has two axes (number lines). The x-axis, across the bottom, shows length in feet. The y-axis, going up, shows weight in ounces.
A point is written (x, y). The across number comes first, then the up number. So in (1, 4), x is 1 and y is 4.
Say what the point means
x = 1 means a length of 1 foot. y = 4 means a weight of 4 ounces. So a pole 1 foot long weighs 4 ounces.
Watch out: do not swap the numbers. “A pole 4 feet long weighs 1 ounce” reads them the wrong way round. That is choice C.
See that this is a unit rate
A unit rate is an amount for 1 unit, here the ounces in 1 foot. The line starts at O, the origin (0, 0), and it is straight. So each extra foot adds the same weight.
$\text{unit rate} = \frac{y}{x} = \frac{4}{1} = 4$
The unit rate is 4 ounces per foot. Watch out: choice D turns this upside down. It gives feet for each ounce, not ounces for each foot.
Rule out the y-intercept
The y-intercept is where the line crosses the y-axis. This line crosses it at O, where y is 0. So the y-intercept is 0, not 4, and choice B is wrong.
So the answer is A, the unit rate is 4 ounces per foot.

Check your answer

Try another point on the line, such as (3, 12). A pole 3 feet long weighs 12 ounces. $12 \div 3 = 4$ , so it is still 4 ounces per foot. The rate stays the same all along the line.

More questions like this

The original DOE question

Question 63 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 63. Open page 200 of the handbook (PDF)

Question 7 of 57

2024 Test B Math, question 7

Question

The number line shows points E, F, G, and H.

$Number line from −7 to 7 with a tick at every integer and arrows at both ends; the ticks are labeled −7, −5, −3, −1, 0, 1, 3, 5 and 7. Point E is on −7, point F is on the unlabeled tick between −3 and −1, point G is on the unlabeled tick between 1 and 3, and point H is on 7.$

Which point represents the sum of $2.5+(-4.5)$ ?

Apoint E
Bpoint F
Cpoint G
Dpoint H

Show the answer

Correct answer: B

Find the sum of $2.5+(-4.5)$ :

$2.5+(-4.5)=2.5-4.5=-2$

Point F is located at −2.

Question analysis

A number line question with points E, F, G and H marked on a line labeled from −7 to 7. Students add 2.5 and negative 4.5 and find which point shows the sum. It tests adding a positive and a negative decimal, then reading positions on a line whose labels skip some integers.

NYS standards: 7.NS.1b rational number addition
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding rational numbers with different signs and showing the sum on a number line is exactly the grade 7 standard. This is one routine sum with simple decimals, a typical classroom exercise.
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
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 1, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a one-line sum followed by a lookup, and most students finish quickly. The four points are placed so that each common sign mistake lands on one of them: −7, 2 and 7 all come from mishandling the negative. Picturing the sum as a move of 4.5 units to the left from 2.5 shows both the size and the direction at once.

Error-prone arithmetic minor Working it out
Adding −4.5 to 2.5 means taking 4.5 away from 2.5, and the larger size of −4.5 makes the sum negative, a sign rule that is easy to apply backwards. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph minor Understanding the question
Only 0 and the odd numbers are labeled, so point F sits at an unlabeled tick between −3 and −1 and must be found by counting. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Rewrite adding −4.5 as subtracting 4.5, work out the sum, then find it on the line by counting tick marks from a labeled number, one tick per integer.

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: weak
Adds the sizes 2.5 and 4.5 and makes the total negative, treating both numbers as negative, which lands on −7 at point E.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Subtracts 2.5 from 4.5 to get the right size, 2, but keeps it positive, missing that the negative number has the larger size.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Ignores the negative sign on 4.5 and adds the two numbers, 2.5 + 4.5 = 7, landing on point H at the far right.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding a negative number: $a + (-b) = a - b$; a is any number, and −b is a negative number. Adding a negative number is the same as subtracting its size. On a number line, it means moving b steps to the left.

Sort out the question
You need to find $2.5 + (-4.5)$ . Then find that sum (the answer to the addition) on the number line.
The number line has a tick at every whole number from −7 to 7. Only 0 and the odd numbers have labels.
Turn the addition into subtraction
$a + (-b) = a - b$
$2.5 + (-4.5) = 2.5 - 4.5$
You are taking 4.5 away from 2.5. You take away more than you have, so the answer will be negative.
Do the subtraction
The two sizes differ by 2, because $4.5 - 2.5 = 2$ . The larger size, 4.5, belongs to the negative number. So the sum is negative.
$2.5 - 4.5 = -2$
Watch out: the answer is −2, not 2. Point G is at 2.
Find −2 on the line
−2 has no label. It is the tick between −3 and −1. Point F sits on that tick.
Watch out: adding 2.5 + 4.5 = 7 lands on point H. Making both numbers negative gives −7, point E.
So the answer is B, point F.

Check your answer

Start at 2.5 on the number line and move 4.5 steps to the left. The first 2.5 steps take you to 0. You still have 2 steps to go. Two more steps left of 0 lands on −2, which is point F.

More questions like this

The original DOE question

Question 64 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 64. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 8 of 57

2024 Test B Math, question 8

Question

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

Show the answer

Correct answer: C

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

$9\bullet 8=72$

$2x\bullet 3y=6xy$

$72=6xy$

$12=xy$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 9 of 57

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 10 of 57

2024 Test B Math, question 10

Question

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

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

Show the answer

Correct answer: B

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

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

Question analysis

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

NYS standards: 6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing fractions is a grade 6 standard, and this is one division with small mixed numbers. Grade 7 work usually adds negative numbers or a word problem around it, so this sits below grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 16+, 5 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 11 of 57

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 12 of 57

2024 Test B Math, question 12

Question

Lena is building a fence. She will need to dig holes to help support the posts that hold up the fence. The holes need to have a depth of $3\frac{1}{3}$ feet below the ground. Each post is 10 feet long. What is the height of the part of the post that is above the ground?

A $6\frac{2}{3}$ feet
B $7\frac{1}{3}$ feet
C10 feet
D $13\frac{1}{3}$ feet

Show the answer

Correct answer: A

The posts are each 10 feet long, and $3\frac{1}{3}$ feet of each post is below the ground, so the height of the part of the posts that is above the ground is the difference between the two amounts:

$10-3\frac{1}{3}=$

$\frac{10}{1}-\frac{10}{3}=$

$\frac{30}{3}-\frac{10}{3}=\frac{20}{3}=6\frac{2}{3}\text{ feet}$

Question analysis

A fence-building word problem: each post is 10 feet long, and 3 1/3 feet of it goes into a hole in the ground. Students find how much of the post stands above ground by subtracting a mixed number from a whole number. Regrouping the whole number into thirds is the step that trips students up.

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 Subtracting a mixed number from a whole number is fraction work from before grade 7, set here in a one-step story. It asks much less than grade 7 problems, which add negative numbers or several steps.
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
Solution: 1 step
Reading level: Question and choices: about grade 2, 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

The situation is plain, a post partly buried in a hole, and the subtraction 10 − 3 1/3 is the whole task. The only real trap is the regrouping, since there are no thirds to take the 1/3 from until one foot is split into three. Checking that the two parts of the post add back to 10 feet catches the slip.

Error-prone arithmetic minor Working it out
Taking 3 1/3 from the whole number 10 requires regrouping 10 as 9 3/3, the step where students subtract the whole numbers and simply carry the 1/3 along. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the buried part from the whole post, rewriting 10 as 9 and 3/3, or as 30/3, so that the thirds can be taken away.

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).
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 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Subtracts the whole numbers, 10 − 3 = 7, then carries the 1/3 along instead of taking it away too, getting 7 1/3.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the post's full 10-foot length, the whole post instead of the part above ground, overlooking the 3 1/3 feet sunk in the hole.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the depth of the hole to the length of the post, 10 + 3 1/3, as if digging the hole made the post taller.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a whole: $\text{above} = \text{whole} - \text{below}$; The whole is the full length of the post. Below is the part in the hole, and above is the part you can see. The two parts add up to the whole, so subtract to find one part.
One whole in thirds: $1 = \frac{3}{3}$; One whole can be cut into 3 equal thirds. So 1 is the same as 3 over 3. Use this when you need to take thirds away from a whole number.

Sort out the question
Each post is 10 feet long. Part of it goes into a hole $3\frac{1}{3}$ feet deep. The rest sticks up above the ground. You need the height of the part above the ground.
Decide to subtract
The part below and the part above make up the whole post. So take the part below away from the whole.
$\text{above} = \text{whole} - \text{below}$
$\text{above} = 10 - 3\frac{1}{3}$
Watch out: do not add. $10 + 3\frac{1}{3} = 13\frac{1}{3}$ , choice D, makes the post longer than it is.
Turn one whole into thirds
The 10 has no thirds to take 1/3 from. So take 1 whole from the 10 and cut it into 3 thirds.
$1 = \frac{3}{3}$
$10 = 9 + \frac{3}{3} = 9\frac{3}{3}$
Now there are thirds to subtract from.
Subtract wholes and thirds
Subtract the whole numbers, then the thirds.
$9\frac{3}{3} - 3\frac{1}{3} = 6\frac{2}{3}$
That is $9 - 3 = 6$ and $\frac{3}{3} - \frac{1}{3} = \frac{2}{3}$ . Watch out: the one third must be taken away too. Doing 10 − 3 = 7 and then keeping the one third gives $7\frac{1}{3}$ , choice B.
So the answer is A, $6\frac{2}{3}$ feet.

Check your answer

Add the two parts back together: $6\frac{2}{3} + 3\frac{1}{3} = 9\frac{3}{3} = 10$ . That is the whole post. Also, the answer must be less than 10 feet, since part of the post is under the ground.

More questions like this

The original DOE question

Question 69 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 69. Open page 202 of the handbook (PDF)

Question 13 of 57

2024 Test B Math, question 13

Question

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

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

Show the answer

Correct answer: B

First substitute 2 for r in the equation.

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

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

Then solve for p.

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

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

$p+4=2p+3$

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

$p+1=2p$

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

$1=p$

The value of p is 1.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 57

2024 Test B Math, question 14

Question

Karen reads at the rate of 60 pages per day. Martina reads 25% more pages per day than Karen. If the two students begin 1,500-page reading assignments on the same day, how many days sooner will Martina complete the assignment than Karen?

Show the answer

Correct answer: B

First, determine the rate at which Martina reads, in pages per day, using the rate at which Karen reads, 60 pages per day.

$60+60\bullet\frac{25}{100}=$

$60+60\bullet 0.25=$

$60+15=75$

Next, determine the number of days it will take each person to complete the 1,500-page reading assignment.

Karen: $\frac{1{,}500\text{ pages}}{k\text{ days}}=\frac{60\text{ pages}}{1\text{ day}}$

$60k=1{,}500$

$(60\div 60)k=1{,}500\div 60$

$k=25\text{ days}$

Martina: $\frac{1{,}500\text{ pages}}{m\text{ days}}=\frac{75\text{ pages}}{1\text{ day}}$

$75m=1{,}500$

$(75\div 75)m=1{,}500\div 75$

$m=20\text{ days}$

Finally, determine the difference between the two numbers of days.

$25-20=5$

So Martina will complete the assignment 5 days sooner than Karen.

Question analysis

A reading-rate word problem: Karen reads 60 pages a day, and Martina reads 25% more pages a day. Both start a 1,500-page assignment on the same day, and students find how many days sooner Martina finishes. It combines a percent increase with rates and times, and asks for a difference in days.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually stop after one increase or decrease. This one turns the percent into a new rate, finds two finishing times and compares them, four steps that go beyond a typical grade 7 task.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Combines ideas moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is easy on its own; the difficulty is in carrying the chain to the end. Three of the choices are numbers a student meets or makes along the way: the extra pages per day, a quotient of two rates, or a rate built on 25 pages instead of 25%. Writing both finishing times down before subtracting keeps the target, a number of days, in view.

Many steps moderate Working it out
Four operations chain together: 25% of 60, Martina's 75 pages, the two finishing times 1,500 ÷ 60 and 1,500 ÷ 75, and finally their difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Combines ideas moderate Working it out
A percent increase in a daily rate has to be turned into a change in days, and 25% more pages a day does not mean 25% fewer days. (Embretson, 1998; Daniel & Embretson, 2010)
Turning words into math minor Understanding the question
“25% more pages per day than Karen” makes Martina's rate 1.25 × 60, and “how many days sooner” asks for a difference of two times, not of two rates. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Increase 60 by 25% to get Martina's daily pages, divide 1,500 by each girl's daily rate to get her number of days, then subtract the two numbers of days.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
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
Divides Karen's 60 pages a day by Martina's 15 extra pages a day, combining two rates without finding how long either girl takes.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds 25 pages instead of 25%, giving Martina 85 pages a day, so she needs 1,500 ÷ 85 ≈ 17.6, or 18 days, and 25 − 18 = 7.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that Martina reads 25% of 60 = 15 more pages each day and stops there, giving a number of pages instead of days.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{old} + \frac{p}{100} \times \text{old}$; Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.
Time from a rate: $\text{days} = \text{total pages} \div \text{pages per day}$; Total pages is the whole assignment. Pages per day is how fast someone reads. Dividing tells how many days of reading fit into the whole job.

Sort out what you know
Karen reads 60 pages per day. Martina reads 25% more pages per day than Karen. Each girl must read 1,500 pages. You need how many days sooner Martina finishes. That is the difference between their numbers of days.
Find Martina’s pages per day
25% more means Karen’s 60 pages plus 25% of 60.
$\text{new} = \text{old} + \frac{p}{100} \times \text{old}$
$60 + \frac{25}{100} \times 60 = 60 + 15 = 75$
Martina reads 75 pages per day. Watch out: 15 is extra pages, not days. Stopping here gives 15, choice D. Adding 25 pages instead of 25% gives 85 pages and leads to 7, choice C.
Find each girl’s number of days
Divide the whole assignment by each girl’s pages per day.
$\text{days} = \text{total pages} \div \text{pages per day}$
$\text{Karen: } 1{,}500 \div 60 = 25$
$\text{Martina: } 1{,}500 \div 75 = 20$
Karen needs 25 days, and Martina needs 20 days.
Subtract the days
“How many days sooner” asks for the difference.
$25 - 20 = 5$
Martina finishes 5 days sooner. Watch out: dividing 60 by 15 mixes two rates and gives 4, choice A.
So the answer is B, 5.

Check your answer

Multiply back. Karen: $25 \times 60 = 1{,}500$ pages. Martina: $20 \times 75 = 1{,}500$ pages. Both read all 1,500 pages, and 25 days is 5 days more than 20 days.

More questions like this

The original DOE question

Question 71 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 71. Open page 202 of the handbook (PDF)

Question 15 of 57

2024 Test B Math, question 15

Question

Claire is running on a circular track. She has already run 650 meters. Each lap around the track is a distance of 120 meters. Her goal is to run at least 2,500 meters in all. How many more laps does she need to complete in order to reach her goal?

Show the answer

Correct answer: D

Claire’s goal is to run at least 2,500 meters, and she has already run 650 meters.

$2{,}500-650=1{,}850$

Claire has at least 1,850 meters left to run. Divide by the distance around the track to find the number of laps she needs to complete:

$1{,}850\div 120\approx 15.417$

Since she wants to run at least 2,500 meters, round 15.417 up to 16. Claire needs to run 16 laps to meet her goal.

Question analysis

A word problem about running laps: Claire has already run 650 meters on a 120-meter track and wants to run at least 2,500 meters in all. Students find how many more laps she must complete. Subtracting and dividing are straightforward; deciding what to do with the leftover part of a lap is the real test.

NYS standards: 7.EE.4b two-step inequalities
6.NS.2 multi-digit division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve word problems that lead to inequalities like px + q ≥ r and interpret the solution in context. This is a typical three-step example, where the solution must be turned into a whole number of 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
A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 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 leads to about 15.4 laps, and the question is entirely about what to do with the fraction of a lap. Rounding to the nearest whole number is a habit that fails here, because the goal must be met or passed. Multiplying back, 15 × 120 = 1,800 and 16 × 120 = 1,920, shows which count actually covers the 1,850 meters still to run.

Turning words into math moderate Understanding the question
“At least 2,500 meters in all” means a part lap must still be run in full, so 1,850 ÷ 120 ≈ 15.4 has to be rounded up, against the usual rounding rule. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
15 and 16 laps are one apart and both sit next to the quotient 15.4, so estimating cannot decide between them; only the condition that the goal be reached can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Dividing 1,850 by 120 by hand does not come out even, and long division by a three-digit number invites slips in placing the quotient. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract what she has run from 2,500, divide by 120, and check your whole number by multiplying it back: those laps must cover all of the remaining distance.

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 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the 650 meters already run by 120 and rounds down to 5, counting laps finished so far instead of laps still needed.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Divides the 650 meters already run by 120, about 5.4, and rounds up as the goal requires, but for the distance done rather than the distance left.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds 1,850 ÷ 120 ≈ 15.4 and rounds down to 15 by the usual rule, but 15 more laps leave her 50 meters short of the goal.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance left: $\text{left} = \text{goal} - \text{done}$; Goal is the distance she wants to reach in all. Done is the distance she has already run. Subtract to find how far she still has to go.
Number of laps: $\text{laps} = \text{distance} \div \text{one lap}$; Distance is how far she still has to run. One lap is the length of the track, 120 meters. Dividing tells how many laps fit into that distance. If it does not come out even, part of a lap is left over.

Sort out what you know
Claire has run 650 meters. One lap is 120 meters. Her goal is “at least 2,500 meters in all”. At least means 2,500 or more, so she must not fall short. You need how many more laps she must run.
Find the distance left
$\text{left} = \text{goal} - \text{done}$
$\text{left} = 2{,}500 - 650 = 1{,}850$
She still needs 1,850 meters. Watch out: divide the distance left, not the 650 meters she already ran. Dividing 650 by 120 leads to 5 or 6, choices A and B.
Divide by the length of a lap
$\text{laps} = \text{distance} \div \text{one lap}$
$1{,}850 \div 120 \approx 15.4$
The sign ≈ means about. Check with whole laps: $15 \times 120 = 1{,}800$ . That leaves $1{,}850 - 1{,}800 = 50$ meters still to go. So 15 laps are not quite enough.
Round up to reach the goal
She needs part of one more lap. She has to complete whole laps, so she runs a 16th lap. Here you round up, even though 15.4 would usually round down. Watch out: 15 laps leave her 50 meters short, so choice C is wrong.
So the answer is D, 16.

Check your answer

Add up her total. With 15 more laps: $650 + 1{,}800 = 2{,}450$ meters, which is short of 2,500. With 16 more laps: $650 + 16 \times 120 = 650 + 1{,}920 = 2{,}570$ meters. That reaches the goal.

More questions like this

The original DOE question

Question 72 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 72. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 16 of 57

2024 Test B Math, question 16

Question

A company with 200 employees grew in size by 25% between 2005 and 2010. The company grew again by 10% between 2010 and 2015. What was the total number of employees in 2015?

A235
B260
C270
D275

Show the answer

Correct answer: D

First, determine the number of employees in 2010:

$200\times 25\%=200\times 0.25=50$

So the number of employees in 2010 is $200+50=250$ .

Next, determine the number of employees in 2015:

$250\times 10\%=250\times 0.10=25$

So the total number of employees in 2015 is $250+25=275$ .

Question analysis

A percent change word problem: a company of 200 employees grows by 25% from 2005 to 2010 and then by another 10% from 2010 to 2015. Students find the number of employees in 2015. Two percent changes in a row are the tricky part, since each percent must be taken of the right amount.

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 Percent increase is a grade 7 topic, and two increases in a row are a common grade 7 multistep problem. The numbers are friendly, so the demand matches typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Turning words into math moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each increase alone is easy, and the whole question is about the base of the second one. Adding the percents into a single 35% feels like a shortcut and lands five employees short, a gap estimation cannot reveal. Working period by period, or multiplying 200 by 1.25 and then by 1.1, keeps each percent attached to the right amount.

A very close wrong answer moderate Choosing the answer
270 and 275 employees are five apart, too close to separate by estimate; only taking the second 10% of 250 rather than of 200 decides between them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math moderate Understanding the question
“Grew again by 10%” refers to the company's size in 2010, so the base for the second percent is the new 250 employees, not the original 200. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two percent increases make four operations, a percent and an addition for each period, and the second period starts from the first one's result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work one period at a time: find the 2010 total by increasing 200 by 25%, then increase that new total by 10%.

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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Adds the two percents to the employee count as if they were numbers of people, 200 + 25 + 10 = 235.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals partial knowledge Expected pull: weak
Finds the 250 employees in 2010 correctly, then adds 10 employees instead of 10% of 250, treating the second percent as a head count.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes both increases from the original 200, 50 + 20 = 70 more, missing that the second 10% applies to the 250 employees of 2010.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{old} + \frac{p}{100} \times \text{old}$; Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.

Sort out what you know
The company has 200 employees at the start, in 2005. From 2005 to 2010 it grows by 25%. From 2010 to 2015 it grows again, by 10%. You need the number of employees in 2015.
Find the number in 2010
$\text{new} = \text{old} + \frac{p}{100} \times \text{old}$
$200 + \frac{25}{100} \times 200 = 200 + 50 = 250$
In 2010 there were 250 employees. Watch out: 25% does not mean 25 people. Adding 25 people and then 10 people gives 235, choice A.
Grow from the new number
The second increase starts in 2010. So take 10% of 250, not 10% of 200.
$250 + \frac{10}{100} \times 250 = 250 + 25 = 275$
In 2015 there were 275 employees.
Avoid the traps and answer
Watch out: taking both percents of the first 200 gives $200 + 50 + 20 = 270$ , choice C. That misses the growth of the 50 new workers. Adding 10 people instead of 10% gives 260, choice B.
So the answer is D, 275.

Check your answer

Growing by 25% is the same as multiplying by 1.25. Growing by 10% is the same as multiplying by 1.1. $200 \times 1.25 \times 1.1 = 250 \times 1.1 = 275$ . The answer should also be a bit more than 270, since the second 10% is taken of a bigger number.

More questions like this

The original DOE question

Question 73 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 73. Open page 203 of the handbook (PDF)

Question 17 of 57

2024 Test B Math, question 17

Question

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

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

Show the answer

Correct answer: A

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

$12=2r$

$r=6$

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

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

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

Question analysis

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

NYS standards: 6.SP.8a uniform probability models
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Simple probability is grade 6 content in New York, and grade 7 problems usually give every count. Working backward from a comparison of two probabilities to a missing count, then forward to a third probability, is more than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Combines ideas moderate
Solution: 3 steps
Reading level: Question and choices: about grade 7, 49 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question withholds the red count, and everything depends on recovering it from the comparison of probabilities. Students who cannot make that step tend to leave red out altogether, while those who find 6 red candies can still slip at the end by using the wrong total or the wrong color. Writing all three counts and the total of 22 before forming any fraction prevents both endings.

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

Strategy: Turn twice as likely into a statement about numbers of candies to find the red count, add all three colors for the total, then put the yellow count over that total and simplify.

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 18 of 57

2024 Test B Math, question 18

Question

A taxi driver recorded the number of fares each day for one month. The box plot displays the data.

$Box plot titled DAILY FARES above a number line labeled Number of Fares, with arrows at both ends and a tick labeled at every even number from 0 to 32. The left whisker starts at 2, the box runs from 12 to 24 with the median line at 16, and the right whisker ends at 32.$

Based on the box plot, which statement is a correct interpretation of the data?

AThe data point 2 is an outlier.
BThe mean of the data is 16.
CThe interquartile range (IQR) of the data is 30.
DOn about 25% of the days, there were between 12 and 16 fares.

Show the answer

Correct answer: D

The amount of data in a box plot is broken into 4 parts, or quartiles, which each represent 25% of the data. So the amount of data represented between Quartile 1 (12) and the median (16) is about 25% of the data.

Question analysis

A statistics question on a box plot of the number of fares a taxi driver recorded each day for one month. Students decide which of four statements is a correct interpretation of the data, using what a box plot does and does not show. The statements draw on several different box plot ideas, so each one must be checked.

NYS standards: 7.SP.1 box plots and interquartile range
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Reading box plots, finding the interquartile range and testing for outliers are all grade 7 standards in New York. This item gathers several of them in one question, but each check is short, typical of 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 understand
What makes it hard: Something you must already know moderate
Every choice must be checked moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 63 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

No calculation here is long; the challenge is knowing what a box plot does and does not show. Two of the wrong statements sound reasonable because they borrow real features of the plot, its long left whisker and its middle line, and give them the wrong meaning. Remembering that the plot splits the days into four groups of about equal size makes the true statement easy to confirm.

Something you must already know moderate Working it out
Each statement draws on a box plot fact the test does not supply: what the middle line marks, how the IQR is found, and when a value counts as an outlier. (Anderson & Krathwohl, 2001)
Every choice must be checked moderate Choosing the answer
The four statements test four separate ideas, an outlier, the mean, the IQR and a quartile's share, so ruling out one says nothing about the next. (Embretson & Wetzel, 1987)
Reading a figure, table or graph minor Understanding the question
The scale is labeled every 2 fares with the numbers crowded together, so the five values 2, 12, 16, 24 and 32 must be read off with care. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the minimum, quartiles, median and maximum off the plot, then test each statement in turn, computing the IQR and the outlier cutoffs rather than judging by eye.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

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
Takes the long left whisker as a sign that 2 is an outlier, but that whisker holds a quarter of the days, and 2 is above the cutoff 12 − 1.5 × 12 = −6.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Takes the median line at 16 for the mean, though a box plot shows the median and quartiles and cannot show the mean.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Subtracts the minimum from the maximum, 32 − 2 = 30, which is the range, not the interquartile range of 24 − 12 = 12.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range (IQR): $\text{IQR} = Q_3 - Q_1$; Q1 is the first quartile, the left edge of the box. Q3 is the third quartile, the right edge. The IQR is the length of the box. It tells how spread out the middle half of the data is.
Low cutoff for an outlier: $\text{low cutoff} = Q_1 - 1.5 \times \text{IQR}$; Q1 and the IQR are found from the box. An outlier is a value far away from the rest. A value below this cutoff counts as an outlier on the low side. A value at or above it does not.

Read the five numbers
A box plot sums up a set of data with five numbers. Here each piece of data is the number of fares on one day.
- Minimum, the smallest value: 2
- First quartile, Q1, the left edge of the box: 12
- Median, the middle value, the line inside the box: 16
- Third quartile, Q3, the right edge of the box: 24
- Maximum, the largest value: 32
Learn what each part holds
These five numbers split the days into four parts. Each part holds about one fourth of the days, or about 25%. For example, about 25% of the days fall between the minimum and Q1. Another 25% fall between Q1 and the median.
Check choice D
Choice D talks about days with between 12 and 16 fares. 12 is Q1, and 16 is the median. That stretch is one of the four parts. So on about 25% of the days, there were between 12 and 16 fares. Choice D is correct.
Rule out choices B and C
Choice B: a box plot shows the median, 16, not the mean (the average). You cannot find the mean from it. Watch out: do not mix up the median and the mean.
$\text{IQR} = Q_3 - Q_1 = 24 - 12 = 12$
Choice C says the IQR is 30. But 30 is the range, $32 - 2$ . Watch out: the IQR uses the edges of the box, not the ends of the whiskers.
Rule out choice A and answer
A long whisker does not make 2 an outlier. Use the cutoff rule.
$\text{low cutoff} = Q_1 - 1.5 \times \text{IQR}$
$\text{low cutoff} = 12 - 1.5 \times 12 = 12 - 18 = -6$
2 is above −6, so 2 is not an outlier. The long whisker just shows that the lowest fourth of the days is spread out.
So the answer is D.

Check your answer

Go back over the four choices. Only D uses what a box plot really shows: four parts of about 25% each. Choice B needs the mean, which the plot does not show. Choice C gives the range, not the IQR. And 2 is not far enough out to be an outlier, so A is wrong.

More questions like this

The original DOE question

Question 75 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 75. Open page 203 of the handbook (PDF)

Question 19 of 57

2024 Test B Math, question 19

Question

Serena was studying the number of water stations for different lengths of races. The table below shows the data she collected.

WATER STATIONS IN RACES

Length of Race (mi)	Number of Water Stations
3	2
12	8
18	12

Which equation represents the relationship between the length of the race, x miles, and the number of water stations, y?

A $y=\frac{2}{3}x$
B $y=\frac{3}{2}x$
C $y=x+6$
D $y=x-1$

Show the answer

Correct answer: A

There is a proportional relationship between the length of the race, x miles, and the number of water stations, y. The constant of proportionality is

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

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

Therefore, the equation $y=\frac{2}{3}x$ represents this relationship.

Question analysis

A table from a study of races shows the number of water stations for races of 3, 12 and 18 miles. Students choose the equation that links the race length x to the number of stations y. It tests reading a relationship from a table and expressing it as an equation in x and y.

NYS standards: 7.RP.2c proportional relationship equations
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an equation for a relationship given in a table is a central grade 7 skill. With three clean rows and whole numbers, this is a typical grade 7 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: A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The table is small and its numbers are friendly, so the question comes down to which way the ratio runs and whether every row is checked. One wrong equation fits only the first row and another fits none, so testing a single row is not enough. Putting the 12-mile race into each equation, which should give 8 stations, leaves only one rule standing.

A very close wrong answer moderate Choosing the answer
The coefficients 2/3 and 3/2 use the same two digits, and only dividing stations by miles, 2 ÷ 3, rather than miles by stations, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
The table lists miles first and stations second, while each equation puts stations, y, alone on the left, so every row must be read as an (x, y) pair. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Test each equation on all three rows of the table, not just the first; only a rule that fits every row represents the relationship.

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).
Checks only one number: Tries one value in each choice and takes the first that works, when that value happens to work for a wrong choice too; a second value would rule 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

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides race length by water stations, 3 ÷ 2, turning the ratio upside down, so the rule gives more stations than miles.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Builds an addition rule from the gap of 6 between 18 and 12 in the last row, with the sign reversed, and it fits no row of the table.
D Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Checks only the first row, where 3 − 1 = 2 works, without testing the 12- and 18-mile races, where it gives 11 and 17.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx$; x is the race length in miles, and y is the number of water stations. k is the constant of proportionality, the number of stations for each mile. Find it with y ÷ x. If every row gives the same k, the table is proportional.

Sort out the table
Each row of the table is a pair of numbers. The first is the race length, x, in miles. The second is the number of water stations, y. You need the equation that works for every row.
Divide y by x in each row
See if every row has the same ratio of stations to miles. Divide y by x.
Miles (x) Stations (y) y ÷ x
3 2 $\frac{2}{3}$
12 8 $\frac{8}{12} = \frac{2}{3}$
18 12 $\frac{12}{18} = \frac{2}{3}$
Every row gives 2/3. So the relationship is proportional: y is always the same number times x.
Write the equation
$y = kx$
Here k is the ratio you found, $\frac{2}{3}$ .
$y = \frac{2}{3}x$
Watch out: divide stations by miles, not miles by stations. 3 ÷ 2 gives 3/2, choice B. That rule would give more stations than miles.
Test the other choices and answer
Watch out: test every row, not just the first one. Choice D, y = x − 1, works for 3 miles. But it gives 11 stations for 12 miles, not 8. Choice C, y = x + 6, gives 9 stations for 3 miles, not 2.
So the answer is A, $y = \frac{2}{3}x$ .

Miles (x)	Stations (y)	y ÷ x
3	2	$\frac{2}{3}$
12	8	$\frac{8}{12} = \frac{2}{3}$
18	12	$\frac{12}{18} = \frac{2}{3}$

Check your answer

Put each race length into $y = \frac{2}{3}x$ . $\frac{2}{3} \times 3 = 2$ , $\frac{2}{3} \times 12 = 8$ and $\frac{2}{3} \times 18 = 12$ . All three match the table.

More questions like this

The original DOE question

Question 76 as printed on page 204 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 204, question 76. Open page 204 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 20 of 57

2024 Test B Math, question 20

Question

Centerville will hold a vote on whether to build a new park. The town council surveyed a random sample of 200 residents to determine whether they would be in favor of building this park. The table shows the results of the survey.

SURVEY RESULTS

Opinion	Number of Residents
Yes	142
No	38
Undecided	20

The mayor estimates that half the undecided voters will vote in favor of building the park. Based on the survey results, if the mayor is correct, how many of the 24,100 voters will be in favor of building the park?

A17,111
B18,316
C19,401
D19,521

Show the answer

Correct answer: B

If half the undecided voters (half of 20) vote in favor of building the park, the total number of “yes” votes from the survey will be 152.

$142+10=152$

There are 24,100 voters, and 200 residents were surveyed. Set up a proportion to solve:

$\frac{152}{200}=\frac{x}{24{,}100}$

$\frac{19}{25}=\frac{x}{24{,}100}$

$(24{,}100)\left(\frac{19}{25}\right)=x$

$18{,}316=x$

Question analysis

A survey question: a random sample of 200 Centerville residents answered yes, no or undecided about building a park, and the mayor expects half the undecided to vote yes. Students use the sample to predict how many of 24,100 voters will favor the park. It joins a proportion from a sample with large-number arithmetic.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.SP.1b samples and populations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Using a random sample to estimate a count in a whole population is a typical grade 7 proportion task. The extra step for the undecided group and the large numbers add care without going beyond grade 7.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 88 words.
Students, last 30 days: 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 a sample-to-population proportion, and the trap is the mayor's assumption in the middle of the stem. Students who scale the 142 yes answers directly, or who count every undecided resident, reach believable totals that sit right among the choices. Once the count of 152 is settled, a rough estimate, about three quarters of 24,000, is enough to confirm the exact product.

Error-prone arithmetic moderate Working it out
Scaling 152 out of 200 up to 24,100 voters means computing 24,100 × 0.76, or 120.5 × 152, by hand with five-digit numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
The mayor's sentence adds a step: “half the undecided voters” turns the 20 undecided residents into 10 more yes votes, giving 152 before any scaling. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table gives Yes, No and Undecided counts, and the half must be taken of the Undecided row, not of the No row just above it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Count the sample's expected yes votes, including half of the undecided, write them as a fraction of 200, and multiply that fraction by 24,100; estimate first to check the choice.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Uses only the 142 yes answers, 142/200 of 24,100 = 17,111, leaving out the undecided residents the mayor expects to vote yes.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Takes half of the 38 No answers instead of the 20 Undecided, reading the wrong row, so 161/200 of 24,100 ≈ 19,401.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Counts all 20 undecided residents as yes votes instead of half of them, so 162/200 of 24,100 = 19,521.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling up a sample: $\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}$; A random sample is a smaller group picked by chance. Its share of yes votes should be about the same as the whole town’s share. So the two fractions are set equal. Use it to scale a survey up to a whole group.

Sort out what you know
The town surveyed 200 people. 142 said Yes, 38 said No, and 20 were Undecided (not sure yet). The mayor thinks half of the undecided people will vote yes. There are 24,100 voters in all. You need how many of them will vote yes.
Count the yes votes in the survey
Half of the 20 undecided people is $20 \div 2 = 10$ . Add them to the yes votes.
$142 + 10 = 152$
So 152 of the 200 people surveyed would vote yes. Watch out: take half of the Undecided row, not the No row. Half of 38 leads to 19,401, choice C.
Find the share who vote yes
$\frac{152}{200} = 0.76$
That is 76% of the people surveyed. The sample was random, so about 76% of all the voters should vote yes too. Watch out: using only the 142 yes votes gives 17,111, choice A. Counting all 20 undecided people gives 19,521, choice D.
Scale up to all the voters
$\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}$
$\frac{152}{200} = \frac{x}{24{,}100}$
$x = 0.76 \times 24{,}100$
Break it into easy parts. 75% is $\frac{3}{4} \times 24{,}100 = 18{,}075$ , and 1% is $24{,}100 \div 100 = 241$ .
$x = 18{,}075 + 241 = 18{,}316$
So the answer is B, 18,316.

Check your answer

Estimate. 76% is a little more than three fourths. Three fourths of 24,000 is 18,000. So the answer should be a little more than 18,000, and 18,316 fits. Also, $18{,}316 \div 24{,}100 = 0.76$ , the same share as the survey.

More questions like this

The original DOE question

Question 77 as printed on page 204 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 204, question 77. Open page 204 of the handbook (PDF)

Question 21 of 57

2024 Test B Math, question 21

Question

$-\frac{1}{49},\ -49,\ -\frac{1}{7},\ -1$

Which number shown above is the greatest?

A $-\frac{1}{49}$
B $-49$
C $-\frac{1}{7}$
D $-1$

Show the answer

Correct answer: A

All the numbers are negative, so they are all less than zero. Consider the absolute value of each number and where it falls on a number line. The negative number closest to 0 will be the greatest.

The negative number with the greatest absolute value is −49, so it has the least value.

The negative number with the next greatest absolute value is −1, so it would fall next in order from least to greatest.

$-49,\ -1$

Compare the absolute values of $-\frac{1}{49}$ and $-\frac{1}{7}$ .

$\frac{1}{49}<\frac{1}{7}$

Therefore, the numbers listed from least to greatest are:

$-49,\ -1,\ -\frac{1}{7},\ -\frac{1}{49}$

So the number $-\frac{1}{49}$ is the greatest.

Question analysis

Four negative numbers are listed: negative 1/49, negative 49, negative 1/7 and negative 1. Students decide which of them is the greatest. Comparing negative fractions with negative whole numbers tests whether students order numbers by their position on the number line or by their size without the sign.

NYS standards: 6.NS.7a inequality as relative position
6.NS.7d absolute value versus order
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Ordering positive and negative rational numbers on a number line is grade 6 content in New York. This is a one-step comparison of four numbers, lighter than typical grade 7 work with signed 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: A very close wrong answer moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

With no arithmetic to do, the question tests how firmly students picture negative numbers on a line. Comparing sizes without the sign picks the number farthest from 0, and comparing the two fractions by habit picks the wrong one of the pair. Placing all four as points left of 0, with −1 between −49 and the fractions, makes the order plain.

A very close wrong answer moderate Choosing the answer
−1/49 and −1/7 differ only in the denominator, and for negative numbers the larger denominator gives the greater value, the reverse of what their sizes suggest. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must bring the number-line fact that a negative number's value falls as its distance from 0 grows, so −49 is the least of the four, not the greatest. (Anderson & Krathwohl, 2001)

Strategy: Picture each number's place on the number line: among negative numbers, the one closest to 0 is the greatest, so compare their distances from 0.

Traps

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

The wrong answers, classified

B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Picks the number with the largest digits, 49, judging its size without the minus sign that puts it farthest below 0.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Knows 1/7 is greater than 1/49 and keeps that order for the negatives, missing that the order reverses on the left of 0.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Places −1/7 and −1/49 below −1, as if dividing a negative number made it more negative, so −1 looks like the greatest.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers: $\text{if } a \lt b \text{, then } -a \gt -b$; a and b are positive numbers. Putting a minus sign on both flips their order. The one with the smaller size is closer to 0, so it is greater. For example, 2 is less than 5, but −2 is greater than −5.

Sort out the question
All four numbers are negative, which means less than 0. You need the greatest one, the one with the highest value. On a number line, greater numbers are farther to the right.
Think about distance from 0
For negative numbers, the one closest to 0 is the greatest. The size of a number without its sign is its absolute value. It is the number’s distance from 0. So look for the smallest size.
$\text{if } a \lt b \text{, then } -a \gt -b$
Compare the sizes
The sizes are 1/49, 49, 1/7 and 1. Both 1/49 and 1/7 have 1 on top. Cutting 1 into 49 pieces gives much smaller pieces than cutting it into 7. So 1/49 is less than 1/7.
$\frac{1}{49} \lt \frac{1}{7} \lt 1 \lt 49$
So 1/49 is the smallest size. That puts $-\frac{1}{49}$ closest to 0.
Put the numbers in order
Flip the order of the sizes to order the negative numbers, from least to greatest.
$-49 \lt -1 \lt -\frac{1}{7} \lt -\frac{1}{49}$
Watch out: −49 has the biggest digits, but it is the least, so choice B is wrong. Watch out: 1/7 is more than 1/49, but −1/7 is less than −1/49. So choice C is wrong.
So the answer is A, $-\frac{1}{49}$ .

Check your answer

Change the fractions to decimals: $-\frac{1}{49} \approx -0.02$ and $-\frac{1}{7} \approx -0.14$ . Think of money you owe. Owing 2 cents is better than owing 14 cents, $1 or $49. So −1/49 is the greatest.

More questions like this

The original DOE question

Question 78 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 2025 Admissions, page 205, question 78. 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 22 of 57

2024 Test B Math, question 22

Question

What decimal is equivalent to $-\frac{13}{11}$ ?

A $-1.18$
B $-1.\overline{118}$
C $-1.\overline{18}$
D $-1.\overline{81}$

Show the answer

Correct answer: C

Calculate the decimal equivalent of $-\frac{13}{11}$ :

$-\frac{13}{11}=-1.18181818...=-1.\overline{18}$

Question analysis

A fractions and decimals question asks which decimal equals negative 13/11. The choices differ in which digits repeat and how the repetition is marked. Students divide 13 by 11, spot the block of digits that repeats, and read the bar notation carefully.

NYS standards: 7.NS.2d fractions to decimals by long division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Converting a fraction to a repeating decimal by long division is a named grade 7 skill. This is one short division, typical of grade 7 work, with care needed only in reading the bar notation.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 15, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Dividing 13 by 11 is quick; the question is really about reading and writing repeating decimals. Three choices differ only in where the bar goes, so students who know the digits run 1.1818… can still pick the wrong bar. Writing out each barred choice for a few digits and comparing it with the quotient is a fast, sure check.

A very close wrong answer moderate Choosing the answer
The bars over 18 and over 81 cover the same two digits, so only checking which digit comes right after the decimal point, 1 before 8, tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Reading bar notation is assumed: the bar marks exactly the block of digits that repeats, and the whole-number 1 never belongs under it. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The long division 13 ÷ 11 has to run until the remainder 2 comes back, which is where the repeating pair is confirmed. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Do the long division of 13 by 11 until a remainder comes back, note which digits repeat from that point, and match that block to the bar in each choice.

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.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
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
Stops the division after two decimal places and treats −1.18 as exact, missing that the digits 1 and 8 go on repeating.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Counts the whole-number 1 into the block, seeing 1, 18, 18 … as 118 repeating, and bars 118, which means −1.118118… rather than −1.1818….
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Bars 81, the pair in the wrong order, which gives −1.8181…, equal to −1 9/11 rather than the −1 2/11 that −13/11 is.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction as division: $\frac{a}{b} = a \div b$; a is the top number, the numerator. b is the bottom number, the denominator. A fraction bar means divide. To turn a fraction into a decimal, divide the top by the bottom.
Bar for repeating digits: $0.\overline{18} = 0.181818\ldots$; The bar sits over the digits that repeat forever, in the order they repeat. The dots mean the pattern goes on and on. Digits that are not under the bar show up only once.

Sort out the question
You need the decimal equal to $-\frac{13}{11}$ . The fraction is negative, so the decimal will be negative too. Work with $\frac{13}{11}$ first. Then put the minus sign back.
Divide 13 by 11
$\frac{a}{b} = a \div b$
Use long division for $13 \div 11$ .
- 11 goes into 13 once, with 2 left over. Write 1 and a decimal point.
- Bring down a 0 to make 20. 11 goes into 20 once, with 9 left over. Write 1.
- Bring down a 0 to make 90. 11 goes into 90 eight times, with 2 left over. Write 8.
So far you have 1.18.
Spot the repeating pattern
The remainder (the amount left over) is 2 again. That is where you were after the first step. So the same digits come back: 1, 8, 1, 8, and so on forever.
$\frac{13}{11} = 1.181818\ldots$
Watch out: −1.18, choice A, stops too soon. The digits never end.
Write it with a bar
$0.\overline{18} = 0.181818\ldots$
A bar goes over only the digits that repeat. Here 18 repeats, starting right after the decimal point. The whole number 1 does not repeat, so it is not under the bar. Put the minus sign back.
$-\frac{13}{11} = -1.\overline{18}$
Watch out: a bar over 81, choice D, means −1.8181…. A bar over 118, choice B, means −1.118118….
So the answer is C, $-1.\overline{18}$ .

Check your answer

Multiply back: $11 \times 1.1818 = 12.9998$ . That is almost exactly 13, and more 18s bring it even closer. Also, 13/11 is a bit more than 1, so a decimal between 1 and 2 makes sense.

More questions like this

The original DOE question

Question 79 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 2025 Admissions, page 205, question 79. Open page 205 of the handbook (PDF)

Question 23 of 57

2024 Test B Math, question 23

Question

Each loaf of bread a baker makes requires $\frac{1}{5}$ ounce of salt. How many loaves can be made if 40 ounces of salt are used?

A8
B20
C45
D200

Show the answer

Correct answer: D

Each loaf of bread requires $\frac{1}{5}$ ounce of salt. Therefore, the baker can make 5 loaves of bread with 1 ounce of salt. Multiply to find the number of loaves that can be made with 40 ounces of salt:

$5\times 40=200$

Question analysis

A baking word problem: each loaf of bread needs 1/5 ounce of salt, and the baker uses 40 ounces. Students find how many loaves that much salt makes, working with a whole number and a unit fraction. Deciding whether to multiply or divide is the whole question.

NYS standards: 6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing a whole number by a fraction in a word problem is a grade 6 standard. This takes one step with a unit fraction, simpler than typical grade 7 rational number problems.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Solution: 1 step
Reading level: Question and choices: about grade 3, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The only decision is whether to multiply or divide, and thinking about one ounce first settles it: one ounce holds five fifths, so it makes five loaves. Students who multiply 40 by 1/5 get a small number, but a tiny amount of salt per loaf should give many loaves, not few. That reasonableness check leaves a single choice.

Turning words into math minor Understanding the question
“How many loaves” from 40 ounces at 1/5 ounce each asks how many fifths fit in 40, a division that many students turn into multiplying by 1/5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work out how many loaves one ounce of salt makes, then scale up to 40 ounces; dividing 40 by 1/5 gives the same count.

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).
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
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 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 40 by 1/5 instead of dividing, finding a fifth of the salt rather than how many fifths of an ounce it holds.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Knows to multiply 40 by 5 but drops a zero, writing the product as 20 instead of 200.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the 5 from 1/5 to the 40 ounces, combining the numbers in the problem without a model of the situation.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal groups: $\text{groups} = \text{total} \div \text{size of one group}$; Total is the whole amount you have. Size of one group is how much each one uses. Dividing tells how many groups fit into the total. Here one group is the salt for one loaf.
Dividing by a unit fraction: $a \div \frac{1}{n} = a \times n$; a is any number. 1 over n is a unit fraction, a fraction with 1 on top. Dividing by it asks how many nths fit into a. Each whole holds n of them, so you multiply by n.

Sort out what you know
Each loaf uses $\frac{1}{5}$ ounce of salt. The baker uses 40 ounces of salt. You need how many loaves that makes. That means how many fifths of an ounce fit into 40 ounces.
Decide to divide
You are splitting the salt into equal groups of 1/5 ounce. Each group makes one loaf. So divide the total salt by the salt for one loaf.
$\text{groups} = \text{total} \div \text{size of one group}$
$\text{loaves} = 40 \div \frac{1}{5}$
Watch out: do not multiply by 1/5. $40 \times \frac{1}{5} = 8$ , choice A, is a fifth of the salt, not a number of loaves.
Divide by the fraction
Each ounce holds 5 fifths, so 1 ounce makes 5 loaves. Dividing by 1/5 is the same as multiplying by 5.
$a \div \frac{1}{n} = a \times n$
$40 \div \frac{1}{5} = 40 \times 5 = 200$
The salt makes 200 loaves. Watch out: keep both zeros. 20, choice B, drops one.
So the answer is D, 200.

Check your answer

Multiply back. 200 loaves at $\frac{1}{5}$ ounce each use $200 \times \frac{1}{5} = 40$ ounces. That is all the salt. Also, each loaf uses less than 1 ounce, so there should be more loaves than ounces. 200 is more than 40.

More questions like this

The original DOE question

Question 80 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 2025 Admissions, page 205, question 80. 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 24 of 57

2024 Test B Math, question 24

Question

What is the value of $\frac{\left(1\frac{5}{7}-1\frac{6}{7}\right)}{\left(3\frac{4}{7}-3\frac{6}{7}\right)}$ ?

A $-\frac{1}{2}$
B $-\frac{1}{7}$
C $\frac{1}{7}$
D $\frac{1}{2}$

Show the answer

Correct answer: D

First, subtract to simplify.

$\frac{1\frac{5}{7}-1\frac{6}{7}}{3\frac{4}{7}-3\frac{6}{7}}=\frac{-\frac{1}{7}}{-\frac{2}{7}}=-\frac{1}{7}\div -\frac{2}{7}$

Follow the process for dividing fractions:

$\left(-\frac{1}{7}\right)\times\left(-\frac{7}{2}\right)=\frac{7}{14}=\frac{1}{2}$

Question analysis

A fractions question: find the value of a quotient whose top is 1 5/7 minus 1 6/7 and whose bottom is 3 4/7 minus 3 6/7. Students subtract mixed numbers that share a denominator and then divide one difference by the other. Keeping track of positive and negative signs through both steps is the main hurdle.

NYS standards: 7.NS.2b rational number division
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students subtract and divide rational numbers, including negatives, and use the sign rules for quotients. This item chains three such operations with easy sevenths, a typical grade 7 exercise.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 3 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 mixed numbers look heavy, but the whole parts cancel and only sevenths remain. What decides the question is sign handling: each difference is negative, and one negative divided by another is positive. Students who stop after the top, subtract instead of divide, or keep a minus sign each find their result among the choices.

Error-prone arithmetic moderate Working it out
Both differences are negative, 1 5/7 − 1 6/7 = −1/7 and 3 4/7 − 3 6/7 = −2/7, and dividing them needs both the sign rule and a reciprocal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Two subtractions must be finished before the division, and a sign lost in either one carries into the final quotient. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out the top and the bottom separately, writing each difference as a signed fraction in sevenths, then divide by multiplying the top by the reciprocal of the bottom and apply the sign rule.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds −1/7 and −2/7 correctly but keeps a negative sign when dividing, forgetting that a negative divided by a negative is positive.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Works out the top, 1 5/7 − 1 6/7 = −1/7, and stops there without dividing it by the bottom difference.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Subtracts the bottom difference from the top one, −1/7 − (−2/7) = 1/7, instead of dividing the two.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.
Dividing two negative numbers: $(-a) \div (-b) = a \div b$; a and b are positive numbers, and b is not zero. A negative number divided by a negative number gives a positive answer. The two minus signs cancel. Multiplying two negatives works the same way.

Sort out the problem
The long fraction bar means divide. So you need the top difference divided by the bottom difference. A difference is the answer to a subtraction. Work out the top and the bottom first, then divide.
Work out the top
Both numbers on top have the same whole number, 1. So the wholes cancel, and you only need $\frac{5}{7} - \frac{6}{7}$ . You take away more than you have, so the answer is negative.
$1\frac{5}{7} - 1\frac{6}{7} = \frac{5}{7} - \frac{6}{7} = -\frac{1}{7}$
Work out the bottom
Both numbers on the bottom have the whole number 3, so those cancel too.
$3\frac{4}{7} - 3\frac{6}{7} = \frac{4}{7} - \frac{6}{7} = -\frac{2}{7}$
Watch out: do not stop after the top. $-\frac{1}{7}$ , choice B, is only part of the problem.
Divide the top by the bottom
Both results are negative. The two minus signs cancel.
$(-a) \div (-b) = a \div b$
$-\frac{1}{7} \div \left(-\frac{2}{7}\right) = \frac{1}{7} \div \frac{2}{7}$
Now flip the second fraction and multiply.
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{1}{7} \times \frac{7}{2} = \frac{7}{14} = \frac{1}{2}$
Avoid the traps and answer
Watch out: keeping a minus sign gives $-\frac{1}{2}$ , choice A. Subtracting the bottom from the top, instead of dividing, gives $\frac{1}{7}$ , choice C.
So the answer is D, $\frac{1}{2}$ .

Check your answer

Think in sevenths. The top is −1 seventh and the bottom is −2 sevenths. That is like $(-1) \div (-2) = \frac{1}{2}$ . Or multiply back: $\frac{1}{2} \times \left(-\frac{2}{7}\right) = -\frac{1}{7}$ , which is the top.

More questions like this

The original DOE question

Question 81 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 2025 Admissions, page 205, question 81. Open page 205 of the handbook (PDF)

Question 25 of 57

2024 Test B Math, question 25

Question

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

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

Show the answer

Correct answer: B

Split the compound inequality into two inequalities.

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

For each inequality, subtract 1 from each side.

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

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

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

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

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

Question analysis

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

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve and graph two-step inequalities one at a time. This compound inequality needs two solved together and a division by a negative number, which reverses the signs, a step many grade 7 classes only touch on.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 12, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 57

2024 Test B Math, question 26

Question

If 1 sind = 0.75 plunk, how many sinds are equivalent to 8 plunks, rounded to the nearest hundredth?

A1.33
B6.00
C7.25
D10.67

Show the answer

Correct answer: D

Set up a proportion:

$\frac{1\text{ sind}}{0.75\text{ plunk}}=\frac{x\text{ sinds}}{8\text{ plunks}}$

$1(8)=0.75x$

$8=\frac{75}{100}x$

$8=\frac{3}{4}x$

$8\left(\frac{4}{3}\right)=x$

$\frac{32}{3}=x$

$10\frac{2}{3}=x$

$10.67\approx x$

Question analysis

A unit conversion question with made-up units: 1 sind equals 0.75 plunk, and students find how many sinds equal 8 plunks, rounded to the nearest hundredth. With unfamiliar unit names there is no everyday sense to lean on, so deciding whether to multiply or divide by 0.75 is the hard part.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with ratio reasoning is grade 6 content, and made-up unit names are a familiar SHSAT twist. Working with a decimal rate and rounding to hundredths put it at about the level of typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
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

Made-up units remove the everyday sense that usually tells students which way a conversion goes, so the relationship must be reasoned out. The most tempting error multiplies by 0.75 and goes the wrong way; another finds the rate for one plunk and stops. A size check settles it: each sind is less than a plunk, so the number of sinds must exceed 8.

Turning words into math moderate Understanding the question
“1 sind = 0.75 plunk” says a sind is smaller than a plunk, so 8 plunks hold more than 8 sinds; seeing that decides between dividing and multiplying by 0.75. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
8 ÷ 0.75 is easiest as 8 × 4/3 = 32/3, and the repeating decimal 10.666… must then be rounded to the nearest hundredth. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Set up a proportion with sinds over plunks on both sides, put the 8 plunks in the matching place, solve for the sinds, and round to the nearest hundredth.

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.
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 how many sinds make one plunk, 1 ÷ 0.75 ≈ 1.33, and stops before scaling up to 8 plunks.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 8 by 0.75 instead of dividing, converting in the wrong direction and finding the plunks in 8 sinds.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts 0.75 from 8, combining the two numbers in the question without setting up a conversion.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a, b, c and d are numbers or letters, and b and d are not zero. When two fractions are equal, multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.
Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b over c is the fraction you divide by. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down. It asks how many of that fraction fit into a.

Sort out what you know
Sinds and plunks are made-up units, like inches and feet. One sind equals 0.75 plunk, so a sind is smaller than a plunk. You need how many sinds equal 8 plunks. Round to the nearest hundredth, two places after the decimal point.
Decide if the answer is more than 8
A sind is only 3/4 of a plunk. So it takes more than 8 sinds to make 8 plunks. The answer must be more than 8. Watch out: $8 \times 0.75 = 6$ , choice B, goes the wrong way. That is the number of plunks in 8 sinds.
Set up a proportion
A proportion says that two ratios are equal. Put sinds on top and plunks on the bottom on both sides. Let x be the number of sinds.
$\frac{1}{0.75} = \frac{x}{8}$
$\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$
$1 \times 8 = 0.75 \times x$
So $0.75x = 8$ .
Divide by 0.75
0.75 is the same as $\frac{3}{4}$ . Divide both sides by it. To divide by a fraction, multiply by its reciprocal.
$a \div \frac{b}{c} = a \times \frac{c}{b}$
$x = 8 \div \frac{3}{4} = 8 \times \frac{4}{3} = \frac{32}{3} = 10.666\ldots$
Watch out: 1.33, choice A, is the number of sinds in just 1 plunk. Keep going to 8 plunks.
Round to the nearest hundredth
Look at the third digit after the point, the thousandths digit. It is 6, which is 5 or more. So round the hundredths digit up, from 6 to 7.
$10.666\ldots \approx 10.67$
So the answer is D, 10.67.

Check your answer

Multiply back. Each sind is 0.75 plunk, so 10.67 sinds are $10.67 \times 0.75 \approx 8.00$ plunks. That matches the 8 plunks. The answer is also more than 8, as it must be.

More questions like this

The original DOE question

Question 83 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 83. Open page 206 of the handbook (PDF)

Question 27 of 57

2024 Test B Math, question 27

Question

Simplify:

$8x-(7+2.5x)+2$

A $5.5x-9$
B $5.5x-5$
C $10.5x-9$
D $10.5x-5$

Show the answer

Correct answer: B

$8x-(7+2.5x)+2$

First, distribute −1 over $7+2.5x$ :

$8x-7-2.5x+2$

Then combine the like terms:

$5.5x-5$

The simplified form of the expression is $5.5x-5$ .

Question analysis

An algebra question asks students to simplify 8x minus the quantity 7 + 2.5x, plus 2. They remove the parentheses and combine the x terms and the constants. Decimal coefficients and a subtraction in front of a group make sign errors likely.

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 Simplifying a linear expression with a subtracted group and decimal coefficients is standard grade 7 work. This is a typical two-step textbook exercise, with the usual sign trap and nothing extra.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 16+, 6 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question checks one skill, removing parentheses that follow a minus sign, and its choices are built from the two ways to get it wrong. A student who applies the minus to too few terms gets 10.5x, and one who extends it past the parentheses gets −9. Handling the x terms and the constants as two separate small checks makes each slip easy to catch.

Error-prone arithmetic moderate Working it out
The minus sign before the parentheses must turn + 2.5x into − 2.5x and 7 into − 7, while the + 2 written after the parentheses keeps its sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
The choices pair two x terms, 5.5x and 10.5x, with two constants, −9 and −5, so each wrong one differs from the right one in a single part. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Change the minus in front of the parentheses into a minus on each term inside, then combine like terms, keeping the + 2 outside the group as it is.

Traps

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

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Carries the minus sign past the parentheses onto the + 2 as well, so the constants become −7 − 2 = −9.
C Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Makes both slips at once, keeping + 2.5x so the x terms add to 10.5x, and subtracting the outside 2 so the constants give −9.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Applies the minus sign to the 7 but not to 2.5x, so the x terms add to 10.5x while the constants come out right.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a group in parentheses: $-(a + b) = -a - b$; a and b are the terms inside the parentheses. A minus sign in front means take away the whole group. So each term inside is taken away, and its sign flips. Terms outside the parentheses keep their own signs.
Combining like terms: $ax + bx = (a + b)x$; a and b are numbers, and x is the letter. Like terms have the same letter part. You add or subtract their numbers and keep the letter. Plain numbers combine only with plain numbers.

Sort out the problem
You need to simplify $8x - (7 + 2.5x) + 2$ . To simplify means to write it in its shortest form. The tricky part is the minus sign in front of the parentheses. It takes away everything inside.
Remove the parentheses
$-(a + b) = -a - b$
$8x - (7 + 2.5x) + 2 = 8x - 7 - 2.5x + 2$
Both 7 and 2.5x get a minus sign. The + 2 is outside the parentheses, so it stays + 2. Watch out: 2.5x must become −2.5x. Leaving it as + 2.5x gives 10.5x, as in choices C and D.
Combine the x terms
Like terms have the same letter part. Put the two x terms together.
$ax + bx = (a + b)x$
$8x - 2.5x = (8 - 2.5)x = 5.5x$
Combine the plain numbers
$-7 + 2 = -5$
Watch out: the + 2 is not inside the parentheses. Making it −2 gives −9, as in choice A.
Put the parts together: $5.5x - 5$ . So the answer is B, $5.5x - 5$ .

Check your answer

Pick x = 2 and try both forms. The first form: $16 - (7 + 5) + 2 = 6$ . Choice B: $5.5 \times 2 - 5 = 11 - 5 = 6$ . They match. Choice D gives $10.5 \times 2 - 5 = 16$ , so D is wrong.

More questions like this

The original DOE question

Question 84 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 84. Open page 206 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 28 of 57

2024 Test B Math, question 28

Question

A community center has two 30-member exercise groups. The director recorded the ages of each group’s members in the box plots.

$Two box plots titled AGES OF GROUPS over one number line from 15 to 33, with a tick at every whole number and labels at 15, 18, 21, 24, 27, 30 and 33. Group P: whiskers from 17 to 33, box from 19 to 27, median line at 24. Group Q: whiskers from 15 to 29, box from 17 to 26, median line at 21.$

Which statement about the two groups can be verified based on the box plots?

ABoth groups have at least one member who is exactly 29 years old.
BGroup P has more members who are at least 17 years old than Group Q has.
CBoth groups have an equal number of members who are between 21 and 24 years old.
DGroup Q has fewer members who are between 19 and 27 years old than Group P has.

Show the answer

Correct answer: B

Groups P and Q have an equal number of members. All the members of Group P are 17 or older, but only about $\frac{3}{4}$ of the members of Group Q are 17 or older.

Question analysis

Box plots show the ages of the members of two 30-member exercise groups at a community center, Group P and Group Q. Students decide which statement comparing the groups can be verified from the plots alone. The challenge is knowing what a box plot guarantees about individual ages and counts, and what it leaves unknown.

NYS standards: 7.SP.1 box plots and interquartile range
7.SP.4 comparative inferences about populations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students read box plots and compare medians and spreads. Deciding which claims two box plots can prove, from what the five key values guarantee about counts, is a harder reasoning task 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 3, strategic thinking · mainly evaluate
What makes it hard: Every choice must be checked major
Something you must already know moderate
Reading a figure, table or graph moderate
Solution: 3 steps
Reading level: Question and choices: about grade 5, 91 words.
Students, last 30 days: 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 question about evidence: the right statement is the only one the box plots guarantee, while the others may or may not be true. Students who treat the plots as pictures of every member, assuming ages fill each whisker or spread evenly through each box, find support for claims that cannot be checked. Asking what must hold for any 30 ages with these summaries separates the verifiable claim from the merely plausible ones.

Every choice must be checked major Choosing the answer
Three of the statements are not false but unverifiable, so each must be tested against what the five-number summaries guarantee, and none can be dismissed at a glance. (Embretson & Wetzel, 1987)
Something you must already know moderate Working it out
Students must know that the minimum and maximum are actual ages, that each quarter of a plot holds about a quarter of the 30 members, and that a whisker shows nothing in between. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph moderate Understanding the question
The scale is labeled only every 3 years, so values such as 17, 19, 26 and 29 have to be read from unlabeled ticks on two separate plots. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List each group's minimum, first quartile, median, third quartile and maximum, then test every statement against only what those five values guarantee about the 30 ages.

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
Sees that Group Q's maximum is exactly 29 and that 29 also lies in Group P's upper whisker, assuming someone in Group P is that age.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the two medians, 21 and 24, and assumes the stretch between them holds the same number of members in both groups.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Knows Group P's box from 19 to 27 holds half its members, then assumes ages spread evenly so Group Q has fewer there, which its plot cannot show.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Members in each part of a box plot: $\text{members in each part} \approx \frac{1}{4} \times n$; n is the number of members in a group, here 30. A box plot splits the data into four parts at the minimum, Q1, the median, Q3 and the maximum. Each part holds about one fourth of the data. It does not show how the ages spread out inside a part.

Read the two box plots
A box plot shows five numbers. The minimum is the youngest age, and the maximum is the oldest. Q1 is the left edge of the box, and Q3 is the right edge. The median is the line inside the box.
Group Min Q1 Median Q3 Max
P 17 19 24 27 33
Q 15 17 21 26 29
Each group has 30 members.
Know what can be verified
To verify a statement means to know for sure that it is true. A box plot shows only five numbers, not every age. The minimum and maximum are real ages of real members. The other ages are hidden inside the box and whiskers.
Check choice B
In Group P, the youngest member is 17. So all 30 members are at least 17. In Group Q, the youngest member is 15, and 15 is under 17. So at most 29 members of Group Q are at least 17.
Group P has 30 such members, and Group Q has 29 or fewer. So choice B is true for sure.
Rule out choices A and C
A: Group Q’s oldest member is 29, so Group Q has a 29-year-old. But in Group P, 29 is inside the right whisker. A whisker does not show single ages. Watch out: 29 is in Group P’s range, but maybe no one in Group P is exactly 29.
C: The plots do not show how many members are between 21 and 24 in either group.
Rule out choice D and answer
Group P’s box, from 19 to 27, is two of its four parts.
$\text{members in each part} \approx \frac{1}{4} \times 30 = 7.5$
So about 15 members of Group P are between 19 and 27. But 19 and 27 fall in the middle of Group Q’s parts, and ages inside a part can bunch up in any way. Group Q could have more or fewer. Watch out: D assumes the ages are spread out evenly, and a box plot never tells you that.
So the answer is B.

Group	Min	Q1	Median	Q3	Max
P	17	19	24	27	33
Q	15	17	21	26	29

Check your answer

Choice B uses only the two minimums, which are real ages. Group P’s youngest member is 17, and Group Q’s is 15. That alone proves it. Each other choice needs ages inside a box or whisker, which a box plot does not show.

More questions like this

The original DOE question

Question 85 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 85. Open page 207 of the handbook (PDF)

Question 29 of 57

2024 Test B Math, question 29

Question

Jar Q contains 12 balls, and Jar R contains 8 balls. The probability of drawing a yellow ball at random from Jar Q is $\frac{1}{3}$ . The probability of drawing a yellow ball at random from Jar R is $\frac{3}{4}$ . All the balls are poured into an empty jar, Jar S. What is the probability of drawing a yellow ball at random from Jar S?

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

Show the answer

Correct answer: C

Jar Q contains 12 balls, and the probability of drawing a yellow ball at random is $\frac{1}{3}$ .

$\frac{1}{3}\times\frac{4}{4}=\frac{4}{12}$

There are 4 yellow balls in Jar Q.

Jar R contains 8 balls, and the probability of drawing a yellow ball at random is $\frac{3}{4}$ .

$\frac{3}{4}\times\frac{2}{2}=\frac{6}{8}$

There are 6 yellow balls in Jar R.

Jar S contains all the balls from Jar Q and Jar R.

$12+8=20$

There are 20 balls in Jar S.

$4+6=10$

There are 10 yellow balls in Jar S.

$\frac{10}{20}\div\frac{10}{10}=\frac{1}{2}$

The probability of drawing a yellow ball at random from Jar S is $\frac{1}{2}$ .

Question analysis

Two jars hold different numbers of balls and give different chances of drawing a yellow ball, and then everything is poured into a third jar. Students find the probability of drawing yellow from the combined jar. The tricky part is that the jars are different sizes, so their probabilities cannot be merged without first counting the yellow balls.

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 Grade 7 probability tasks usually give the counts and ask for a fraction. Here each jar's probability must be turned back into a count of yellow balls before the jars are pooled, a reverse step that typical classwork rarely asks for.
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 minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 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

The arithmetic is light; the hurdle is seeing that chances from jars of different sizes cannot be combined as fractions. Every wrong choice works on 1/3 and 3/4 directly, by multiplying them, by keeping one, or by adding tops and bottoms, while the right path runs through counts of balls. Asking how many yellow balls each jar holds before anything is poured avoids all three.

Turning words into math moderate Understanding the question
Each jar's probability, 1/3 of 12 balls and 3/4 of 8 balls, has to be turned back into a count of yellow balls, 4 and 6, before the two jars can be combined. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The straight-across 4/7, about 0.57, sits just above the right 1/2, and both lie between the jars' chances of 1/3 and 3/4, so a rough estimate does not rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Turn each jar's probability into a number of yellow balls by multiplying it by that jar's total, then divide all the yellow balls by all the balls.

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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Multiplies the jars' chances, 1/3 × 3/4 = 1/4, the rule for two draws that must both be yellow, though only one ball is drawn from the mixture.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Repeats Jar Q's 1/3 from the stem, as if the bigger jar's chance carried over to the mixture unchanged.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the fractions straight across, (1 + 3)/(3 + 4) = 4/7, treating 1/3 and 3/4 as if they were counts of balls rather than chances.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{yellow}) = \frac{\text{yellow balls}}{\text{all balls}}$; P(yellow) is the probability, or chance, of drawing a yellow ball. It is a number from 0 to 1. The top counts the yellow balls. The bottom counts all the balls. It works because every ball is equally likely to be drawn.

Sort out what you know
Jar Q holds 12 balls. The probability, or chance, of drawing yellow from it is $\frac{1}{3}$ .
Jar R holds 8 balls. The chance of drawing yellow from it is $\frac{3}{4}$ .
All the balls are poured into Jar S. You need the chance of drawing yellow from Jar S.
Count the yellow balls in Jar Q
The jars are different sizes, so you can’t mix their chances directly. Count the yellow balls instead. The chance tells what fraction of the balls are yellow. So multiply it by the number of balls.
$\text{yellow balls} = \frac{1}{3} \times 12 = 4$
Jar Q has 4 yellow balls.
Count the yellow balls in Jar R
Do the same for Jar R. Three fourths of its 8 balls are yellow.
$\text{yellow balls} = \frac{3}{4} \times 8 = 6$
Jar R has 6 yellow balls.
Count the balls in Jar S
Jar S gets every ball from both jars. Add all the balls. Then add the yellow balls.
$12 + 8 = 20$
$4 + 6 = 10$
Jar S has 20 balls, and 10 of them are yellow.
Find the chance for Jar S
Use the probability formula with the Jar S counts.
$P(\text{yellow}) = \frac{10}{20} = \frac{1}{2}$
Watch out: Adding the tops and the bottoms, 1 + 3 over 3 + 4, gives $\frac{4}{7}$ , choice D. Those numbers are parts of fractions, not balls. Multiplying $\frac{1}{3} \times \frac{3}{4}$ gives $\frac{1}{4}$ , choice A. That is the chance of two yellow draws, one from each jar.
So the answer is C, $\frac{1}{2}$ .

Check your answer

Half of the 20 balls is 10, and 4 + 6 = 10 yellow balls. That matches. Also, the mixed jar’s chance should land between the two jars’ chances, $\frac{1}{3}$ and $\frac{3}{4}$ . One half does.

More questions like this

The original DOE question

Question 86 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 86. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 57

2024 Test B Math, question 30

Question

GIA’S MIX

Snack	Number in Bag
Pretzels	12
Raisins	9

Gia and her friends counted the numbers of pretzels and raisins in their snack mix. The table above shows the numbers in Gia’s mix. Which other mix has a ratio of pretzels to raisins in the same proportional relationship as Gia’s mix?

A
ADELE’S MIX

Snack Number in Bag
Pretzels 6
Raisins 18
B
CARL’S MIX

Snack Number in Bag
Pretzels 15
Raisins 12
C
ISABEL’S MIX

Snack Number in Bag
Pretzels 18
Raisins 15
D
TREVON’S MIX

Snack Number in Bag
Pretzels 16
Raisins 12

Snack	Number in Bag
Pretzels	6
Raisins	18

Snack	Number in Bag
Pretzels	15
Raisins	12

Snack	Number in Bag
Pretzels	18
Raisins	15

Snack	Number in Bag
Pretzels	16
Raisins	12

Show the answer

Correct answer: D

The ratio of pretzels to raisins in Gia’s bag of snack mix, $12:9$ , is equivalent to the ratio of pretzels to raisins in Trevon’s bag, $16:12$ .

$\frac{12}{9}=\frac{16}{12}=\frac{4}{3}$

Question analysis

Gia's snack mix table lists how many pretzels and raisins are in her bag, and four friends' tables list theirs. Students pick the mix whose ratio of pretzels to raisins matches Gia's. It tests equivalent ratios and proportional relationships, and some mixes look close at a glance, so each ratio has to be tested exactly.

NYS standards: 7.RP.2a testing for proportionality
6.RP.3a tables of equivalent ratios
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Testing whether two ratios are equivalent is core grade 7 proportional reasoning. The numbers are small whole numbers in four short tables, so the work is routine, though every table has to be checked against Gia's.
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 minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 77 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

A short ratio check made harder only by an additive pull: Carl's and Isabel's mixes keep the gap of 3 between pretzels and raisins that Gia's mix has, so students who compare differences instead of ratios see two matches. Reducing every table to lowest terms, Gia's 12:9 to 4:3, shows that exactly one mix keeps the ratio.

Every choice must be checked minor Choosing the answer
Each of the four tables has to be reduced in turn, to 1:3, 5:4, 6:5 and 4:3, because none of them can be ruled out at a glance. (Embretson & Wetzel, 1987)
Reading a figure, table or graph minor Understanding the question
The counts sit in five separate tables, and pretzels must stay first in every ratio, in the same order as Gia's 12 pretzels to 9 raisins. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Simplify the pretzels-to-raisins ratio in Gia's table to lowest terms, then simplify each other mix's ratio in the same order and compare.

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: weak
Halves Gia's 12 pretzels and doubles her 9 raisins, which keeps the product 108 the same but changes the ratio from 4:3 to 1:3.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds 3 to both of Gia's counts, 12 + 3 and 9 + 3, keeping the difference of 3 instead of the ratio; 15:12 reduces to 5:4.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds 6 to both counts, giving 18 and 15, so pretzels still outnumber raisins by 3, but 18:15 reduces to 6:5, not 4:3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent ratios: $a : b = (a \div k) : (b \div k)$; a and b are the two amounts being compared, in order. k is any number that divides both evenly. Dividing both amounts by the same number keeps the ratio the same. When no number but 1 divides both, the ratio is in simplest form.

Sort out the question
A ratio compares two amounts. Gia’s mix has 12 pretzels and 9 raisins. So her ratio of pretzels to raisins is 12 : 9.
The “same proportional relationship” means the ratios are equal once you simplify them. For example, 2 : 1 and 6 : 3 are equal. Always keep pretzels first and raisins second.
Simplify Gia’s ratio
Both 12 and 9 can be divided by 3. So divide both by 3.
$12 : 9 = (12 \div 3) : (9 \div 3) = 4 : 3$
Gia’s mix has 4 pretzels for every 3 raisins.
Simplify each other mix
Do the same for each choice. Divide both numbers by the biggest number that goes into both.
Choice Pretzels : raisins Divide by Simplest form
A, Adele 6 : 18 6 1 : 3
B, Carl 15 : 12 3 5 : 4
C, Isabel 18 : 15 3 6 : 5
D, Trevon 16 : 12 4 4 : 3
Pick the matching mix
Only Trevon’s mix simplifies to 4 : 3, the same as Gia’s.
Watch out: Carl’s and Isabel’s mixes each have 3 more pretzels than raisins, just like Gia’s. But adding the same number to both amounts changes the ratio. Equal ratios come from multiplying or dividing. That trap leads to choice B or C.
So the answer is D, Trevon’s mix.

Choice	Pretzels : raisins	Divide by	Simplest form
A, Adele	6 : 18	6	1 : 3
B, Carl	15 : 12	3	5 : 4
C, Isabel	18 : 15	3	6 : 5
D, Trevon	16 : 12	4	4 : 3

Check your answer

Go the other way. Multiply both parts of 4 : 3 by 4. You get 16 : 12, which are Trevon’s exact numbers. You can also divide pretzels by raisins. Both $12 \div 9$ and $16 \div 12$ equal $\frac{4}{3}$ .

More questions like this

The original DOE question

Question 87 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 87. Open page 208 of the handbook (PDF)

Question 31 of 57

2024 Test B Math, question 31

Question

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

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

Show the answer

Correct answer: C

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

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

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

Then simplify the expression:

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

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

$12x+24x+10-x$

$12x+24x-x+10$

$35x+10$

The simplified expression is $35x+10$ .

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. Terms are the parts joined by plus or minus signs. Multiply a by each term inside. It works because 2 groups of (12x + 5) hold 2 of the 12x and 2 of the 5.

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

Check your answer

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

More questions like this

The original DOE question

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

Question 32 of 57

2024 Test B Math, question 32

Question

Two circular plates have the dimensions shown.

$Two circles side by side. The larger circle has a dashed diameter labeled 18 in., and the smaller circle has a dashed diameter labeled 12 in.$

What is the difference between the areas of the two plates, in square inches?

A $6\pi$
B $9\pi$
C $45\pi$
D $180\pi$

Show the answer

Correct answer: C

First, determine the area of each circle using the formula $A=\pi r^{2}$ :

A circle with a diameter of 18 inches has a radius of 9 inches, so the area is $\pi(9)^{2}=81\pi$ .

A circle with a diameter of 12 inches has a radius of 6 inches, so the area is $\pi(6)^{2}=36\pi$ .

The difference between the two areas is $81\pi-36\pi=45\pi$ .

Question analysis

Two circular plates are drawn with their diameters marked in inches, and students find the difference between their areas in square inches, in terms of pi. It tests the area of a circle. The diameters, not the radii, are marked, so a step comes before the area formula, which the test does not provide.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a circle's area is a grade 7 standard, and this asks for two areas and their difference. It matches typical classwork: the formula must be recalled and each diameter halved first, with no further twist.
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
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two area calculations and a subtraction, kept in terms of π, make this routine for students who know the area formula. Points are lost by using the labeled diameters as radii, by subtracting circumferences, or by subtracting the radii before squaring, and each of these has a listed answer. Writing both areas, 81π and 36π, before subtracting avoids all three.

Something you must already know moderate Working it out
The area formula A = πr² is not given, and the figure labels each plate by its diameter, so 18 and 12 must be halved to radii of 9 and 6 first. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The dashed lines run all the way across each circle, which marks them as diameters; reading the 18 in. and 12 in. as radii leads to a listed choice. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Halve each diameter to get the radius, find each plate's area as pi times the radius squared, and subtract the smaller area from the larger.

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

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Subtracts the circumferences, 18π − 12π = 6π, using πd, the distance around a circle, in place of the area formula.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the radii first, 9 − 6 = 3, and then squares, π(3)² = 9π, as if the difference of two squares were the square of the difference.
D Wrong formula (King et al., 2004)
Reveals a slip Expected pull: strong
Uses the labeled diameters as radii, π(18)² − π(12)² = 324π − 144π = 180π, forgetting to halve them first.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle: $A = \pi r^2$; A is the area, the flat space inside the circle. r is the radius, the distance from the center to the edge. It is half the diameter. π (pi) is a fixed number, about 3.14. r² means r × r. Use it for the space inside any circle.

Sort out the question
Each dashed line goes all the way across its circle, through the center. A line like that is a diameter. The big plate has a diameter of 18 inches. The small plate has a diameter of 12 inches.
You need the difference between the two areas. That means the big area minus the small area, in square inches.
Find each radius
The area formula uses the radius, not the diameter. The radius is half the diameter.
$18 \div 2 = 9$
$12 \div 2 = 6$
The radii are 9 inches and 6 inches. Watch out: If you use 18 and 12 as the radii, you get $180\pi$ , choice D.
Find the big plate’s area
$A = \pi r^2 = \pi \times 9^2 = \pi \times 81 = 81\pi$
The big plate’s area is $81\pi$ square inches. Keep π as a symbol, since all the choices use π.
Find the small plate’s area
$A = \pi \times 6^2 = \pi \times 36 = 36\pi$
The small plate’s area is $36\pi$ square inches. Squaring means multiplying a number by itself, so $6^2 = 6 \times 6 = 36$ .
Subtract the two areas
$81\pi - 36\pi = 45\pi$
Subtract the numbers in front of π, just as 81 apples minus 36 apples is 45 apples. The difference is $45\pi$ square inches.
Watch out: Don’t subtract the radii first. Squaring 9 − 6 = 3 gives $9\pi$ , choice B. Subtracting the distances around the plates, $18\pi - 12\pi$ , gives $6\pi$ , choice A. That uses the wrong formula.
So the answer is C, $45\pi$ .

Check your answer

Estimate with π about 3. The big plate is about $3 \times 81 = 243$ square inches. The small one is about $3 \times 36 = 108$ . The difference is about 135, and $45 \times 3 = 135$ . It matches.

More questions like this

The original DOE question

Question 89 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 89. Open page 208 of the handbook (PDF)

Question 33 of 57

2024 Test B Math, question 33

Question

If $x=-4$ , what is the sum of x and 10% of $|x|$ ?

A−4.4
B−3.6
C0.4
D4.4

Show the answer

Correct answer: B

First, determine 10% of $|x|$ .

$0.10\bullet|x|=$

$0.10\bullet|-4|=$

$0.10\bullet 4=0.4$

Then determine the sum of x and 10% of $|x|$ .

$-4+0.4=-3.6$

The sum is −3.6.

Question analysis

With x equal to a negative number, students find the sum of x and 10 percent of the absolute value of x. It tests absolute value, a simple percent and adding numbers with different signs. The tricky part is keeping track of which term is negative and which is not.

NYS standards: 7.NS.1b rational number addition
6.NS.7c absolute value as distance
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 add positive and negative decimals and find simple percents. This joins those skills with an absolute value in two short steps, so it matches typical practice, with the sign of each term as the main hazard.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

The question is short, but it asks students to treat the same number two ways: x keeps its negative sign while |x| drops it. Each wrong choice gets the sign wrong in one place, by taking 10% of −4, by making both terms positive, or by stopping at the positive 0.4. Writing the two terms separately, −4 and +0.4, before adding prevents each slip.

Error-prone arithmetic moderate Working it out
The sum mixes a negative integer with a positive decimal, −4 + 0.4, and the result has to keep the negative sign while moving 0.4 toward zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must know that |−4| = 4, so the 10% is taken of a positive number even though x itself is negative. (Anderson & Krathwohl, 2001)

Strategy: Evaluate the absolute value first, take 10 percent of that positive number, and then add the result to x itself, keeping x's negative 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.)
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Takes 10% of x instead of |x|, getting −0.4, and adds it to −4 to reach −4.4, overlooking the absolute value bars.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds 10% of |x| = 0.4 correctly and stops there, without adding x as the word sum requires.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Uses 4 for x as well as for |x|, so adds 4 + 0.4 = 4.4 and loses the negative sign of x itself.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{distance from } a \text{ to } 0$; The two bars mean absolute value. a is any number. Its absolute value is how far it is from 0 on the number line. A distance is never negative, so |−4| = 4.
Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get. Use it to find a percent of any number.

Sort out the question
You know that x = −4. You need the sum of two things: x itself, and 10% of |x|. A sum means you add.
The bars around x mean absolute value. So the 10% is taken of |x|, not of x.
Find the absolute value
−4 is 4 steps from 0 on the number line. So its absolute value is 4. An absolute value is never negative.
$|x| = |-4| = 4$
Take 10 percent of it
$\text{part} = \frac{10}{100} \times 4 = 0.4$
So 10% of |x| is 0.4.
Watch out: If you take 10% of x instead, you get −0.4. Then −4 + (−0.4) = −4.4, choice A. The bars make the number positive first.
Add x and the result
Now add x, which is still −4. Its own minus sign stays.
$-4 + 0.4 = -3.6$
Start at −4 on the number line. Move 0.4 to the right, toward 0. You land on −3.6.
Watch out: Stopping at 0.4 gives choice C. Using 4 in place of x gives 4.4, choice D.
So the answer is B, −3.6.

Check your answer

Adding a small positive number should move −4 a little toward 0. The answer −3.6 is just to the right of −4, so that fits. Also, $-3.6 - (-4) = 0.4$ , the amount you added.

More questions like this

The original DOE question

Question 90 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 90. Open page 209 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 34 of 57

2024 Test B Math, question 34

Question

The volume of a cube is 512 cubic centimeters. What is the length, in centimeters, of one edge of this cube?

A8
B $42\frac{2}{3}$
C $85\frac{1}{3}$
D128

Show the answer

Correct answer: A

The volume of a cube is $V=s^{3}$ , where s is the length of one edge.

So $512=s^{3}$ , and $s=\sqrt[3]{512}$ .

If you are not sure how to find the cube root, you can estimate.

We know $10\times 10\times 10=10^{3}=1{,}000$ , so $\sqrt[3]{1{,}000}=10$ .

Since 512 is less than 1,000, the answer must be less than 10. The only given option less than 10 is 8. Double-check to make sure this is correct: $8^{3}=8\times 8\times 8=512$ .

Question analysis

A cube has a volume of 512 cubic centimeters, and students work backward to find the length of one edge. It tests the volume formula for a cube and the idea of a cube root. The tricky part is going from a volume to a length, the reverse of the usual volume problem.

NYS standards: 8.EE.2 square and cube roots
6.G.5 perfect squares and perfect cubes
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 volume work goes from edge lengths to a volume. Going backward needs a cube root, a grade 8 idea, and 512 is beyond the perfect cubes students are expected to know, though checking a guess by multiplying keeps it short.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: 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

This is a one-fact question: a cube's volume is its edge used three times as a factor. Students who remember that need only the number whose cube is 512, and the choices help, because only one is under 10 while 10 × 10 × 10 = 1,000. The wrong choices divide 512 by numbers tied to other measures of a cube or a square, so they catch students who do not recall the formula.

Something you must already know moderate Working it out
Neither V = s³ nor the cube root of 512 is given, so students must recall the formula and know, or find by trial, that 8 × 8 × 8 = 512. (Anderson & Krathwohl, 2001)

Strategy: Write the volume as edge × edge × edge, then find the number that, multiplied by itself three times, gives 512, and confirm it by multiplying.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 512 by 12, the number of edges, as if the volume were the total length of all the edges, giving 42 2/3.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 512 by 6, the number of faces, mixing up volume with surface area, giving 85 1/3.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 512 by 4, as when a square's side is found from its perimeter, instead of finding the number whose cube is 512.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube: $V = s^3$; V is the volume, the space inside, in cubic units. s is the length of one edge. s³ means s × s × s. A cube’s length, width and height are all s, so length × width × height becomes s × s × s.

Sort out the question
The cube’s volume is 512 cubic centimeters. Volume is the space inside a solid. You need the length of one edge.
A cube has 12 edges, and they are all the same length. So one edge tells you all of them.
Write the volume formula
$V = s^3$
$512 = s \times s \times s$
You need a number that gives 512 when you multiply it by itself three times. That number is called the cube root of 512.
Try numbers to find the edge
Start with an easy one: $10 \times 10 \times 10 = 1{,}000$ . That is too big, so the edge is less than 10. Try 8.
$8 \times 8 \times 8 = 64 \times 8 = 512$
It works. Each edge is 8 centimeters.
Rule out the other choices
Watch out: The volume is not shared out among the faces or the edges. Dividing 512 by the 6 faces gives $85\frac{1}{3}$ , choice C. Dividing by the 12 edges gives $42\frac{2}{3}$ , choice B. Dividing by 4 gives 128, choice D.
So the answer is A, 8.

Check your answer

Test the biggest choice. $128 \times 128 \times 128$ is over 2 million, far more than 512. Only an edge less than 10 can work, and 8 is the only choice less than 10. And $8^3 = 512$ .

More questions like this

The original DOE question

Question 91 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 91. Open page 209 of the handbook (PDF)

Question 35 of 57

2024 Test B Math, question 35

Question

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

A4.0° F
B8.0° F
C8.4° F
D8.8° F

Show the answer

Correct answer: B

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

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

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

Therefore, the temperature on Monday was 80° F.

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually apply a percent to a known amount. Here Monday's temperature must first be recovered from a percent increase and then a second percent applied to it, a reverse step most grade 7 students find unfamiliar.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 54 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 36 of 57

2024 Test B Math, question 36

Question

The integers $(n-1)$ , n, and $(n+1)$ are factors of 168. What is the greatest possible value of n?

Show the answer

Correct answer: B

First, determine all possible factors of 168. One method of doing this is to use a table to list pairs of factors.

Factors of 168
1	168
2	84
3	56
4	42
6	28
7	24
8	21
12	14

Then determine the group of three sequential factors $(n-1)$ , n, and $(n+1)$ where n is as large as possible. This group is 6, 7, and 8, where $n=7$ .

Question analysis

Three consecutive integers, written as n − 1, n and n + 1, must all be factors of 168, and students find the greatest possible value of n. It tests factors and consecutive integers. The catch is that a number can be a factor while its neighbors are not, so every member of the run needs checking.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Listing factors is grade 6 work, but here students must read n − 1, n and n + 1 as three consecutive integers and search the factor list for the largest such run. Pairing algebraic notation with factors matches grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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

Once the factors of 168 are listed in order, 1, 2, 3, 4, 6, 7, 8, 12 and up, the runs of three in a row stand out, and the last of them is 6, 7, 8. Students lose points by stopping at the first run they see or by checking only n itself. Testing each choice with its two neighbors, starting from the largest, settles it in under a minute.

Variables instead of numbers minor Understanding the question
The stem names the numbers only as (n − 1), n and (n + 1), which has to be read as three whole numbers in a row that all divide 168. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Each candidate must pass three divisibility tests at once, for itself and for both neighbors, while the search continues for the greatest such n. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: List the factor pairs of 168 in order, then look along the list for three whole numbers in a row; or test each choice together with its two neighbors.

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: strong
Finds the run 2, 3, 4 among the small factors and stops searching, missing the later run 6, 7, 8 that 168 also contains.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Checks only that 14 divides 168, since 12 × 14 = 168, without testing its neighbors; neither 13 nor 15 is a factor.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Takes the largest factor below 168 itself, 84, for the greatest n, ignoring that 83 and 85 would also have to divide 168.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair: $a \times b = 168$; a and b are whole numbers. If a times b equals 168, then a and b are both factors of 168. A factor divides a number with no remainder. Listing every pair, from small a upward, finds every factor.

Sort out the question
A factor of 168 divides 168 with no remainder. The numbers n − 1, n and n + 1 are three whole numbers in a row, such as 6, 7 and 8. The middle one is n.
All three must be factors of 168. You need the greatest n that works.
List the factor pairs of 168
Try each whole number from 1 up. When it divides 168, write it with its partner. Stop when the pairs meet, after 12 and 14.
Small factor Partner
1 168
2 84
3 56
4 42
6 28
7 24
8 21
12 14
Look for three in a row
Here are all the factors in order:
1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168
The runs of three in a row are 1, 2, 3, then 2, 3, 4, then 6, 7, 8. No other three factors sit side by side.
Pick the greatest middle number
The middle numbers of the runs are 2, 3 and 7. The greatest is 7, so n = 7.
Watch out: Stopping at the run 2, 3, 4 gives 3, choice A. Keep looking. Also, 14 and 84 are factors, but their neighbors are not. Neither 13 nor 15 divides 168, so 14, choice C, fails. For 84, choice D, neither 83 nor 85 is a factor.
So the answer is B, 7.

Check your answer

Divide 168 by each number in the run: $168 \div 6 = 28$ , $168 \div 7 = 24$ and $168 \div 8 = 21$ . All three are whole numbers, so 6, 7 and 8 are all factors.

More questions like this

The original DOE question

Question 93 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 93. Open page 209 of the handbook (PDF)

Question 37 of 57

2024 Test B Math, question 37

Question

Raoul gave two musical performances. His second performance occurred 86 days after his first performance. If Raoul’s second performance was on a Tuesday, on what day of the week was his first performance?

ASunday
BMonday
CWednesday
DThursday

Show the answer

Correct answer: A

The total number of days (86) is divided by the number of days in a week (7).

$86\div 7=12\text{, with a remainder of }2$

The 2 days are counted back from Tuesday, so the first performance was on a Sunday.

Question analysis

Raoul's second musical performance came 86 days after his first and fell on a Tuesday. Students work out the day of the week of the first performance. It tests division with remainders in a repeating weekly cycle, and the traps are counting in the wrong direction or miscounting the leftover days.

NYS standards: 6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Dividing with a remainder is grade 6 arithmetic, but using the remainder to step through a seven-day cycle, and counting backward, is less routine than grade 7 classwork. It needs care with direction rather than new content.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
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

Counting 86 days one at a time is hopeless in the time allowed, so the question rewards seeing that whole weeks return to the same weekday. Two decisions remain after that, and each has a listed wrong answer: which way to count, and whether Tuesday itself is one of the 2 days. Saying the days aloud, Monday is 1 back and Sunday is 2 back, avoids both.

No obvious method moderate Working it out
Nothing in the stem says to divide by 7; the key is that 84 of the 86 days are exactly 12 weeks, which leaves only 2 days to count. (Schoenfeld, 1985)
Turning words into math minor Understanding the question
"His second performance occurred 86 days after his first" puts the unknown day before the Tuesday, so the leftover days are counted backward. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Remove whole weeks from the 86 days, since each full week lands on the same weekday, then count the leftover days backward from Tuesday.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts Tuesday itself as the first of the 2 leftover days, so moves back only one day, to Monday.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Counts forward from Tuesday instead of back and also counts Tuesday as one of the 2 days, landing on Wednesday.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Counts the 2 leftover days forward from Tuesday, finding the day 86 days after the second performance instead of before it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Weeks and leftover days: $\text{days} = 7 \times \text{weeks} + \text{leftover}$; days is the total number of days. weeks is the number of full weeks inside it. leftover is the days that remain, from 0 to 6. The day of the week repeats every 7 days, so only the leftover days change it.

Sort out the question
The second show was on a Tuesday. It came 86 days after the first show. So the first show was 86 days before that Tuesday.
You need the day of the week of the first show. So you count back 86 days from Tuesday.
Take out the full weeks
Counting back 86 days one at a time is slow. Here is a shortcut. Every 7 days, the same weekday comes back. So divide 86 by 7.
$86 \div 7 = 12 \text{ remainder } 2$
$86 = 7 \times 12 + 2$
That is 12 full weeks and 2 extra days. Going back 12 weeks from a Tuesday lands on a Tuesday again.
Count back the 2 extra days
Start on Tuesday and go back one day at a time. One day back is Monday. Two days back is Sunday.
Watch out: Do not count Tuesday itself as one of the 2 days. That lands on Monday, choice B. And count backward, not forward. Counting forward 2 days gives Thursday, choice D.
So the answer is A, Sunday.

Check your answer

Count forward to be sure. Start on a Sunday. After 84 days, which is 12 weeks, it is Sunday again. Two more days are Monday and Tuesday. That is day 86, a Tuesday. It matches.

More questions like this

The original DOE question

Question 94 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 94. Open page 210 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 57

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 39 of 57

2024 Test B Math, question 39

Question

$A closed rectangular box drawn in three dimensions. Its height is labeled 20 cm, the bottom edge running back on the left is labeled 40 cm, and the bottom front edge is labeled 50 cm.$

What is the total surface area, in square centimeters, of the closed box shown above?

A3,800
B5,600
C7,600
D40,000

Show the answer

Correct answer: C

To find surface area, use the formula $A=2(hw+hl+lw)$ :

$A=2[(20)(40)+(20)(50)+(50)(40)]$

$A=2(800+1{,}000+2{,}000)$

$A=2(3{,}800)=7{,}600$

Question analysis

A drawing of a closed box gives its length, width and height in centimeters, and students find its total surface area in square centimeters. It tests surface area of a rectangular prism. The work is short, but every face of the box has to be accounted for, and surface area must be kept apart from volume.

NYS standards: 7.G.6 area, surface area, and volume
6.G.4 nets and surface area
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Surface area of a rectangular prism is a grade 7 standard, and this box has whole-number dimensions shown in a drawing. Three face areas, a sum and the opposite faces make it typical practice rather than a stretch.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Reading a figure, table or graph minor
Solution: 3 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

Finding the three face areas, 800, 1,000 and 2,000, is quick; the question tests whether students account for all six faces. The wrong choices are what the box would give with only three faces, with the top left off, or with the dimensions multiplied into a volume. Listing the faces in pairs before adding is the safest route.

Something you must already know minor Working it out
No surface-area formula is given; students must know that a box's six faces come in three matching pairs, 20 × 40, 20 × 50 and 40 × 50. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The drawing shows only three faces and labels 20, 40 and 50 cm on separate edges, so the hidden faces and their dimensions have to be pictured. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the areas of the three different faces from the labeled length, width and height, add them, and double the sum to cover the opposite faces.

Traps

Stops one step early: A choice is a number from partway through the solution.
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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Adds the three different face areas, 800 + 1,000 + 2,000 = 3,800, and stops before doubling for the matching opposite faces.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Leaves out the 40-by-50 top, as for an open box, 7,600 − 2,000 = 5,600, though the stem says the box is closed.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies the three dimensions, 20 × 40 × 50 = 40,000, which is the box's volume in cubic centimeters, not its surface area.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Surface area of a box: $SA = 2(lw + lh + wh)$; SA is the surface area, the total area of all six faces. l is the length, w is the width and h is the height. lw, lh and wh are the areas of the three different faces. You double because each face has a matching face on the opposite side.

Sort out the question
The box is a rectangular prism, a box shape with six rectangle faces. Its length is 50 cm, its width is 40 cm and its height is 20 cm.
The box is closed, so it has a top. Surface area is the total area of all the outside faces, in square centimeters.
Find the three different faces
The faces come in matching pairs. Find the area of one face from each pair. Multiply its two edges.
Faces Edges Area of one (sq cm)
Front and back 50 × 20 1,000
Left and right 40 × 20 800
Top and bottom 50 × 40 2,000
Add the faces and double
Use the formula. Add the three areas. Then double the sum to cover the matching faces you can’t see.
$SA = 2(50 \times 40 + 50 \times 20 + 40 \times 20)$
$SA = 2(2{,}000 + 1{,}000 + 800) = 2 \times 3{,}800 = 7{,}600$
The surface area is 7,600 square centimeters.
Avoid the traps
Watch out: Stopping at 3,800 counts only three faces, choice A. Leaving off the top, as for an open box, gives 5,600, choice B. Multiplying all three edges gives 40,000, choice D. That is the volume, in cubic centimeters.
So the answer is C, 7,600.

Faces	Edges	Area of one (sq cm)
Front and back	50 × 20	1,000
Left and right	40 × 20	800
Top and bottom	50 × 40	2,000

Check your answer

Add all six faces one by one: $1{,}000 + 1{,}000 + 800 + 800 + 2{,}000 + 2{,}000 = 7{,}600$ . Also check the units. Area multiplies two lengths, so it is in square centimeters. Multiplying three lengths gives cubic units, which is volume.

More questions like this

The original DOE question

Question 96 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 96. Open page 210 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 57

2024 Test B Math, question 40

Question

The value of a particular car decreases at a constant rate. If the car is worth $15,000 three years after its original purchase date and $10,000 five years after its original purchase date, what was the value of the car on its original purchase date?

A$18,000
B$20,000
C$22,500
D$25,000

Show the answer

Correct answer: C

The value of the car decreased $5,000 in two years: $15,000 to $10,000.

$5{,}000\div 2=2{,}500$

Therefore, the value of the car decreases $2,500 each year. After the first 3 years, the car’s value had decreased $7,500.

So $7{,}500+15{,}000=22{,}500$ .

Question analysis

A car loses value at a constant rate, and its worth three years and five years after purchase is given. Students find what the car was worth on the day it was bought. It tests rate of change and initial value in a linear relationship, and the tricky part is counting the right number of years when working back.

NYS standards: 8.F.4 modeling with linear functions
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Finding a starting value from two later values of a steady change is grade 8 linear-function work. A grade 7 student can reason with a yearly rate, but extending it back to the purchase date goes beyond typical grade 7 problems.
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 10, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is a straight-line pattern read backward, and its numbers are friendly; the difficulty is counting years correctly. The $5,000 drop happens over 2 years, not 5, and the purchase lies 3 years before the $15,000 value, not 2. Each miscount gives a listed answer, so sketching a timeline from year 0 to year 5 before computing pays off.

Turning words into math moderate Understanding the question
The phrases "three years after its original purchase date" and "five years after" must become points at year 3 and year 5 on a timeline whose start, year 0, holds the unknown value. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three moves are chained: the $5,000 drop over 2 years, the $2,500 yearly rate, and three years of that rate added back to $15,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the yearly drop from the change between the two given years, then add that drop back once for each year between the purchase and year 3.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $5,000 drop by 5, the year of the second value, instead of the 2 years between year 3 and year 5, then adds back 3 × $1,000.
B Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Adds back the $5,000 two-year drop once, which gives the value at year 1, one year short of the 3 years back to the purchase.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two given values, $15,000 + $10,000, combining numbers from the question without a model of the steady decrease.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate of change: $\text{rate} = \frac{\text{change in value}}{\text{change in time}}$; change in value is how many dollars the value went up or down. change in time is how many years passed. rate is the change for one year. A constant rate means the change is the same every year.

Sort out the question
The car’s value “decreases at a constant rate.” That means the car loses the same number of dollars each year. Year 0 is the day it was bought.
Year Value
0 ?
3 $15,000
5 $10,000
You need the value at year 0.
Find the drop for one year
From year 3 to year 5 is 2 years. In that time, the value fell by $15{,}000 - 10{,}000 = 5{,}000$ dollars.
$\text{rate} = \frac{5{,}000}{2} = 2{,}500$
The car loses $2,500 each year. Watch out: Divide by the 2 years between the values, not by 5. Dividing by 5 leads to $18,000, choice A.
Go back to year 0
From year 0 to year 3 is 3 years. The car lost $2,500 in each of them. So add that amount back 3 times.
$15{,}000 + 3 \times 2{,}500 = 15{,}000 + 7{,}500 = 22{,}500$
Watch out: Adding back only one $5,000 gives $20,000, choice B. That is the value at year 1, not year 0.
So the answer is C, $22,500.

Year	Value
0	?
3	$15,000
5	$10,000

Check your answer

Count forward from $22,500, taking off $2,500 each year. Year 1 is $20,000. Year 2 is $17,500. Year 3 is $15,000. Year 4 is $12,500. Year 5 is $10,000. Both given values match.

More questions like this

The original DOE question

Question 97 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 97. Open page 210 of the handbook (PDF)

Question 41 of 57

2024 Test B Math, question 41

Question

A bag contains 2 green, 4 blue, and 3 yellow disks. Alana selects 4 disks from the bag at random, one at a time, without replacing them. What is the probability that she selects all 4 blue disks?

A $\frac{8}{2{,}187}$
B $\frac{1}{126}$
C $\frac{256}{6{,}561}$
D $\frac{16}{9}$

Show the answer

Correct answer: B

If Alana picks 4 blue disks in a row, the number of blue disks and the total number of disks will both decrease by 1 after each selection.

For the first selection, there are 4 blue disks out of 9 total disks, so the probability of selecting the first blue disk is $\frac{4}{9}$ .

The probabilities of selecting blue for the next three selections are $\frac{3}{8}$ , $\frac{2}{7}$ , and $\frac{1}{6}$ .

$\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}=\frac{24}{3{,}024}=\frac{1}{126}$

Question analysis

A bag holds green, blue and yellow disks, and Alana draws four of them one at a time without putting any back. Students find the probability that all four disks are blue. It tests the probability of dependent events, and the difficulty is keeping track of how the bag changes after each draw.

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 lists or tree diagrams for two stages. Four dependent draws, each changing the counts, are too many to list and need the multiplication rule, which grade 7 does not formally teach.
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
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 5, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Students who know the multiplication rule for successive draws still have two adjustments to make on every draw, and each wrong choice skips at least one of them. Leaving the bag at 9 disks, or putting each disk back, both produce fractions over 6,561 that look as exact as the right one. Writing the four fractions in a row before multiplying, and cancelling as you go, keeps the arithmetic small.

Something you must already know moderate Working it out
The rule for successive draws without replacement is assumed: each draw's chance is multiplied in, with both the blue count and the total lowered by one. (Anderson & Krathwohl, 2001)
Error-prone arithmetic moderate Working it out
The product 4/9 × 3/8 × 2/7 × 1/6 = 24/3,024 has to be reduced to 1/126, and two wrong choices hide behind 9 to the fourth power, 6,561. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the chance of blue for each of the four draws in turn, lowering both the blue count and the total by one each time, and multiply.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
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 Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: moderate
Lowers the blue count on each draw but keeps all 9 disks in the bag, 4 × 3 × 2 × 1 over 9 × 9 × 9 × 9, which is 24/6,561 = 8/2,187.
C Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Uses 4/9 for every draw, (4/9) to the fourth power = 256/6,561, as if each disk were put back before the next draw.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Adds the first draw's 4/9 four times, or multiplies it by 4, instead of multiplying the draws' chances, and misses that a probability cannot exceed 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(A \text{ then } B) = P(A) \times P(B \text{ after } A)$; A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

Sort out the question
The bag holds 2 green, 4 blue and 3 yellow disks. That is 9 disks in all. Alana takes 4 disks, one at a time.
“Without replacing them” means a disk she takes does not go back in the bag. You need the chance that all 4 disks are blue.
Find the chance for each pick
For all 4 to be blue, each pick must be blue. After each blue pick, one blue disk is gone. One disk is gone from the total, too. So the top and the bottom each go down by 1.
Pick Blue disks left Disks left Chance of blue
1st 4 9 $\frac{4}{9}$
2nd 3 8 $\frac{3}{8}$
3rd 2 7 $\frac{2}{7}$
4th 1 6 $\frac{1}{6}$
Multiply the four chances
All four picks must go right, one after another. Use the rule for events in a row and multiply.
$\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{4 \times 3 \times 2 \times 1}{9 \times 8 \times 7 \times 6} = \frac{24}{3{,}024}$
Simplify the fraction
Divide the top and the bottom by 24. That keeps the value the same.
$\frac{24 \div 24}{3{,}024 \div 24} = \frac{1}{126}$
Watch out: Using $\frac{4}{9}$ for every pick gives $\frac{256}{6{,}561}$ , choice C. That is right only if each disk goes back in. Keeping 9 on the bottom each time gives $\frac{8}{2{,}187}$ , choice A. Choice D is more than 1, and no chance can be more than 1.
So the answer is B, $\frac{1}{126}$ .

Pick	Blue disks left	Disks left	Chance of blue
1st	4	9	$\frac{4}{9}$
2nd	3	8	$\frac{3}{8}$
3rd	2	7	$\frac{2}{7}$
4th	1	6	$\frac{1}{6}$

Check your answer

Cancel before you multiply. On top, $4 \times 2 = 8$ cancels the 8 on the bottom. That leaves $\frac{3}{9 \times 7 \times 6} = \frac{3}{378}$ . Divide the top and bottom by 3 to get $\frac{1}{126}$ again.

More questions like this

The original DOE question

Question 98 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 98. Open page 211 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 57

2024 Test B Math, question 42

Question

Jevon earns $\frac{1}{2}\%$ per year on the money in his savings account. Which expression is not another way to write $\frac{1}{2}\%$ ?

A $\frac{0.005}{100}$
B $\frac{\frac{1}{2}}{100}$
C $\frac{5}{1{,}000}$
D $\frac{1}{200}$

Show the answer

Correct answer: A

Rewrite the percent as a decimal:

$\frac{1}{2}\%=0.5\%=0.005$

So the correct answer must not equal 0.005:

Option A

$\frac{0.005}{100}=0.00005$

Option B

$\frac{\frac{1}{2}}{100}=0.5\div 100=0.005$

Option C

$\frac{5}{1{,}000}=0.005$

Option D

$\frac{1}{200}=0.005$

All answer options except Option A equal 0.005. Therefore, Option A is the correct answer.

Question analysis

Jevon's savings account earns one-half percent a year, and students must find the expression that is not another way to write one-half percent. It tests converting a percent less than 1 into fractions and decimals. Reading the word not carefully, and evaluating each stacked or unfamiliar fraction correctly, are what make it tricky.

NYS standards: 6.RP.3c percent as a rate per 100
7.NS.2d fractions to decimals by long division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Converting between percents, fractions and decimals is grade 6 and 7 work. A percent smaller than 1 is less familiar, and the question asks which form is not equal, so all four must be checked, which keeps it at typical grade 7 demand.
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: A question in the negative moderate
Every choice must be checked moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 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

Every expression is built from the same idea, so the question tests whether students can convert a percent smaller than 1 without losing track of decimal places. The odd one out, 0.005 over 100, looks like a correct form, because 0.005 is one-half percent and dividing by 100 seems to turn it into a percent. Converting one-half percent to 0.005 once and comparing each expression with it beats reasoning about each form.

A question in the negative moderate Understanding the question
The bold word not asks for the one expression unequal to one-half percent, so the three expressions that check out are exactly the ones to reject. (Haladyna, Downing & Rodriguez, 2002)
Every choice must be checked moderate Choosing the answer
Three of the four forms are equal, so each has to be converted to a decimal and compared with 0.005 before the odd one out can be named. (Embretson & Wetzel, 1987)
Error-prone arithmetic minor Working it out
One-half percent is 0.5 with the point moved two places, 0.005, and dividing that by 100 moves it two more places, to 0.00005. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write one-half percent as a single decimal by dividing one half by 100, then evaluate each expression as a decimal and compare it with that value.

Traps

NOT, LEAST or EXCEPT: The question asks for the exception, the least, or what is not true.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Misses the bold not and picks the expression that equals one-half percent most plainly, one half over 100, the literal meaning of percent.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Believes a percent must be written over 100, so 5 over 1,000 looks like a different number, though both equal 0.005.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Divides 1 by 200 and misplaces the decimal point, getting 0.05 instead of 0.005, so 1/200 seems not to equal one-half percent.

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 any number. The % sign means out of 100. So p percent is p divided by 100. Dividing by 100 moves the decimal point two places to the left.

Sort out the question
The question asks which expression is not equal to $\frac{1}{2}\%$ . So three choices are equal to it, and one is different. You need the odd one out.
Write the percent as a decimal
One half is 0.5. A percent means out of 100, so divide by 100.
$\frac{1}{2}\% = 0.5\% = \frac{0.5}{100} = 0.005$
So $\frac{1}{2}\%$ is 0.005. It is half of one percent, a very small amount.
Test each choice
Turn each choice into a decimal. Then compare it with 0.005.
Choice Work it out Equal to 0.005?
A $0.005 \div 100 = 0.00005$ No
B $0.5 \div 100 = 0.005$ Yes
C $5 \div 1{,}000 = 0.005$ Yes
D $1 \div 200 = 0.005$ Yes
Pick the one that is different
Choice A divides by 100 twice. The 0.005 was already divided by 100 once. Dividing again makes it 100 times too small.
Watch out: Choice B is the plainest way to write $\frac{1}{2}\%$ : one half over 100. Picking it means missing the word “not.” Also, 1 ÷ 200 is 0.005, not 0.05, so D is equal too.
So the answer is A, $\frac{0.005}{100}$ .

Choice	Work it out	Equal to 0.005?
A	$0.005 \div 100 = 0.00005$	No
B	$0.5 \div 100 = 0.005$	Yes
C	$5 \div 1{,}000 = 0.005$	Yes
D	$1 \div 200 = 0.005$	Yes

Check your answer

Use fractions instead. Half of one hundredth is $\frac{1}{2} \times \frac{1}{100} = \frac{1}{200}$ , which is choice D. And $\frac{5}{1{,}000}$ simplifies to $\frac{1}{200}$ too. Only A is a different amount.

More questions like this

The original DOE question

Question 99 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 99. Open page 211 of the handbook (PDF)

Question 43 of 57

2024 Test B Math, question 43

Question

A community garden had 50 members in 2001. In each of the next three years, the number of members increased by 40% over the number of members from the previous year. How many members were there in 2004? (Round to the nearest whole number.)

A70
B98
C110
D137

Show the answer

Correct answer: D

To find out how many members there were in 2002, calculate 40% of 50 and then add it to 50.

$50\times 0.40=20$

$20+50=70\text{ members}$

To find out how many members there were in 2003, calculate 40% of 70 and then add it to 70.

$70\times 0.40=28$

$28+70=98\text{ members}$

To find out how many members there were in 2004, calculate 40% of 98 and then add it to 98.

$98\times 0.40=39.2$

The result is 39.2, which is rounded to 39 to keep members to the nearest whole number.

$39+98=137\text{ members}$

Question analysis

A community garden has 50 members in 2001, and membership grows by 40 percent each year over the year before. Students find the number of members in 2004, rounded to a whole number. It tests repeated percent increase, where each of the three rounds of growth must start from the previous year's total.

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 is grade 7 content, and each step here is a routine one. Repeating it three times, each time on the new total, adds length and bookkeeping but no new idea, so it matches 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: Many steps moderate
Turning words into math minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 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

Compound growth is the whole question: each year's 40% is taken of a larger number, so the increases grow from 20 to 28 to about 39. Students who add 20 three times treat the growth as simple, and those who lose count of the years stop at an earlier total that is also listed. Multiplying by 1.4 three times, or checking that the third increase is the largest, separates the right count from the rest.

Many steps moderate Working it out
Three yearly increases have to be applied in turn, 50 to 70 to 98 to 137.2, each taking 40% of the new total rather than of the original 50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The phrase "increased by 40% over the number of members from the previous year" sets a new base each year, the wording that separates compound from simple growth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
The third year's 40% of 98 is 39.2, the first decimal in the chain, and the rounding instruction applies only to the final count. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply the membership by 1.4 once for each year from 2002 through 2004, carrying the unrounded value forward, and round only at the end.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Applies the 40% increase once, 50 + 20 = 70, which is the 2002 membership, and stops two years short of 2004.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Applies two increases, reaching 98 members in 2003, and stops one year short of the 2004 total the question asks for.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Adds 40% of the original 50, that is 20, in each of the three years, 50 + 60 = 110, instead of 40% of the previous year's total.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{old} + \frac{p}{100} \times \text{old}$; old is last year’s number of members. p is the percent increase, here 40. new is this year’s number. You add p percent of the old amount to the old amount. Each year, the new number becomes next year’s old number.

Sort out the question
In 2001 the garden had 50 members. In each of the next three years, 2002, 2003 and 2004, the members grew by 40%.
The 40% is of last year’s members, so the base changes every year. You need the number for 2004, rounded to the nearest whole number.
Find the members in 2002
40% of 50 is 20. Add it to 50.
$\text{new} = 50 + \frac{40}{100} \times 50 = 50 + 20 = 70$
There were 70 members in 2002.
Find the members in 2003
Now 70 is the old number. Take 40% of 70, not of 50.
$\text{new} = 70 + \frac{40}{100} \times 70 = 70 + 28 = 98$
There were 98 members in 2003.
Find the members in 2004
Now 98 is the old number.
$\text{new} = 98 + \frac{40}{100} \times 98 = 98 + 39.2 = 137.2$
Round to the nearest whole number. 137.2 is closer to 137 than to 138, so there were 137 members.
Watch out: Adding 20 members every year uses 40% of 50 each time. That gives 110, choice C. Stopping after one or two years gives 70 or 98, choices A and B.
So the answer is D, 137.

Check your answer

A 40% rise is the same as multiplying by 1.4. So $50 \times 1.4 \times 1.4 \times 1.4 = 50 \times 2.744 = 137.2$ , which rounds to 137. It matches.

More questions like this

The original DOE question

Question 100 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 100. Open page 211 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 57

2024 Test B Math, question 44

Question

Which inequality is equivalent to $6x-20<5x-2(5-3x)$ ?

A $x<16$
B $x<\frac{5}{2}$
C $x<-2$
D $x>-2$

Show the answer

Correct answer: D

$6x-20<5x-2(5-3x)$

Distribute the coefficient −2 on the right-hand side:

$6x-20<5x+(-2)(5)+(-2)(-3x)$

$6x-20<5x-10+6x$

Subtract $6x$ from both sides:

$-20<5x-10$

Add 10 to both sides:

$-10<5x$

Divide both sides by 5:

$-2<x$

This is equivalent to

$x>-2$

Question analysis

Students choose the simplest equivalent form of a linear inequality that has x on both sides and a negative number multiplied into parentheses. It tests distributing, combining like terms and solving inequalities. Sign errors in the distribution, and reading the direction of the final inequality, are where most mistakes happen.

NYS standards: 7.EE.4b two-step inequalities
8.EE.7b linear equations with rational coefficients
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 inequalities have the form px + q < r. This one has x on both sides and a negative factor to distribute across a subtraction, steps that belong to grade 8 equation work, so it is harder than typical grade 7 practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 16+, 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

Every move is standard algebra, but the inequality is built so that small slips change the answer completely. The −2 has to reach both terms in the parentheses, and once the boundary −2 is found its direction still has to be read correctly, since the variable ends up on the right as −2 < x. Rewriting that as x > −2, or substituting x = 0 into the original, catches the trap.

Error-prone arithmetic moderate Working it out
Distributing −2 over (5 − 3x) gives −10 + 6x, two sign decisions in one move, and the x terms then sit on both sides of the inequality. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
x < −2 and x > −2 share the boundary and differ only in direction, which depends on reading −2 < x correctly or flipping the sign when dividing by a negative. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
Four moves follow one another, distributing, collecting 5x + 6x, moving the x terms and dividing by 5, and a slip at any of them changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Distribute the −2 to both terms in the parentheses, combine the x terms on the right, then gather the x terms on the side that keeps their coefficient positive.

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Treats 5 − 3x as 2, combining a number with an x term, so the right side becomes 5x − 4 and the solution x < 16.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Multiplies the −2 by the 5 but not by the −3x, writing 5x − 10 − 3x, which leads to 4x < 10 and x < 5/2.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the boundary −2 but points the inequality the wrong way, by misreading −2 < x or by not flipping the sign after dividing by −5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. Multiply a by each one. A minus sign inside counts as a negative term, so 5 − 3x is 5 + (−3x). A negative times a negative is positive.
Keeping an inequality true: $\text{if } a < b \text{, then } a + c < b + c$; a and b are the two sides, and c is any number. Adding or subtracting the same number on both sides keeps the inequality true. Dividing both sides by a positive number also keeps it true. Only multiplying or dividing by a negative number flips the sign.

Sort out the question
An inequality compares two sides with a sign like <, which means is less than. To solve it, you get x alone on one side.
You need the short inequality that has the same answers as the long one.
Distribute the −2
Multiply −2 by both terms inside the parentheses, 5 and −3x.
$-2(5 - 3x) = (-2) \times 5 + (-2) \times (-3x) = -10 + 6x$
A negative times a negative is positive, so the second part is +6x. Watch out: If the −2 multiplies only the 5, you end up with $x < \frac{5}{2}$ , choice B.
Combine the x terms
The right side is now 5x − 10 + 6x. Put the x terms together: 5x + 6x = 11x.
$6x - 20 < 11x - 10$
Gather the x terms on one side
Subtract 6x from both sides. That leaves the x terms on the right, where they stay positive.
$6x - 20 - 6x < 11x - 10 - 6x$
$-20 < 5x - 10$
Get x by itself
Add 10 to both sides. Then divide both sides by 5.
$-10 < 5x$
$-2 < x$
You divided by a positive 5, so the sign does not flip.
Read it with x first
$-2 < x$ says −2 is less than x. That is the same as saying x is greater than −2, just as 3 < 5 means 5 > 3.
$x > -2$
Watch out: $x < -2$ , choice C, has the same number but points the wrong way.
So the answer is D, $x > -2$ .

Check your answer

Test x = 0, which is greater than −2. The left side is $6(0) - 20 = -20$ . The right side is $5(0) - 2(5 - 0) = -10$ . And −20 < −10 is true. Now test x = −3. The left side is −38 and the right side is −43. That is false, as it should be.

More questions like this

The original DOE question

Question 101 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 101. Open page 211 of the handbook (PDF)

Question 45 of 57

2024 Test B Math, question 45

Question

A box contains a total of 30 paper clips with colors in the ratio $\text{red}:\text{white}:\text{blue}=2:5:3$ . Suppose that one paper clip of each color is removed from the box and not replaced. What is the probability that the next paper clip chosen will be blue?

A $\frac{5}{27}$
B $\frac{2}{7}$
C $\frac{8}{27}$
D $\frac{3}{10}$

Show the answer

Correct answer: C

The ratio is $\text{red}:\text{white}:\text{blue}=2:5:3$ . Find the number of each color in a box of 30 paper clips.

$2x+3x+5x=30$

$10x=30$

$x=3$

$\text{Red: }2x=2(3)=6$

$\text{White: }5x=5(3)=15$

$\text{Blue: }3x=3(3)=9$

If one paper clip of each color is removed (3 total), there will be 8 blue paper clips out of 27 in the box. The probability that the next paper clip chosen will be blue is $\frac{8}{27}$ .

Question analysis

A box of 30 paper clips has red, white and blue clips in the ratio 2 to 5 to 3, and one clip of each color is taken out. Students find the probability that the next clip drawn is blue. It tests ratios and simple probability; the ratio gives shares rather than counts, and the removals change both the blue count and the total.

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 Splitting a total by a three-part ratio and finding a simple probability are both familiar skills. Combining them, then adjusting the counts after the removals, adds steps but stays within 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: Many steps moderate
A very close wrong answer moderate
Turning words into math minor
Solution: 4 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

The question chains a ratio problem into a probability problem, with a small change in between. Students who skip the removal get 3/10, a choice that is almost the right value and looks natural because it matches the ratio. The cleanest route lists the actual counts, 6, 15 and 9, takes one from each, and writes blue's 8 over the remaining 27.

Many steps moderate Working it out
The ratio 2:5:3 must become counts of 6, 15 and 9 clips, one clip of each color must come out, and only then can blue's 8 be put over 27. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
3/10 = 0.3 and 8/27, about 0.296, are almost equal, so an estimate cannot separate them; only subtracting the three removed clips does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
The removal of one clip of each color, not replaced, happens before the draw, so it changes both the blue count and the total in the box. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide 30 by the sum of the ratio parts to find each color's count, take one clip away from each color, then put the blue count over the new total.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
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 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Works out the counts correctly but puts red's remaining 5 clips over the 27 left, answering for the wrong color.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the ratio terms as actual counts and removes one from each, so blue is 3 − 1 = 2 of 10 − 3 = 7 clips.
D Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Gives blue's share of the full box, 9 of 30 or 3 of 10 ratio parts, ignoring the three clips removed before the draw.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Size of one ratio part: $\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}}$; The ratio numbers tell how many parts each color gets. Their sum is the number of parts in all. total is the number of clips. Dividing gives the size of one part. Each color is then its ratio number times one part.
Probability: $P(\text{blue}) = \frac{\text{blue clips}}{\text{all clips}}$; P(blue) is the chance of picking a blue clip. The top counts the blue clips in the box. The bottom counts all the clips in the box. It works because each clip is equally likely to be picked.

Sort out the question
The box has 30 clips. The ratio red : white : blue = 2 : 5 : 3 means that for every 2 red clips, there are 5 white clips and 3 blue clips.
Then one clip of each color is taken out and not put back. You need the chance that the next clip picked is blue.
Find the size of one part
The ratio has $2 + 5 + 3 = 10$ parts in all. The 30 clips are shared among those 10 parts.
$\text{one part} = \frac{30}{10} = 3$
Each part is 3 clips.
Count the clips of each color
Multiply each ratio number by the size of one part.
$\text{red} = 2 \times 3 = 6$
$\text{white} = 5 \times 3 = 15$
$\text{blue} = 3 \times 3 = 9$
Check: 6 + 15 + 9 = 30 clips. That matches the total.
Take out one of each color
One red, one white and one blue clip come out. Each color drops by 1, and the total drops by 3.
Color Before After
Red 6 5
White 15 14
Blue 9 8
All 30 27
Find the chance of blue
$P(\text{blue}) = \frac{8}{27}$
Watch out: $\frac{9}{30} = \frac{3}{10}$ , choice D, ignores the 3 clips taken out. $\frac{5}{27}$ , choice A, is the chance of red. Taking 1 from each ratio number, as if the box held only 10 clips, gives $\frac{2}{7}$ , choice B.
So the answer is C, $\frac{8}{27}$ .

Color	Before	After
Red	6	5
White	15	14
Blue	9	8
All	30	27

Check your answer

Add the clips left: 5 + 14 + 8 = 27. That matches. Also, $\frac{8}{27}$ is about 0.296, a bit less than blue’s share of the full box, 0.3. That makes sense, since one blue clip came out.

More questions like this

The original DOE question

Question 102 as printed on page 212 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 102. Open page 212 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 57

2024 Test B Math, question 46

Question

The least positive integer evenly divisible by the first 6 positive integers is 60. What is the least positive integer evenly divisible by the first 8 positive integers?

A1,680
B840
C480
D420

Show the answer

Correct answer: B

Each option is divisible by 1, 2, 3, 4, 5, and 6.

Option A:

1,680 is evenly divisible by the first 8 positive integers.

Option B:

840 is evenly divisible by the first 8 positive integers and is less than 1,680.

Option C:

480 is not evenly divisible by 7.

$480\div 7\approx 68.57$

Option D:

420 is not evenly divisible by 8.

$420\div 8=52.5$

Option B (840) is the correct answer.

Question analysis

Knowing that 60 is the smallest number divisible by each of 1 through 6, students find the smallest positive integer divisible by each of 1 through 8. It tests least common multiples and prime factors. The challenge is adding only the factors that 7 and 8 really need, so the result works for all eight numbers and is still the least.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Least common multiples are grade 6 content, usually for two small numbers. Extending the given result from six numbers to eight means deciding what 7 and 8 add, more reasoning than typical classwork but no new content.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Every choice must be checked moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 13, 25 words.
Corrected from the DOE handbook: A rounded decimal in the explanation is now marked as approximate (≈ instead of =).
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The given fact, 60 for the first six integers, is a head start, but only for students who see what 7 and 8 add. Multiplying 60 by 7 is needed; multiplying by all of 8 is not, since 60 already supplies two of its three factors of 2. Students who multiply by the new numbers whole get answers that are too large or that miss one divisor, so checking the smaller choices first is the efficient route.

Every choice must be checked moderate Choosing the answer
Both 1,680 and 840 are divisible by 7 and by 8, so testing the choices is safe only from the smallest up; 420 fails for 8 and 480 for 7. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
Building on 60 needs prime factors: 7 adds a new prime, while 8 = 2 × 2 × 2 needs only one more 2, since 60 already holds two. (Anderson & Krathwohl, 2001)

Strategy: Start from 60 and ask which extra prime factors 7 and 8 require, or test the choices from the smallest up for divisibility by both 7 and 8.

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

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Multiplies 60 by 7 and by 4 to cover 8, getting 1,680, which all of 1 to 8 divide, but overlooks that the question wants the least.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Multiplies 60 by 8 to bring in the 8 but never checks divisibility by 7, which 480 fails.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Multiplies 60 by 7 to bring in the 7 and stops, but 420 has only two factors of 2, so 8 does not divide it.

What was corrected from the DOE handbook, and why

Explanation: To show that 480 is not evenly divisible by 7, the handbook writes 480 ÷ 7 = 68.57. The quotient is really 68.571428…, a decimal that never ends, so 68.57 is only a rounded value. The site writes 480 ÷ 7 ≈ 68.57, as the handbook itself does for other rounded results in this test. The point of the step is unchanged: the quotient is not a whole number, so 480 is not evenly divisible by 7 and is not divisible by all of the first 8 positive integers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Evenly divisible: $a \div b = \text{a whole number}$; a is the number you test, and b is the number you divide by. If the answer is a whole number, with no remainder, a is evenly divisible by b. Then b is a factor of a.

Sort out the question
The first 8 positive integers are 1, 2, 3, 4, 5, 6, 7 and 8. “Evenly divisible” means dividing leaves no remainder.
You need the least, or smallest, number that all 8 of them divide. You are told that 60 is the smallest for 1 to 6. So build on 60.
Bring in the 7
60 is not divisible by 7. And 7 is prime, which means its only factors are 1 and 7. So the new number needs a factor of 7. Multiply 60 by 7.
$60 \times 7 = 420$
Bring in the 8
8 is $2 \times 2 \times 2$ , so the number needs three 2s as factors. But 420 has only two: $420 = 2 \times 2 \times 105$ , and 105 is odd. Multiply by one more 2.
$420 \times 2 = 840$
Now $840 \div 8 = 105$ and $840 \div 7 = 120$ . Both are whole numbers.
Test the choices from smallest up
Choice D: $420 \div 8 = 52.5$ . That is not a whole number, so 420 fails.
Choice C: $480 \div 7 \approx 68.57$ . The ≈ sign means about, since the decimal never ends. It is not a whole number, so 480 fails.
Choice B: 840 works for 7 and 8. It also works for 1 to 6, since $840 = 60 \times 14$ .
Watch out: 1,680, choice A, works too. But it is twice 840, so it is not the least.
So the answer is B, 840.

Check your answer

Divide 840 by each number from 1 to 8. You get 840, 420, 280, 210, 168, 140, 120 and 105. Every answer is a whole number, so 840 works for all eight.

More questions like this

The original DOE question

Question 103 as printed on page 212 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 103. Open page 212 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 47 of 57

2024 Test B Math, question 47

Question

Three different colors of paint are available to paint the window frame, the door, and the wall of a house. If no two parts of the house may be painted the same color, how many different ways are there to paint the three parts of the house?

Show the answer

Correct answer: B

One method of solving the problem is to use a table to represent the sample space. Let the three colors be red, blue, and green.

PAINT COLORS

Window Frame	Door	Wall
Red	Blue	Green
Red	Green	Blue
Blue	Red	Green
Blue	Green	Red
Green	Red	Blue
Green	Blue	Red

There are 6 possible combinations of colors for the window frame, the door, and the wall where no two parts are painted the same color.

Question analysis

Three paint colors are available for a house's window frame, door and wall, and no two of the three parts may be painted the same color. Students count the different ways to paint them. It tests counting arrangements with a sample space, and the rule against repeating a color is what shapes the count.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Counting arrangements with an organized list is grade 7 work, and this one has only three parts and three colors. It is typical practice, with the no-repeat rule as the one condition to keep in mind.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 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

With three colors and three parts, the six arrangements can be listed in well under a minute, so the question rewards either a quick organized list or the counting principle. The main loss comes from ignoring the rule that no two parts match, which turns 3 × 2 × 1 into 3 × 3 × 3. The other wrong choices simply combine the two 3s in the stem.

Something you must already know minor Working it out
The counting principle is assumed: 3 color choices for the window frame, 2 left for the door, then 1 for the wall, multiplied together. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The rule that "no two parts of the house may be painted the same color" means each color is used exactly once, which removes one choice at every step. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the color choices for the window frame, then for the door with one color already used, then for the wall, and multiply; or list the arrangements.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
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 number of colors, 3, as the number of ways, without arranging the colors across the three parts.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Multiplies the 3 colors by the 3 parts, 3 × 3 = 9, a product of the stem's numbers rather than a count of arrangements.
D Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Allows any color on each part, 3 × 3 × 3 = 27, ignoring the rule that a color used once cannot be used again.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total ways} = a \times b \times c$; a is the number of choices for the first part, b for the second and c for the third. Multiply them to get every possible way. It works because each choice for one part can go with each choice for the next part.

Sort out the question
There are 3 colors and 3 parts to paint: the window frame, the door and the wall. No two parts may be the same color. So each color is used exactly once.
You need the number of different ways to paint the three parts.
Count the choices for each part
Paint the parts in order. Each color you use is gone for the next part.
- Window frame: any of the 3 colors, so 3 choices.
- Door: one color is used, so 2 choices are left.
- Wall: only 1 color is left, so 1 choice.
Multiply the choices
Use the counting principle. Each choice for the window frame goes with each choice for the door, and so on. So you multiply.
$\text{total ways} = 3 \times 2 \times 1 = 6$
Watch out: $3 \times 3 \times 3 = 27$ , choice D, lets a color be used twice. $3 \times 3 = 9$ , choice C, just multiplies the numbers in the question.
So the answer is B, 6.

Check your answer

List them. Use R for red, B for blue and G for green, in the order frame, door, wall: RBG, RGB, BRG, BGR, GRB and GBR. That is 6 ways, and each one uses all three colors once.

More questions like this

The original DOE question

Question 104 as printed on page 212 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 104. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 48 of 57

2024 Test B Math, question 48

Question

$Two triangles. Triangle MNP has N at the top and base MP labeled 10 cm; sides MN and NP each have a tick mark, showing they are equal. The smaller triangle XYZ has Y at the top, base XZ labeled 6 cm, and sides XY and YZ each labeled 9 cm.$

Triangle MNP is similar to triangle XYZ, and $\mathrm{MN}=\mathrm{NP}$ . What is the perimeter of triangle MNP?

A25 cm
B30 cm
C36 cm
D40 cm

Show the answer

Correct answer: D

Similar geometric figures have sides that are proportional.

$\frac{10}{6}=\frac{x}{9}$

$\frac{5}{3}=\frac{x}{9}$

$x=\frac{(5)(9)}{3}=15$

Because $\mathrm{MN}=\mathrm{NP}$ , both unknown sides are equal to 15. Therefore, the three sides of MNP are 15, 15, and 10. The perimeter of a triangle is the sum of its three side lengths.

$15+15+10=40$

Question analysis

Two isosceles triangles are drawn, one with all its side lengths labeled and the other with only its base labeled, and the triangles are similar. Students find the perimeter of triangle MNP in centimeters. It tests similar figures and proportional sides, and the catch is that two sides must be found from the scale and the equal-sides marks.

NYS standards: 8.G.4 similarity and dilations
7.G.1 scale drawings
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Similar triangles are grade 8 geometry, though grade 7 scale drawings use the same proportional reasoning. Students also need the equal-sides marks to find a missing side, a combined step beyond routine grade 7 scale 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: Reading a figure, table or graph moderate
Something you must already know minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question chains a scale factor to a perimeter, and both links have a trap. Some students scale by adding, since the base grew by 4, and others find the 15 cm leg correctly but then add only some of the sides. Writing all three sides of the large triangle, 15, 15 and 10, before adding is what the question rewards.

Reading a figure, table or graph moderate Understanding the question
The figure carries every number: tick marks show MN = NP, the base MP is 10 cm, and triangle XYZ has 9 cm legs on a 6 cm base. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Students must know that corresponding sides of similar triangles share one scale factor, here 10/6 = 5/3, though the stem never says so. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The scale factor 5/3 is a fraction, and 9 × 5/3 = 15 is easier to get right by dividing 9 by 3 first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Compare the two bases to get the scale factor, multiply a leg of the small triangle by it, then use the equal sides to add up the perimeter.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds a scaled leg of 15 cm but adds only it and the 10 cm base, 15 + 10 = 25, leaving out the second 15 cm side.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals a slip Expected pull: weak
Finds both legs, 15 cm each, but adds only 15 + 15 and leaves out the 10 cm base; calling all three sides 10 cm also gives 30.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds 4 cm to each leg because the base grew from 6 to 10, making the legs 13 cm: 13 + 13 + 10 = 36.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Similar triangles: $\frac{\mathrm{MN}}{\mathrm{XY}} = \frac{\mathrm{MP}}{\mathrm{XZ}}$; Similar triangles have the same shape but can differ in size. MN matches XY, and MP matches XZ. Each side of the big triangle is the same number of times as long as its matching side. So these two fractions are equal.
Perimeter of a triangle: $P = a + b + c$; P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. Add them to go once around the shape.

Sort out the question
In triangle MNP, the base MP is 10 cm. The tick marks on MN and NP show those two sides are equal. In triangle XYZ, the base XZ is 6 cm, and XY and YZ are each 9 cm.
Similar means the same shape, with every side scaled by the same amount. You need the perimeter of MNP, the sum of its three sides.
Match the sides
The letters line up in order: M with X, N with Y, and P with Z. So the base MP matches the base XZ. The side MN matches XY, and NP matches YZ.
Find the scale factor
The scale factor is how many times bigger MNP is. Compare the two bases.
$\frac{\mathrm{MP}}{\mathrm{XZ}} = \frac{10}{6} = \frac{5}{3}$
Every side of MNP is $\frac{5}{3}$ times as long as its matching side in XYZ.
Find the equal sides
Use the similar triangles formula with XY = 9.
$\frac{\mathrm{MN}}{9} = \frac{5}{3}$
$\mathrm{MN} = 9 \times \frac{5}{3} = 3 \times 5 = 15$
Divide 9 by 3 first to keep the numbers small. MN is 15 cm. Since MN = NP, NP is 15 cm too.
Watch out: The base grew by 4 cm, but adding 4 cm to each side is wrong. That gives 13 cm sides and 36, choice C. Similar shapes grow by multiplying.
Add the three sides
$P = 15 + 15 + 10 = 40$
The perimeter is 40 cm. Watch out: Leaving out one 15 cm side gives 25, choice A. Leaving out the base gives 30, choice B.
So the answer is D, 40 cm.

Check your answer

The small triangle’s perimeter is 9 + 9 + 6 = 24 cm. Perimeters grow by the same scale factor as the sides. So $24 \times \frac{5}{3} = 40$ . It matches.

More questions like this

The original DOE question

Question 105 as printed on page 212 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 105. Open page 212 of the handbook (PDF)

Question 49 of 57

2024 Test B Math, question 49

Question

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

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

Show the answer

Correct answer: C

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

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

y can be a positive or a negative number.

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 50 of 57

2024 Test B Math, question 50

Question

Chris has 1 red shirt, 2 white shirts, and 3 black shirts. Chris picks a shirt at random. Without putting the first shirt back, he picks a second shirt at random. What is the probability that the first shirt is black and the second shirt is red?

A $\frac{1}{12}$
B $\frac{1}{10}$
C $\frac{1}{5}$
D $\frac{7}{10}$

Show the answer

Correct answer: B

Use the joint probability formula: P (black shirt first) • P (red shirt second, given black shirt first).

Initially there are 3 black shirts out of 6 total shirts, so the probability of picking a black shirt first is $\frac{3}{6}$ .

Once a black shirt is picked first, there is 1 red shirt out of 5 total shirts remaining.

So the probability of picking a red shirt second, given that a black shirt was picked first, is $\frac{1}{5}$ .

Then P (black shirt first) • P (red shirt second, given black shirt first)
$=\frac{3}{6}\bullet\frac{1}{5}=\frac{3}{30}=\frac{1}{10}$ .

Question analysis

Chris has red, white and black shirts and picks two at random, one after the other, without putting the first back. Students find the probability that the first shirt is black and the second is red. It tests the probability of two dependent events, where the second pick has one fewer shirt to choose from.

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. Only two picks are involved, but adjusting the second pick for the shirt already taken goes beyond typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 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

Two dependent picks with small numbers make this a textbook question, but the choices punish each half-step. Students who leave the first shirt in the drawer get a fraction just below the right one, students who find only the second pick's chance stop at a listed value, and students who add the two chances get a probability far too large. Writing both fractions, 3/6 and 1/5, and multiplying is the whole method.

Something you must already know moderate Working it out
The rule for two dependent picks, the first pick's chance times the second's given the first, is assumed, along with the drop to 5 shirts for the second pick. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
1/10 and 1/12 differ only in the second pick's denominator, 5 shirts or 6, so estimating cannot separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
"Without putting the first shirt back" sets the second pick's total at 5 shirts, and the order, black first then red, fixes which fraction comes first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the chance of black on the first pick, then the chance of red on the second pick with one shirt fewer, and multiply the two.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Keeps all 6 shirts for the second pick, 3/6 × 1/6 = 1/12, as if the black shirt had been put back.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Gives only the chance that the second shirt is red once a black one is gone, 1/5, leaving out the first pick's 3/6.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds the two chances, 1/2 + 1/5 = 7/10, instead of multiplying them for two events that must both happen.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(A \text{ then } B) = P(A) \times P(B \text{ after } A)$; A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

Sort out the question
Chris has 1 red, 2 white and 3 black shirts. That is 6 shirts. He picks one and keeps it out. Then he picks a second shirt.
You need the chance that the first shirt is black and the second is red. Both must happen, in that order.
Find the chance of black first
3 of the 6 shirts are black. Each shirt is equally likely to be picked.
$P(\text{black first}) = \frac{3}{6}$
That is one half.
Find the chance of red second
A black shirt is now out and not put back. That leaves 5 shirts: 1 red, 2 white and 2 black.
$P(\text{red second}) = \frac{1}{5}$
Watch out: The bottom is 5, not 6. Using 6 gives $\frac{1}{12}$ , choice A, as if the shirt went back.
Multiply the two chances
Use the rule for events in a row. The second pick must go right too, inside the half of the time the first one did.
$P(\text{black, then red}) = \frac{3}{6} \times \frac{1}{5} = \frac{3}{30} = \frac{1}{10}$
Watch out: Adding the chances gives $\frac{7}{10}$ , choice D. Stopping at $\frac{1}{5}$ gives choice C.
So the answer is B, $\frac{1}{10}$ .

Check your answer

Count every ordered pair of picks. There are 6 choices for the first and 5 for the second, so $6 \times 5 = 30$ pairs. Pairs with black then red: 3 black shirts times 1 red shirt is 3. So the chance is $\frac{3}{30} = \frac{1}{10}$ .

More questions like this

The original DOE question

Question 107 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 107. Open page 213 of the handbook (PDF)

Question 51 of 57

2024 Test B Math, question 51

Question

Rita used 125 small solid cubes to build a large solid cube. The measure of each edge of the small cubes equals 1 inch. What is the measure of each edge of the large cube, in inches?

A5
B25
C41.33
D125

Show the answer

Correct answer: A

The volume of a cube with a side length of s is

$V=s^{3}$

Since 125 small cubes were used to build the large cube, $125=s^{3}$ .

Each side of the large cube is made up of 5 small cubes.

$125=5^{3}$

Since 5 small cubes make up each edge of the large cube, and the edge of each small cube measures 1 inch, the measure of each edge of the large cube is $1\times 5=5\text{ inches}$ .

Question analysis

Rita stacks 125 one-inch cubes into a single large solid cube. Students find the length of each edge of the large cube in inches. It tests volume and perfect cubes with a building-block model, and the care needed is in telling apart the cubes along an edge, across a face and in the whole solid.

NYS standards: 6.G.5 perfect squares and perfect cubes
6.G.2 volume with fractional edge lengths
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Building a large cube from unit cubes is the model grade 6 uses to introduce perfect cubes. It is a single step with small numbers, easier than typical grade 7 geometry, which works with composite solids and formulas.
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 3, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is one of the easiest questions on the test, since 125 is a familiar cube and the small cubes make the large one easy to picture as 5 rows of 5 in 5 layers. The wrong choices name a layer instead of an edge, a division by 3 for three dimensions, or the total count itself, and multiplying 5 × 5 × 5 as a check rules all of them out.

Something you must already know minor Working it out
Students must see that 125 one-inch cubes make a volume of 125 cubic inches, and know or find that 5 × 5 × 5 = 125. (Anderson & Krathwohl, 2001)

Strategy: Think of the large cube as equal rows, columns and layers of small cubes, and find the number that, used three times as a factor, gives 125.

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

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Gives the number of small cubes in one layer or on one face, 5 × 5 = 25, instead of the number along one edge.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 125 by 3 for the cube's three dimensions instead of taking a cube root; that gives 41.67, and 41.33 is the only choice near it.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives back the 125 from the stem, the count of small cubes and the volume in cubic inches, rather than an edge length.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube: $V = s^3$; V is the volume, the space inside, in cubic units. s is the length of one edge. s³ means s × s × s. A cube’s length, width and height are all s, so length × width × height becomes s × s × s.

Sort out the question
Rita stacks 125 small cubes into one big solid cube. Each small cube has edges of 1 inch. You need the length of one edge of the big cube, in inches.
Find the big cube’s volume
Each small cube is $1 \times 1 \times 1 = 1$ cubic inch. The big cube is solid, with no gaps. So its volume is 125 cubic inches.
Find the edge length
$V = s^3$
$125 = s \times s \times s$
Find the number that makes 125 when you use it three times. Try 5.
$5 \times 5 \times 5 = 25 \times 5 = 125$
So 5 small cubes sit along each edge. Each one is 1 inch long, so the edge is 5 inches.
Rule out the other choices
Watch out: 25, choice B, is the number of small cubes in one layer, 5 rows of 5. 125, choice D, is the number of cubes, or the volume. Dividing 125 by 3 does not give an edge either. That is about 41.67, near choice C.
So the answer is A, 5.

Check your answer

Picture building it. A row of 5 cubes is 5 inches long. Five rows make a layer of 25 cubes. Five layers make $25 \times 5 = 125$ cubes. That matches.

More questions like this

The original DOE question

Question 108 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 108. Open page 213 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 52 of 57

2024 Test B Math, question 52

Question

$Two open boxes, labeled A and B. Box A holds two balls, numbered 2 and 6. Box B holds three balls, numbered 3, 4 and 5.$

Five numbered balls are placed in two boxes as shown in the figure above. Davina picks one ball at random from Box B and places it in Box A. Then Yusuf picks one ball from Box A without looking. What is the probability that Yusuf will pick a ball with an odd number?

A $\frac{2}{9}$
B $\frac{1}{3}$
C $\frac{2}{5}$
D $\frac{2}{3}$

Show the answer

Correct answer: A

The probability of Davina picking an odd-numbered ball from Box B is $\frac{2}{3}$ . If Davina places an odd-numbered ball in Box A, then the probability of Yusuf picking an odd-numbered ball from Box A is $\frac{1}{3}$ . The probability of both of these events happening is the product of the two probabilities.

$\frac{2}{3}\times\frac{1}{3}=\frac{2}{9}$

Question analysis

Two boxes hold five numbered balls, and Davina moves one ball at random from Box B into Box A before Yusuf draws one from Box A. Students find the probability that Yusuf's ball shows an odd number. It tests two-stage probability, and the tricky part is that the second draw depends on which ball was moved.

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 lists outcomes for independent stages such as coins or spinners. Here the first random move changes what is in the second box, so students must follow two linked stages, a dependence typical grade 7 work does not include.
Estimated difficulty: Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Thinking it demands: Depth of knowledge 3, strategic thinking · mainly analyze
What makes it hard: No obvious method major
Something you must already know minor
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 57 words.
Students, last 30 days: 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 chance events are hidden in what reads like a question about a single draw. Every listed wrong answer is a reasonable single-stage answer, either one of the two factors on its own or a count of odd balls across both boxes. Listing the nine equally likely pairs, Davina's ball and then Yusuf's, shows that only two of them end with an odd ball and makes the structure visible without any formula.

No obvious method major Working it out
The stem asks one question about Yusuf's draw, but the answer depends on Davina's earlier random pick; the key is that Box A starts with only even balls, so Yusuf can draw odd only by drawing the ball she moved. (Schoenfeld, 1985)
Something you must already know minor Working it out
Multiplying the chances of the two stages, 2/3 for an odd ball moving and 1/3 for Yusuf then drawing it, is a rule students must bring themselves. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The numbers appear only in the figure, Box A holding 2 and 6 and Box B holding 3, 4 and 5, so the odd balls have to be spotted there. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List every equally likely pair of a ball moved by Davina and a ball picked by Yusuf, and count the pairs that end with an odd ball.

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 Yusuf's chance of drawing the odd ball once one is in Box A, 1 out of 3, leaving out the chance that Davina moves an odd ball at all.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Counts the odd balls in the figure, 3 and 5, over all five balls, 2/5, ignoring that Yusuf draws only from Box A.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Gives the chance that Davina moves an odd ball, 2 out of 3, the first stage only, without multiplying by Yusuf's 1/3 chance of drawing it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(A \text{ then } B) = P(A) \times P(B \text{ after } A)$; A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

Read the figure
Box A holds balls numbered 2 and 6. Box B holds balls numbered 3, 4 and 5.
An odd number does not divide evenly by 2. So the only odd balls are 3 and 5, and both start in Box B. Box A starts with no odd balls at all.
Spot the two chance events
Two things happen by chance, one after the other:
1. Davina picks one of the 3 balls in Box B and moves it to Box A.
2. Then Yusuf picks one of the 3 balls now in Box A.
Yusuf gets an odd ball only if both go right. Davina must move an odd ball. Then Yusuf must pick that very ball.
The question asks only about Yusuf. But his chance depends on Davina’s pick, which comes first.
Find the chance for Davina’s pick
Box B has 3 balls, and 2 of them, the 3 and the 5, are odd.
$P(\text{odd ball moved}) = \frac{2}{3}$
If she moves the 4 instead, Box A holds only even balls. Then Yusuf can’t get an odd ball at all.
Find the chance for Yusuf’s pick
Say Davina moved an odd ball. Box A now holds 3 balls: the 2, the 6 and that one odd ball.
$P(\text{Yusuf picks it}) = \frac{1}{3}$
Watch out: $\frac{1}{3}$ alone is choice B. It forgets that Davina might move the 4.
Multiply the two chances
Both events must happen, so use the rule for events in a row.
$P(\text{odd}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$
Watch out: $\frac{2}{3}$ alone, choice D, stops after Davina’s pick. $\frac{2}{5}$ , choice C, counts 2 odd balls out of all 5. But Yusuf picks only from Box A.
So the answer is A, $\frac{2}{9}$ .

Check your answer

List all 9 equally likely outcomes. Davina can move 3, 4 or 5. Then Yusuf picks one of 3 balls.

Ball moved	Box A then	Odd picks
3	2, 6, 3	1
4	2, 6, 4	0
5	2, 6, 5	1

That is 2 odd picks out of 9, or $\frac{2}{9}$ .

More questions like this

The original DOE question

Question 109 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 109. Open page 214 of the handbook (PDF)

Question 53 of 57

2024 Test B Math, question 53

Question

Which graph represents the solution to $|x+1|\geq 2$ ?

A $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1. The line is shaded from −3 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.$
B $Number line from −3 to 3 with a tick at every integer. Closed dots at −1 and 1. The line is shaded from −1 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.$
C $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1, with the segment between them shaded.$
D $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 3. The line is shaded from −3 to the left, ending in an arrow, and from 3 to the right, ending in an arrow.$

Show the answer

Correct answer: A

Since $|x+1|\geq 2$ , either $x+1\leq -2$ or $x+1\geq 2$ .

Solving $x+1\leq -2$ for x gives $x\leq -3$ .

Solving $x+1\geq 2$ for x gives $x\geq 1$ .

A filled circle on a number line represents “or equal to.” So the correct number line representation has filled circles at −3 and 1, with the number line filled to the left of −3 for “less than” and to the right of 1 for “greater than.”

Question analysis

Students choose the number line graph that shows every solution of an inequality in which x + 1 sits inside absolute value bars. It tests absolute value as distance and graphing inequalities. The tricky part is turning one absolute value statement into two pieces and placing each boundary point and shaded ray correctly.

NYS standards: 7.NS.1c rational number subtraction
7.EE.4b two-step inequalities
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above 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: Something you must already know moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 7 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Reading |x + 1| as the distance from x to −1 makes this quick: the points 2 units away are −3 and 1, and greater than or equal means outside them. Students who solve only one case and mirror it across 0 land on graphs centered on 0, and students who find both endpoints may still shade between them. The graphs differ in just one dot or one shading choice, so each must be read carefully.

Something you must already know moderate Working it out
Solving |x + 1| ≥ 2 needs two cases, x + 1 ≤ −2 or x + 1 ≥ 2, joined by or; absolute value inequalities go beyond grade 7, and the stem gives no hint. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
The right graph and the segment from −3 to 1 share both filled endpoints, so only deciding between outside and between the boundaries separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
Four number lines must be read for filled dots and arrow directions, and two of them differ from the right one only in where a single dot sits. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the absolute value as distance from −1 on the number line, find the points exactly 2 units away, and decide whether the solutions lie between or outside them.

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Solves x + 1 ≥ 2 to get x ≥ 1, then mirrors it across 0 to x ≤ −1, as if the solution were centered on 0 rather than on −1.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Finds −3 and 1 but shades between them, the solution of |x + 1| ≤ 2, reversing which side of the boundaries lies at least 2 units from −1.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Solves x + 1 ≤ −2 to get x ≤ −3, then mirrors it across 0 to x ≥ 3, assuming the boundaries are symmetric about 0.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value as distance: $|x - a| = \text{distance from } x \text{ to } a$; x and a are points on the number line. The bars mean absolute value. |x − a| is how far apart the two points are, which is never negative. Since x + 1 = x − (−1), |x + 1| is the distance from x to −1.

Sort out the question
You need the graph of every x that makes $|x + 1| \geq 2$ true. The sign ≥ means greater than or equal to. The bars mean absolute value.
On a number line, a filled dot means that point counts. A thick line with an arrow means all the numbers that way count too.
Read it as a distance
Adding 1 is the same as subtracting −1, so $x + 1 = x - (-1)$ . By the formula, $|x + 1|$ is the distance from x to −1.
So the inequality says that x is at least 2 units away from −1.
Find the two boundary points
Start at −1 and go 2 units each way.
$-1 - 2 = -3$
$-1 + 2 = 1$
At −3 and at 1, the distance is exactly 2. The words or equal to let them count, so they get filled dots.
Decide which side to shade
Being at least 2 away means being farther out. Numbers between −3 and 1 are too close to −1. For example, 0 is only 1 unit away. So shade to the left of −3 and to the right of 1.
In algebra, that is $x + 1 \leq -2$ or $x + 1 \geq 2$ . These give $x \leq -3$ or $x \geq 1$ .
Match the graph
Choice A has filled dots at −3 and 1, with arrows going outward. That matches.
Watch out: Choice C shades between −3 and 1. Those are the points within 2 of −1, the answer to $|x + 1| \leq 2$ . Choices B and D are centered on 0, but the center here is −1.
So the answer is A, the graph shaded outward from −3 and 1.

Check your answer

Test a point from each part. For x = 2, $|2 + 1| = 3$ , and 3 is at least 2. For x = 0, $|0 + 1| = 1$ , which is less than 2. For x = −3, $|-3 + 1| = 2$ , which counts. Graph A agrees with all three.

More questions like this

The original DOE question

Question 110 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 110. Open page 214 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 54 of 57

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 55 of 57

2024 Test B Math, question 55

Question

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

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

Show the answer

Correct answer: A

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

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

$30{,}000=5x$

$6{,}000=x$

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

$6{,}000=3y$

$2{,}000=y$

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 56 of 57

2024 Test B Math, question 56

Question

A rectangular field had a length of 80 meters and a width of 50 meters. The field was recently enlarged, with the length increased by 15% and the width increased by 10 meters. By what percentage was the area of the field increased?

A26.5%
B35%
C38%
D42.5%

Show the answer

Correct answer: C

The original area of the field was 4,000 square meters.

$80\times 50=4{,}000$

When the field was enlarged, the length increased by 15%, so the new length is 92 meters.

$80\times 15\%$

$80\times 0.15=12$

$80+12=92$

The width increased by 10 meters, so the new width is 60 meters.

$50+10=60$

The new area of the field is 5,520 square meters.

$92\times 60=5{,}520$

The percent increase is found by finding the difference between the two areas and dividing the increase by the original area.

$5{,}520-4{,}000=1{,}520$

$1{,}520\div 4{,}000=0.38=38\%$

Question analysis

A rectangular field 80 meters long and 50 meters wide has its length increased by 15 percent and its width by 10 meters. Students find the percent by which the field's area grew. It tests percent increase with rectangle area, and the tricky part is that one side changes by a percent and the other by a fixed length.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.G.1 area of polygons
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually apply one percent change to one quantity. This changes a length by a percent and a width by a fixed amount, then asks for the percent change in area, a five-step chain with two kinds of change.
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
Many steps moderate
A very close wrong answer minor
Solution: 5 steps
Reading level: Question and choices: about grade 6, 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 ordinary, but the two changes are stated in different units, one a percent and one in meters, which tempts students to force them into the same form. Adding the two percents is the most natural shortcut and lands close to the right value, because it misses only the small extra area where both changes overlap. Computing the old and new areas directly is slower but leaves nothing to chance.

Combines ideas moderate Working it out
Percent change has to be combined with area: the length grows by a percent and the width by meters, and the question asks for the percent change of their product. (Embretson, 1998; Daniel & Embretson, 2010)
Many steps moderate Working it out
Five computations are chained: the new length 92, the new width 60, the areas 4,000 and 5,520, and the increase 1,520 as a share of 4,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The 35% from adding 15% and 20% sits only 3 points below 38%, so a rough estimate of the new area cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compute the new length and the new width, find the old and new areas, and divide the increase in area by the original area.

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

The wrong answers, classified

A Units or scale (King et al., 2004)
Reveals a slip Expected pull: moderate
Treats the 10-meter widening as a 10% increase, so multiplies 1.15 × 1.10 = 1.265 and reports a 26.5% rise.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Adds the length's 15% to the width's 20% increase, as if percent increases in two dimensions simply add, missing the 3% where they overlap.
D Units or scale (King et al., 2004)
Reveals a slip Expected pull: weak
Treats the 15% as 15 meters, making the field 95 by 60 meters, an area of 5,700 square meters and a 42.5% increase.

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.
Percent increase: $\text{percent increase} = \frac{\text{new} - \text{old}}{\text{old}} \times 100$; old is the starting amount, and new is the amount after. new − old is how much it grew. Dividing by old compares the growth with the start. Multiplying by 100 turns that into a percent.

Sort out the question
The field was 80 meters long and 50 meters wide. The length grew by 15%, a percent. The width grew by 10 meters, a number of meters. These are two different kinds of increase.
You need the percent by which the area grew.
Find the old area
You need the old area to compare with the new one.
$A = 80 \times 50 = 4{,}000$
The old field covered 4,000 square meters.
Find the new length and width
15% of 80 is $0.15 \times 80 = 12$ . So the new length is $80 + 12 = 92$ meters.
The new width is $50 + 10 = 60$ meters.
Watch out: 10 meters is not 10%. Treating it as 10% leads to 26.5%, choice A. Treating 15% as 15 meters leads to 42.5%, choice D.
Find the new area
Multiply the new length by the new width.
$A = 92 \times 60 = 5{,}520$
The new field covers 5,520 square meters. That is $5{,}520 - 4{,}000 = 1{,}520$ square meters more.
Find the percent increase
$\text{percent increase} = \frac{5{,}520 - 4{,}000}{4{,}000} \times 100 = \frac{1{,}520}{4{,}000} \times 100 = 38$
The area grew by 38%. Watch out: The width grew by 20%, since 10 is 20% of 50. Adding 15% + 20% = 35% gives choice B. That misses the extra corner piece where both new strips meet.
So the answer is C, 38%.

Check your answer

Use growth factors. The length is multiplied by 1.15. The width goes from 50 to 60, so it is multiplied by 1.2. The area is multiplied by $1.15 \times 1.2 = 1.38$ , which is 38% more.

More questions like this

The original DOE question

Question 113 as printed on page 215 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 113. Open page 215 of the handbook (PDF)

Question 57 of 57

2024 Test B Math, question 57

Question

If x and y are both negative integers with $x<y$ , which expression has the least value?

A $x^{2}y$
B $xy^{2}$
C $-x^{2}y$
D $-xy^{2}$

Show the answer

Correct answer: A

Since x and y are both negative integers, both $x^{2}$ and $y^{2}$ are positive.

Therefore, both $x^{2}y$ and $xy^{2}$ are negative, and both $-x^{2}y$ and $-xy^{2}$ are positive.

Because $x<y$ and both x and y are negative, $|y|<|x|$ and $y^{2}<x^{2}$ . So $|y^{2}|<|x^{2}|$ .

Since $|y|<|x|$ and x and y are integers, $|xy^{2}|<|x^{2}y|$ .

So $x^{2}y<xy^{2}$ .

$-xy^{2}=|xy^{2}|\text{ and }-x^{2}y=|x^{2}y|$

Therefore, $x^{2}y<xy^{2}<-xy^{2}<-x^{2}y$ .

Question analysis

Two negative integers x and y, with x less than y, appear in four expressions built from squares and minus signs. Students find the expression with the least value. It tests multiplying negative numbers and comparing their sizes, and the difficulty is tracking how each minus sign and each square changes both the sign and the size of the result.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7d absolute value versus order
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students multiply negative numbers, but this asks them to compare four variable expressions for any pair of negative integers. Deciding how the squares compare when x is less than y goes beyond typical grade 7 sign practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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

Two facts settle the question: squares of negatives are positive, and the more negative number has the larger square. Students who track the minus signs carelessly think a leading minus makes an expression negative, and students who assume the larger y has the larger square choose the wrong negative expression. Substituting x = −3 and y = −2 gives −18, −12, 18 and 12, which orders all four at once.

Variables instead of numbers moderate Understanding the question
No values are given, only that x and y are negative integers with x < y, so students must reason about signs or invent a pair such as −3 and −2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
x²y and xy² use the same letters with the exponent moved; both are negative, and only comparing x² with y² separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
All four expressions have to be placed in order, because the minus signs in front of the last two turn those expressions positive. (Embretson & Wetzel, 1987)

Strategy: Pick two negative integers with x less than y, such as −3 and −2, evaluate all four expressions, and confirm the order with a second pair.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Knows x²y and xy² are both negative but takes y² as the larger square because y > x, though the more negative x has the bigger square.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads the leading minus as making −x²y negative, missing that y is negative, so −x²y is in fact positive.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Assumes the leading minus makes −xy² negative, missing that x is negative, so −xy² is in fact positive.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.
Squaring a number: $x^2 = x \times x$; x is any number. x² (x squared) means x times itself. The square of a number other than 0 is always positive, because two negatives or two positives multiply to a positive.

Sort out the question
x and y are negative integers, such as −3 and −2. Since x < y, x is to the left of y on the number line. So x is more negative, farther from 0.
You need the expression with the least value. That is the most negative one.
Find the sign of each expression
Squares are positive, so $x^2$ and $y^2$ are positive.
- $x^2 y$ is a positive times a negative, so it is negative.
- $x y^2$ is a negative times a positive, so it is negative.
- $-x^2 y$ and $-x y^2$ are the opposites of these, so they are positive.
So the least value must be A or B. Watch out: The minus sign in front of C and D does not make them negative. It turns a negative into a positive.
Compare A and B with numbers
Try x = −3 and y = −2, since −3 < −2.
$x^2 y = (-3)^2 \times (-2) = 9 \times (-2) = -18$
$x y^2 = (-3) \times (-2)^2 = (-3) \times 4 = -12$
Since −18 < −12, A is smaller.
See why it always works
Both expressions contain x × y, which is positive. $x^2 y = (xy) \times x$ and $x y^2 = (xy) \times y$ . Multiplying the same positive number by the more negative x gives the more negative result.
Watch out: y is bigger than x, but $y^2$ is smaller than $x^2$ : 4 is less than 9. Thinking y² is bigger leads to choice B.
So the answer is A, $x^2 y$ .

Check your answer

Try another pair, x = −5 and y = −1. Then A is $25 \times (-1) = -25$ , B is $(-5) \times 1 = -5$ , C is 25 and D is 5. A is still the least.

More questions like this

The original DOE question

Question 114 as printed on page 215 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 114. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Small factor	Partner
1	168
2	84
3	56
4	42
6	28
7	24
8	21
12	14