Practice Test

6.G geometry

Question 1 of 25

2026 Practice Test C Math, question 23

Question

$Parallelogram PQRS, with Q and R at the top and P and S at the bottom; it leans to the right. A vertical segment QT is drawn from Q down to point T on side PS, with a right-angle mark at T, and labeled 18 cm. Side PS is labeled 34 cm, and side RS is labeled 22 cm.$

On parallelogram PQRS above, the length of $\overline{QT}$ is 18 centimeters. What is the area of the parallelogram?

A112 sq cm
B396 sq cm
C612 sq cm
D748 sq cm

Show the answer

Correct answer: C

PQRS is a parallelogram in which PS is a base and QT is an altitude. The base is 34 cm in length, and the height is the length of the altitude, so the height is 18 cm.

To determine the area of a parallelogram, multiply the base × the height:

$34\text{ cm} \times 18\text{ cm} = 612\text{ cm}^{2}$

Question analysis

Parallelogram PQRS is drawn with a 34 cm base PS, a 22 cm slanted side and an 18 cm segment QT at a right angle to the base, and students find its area. It tests the area formula for a parallelogram. One labeled length is not needed, so the base and height that go together have to be chosen.

NYS standards: 6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The area of a parallelogram is grade 6 content, used here once with the height already drawn and labeled. Grade 7 area tasks usually involve circles, composite shapes or scale drawings, so this is easier.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 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

One multiplication answers the question once the right pair of lengths is chosen. Each wrong choice pairs or combines the labeled lengths in another way: the two sides as for a rectangle, the height with the slanted side, or all four sides as a perimeter. Following the right-angle mark at T from QT down to PS identifies the base and height that belong together.

Reading a figure, table or graph minor Understanding the question
Three lengths are labeled, 34 cm, 18 cm and 22 cm, but only the base PS and the height QT meet at the right-angle mark; the slanted side RS is not needed. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The formula, base times height, is not given, and the height must be the perpendicular distance between two sides, not the length of a slanted side. (Anderson & Krathwohl, 2001)

Strategy: Multiply the base by the height measured at a right angle to it, and check that the length you use as the height meets that base at a right-angle mark.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds all four sides, 2 × (34 + 22) = 112, finding the perimeter of the parallelogram instead of its area.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Multiplies the 18 cm height by the 22 cm slanted side, a pair that does not meet at a right angle, getting 396.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies the two side lengths, 34 × 22 = 748, as if a parallelogram's area were side times side like a rectangle's.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a parallelogram: $A = b \times h$; A is the area. b is the base, the side the shape sits on. h is the height, the straight-up distance from the base to the opposite side, at a right angle to the base. It works because you can cut a triangle off one end, move it to the other end, and make a rectangle.

Sort out the question
PQRS is a parallelogram, a four-sided shape whose opposite sides are parallel. The figure shows three lengths. Side PS, the bottom, is 34 cm. Segment QT goes straight down from Q to PS and is 18 cm. Side RS is slanted and is 22 cm. You need the area, in square centimeters.
Pick the base and the height
The base is PS, 34 cm. The height must meet the base at a right angle. The small square at T shows that QT does. So the height is QT, 18 cm.
Watch out: the 22 cm side is slanted, so it is not the height. Multiplying 22 × 18 gives 396, choice B.
Multiply the base by the height
Use the area formula:
$A = 34 \times 18$
Break it up to multiply:
$34 \times 18 = 34 \times 20 - 34 \times 2 = 680 - 68 = 612$
The area is 612 square centimeters.
Choose the answer
Watch out: 34 × 22 = 748, choice D, treats the shape like a rectangle. Adding all four sides gives 112, choice A. That is the perimeter, the distance around.
So the answer is C, 612 sq cm.

Check your answer

Estimate: 34 is about 30 and 18 is about 20, and $30 \times 20 = 600$ . 612 is close. Also, the area must be less than $34 \times 22 = 748$ , because the height is shorter than the slanted side.

More questions like this

The original DOE question

Question 23 as shown on the DOE’s digital test, page 96 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 96, question 23.

Question 2 of 25

2026 Practice Test C Math, question 42

Question

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

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 7,500

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

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

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

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

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

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

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

Question analysis

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

NYS standards: 7.G.6 area, surface area, and volume
6.G.2 volume with fractional edge lengths
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the volume of a solid made from rectangular prisms, here by taking one away, is named in the grade 7 standards. This is a direct case with whole numbers, two volumes and one subtraction.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Reading a figure, table or graph minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a direct composite-volume problem with whole numbers: one cube, one hole, one subtraction. What it checks is that both volumes are real volumes, the hole as tall as the cube, and that the subtraction is done at the end. Students who stop at the hole’s volume, or who subtract only the square face, give amounts that are easy to type and hard to doubt without choices.

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

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

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 3 of 25

2024 Test A Math, question 8

Question

$Square PQRS, with Q at the top left, R at the top right, P at the bottom left and S at the bottom right, and a right-angle mark at S. Point S is marked as the center of a circle that passes through P and R. The quarter of the circle inside the square, bounded by sides SP and SR and the arc from P to R, is shaded.$

In the figure above, PQRS is a square. Point S is the center of the circle, and points P and R are on the circle. If the area of the square is 4 square centimeters, what is the area, in square centimeters, of the shaded quarter of the circle?

A $\frac{\pi}{4}$
B $\pi$
C $2\pi$
D $4\pi$

Show the answer

Correct answer: B

The area of square PQRS is 4 square centimeters. So

$s^{2} = 4$

$\sqrt{s^{2}} = \sqrt{4}$

$s = 2\text{ cm}$

Sides PS and RS are each a radius of the circle. So the radius of the circle is 2 cm.

The shaded region is $\frac{1}{4}$ of the circle. The area of a circle is $A = \pi r^{2}$ . So the area of the shaded region, in square centimeters, is

$\frac{1}{4}A = \frac{1}{4}\pi r^{2} =$

$\frac{1}{4}\pi(2^{2}) =$

$\frac{1}{4}\pi(4) = \pi$

Question analysis

A geometry question with a figure: square PQRS has one corner at the center of a circle, and the quarter of the circle inside the square is shaded. Given the area of the square, students find the area of the shaded quarter circle. The key is connecting the square's side to the circle's radius.

NYS standards: 7.G.4 circle area and circumference
6.G.5 perfect squares and perfect cubes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Circle area is grade 7 content, and this question adds only one linking step from the square's area to the radius. Typical grade 7 circle work gives the radius directly; here it takes one short extra inference.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 52 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question chains three short facts: from the square's area to its side, from the side to the radius, and from the whole circle to a quarter of it. None is hard, but each wrong choice marks a broken link, most often the whole circle's area, which looks like a finished answer. Pointing to S as the center before computing anything settles the radius.

Reading a figure, table or graph moderate Understanding the question
The radius is never labeled: students must see that S is both a corner of the square and the circle's center, so sides PS and RS are radii. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Neither the circle area formula, π times the radius squared, nor the step from a square's area to its side length is given. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
Square and circle are linked in one chain: the area 4 gives a side of 2, the side is the radius, and the shaded part is a quarter of π × 2². (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Take the square root of the square's area to get its side, which is also the radius, then take one quarter of pi times the radius squared.

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

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Halves the square's side as if it were a diameter, so it uses a radius of 1, and one quarter of π × 1² gives π/4.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Misremembers the circle's area as 2πr², blending it with the circumference 2πr, so one quarter of 2π × 2² gives 2π.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the whole circle's area, π × 2² = 4π, and stops before taking the quarter that is shaded.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s^2$; A is the area, the space inside. s is the length of one side, and s² means s × s. All four sides of a square are equal, so its area is side times side.
Area of a circle: $A = \pi r^2$; A is the area of the whole circle. r is the radius: the distance from the center to the edge. π, called pi, is a number a little more than 3, about 3.14. Multiply π by r × r. This works for every circle.

Sort out the figure
PQRS is a square with an area of 4 sq cm. Point S is the center of the circle. Points P and R are on the circle. The shaded part is a quarter of the circle, inside the square. You need the shaded area, in square centimeters. The answer will have π in it.
Find the side of the square
Use the square formula. The area is 4, so find the number that times itself gives 4.
$A = s^2$
$4 = s^2$
$s = 2$
Each side is 2 cm long, because $2 \times 2 = 4$ .
See that the side is the radius
Side SP goes from the center, S, to point P on the circle. So SP is a radius. So is side SR. The radius is 2 cm. Watch out: the side is a radius, not a diameter, the full width of the circle. Using a radius of 1 leads to $\frac{\pi}{4}$ , choice A.
Find the area of the whole circle
$A = \pi r^2$
$A = \pi \times 2^2 = \pi \times 4 = 4\pi$
The whole circle has an area of 4π sq cm. Watch out: this is the whole circle. Stopping here gives choice D.
Take one quarter
The shaded part is one quarter of the circle, so take one quarter of 4π.
$\frac{1}{4} \times 4\pi = \pi$
The shaded area is π sq cm. So the answer is B, $\pi$ .

Check your answer

π is about 3.14, so the shaded part is about 3.14 sq cm. The whole square is 4 sq cm. In the figure, the shaded part fills most of the square, but not all. So a bit less than 4 makes sense. The choice 4π is about 12.6, far too big.

More questions like this

The original DOE question

Question 65 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 4 of 25

2024 Test A Math, question 21

Question

$A coordinate grid with origin O, the x-axis and y-axis each marked 5, 10, 15 and 20, and grid lines every 5 units. One line segment joins the points (0, 20) and (20, 0), and another joins (0, 15) and (10, 0); all four points are marked with dots. The region between the two segments, with corners at (0, 15), (0, 20), (20, 0) and (10, 0), is shaded.$

What is the area, in square units, of the shaded region shown in the figure above?

A75
B125
C150
D200

Show the answer

Correct answer: B

The area of the shaded region, in square units, is the area of the triangle created by the shaded and unshaded regions minus the area of the triangle created by the unshaded region.

The area of the triangle created by the shaded and unshaded regions is 200 square units.

$\frac{1}{2}bh = \frac{1}{2}(20)(20) = \frac{1}{2}(400) = 200$

The area of the triangle created by the unshaded region is 75 square units.

$\frac{1}{2}bh = \frac{1}{2}(10)(15) = \frac{1}{2}(150) = 75$

Subtract to find the area of the shaded region.

$200 - 75 = 125$

Question analysis

A coordinate geometry question: two line segments cross a grid, one from (0, 20) to (20, 0) and one from (0, 15) to (10, 0), and the region between them is shaded. Students find the shaded area. The shape is not a standard figure, so they need to see it as one triangle with another removed.

NYS standards: 6.G.1 area of polygons
6.G.3 polygons in the coordinate plane
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Area of triangles is grade 6 content; finding a composite area by subtraction is the kind of task grade 7 students meet regularly. Points on the axes make every length easy to read.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
No obvious method minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The shaded shape has no name, so the question is about seeing it as a difference of two right triangles whose legs lie on the axes. Once that is seen, both areas are one-line calculations. The wrong choices come from stopping after either triangle or from pairing a base from one segment with a height from the other.

Reading a figure, table or graph moderate Understanding the question
All four lengths come from where the two segments meet the axes, at 20 and 15 on the y-axis and at 10 and 20 on the x-axis, read against grid lines 5 units apart. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
No obvious method minor Working it out
The shaded region is a four-sided shape with no formula of its own; it has to be seen as the large triangle with the small one removed. (Schoenfeld, 1985)
Something you must already know minor Working it out
The area of a triangle, one half of base times height, is assumed and not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of the large right triangle under the outer segment and subtract the area of the smaller triangle under the inner segment, reading the legs from the axes.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the area of the small unshaded triangle, 1/2 × 10 × 15 = 75, and gives that region instead of the shaded one.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Pairs the outer segment's x-intercept, 20, with the inner segment's y-intercept, 15, as if they bounded one triangle: 1/2 × 20 × 15 = 150.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the area of the large triangle, 1/2 × 20 × 20 = 200, and stops before subtracting the unshaded triangle inside it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

Sort out the figure
The grid lines are 5 units apart. Two slanted segments run from the y-axis down to the x-axis. The outer one goes from 20 on the y-axis to 20 on the x-axis. The inner one goes from 15 on the y-axis to 10 on the x-axis. The shaded region is the strip between them. You need its area, in square units.
See the shape as two triangles
The shaded region has four sides, and it has no simple formula. But look at the big triangle under the outer segment, with its corner at O. It is the shaded strip plus the small white triangle. So the shaded area is the big triangle minus the small one.
Find the area of the big triangle
The big triangle has a right angle at O. Its base runs along the x-axis from 0 to 20, so it is 20 units. Its height runs up the y-axis from 0 to 20, so it is 20 units.
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 20 \times 20 = 200$
Watch out: this is the whole big triangle. Stopping here gives 200, choice D.
Find the area of the small triangle
The small triangle also has its right angle at O. Its base is 10 units, from 0 to 10 on the x-axis. Its height is 15 units, from 0 to 15 on the y-axis.
$A = \frac{1}{2} \times 10 \times 15 = 75$
Watch out: 75 is the white triangle, choice A, not the shaded part. Also, pair the base and height of the same triangle. Mixing 20 and 15 gives 150, choice C.
Subtract to find the shaded area
Take the small white triangle away from the big triangle.
$200 - 75 = 125$
The shaded area is 125 square units. So the answer is B, 125.

Check your answer

In the figure, the shaded strip looks bigger than the white triangle. So the shaded area should be more than half of the big triangle, which is $200 \div 2 = 100$ . 125 is more than 100. And $125 + 75 = 200$ , the whole big triangle.

More questions like this

The original DOE question

Question 78 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 81, question 78. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 5 of 25

2024 Test A Math, question 26

Question

$A parallelogram with horizontal top and bottom sides; its left and right sides both slant down to the right. The top side is labeled 30 ft and the slanted left side is labeled 25 ft. A vertical double-headed arrow from the top side down to the bottom side, with a right-angle mark where it meets the bottom side, is labeled 20 ft.$

What is the area of the parallelogram shown above?

A750 sq ft
B600 sq ft
C500 sq ft
D300 sq ft

Show the answer

Correct answer: B

To find the area of a parallelogram, multiply the length of a side by the altitude for that side.

One property of parallelograms is that opposite sides are the same length.

So the bottom side of the parallelogram is 30 feet long and has an altitude of 20 feet.

Therefore, the area of the parallelogram, in square feet, is $A = bh = 30 \times 20 = 600$ .

Question analysis

A geometry question with a figure: a parallelogram is labeled with a 30-foot top side, a 25-foot slanted side and a 20-foot height drawn at a right angle. Students find its area. The figure gives one length that is not needed, so students must pick the base and height that belong together.

NYS standards: 6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Area of a parallelogram is grade 6 content, and this is a one-step use of the formula. It is easier than typical grade 7 area work, which involves composite figures, circles or scale drawings.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 2, 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

This is a one-step formula question with whole numbers, made slightly harder by one length too many. A student who multiplies the two side lengths, as for a rectangle, or brings in a triangle's one half, finds a listed choice. Matching the base to the height marked with a right angle is all it takes.

Reading a figure, table or graph minor Understanding the question
The figure labels three lengths, but only the 30-foot side and the 20-foot height drawn at a right angle to it belong together; the slanted 25-foot side is extra. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The parallelogram area formula, base times height, is not given, and its height must be the perpendicular distance, not the slanted side. (Anderson & Krathwohl, 2001)

Strategy: Use base times height, where the height is the perpendicular distance between the two parallel sides chosen as bases.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies the two side lengths, 30 × 25 = 750, as if a parallelogram's area were side times side like a rectangle's.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Pairs the 20-foot height with the slanted 25-foot side, 25 × 20 = 500, though the height is drawn at a right angle to the 30-foot sides.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses the triangle formula, 1/2 × 30 × 20 = 300, halving the base times the height.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a parallelogram: $A = b \times h$; A is the area. b is the base, one of the two parallel sides. h is the height: the straight distance between the two parallel sides, at a right angle to the base. It works because you can cut a triangle off one end, move it to the other end and make a rectangle.

Sort out the figure
A parallelogram is a four-sided shape whose opposite sides are parallel. Parallel means they never meet. The figure shows three lengths. The top side is 30 ft. The slanted left side is 25 ft. The arrow from top to bottom is 20 ft, and the small square shows a right angle. You need the area, in square feet.
Choose the base and height
The height must make a right angle with the base. The 20 ft arrow meets the bottom side at a right angle. Opposite sides of a parallelogram are equal, so the bottom side is 30 ft too. So the base is 30 ft and the height is 20 ft. Watch out: the slanted 25 ft side is not needed. Using $25 \times 20 = 500$ gives choice C.
Find the area
$A = b \times h$
$A = 30 \times 20 = 600$
The area is 600 sq ft. Watch out: do not multiply the two sides, $30 \times 25 = 750$ . That is choice A. Do not take half either, which is for triangles and gives 300, choice D. So the answer is B, 600 sq ft.

Check your answer

Picture cutting the triangle off the left end and sliding it to the right end. The shape becomes a rectangle 30 ft long and 20 ft tall, with an area of 600 sq ft. 750 is too big, because the shape is only 20 ft tall, not 25 ft.

More questions like this

The original DOE question

Question 83 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 6 of 25

2024 Test A Math, question 31

Question

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

Show the answer

Correct answer: A

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

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

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

$A = 15^{2} = 225$

Subtract to determine the difference between the areas.

$225 - 169 = 56$

The difference, in square inches, is 56.

Question analysis

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

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Area of a square is elementary content, and squaring 13 and 15 is grade 6 exponent work. Typical grade 7 area problems involve circles, composite shapes or scale drawings; this one has two familiar areas and a subtraction.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is short, but the question invites a shortcut: the sides differ by 2, so it feels as if the areas should differ by 2 squared. Two choices reward working with the side lengths and one rewards perimeters. Squaring each side before subtracting avoids all three, and a sketch of the 13-inch square inside the 15-inch one shows the gap is an L-shaped strip, far more than 4 square inches.

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

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

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 7 of 25

2024 Test A Math, question 39

Question

$Triangle with its two slanted sides each labeled 13 in. and its base labeled 10 in. Beside it, a vertical arrow as tall as the triangle, from its base to its top vertex, is labeled 12 in.$

Raquel is cutting out pieces of cardboard to make a pyramid. She will use a square piece for the base and identical triangular pieces for the sides. The figure above shows the dimensions of the triangle for each side. What will be the total surface area, in square inches, of the pyramid, including the square base?

A280
B295
C340
D360

Show the answer

Correct answer: C

Surface area is found by calculating the area of each side, including the base, and adding them all together.

The base of the pyramid measures 10 inches on each side, so the area of the base is

$10 \times 10 = 100$

Each triangular face has a base of 10 inches and a height of 12 inches, so the area of each face is

$\frac{1}{2} \times 10 \times 12 = 60$

There are 4 triangular faces (because the square base has 4 edges, each of which will have a triangular face). So the total surface area of the pyramid is the sum of the areas of 4 triangles and 1 square.

$(4 \times 60) + 100 = 240 + 100$

So the total surface area is 340 square inches.

Question analysis

This surface area word problem shows one triangular face, with sides of 13 inches, a base of 10 inches and a height of 12 inches, that Raquel will use to build a cardboard pyramid with a square base. Students find the total area of the base and all the triangular sides. Picturing the solid and choosing the right measurements from the figure are the tricky parts.

NYS standards: 7.G.6 area, surface area, and volume
6.G.4 nets and surface area
6.G.1 area of polygons
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students find the surface area of a pyramid from a drawn net or solid. Here only one triangular face is shown, so students must picture the whole pyramid, count its faces, and choose which labeled length belongs in the area formula.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Combines ideas moderate
Something you must already know minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 56 words.
Corrected from the DOE handbook: A wrong unit in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Students have to build the pyramid in their heads from one flat triangle, then choose the right measurement from three labels. The 13-inch sides look like the natural height, and using them gives a listed total; picturing a pyramid with three sloping faces gives another. Sketching the square base with a triangle on each edge, and marking 12 as the height, settles both the face count and the formula.

Reading a figure, table or graph moderate Understanding the question
The figure gives one triangle with three length labels, 13, 13 and 10, and puts its height, 12, on a separate arrow beside it, where it is easy to overlook. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
The solid is never drawn: students must see that the square's side equals the triangle's 10-inch base, count four triangles, and add triangle and square areas. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
A triangle's area uses its perpendicular height, not a slanted side, and a pyramid on a square base has four triangular faces; neither fact is given. (Anderson & Krathwohl, 2001)

Strategy: Sketch the pyramid: a square base whose side is the triangle's base, and one triangular face on each side of the square. Find each area with the matching formula, then add them all.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Information you don't need: The question gives numbers or facts that play no part in the answer.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Counts only three triangular faces, as if the pyramid had a triangular base, so 3 × 60 + 100 = 280.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Counts three faces and also uses the 13-inch side as each triangle's height: 3 × 65 + 100 = 295.
D Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: strong
Takes the 13-inch slanted side as each triangle's height instead of the 12-inch arrow beside it, so 4 × 65 + 100 = 360.

What was corrected from the DOE handbook, and why

Explanation: The handbook ends “So the total surface area is 340 square feet.” The triangle’s dimensions are in inches and the question asks for square inches, so the site now reads “340 square inches.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.
Area of a square: $A = s \times s$; A is the area and s is the length of one side. All four sides of a square are equal, so the area is side times side.
Surface area: $\text{surface area} = \text{sum of the areas of all the faces}$; A face is one flat side of a solid shape. The surface area is the total area of the outside. Picture unfolding the solid into flat pieces. Then add up the area of every piece.

Sort out the question
Raquel’s pyramid has a square base. Each side is a triangle like the one in the figure. The triangle’s base is 10 in. Its two slanted sides are 13 in. Its height, the arrow beside it, is 12 in. You need the total surface area, including the square base, in square inches.
Picture the pyramid
Surface area is the total area of all the flat sides, called faces. Each triangle stands on one edge of the square. So each edge of the square is 10 inches, the same as the triangle’s base. A square has 4 edges, so the pyramid has 4 triangles.
$\text{surface area} = \text{sum of the areas of all the faces}$
Here that is 4 triangles plus 1 square. Watch out: counting only 3 triangles gives 280, choice A.
Find the area of one triangle
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 10 \times 12 = 60$
Each triangle is 60 square inches. Watch out: the height is 12 in., the straight-up arrow. It is not 13 in., the slanted side. Using 13 gives 65 for each triangle and 360 in all, choice D.
Find the area of the base
The base is a square with sides of 10 inches, the same as each triangle’s base.
$A = s \times s$
$A = 10 \times 10 = 100$
The square base is 100 square inches.
Add all five faces
$\text{surface area} = 4 \times 60 + 100 = 240 + 100 = 340$
The four triangles and the square together cover 340 square inches.
So the answer is C, 340 square inches.

Check your answer

Each triangle is half of a 10 by 12 rectangle, which has an area of 120. So two triangles make 120, and four make 240. Add the base: $240 + 100 = 340$ . The units fit too: inches times inches gives square inches.

More questions like this

The original DOE question

Question 96 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 96. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 8 of 25

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 9 of 25

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 10 of 25

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 11 of 25

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 12 of 25

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 13 of 25

2023 Test A Math, question 37

Question

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

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

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

Show the answer

Correct answer: B

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

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

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

$r^{2} = 144$

$r = 12$

Use r to calculate the circumference of the circle:

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

The circumference is $24\pi$ feet.

Question analysis

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

NYS standards: 7.G.4 circle area and circumference
6.G.5 perfect squares and perfect cubes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 circle problems usually give the radius and ask for area or circumference. This one runs backward from the area of a 45° sector, needing the sector's share of the circle, a square root and a switch to circumference: four linked steps.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 94 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 94. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 14 of 25

2023 Test A Math, question 50

Question

A rectangular concrete driveway is 30 feet long, 8 feet wide, and 6 inches thick. What is the volume of the concrete?

A44 cu ft
B48 cu ft
C120 cu ft
D240 cu ft

Show the answer

Correct answer: C

First convert 6 inches to 0.5 foot. Then, calculate the volume, in cubic feet:

$30 \times 8 \times 0.5 = 120$

Question analysis

This volume word problem describes a concrete driveway 30 feet long, 8 feet wide and 6 inches thick, and asks for the volume of the concrete in cubic feet. Students convert inches to feet and multiply three dimensions. Mixing units, or stopping at the area of the driveway's surface, is the main way to go wrong.

NYS standards: 6.G.2 volume with fractional edge lengths
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Volume of a rectangular prism with a fractional edge is grade 6 content. The only extra step is changing inches to feet before multiplying, so it is lighter than grade 7 volume work with composite or triangular prisms.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 2, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Volume of a box is familiar, so the question is decided by the mix of units: two dimensions in feet and one in inches. Multiplying the three numbers as printed gives 1,440, which is not a choice, and that should send students back to convert. The listed wrong answers instead come from leaving out the thickness or combining the numbers without the volume rule.

Something you must already know minor Working it out
The stem gives no formula and no conversion, so students must bring both the volume rule, length times width times height, and the 12 inches in a foot. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The thickness becomes the fraction 1/2 foot, so the product is 30 × 8 × 1/2, or, if the inches are kept, 1,440 divided by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Change 6 inches to a fraction of a foot first so that all three measurements share a unit, then multiply length, width and thickness.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the three numbers in the stem, 30 + 8 + 6 = 44, instead of multiplying the dimensions.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies only the width and the thickness, 8 × 6 = 48, combining two of the stem's numbers and leaving out the 30-foot length.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the area of the driveway's surface, 30 × 8 = 240, and stops before multiplying by the half-foot thickness.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Inches to feet: $\text{feet} = \text{inches} \div 12$; There are 12 inches in 1 foot. So to change inches into feet, divide the number of inches by 12.
Volume of a rectangular box: $V = l \times w \times h$; V is the volume, the space inside a box shape. l is the length, w is the width and h is the height, here the thickness. All three must use the same unit. The answer is in cubic units, such as cubic feet.

Sort out the question
The driveway is a flat box of concrete. It is 30 feet long, 8 feet wide and 6 inches thick. You need its volume, the space the concrete fills, in cubic feet. The thickness is in inches, so change it to feet first.
Change the thickness to feet
$\text{feet} = \text{inches} \div 12$
$6 \div 12 = \frac{1}{2}$
So 6 inches is half a foot. Now all three measures are in feet.
Multiply the three measures
$V = l \times w \times h$
$V = 30 \times 8 \times \frac{1}{2} = 240 \times \frac{1}{2} = 120$
The volume is 120 cubic feet. Watch out: 30 × 8 = 240 is only the area of the top surface, in square feet. Stopping there gives choice D. Adding 30 + 8 + 6 = 44 gives choice A.
So the answer is C, 120 cu ft.

Check your answer

Think of it another way. A slab 30 feet by 8 feet and a full foot thick would hold 240 cubic feet. This one is only half a foot thick, so it holds half as much, 120 cubic feet. Using 6 instead of one half would give 1,440, far too big.

More questions like this

The original DOE question

Question 107 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 107. Open page 89 of the handbook (PDF)

Question 15 of 25

2023 Test B Math, question 18

Question

$A rectangle whose bottom side is labeled 16 ft and whose left side is labeled 8 ft. It is shaded gray except for a white triangle inside it: the triangle’s vertex is at the middle of the left side, and its base lies along the right side and is labeled 4 ft.$

The diagram above shows a rectangle with an inscribed triangle. The triangle’s vertex and base touch the sides of the rectangle as shown. What is the total area of the shaded portion?

A32 sq ft
B64 sq ft
C96 sq ft
D128 sq ft

Show the answer

Correct answer: C

First, find the area of the rectangle, in square feet:

$16 \times 8 = 128$

Then, find the area of the triangle:

$\frac{1}{2} \times 4 \times 16 = 32$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle:

$128 - 32 = 96\text{ sq ft}$

Question analysis

A 16-foot by 8-foot rectangle has a white triangle inside it, with one vertex at the middle of the left side and a 4-foot base along the right side. Students find the total area of the shaded part. They subtract the triangle's area, whose height is not labeled and must be read from the figure.

NYS standards: 6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Composite areas built from rectangles and triangles appear in grades 6 and 7. This one is typical, but the triangle lies on its side, so students must see which length is its height rather than read it from a label.
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
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The shaded region is a rectangle with a triangle cut out, so the plan is standard, but the triangle is turned so that its height runs across the figure. Students who see that the height is the 16-foot side have two short areas to find and one subtraction. The listed wrong answers stop at either area or leave out the one half.

Reading a figure, table or graph moderate Understanding the question
The triangle lies on its side: its 4-foot base runs along the right edge, so its height is the rectangle’s full 16-foot length, which the figure labels only as the bottom side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
The rectangle’s area, the triangle’s area and their difference are three separate results, and the first two, 128 and 32, are themselves listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the rectangle's area, find the triangle's area from its base on the right side and its height across the rectangle, then subtract.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
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 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the triangle’s area, ½ × 4 × 16 = 32, and gives the white region instead of the shaded part around it.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Leaves out the ½ in the triangle’s area, 4 × 16 = 64, and subtracts that from 128, which leaves 64.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the rectangle’s area, 16 × 8 = 128, and stops without subtracting the white triangle.

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 space inside a flat shape, in square units. l is the length and w is the width. It works because the rectangle is filled by w rows of unit squares, with l squares in each row.
Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side you measure from. h is the height, the distance from the base to the opposite corner, at a right angle to the base. You take half, because a triangle is half of a rectangle with the same base and height.

Sort out the figure
The figure is a rectangle 16 feet long and 8 feet tall. Inside it is a white triangle. Its tip touches the middle of the left side. Its base lies along the right side and is 4 feet long. The shaded part is the gray area around the triangle. You need its area.
Plan: find the rectangle’s area, then take away the triangle’s area.
Find the area of the rectangle
$A = l \times w$
$A = 16 \times 8 = 128$
The rectangle’s area is 128 square feet. Watch out: this includes the white triangle. Stopping here gives choice D.
Find the triangle’s height
The triangle lies on its side. Its base is the 4-foot piece on the right side, which runs up and down. The height must meet the base at a right angle, a square corner. So the height runs straight across, from the base to the tip on the left side. That is the full length of the rectangle, 16 feet.
Find the area of the triangle
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 4 \times 16 = 32$
The triangle’s area is 32 square feet. Watch out: don’t forget the $\frac{1}{2}$ . Using $4 \times 16 = 64$ leaves 64 for the shaded part, choice B.
Subtract to find the shaded area
$128 - 32 = 96$
The shaded part is 96 square feet. Watch out: 32 is the white triangle, not the shaded part, so choice A is wrong. So the answer is C, 96 sq ft.

Check your answer

The triangle’s base, 4 feet, is half the rectangle’s height of 8 feet. Its height is the full 16 feet. So the triangle is half of half the rectangle, or one fourth of it. One fourth of 128 is 32, and $128 - 32 = 96$ .

More questions like this

The original DOE question

Question 75 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 75. Open page 196 of the handbook (PDF)

Question 16 of 25

2023 Test B Math, question 24

Question

A prism is made up of 9 congruent cubes. Each of the cubes has side lengths measuring 4 units. What is the volume, in cubic units, of the prism?

A108
B144
C480
D576

Show the answer

Correct answer: D

First, determine the volume of each congruent cube. All the side lengths of the cubes are equal, so the length, width, and height of each cube is 4 units. To determine the volume of each cube, multiply $\text{length} \times \text{width} \times \text{height}$ :

$4 \times 4 \times 4 = 64$

So, the volume of each cube is 64 cubic units.

$Left: a prism built from 9 cubes, three rows of three, one cube deep, with one cube’s height, width and depth each labeled 4. Right: a single cube whose height, width and depth are each labeled 4.$

The prism is comprised of 9 of these cubes, so multiply the volume of each cube (64) by the total number of cubes (9) to determine the volume, in cubic units, of the prism:

$64 \times 9 = 576$

Question analysis

A prism is built from 9 identical cubes, and each cube has edges 4 units long. Students find the volume of the prism in cubic units. It tests the volume of a cube and how volumes add when solids are put together, and it rewards keeping area, surface area and volume apart.

NYS standards: 6.G.2 volume with fractional edge lengths
6.G.5 perfect squares and perfect cubes
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Volume of cubes and rectangular prisms is grade 5 and 6 content. This item needs one cube's volume and a multiplication by 9, two short steps, so it is below typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 27 words.
Corrected from the DOE handbook: A letter x printed in place of a multiplication sign in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question needs one formula and one multiplication, and the stem does not even say how the cubes are arranged, because the volume does not depend on it. The wrong choices come from mishandling the cube’s volume: the exponent multiplied instead of applied, the area of one face, or a surface area. Writing the volume out as edge times edge times edge, in cubic units, rules all three out.

Something you must already know minor Working it out
A cube’s volume, edge × edge × edge, is not given; with it, each 4-unit cube holds 64 cubic units, and the arrangement of the 9 cubes does not matter. (Anderson & Krathwohl, 2001)

Strategy: Find the volume of one cube by multiplying its three edges, then multiply by the number of cubes; how the cubes are arranged does not matter.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Takes 4³ as 4 × 3 = 12, multiplying the edge by the exponent, so each cube seems to hold 12 cubic units and 9 × 12 = 108.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses the area of one face, 4 × 4 = 16 square units, as each cube’s volume, so 9 × 16 = 144.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Pictures the 9 cubes as a 3-by-3 slab, 12 by 12 by 4 units, and finds its surface area, 480 square units, instead of its volume.

What was corrected from the DOE handbook, and why

Explanation: The handbook gives the volume of a cube as “length x width x height”, with the letter x where it means “times”. In a math explanation the letter x reads as a variable, so the site writes length × width × height, the sign the handbook itself uses on the next line (4 × 4 × 4 = 64). Nothing else changes: each cube is 64 cubic units, and the 9 cubes make 64 × 9 = 576.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube: $V = s \times s \times s$; V is the volume, the space inside a solid, in cubic units. s is the length of one edge. All the edges of a cube are equal, so its length, width and height are all s. Multiplying them counts how many unit cubes fit inside.

Sort out the question
A prism, a solid with flat faces, is made of 9 congruent cubes. Congruent means exactly the same size and shape. Each cube has sides of 4 units. You need the volume of the whole prism, in cubic units.
The question doesn’t say how the cubes are stacked. You don’t need to know. Volume is the space inside, and 9 cubes fill the same space however they are arranged.
Find the volume of one cube
$V = s \times s \times s$
$V = 4 \times 4 \times 4 = 64$
Each cube has a volume of 64 cubic units. Watch out: $4^3$ means $4 \times 4 \times 4$ , not $4 \times 3 = 12$ . That slip leads to 108, choice A. Also, $4 \times 4 = 16$ is only the area of one face. It leads to 144, choice B.
Multiply by the number of cubes
The prism is 9 of these cubes, so its volume is 9 times as big.
$64 \times 9 = 576$
The prism’s volume is 576 cubic units.
Give the answer
Watch out: volume counts the space inside, not the outside faces. Suppose the cubes formed one flat 3-by-3 layer. Adding up the areas of its outside faces gives 480, choice C. That is surface area, not volume. So the answer is D, 576.

Check your answer

Estimate: $64 \times 10 = 640$ , and one cube fewer is $640 - 64 = 576$ . You can also test one arrangement. A flat 3-by-3 layer is 12 by 12 by 4 units, and $12 \times 12 \times 4 = 576$ . Same answer.

More questions like this

The original DOE question

Question 81 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 198, question 81. Open page 198 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 17 of 25

2023 Test B Math, question 36

Question

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

Show the answer

Correct answer: B

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

The area of square S is:

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

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

$x-1=3$

$x=4$

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

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

Question analysis

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

NYS standards: 6.G.5 perfect squares and perfect cubes
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The area of a square and recognizing 9 as a perfect square are grade 6 ideas. This takes three short steps with small whole numbers, lighter than grade 7 work on composite figures, though the first step runs backward from area to side.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 93 as printed on page 201 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 93. Open page 201 of the handbook (PDF)

Question 18 of 25

2023 Test B Math, question 43

Question

A certain square puzzle has an area of 64 square inches, and a cubic block has a volume of 64 cubic inches. What is the ratio of the length of a side of the puzzle to the length of a side of the block?

A $1:3$
B $1:2$
C $2:3$
D $2:1$

Show the answer

Correct answer: D

The length of one side of the puzzle is $\sqrt{64}=8$ inches. The length of one side of the block is $\sqrt[3]{64}=4$ inches.

The ratio of the side of the puzzle to the side of the block is $8:4$ , which reduces to $2:1$ .

Question analysis

A square puzzle has an area of 64 square inches, and a cube-shaped block has a volume of 64 cubic inches. Students find the ratio of a side of the puzzle to a side of the block. Both figures are described by the number 64, but turning an area and a volume back into lengths takes different steps, and the order of the ratio matters.

NYS standards: 8.EE.2 square and cube roots
6.G.5 perfect squares and perfect cubes
6.RP.1 ratio concepts and language
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 area and volume problems go from side lengths to an area or a volume. Going backward needs a square root and a cube root, and cube roots are grade 8 content, although 64 is both a perfect square and a perfect cube.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Because 64 is both a perfect square and a perfect cube, the computation is easy once students know to undo squaring and cubing. The risk is at the end: with sides of 8 and 4 inches in hand, writing the block first turns the ratio upside down. Students who never find the side lengths are left with ideas such as two dimensions against three.

Something you must already know moderate Working it out
Neither formula is given: the puzzle’s side is the number whose square is 64 and the block’s edge the number whose cube is 64, and cube roots are grade 8 work. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
A square and a cube share the number 64 in different units, so two different inverse operations feed one ratio, which must list the puzzle’s side first. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the side of the square from its area and the edge of the cube from its volume by testing small whole numbers, then write the puzzle's side first and simplify.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 1 : 3; it makes the puzzle’s side the shorter one, the reverse of 8 and 4 inches, so it mostly draws guesses.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Finds both lengths correctly, 8 and 4 inches, but writes the block’s side first, turning the ratio upside down to 1 : 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the square to 2 and the cube to 3, the powers behind square inches and cubic inches, and writes 2 : 3 without finding any length.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s \times s$; A is the area, in square units. s is the length of one side. All sides of a square are equal, so the area is the side times itself. To find the side from the area, ask which number times itself gives the area.
Volume of a cube: $V = s \times s \times s$; V is the volume, the space inside, in cubic units. s is the length of one edge. A cube’s length, width and height are all equal, so the volume is the edge used three times. To go backward, ask which number used three times gives the volume.

Sort out the shapes
The puzzle is a square with an area of 64 square inches. The block is a cube with a volume of 64 cubic inches. A cube is a box with every edge the same length. You need the ratio of the puzzle’s side to the block’s side, with the puzzle first.
Find the puzzle’s side
Use the area formula backward. Which number times itself makes 64?
$8 \times 8 = 64$
The puzzle’s side is 8 inches. This number is the square root of 64.
Find the block’s side
Use the volume formula backward. Which number used three times makes 64? Try small numbers. 3 × 3 × 3 = 27 is too small.
$4 \times 4 \times 4 = 64$
The block’s side is 4 inches. This number is the cube root of 64.
Write and simplify the ratio
Put the puzzle first, as the question asks. Then divide both parts by 4.
$8 : 4 = (8 \div 4) : (4 \div 4) = 2 : 1$
Watch out: Writing the block’s side first gives 1 : 2, choice B. The order in a ratio matters.
So the answer is D, 2 : 1.

Check your answer

Check both sides. 8 × 8 = 64 square inches, and 4 × 4 × 4 = 64 cubic inches. The puzzle’s side is longer, so its number in the ratio must be bigger. Only choice D puts the bigger number first.

More questions like this

The original DOE question

Question 100 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 100. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 19 of 25

2023 Test B Math, question 49

Question

A rectangular swimming pool of uniform depth is 20 yards long and 14 yards wide. If the water is 6 feet deep, what is the volume of water in the pool, in cubic yards?

A40
B286
C560
D1,680

Show the answer

Correct answer: C

First, change 6 feet to yards:

$6\text{ feet}\times\frac{1\text{ yard}}{3\text{ feet}}=2\text{ yards}$

Then, multiply the dimensions of the pool to find the volume, in cubic yards:

$20\times 14\times 2=560$

Question analysis

A rectangular swimming pool is 20 yards long and 14 yards wide, and the water is 6 feet deep. Students find the volume of water in cubic yards. It tests the volume of a rectangular prism, and the depth is in a different unit from the other measurements, so it has to be converted first.

NYS standards: 6.G.2 volume with fractional edge lengths
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The volume of a rectangular prism is grade 5 and 6 content, and the conversion here gives a whole number. Grade 7 volume work is usually longer, with composite or triangular prisms; here the only extra step is one unit conversion.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas minor
Something you must already know minor
Solution: 2 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 only real trap is the mixed units: 20 and 14 are yards, 6 is feet, and the answer is wanted in cubic yards. Multiplying the three numbers as printed gives a listed choice three times too large. Underlining the unit of each measurement before multiplying, and converting the odd one out, takes seconds.

Combines ideas minor Working it out
The length and width are in yards but the depth is in feet, so a conversion, 6 feet = 2 yards, must come before the volume formula is used. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Neither 3 feet = 1 yard nor volume = length × width × depth is given; both are assumed in a stem that asks for cubic yards. (Anderson & Krathwohl, 2001)

Strategy: Change the depth from feet to yards first, so that all three measurements are in yards, then multiply the length, width and depth.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the three measurements, 20 + 14 + 6 = 40, with no conversion and no volume formula, combining the stem’s numbers.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Finds the area of the water’s surface, 20 × 14 = 280, then adds the depth of 6 instead of multiplying by it.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Multiplies 20 × 14 × 6 without changing 6 feet to 2 yards, mixing yards with feet and getting a volume three times too large.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a rectangular box: $V = l \times w \times h$; V is the volume, the space inside, in cubic units. l is the length, w is the width, and h is the height or depth. All three must be in the same unit. Multiplying them counts the unit cubes that fill the box.
Feet to yards: $\text{yards} = \text{feet} \div 3$; There are 3 feet in 1 yard. So to change feet to yards, divide the number of feet by 3. A yard is longer than a foot, so you get a smaller number.

Sort out the measurements
The pool is shaped like a box. Uniform depth means the water is equally deep everywhere. The pool is 20 yards long and 14 yards wide. The water is 6 feet deep. You need the volume, the space the water fills, in cubic yards.
Change the depth to yards
Two sides are in yards, but the depth is in feet. All three must use the same unit before you multiply. The answer is in cubic yards, so change feet to yards.
$6 \div 3 = 2$
The water is 2 yards deep.
Find the volume
Use the volume formula with all three measurements in yards.
$V = 20 \times 14 \times 2 = 560$
The pool holds 560 cubic yards of water.
Give the answer
Watch out: Multiplying 20 × 14 × 6 mixes yards with feet. It gives 1,680, choice D, which is three times too big.
So the answer is C, 560 cubic yards.

Check your answer

Check the units. Yards times yards times yards gives cubic yards, just what the question asks for. Also, 1,680 ÷ 3 = 560. So using 6 feet unchanged makes the answer exactly three times too big.

More questions like this

The original DOE question

Question 106 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 106. 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 20 of 25

2023 Test B Math, question 56

Question

$A coordinate grid with the origin O, the x-axis marked 1, 2, and 3, and the y-axis marked 1 through 5. Line t passes through the point (0, 1) and line s passes through the point (1, 0); they intersect at P (3, 5), and dashed lines run from P across to 5 on the y-axis and down to 3 on the x-axis. The shaded region is the four-sided figure with vertices O, (0, 1), P (3, 5), and (1, 0).$

Lines s and t intersect at point P (3, 5). What is the area, in square units, of the shaded region in the graph above?

Show the answer

Correct answer: C

First, find the area of the rectangle in the graph:

$3\times 5=15$

Then, find the areas of the two right triangles that are not shaded:

Upper triangle:

$\frac{1}{2}\times 4\times 3=6$

Lower triangle:

$\frac{1}{2}\times 2\times 5=5$

The area of the shaded region, in square units, is calculated by subtracting the total area of the two right triangles from the area of the rectangle:

$15-(6+5)=15-11=4$

Question analysis

Two lines meet at the point (3, 5) on a coordinate grid, and the shaded region between them and the axes is a four-sided shape. Students find its area in square units. The shape is not a standard figure, so no single area formula applies to it directly.

NYS standards: 6.G.1 area of polygons
6.G.3 polygons in the coordinate plane
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Areas of triangles and rectangles are grade 6 content, and grade 7 students often find composite areas. This region is an irregular four-sided shape with two slanted sides, so students need a plan, and the lengths come from coordinates rather than labels.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Reading a figure, table or graph moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards a plan more than computation: the region has no name, so students must see it inside the dashed rectangle from O to P, with two right triangles cut away. Reading each triangle’s legs from the axes is the second hurdle, since the 1-unit intercepts shorten one leg of each, to 4 instead of 5 and to 2 instead of 3. Only one wrong choice, from using one coordinate as both heights, has a clear working behind it, so most errors here are guesses after a failed plan.

No obvious method moderate Working it out
The shaded shape has four sides, two of them slanted, and no formula of its own; it must be boxed in the 3-by-5 rectangle with two triangles removed, or cut along OP. (Schoenfeld, 1985)
Reading a figure, table or graph moderate Understanding the question
No lengths are labeled; every one comes from the coordinates of P (3, 5), the y-intercept 1 of line s and the x-intercept 1 of line t. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
The rectangle, two triangle areas and a subtraction make four computations, 15, 6 and 5, then 15 − 11, each built from different coordinates. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Draw the rectangle from O to P, find the areas of the two unshaded right triangles inside it from the coordinates, and subtract both from the rectangle's area.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2; no rectangle or triangle built from the coordinates leaves that area, so it mostly draws students who judge the narrow region by eye.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Cuts the region along OP into two triangles with bases of 1 but uses 3, the x-coordinate of P, as the height of both: 1.5 + 1.5 = 3.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 7; half of the 3-by-5 rectangle, 7.5, comes closest, so it mostly draws students who guess the region fills about half.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle: $A = l \times w$; A is the area, in square units. l is the length and w is the width. Multiplying them counts the unit squares that fill the rectangle.
Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, one side. h is the height: the distance from the base straight up to the far corner. A right triangle has a square corner, and the two sides that meet there are the base and height. A triangle is half of a rectangle with the same base and height.

Sort out the shaded region
The shaded region has four corners: O at (0, 0), (0, 1) on the y-axis, P at (3, 5), and (1, 0) on the x-axis. Line t meets the y-axis at (0, 1), and line s meets the x-axis at (1, 0). There is no easy formula for this shape. But the dashed lines make a rectangle around it. So find the rectangle’s area and take away the white parts.
Find the rectangle’s area
The rectangle has corners O, (3, 0), P and (0, 5). It is 3 units wide and 5 units tall.
$A = 3 \times 5 = 15$
The rectangle’s area is 15 square units.
Find the upper triangle’s area
The white triangle at the top left has corners (0, 1), (0, 5) and P. Its square corner, or right angle, is at (0, 5). One side runs up the y-axis from 1 to 5, so it is 4 units. The other runs across from 0 to 3, so it is 3 units.
$A = \frac{1}{2} \times 4 \times 3 = 6$
Its area is 6 square units.
Find the lower triangle’s area
The white triangle at the bottom right has corners (1, 0), (3, 0) and P. Its square corner is at (3, 0). One side runs along the x-axis from 1 to 3, so it is 2 units. The other goes up from 0 to 5, so it is 5 units.
$A = \frac{1}{2} \times 2 \times 5 = 5$
Its area is 5 square units.
Watch out: The lines cross the axes at 1, not at O. So these sides are 4 and 2, not 5 and 3.
Subtract the white triangles
Take both triangles away from the rectangle.
$15 - (6 + 5) = 15 - 11 = 4$
The shaded area is 4 square units.
So the answer is C, 4.

Check your answer

Cut the shaded region along a line from O to P. One piece has a base of 1 on the x-axis and a height of 5, since P is 5 units up. Its area is 2.5. The other has a base of 1 on the y-axis and a height of 3, since P is 3 units across. Its area is 1.5. Then 2.5 + 1.5 = 4. Using 3 for both heights gives 3, choice B.

More questions like this

The original DOE question

Question 113 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 113. Open page 208 of the handbook (PDF)

Question 21 of 25

2018 Test A Math, question 12

Question

$Square pyramid with apex R over the square base MNPQ; corner N is at the back, and its edges are dashed as hidden. Segment RS runs down face QRP from R to point S on base edge QP, with a right-angle mark at S.$

In the pyramid above, each triangular face has the same area, and the base MNPQ is a square that measures 8 centimeters on each side. If the length of $\overline{\mathrm{RS}}$ = 6 centimeters, what is the surface area of the pyramid excluding the base?

A48 sq cm
B96 sq cm
C128 sq cm
D160 sq cm

Show the answer

Correct answer: B

Line segment $\overline{\mathrm{RS}}$ is the altitude, or height, of triangle QRP. The length of $\overline{\mathrm{QP}}$ is 8 cm. Use that information to find the area of triangle QRP:

$A=\frac{1}{2}bh=\frac{1}{2}(8)(6)=24$ sq cm.

There are 4 congruent triangles in the pyramid, so the surface area is

4 × 24 = 96 sq cm

Question analysis

A geometry question about the surface area of a square pyramid, leaving out its base. The base edges are 8 cm and segment RS, the height of one triangular face drawn from the top to a base edge, is 6 cm. Students find one face's area and multiply by the number of faces, reading excluding the base carefully.

NYS standards: 7.G.6 area, surface area, and volume
6.G.4 nets and surface area
6.G.1 area of polygons
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Surface area of a right pyramid from its faces is named in the grade 7 standards, and this is a direct two-step case. It matches typical grade 7 work, though the drawing's slant height must be read correctly.
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
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

Once RS is read as the height of a face, the computation is short: one triangle, then four of them. Points go to the figure and to one word. Students who count only the faces drawn in front, or who include the square base, reach listed choices, and the bold "excluding" is all that separates the lateral area from the total. The statement that every face has the same area is what makes multiplying by four valid.

Reading a figure, table or graph moderate Understanding the question
The drawing shows two lateral faces in front and two behind dashed edges, and RS, with its right-angle mark at S, is the height of face QRP, not of the pyramid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The triangle area formula, one half times base times height, and the fact that a square pyramid has four lateral faces are both assumed, not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of one triangular face with one half times base times height, using RS as the height, then multiply by the four congruent faces.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: moderate
Finds one face as ½ × 8 × 6 = 24 but counts only the two faces drawn in front, 2 × 24 = 48, missing the two hidden ones.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Uses only the 8 cm edge, as both base and height of each face, 4 × ½ × 8 × 8 = 128, and never uses RS.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Finds the four faces, 96 sq cm, then adds the 8 × 8 = 64 sq cm base that the question says to exclude.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

Sort out the pyramid
The pyramid sits on a square base, MNPQ. Each side of the square is 8 cm. Four triangle faces rise from the base and meet at the top point, R. Every face has the same area. Segment RS is 6 cm. You need the area of the four faces, “excluding the base”. Excluding means leaving it out.
See what RS measures
RS runs from R down face QRP to point S on edge QP. The small square at S marks a right angle. So RS is the height of triangle QRP. The base of that triangle is QP, one side of the square, so it is 8 cm.
Find the area of one face
Use the triangle formula with base 8 cm and height 6 cm.
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 8 \times 6 = 24$
One face has an area of 24 sq cm.
Add the four faces
All four faces are the same, so multiply by 4.
$4 \times 24 = 96$
Watch out: only two faces are drawn in front. The other two hide behind the dashed lines. Counting just two faces gives 48, choice A. Watch out: don’t add the base, $8 \times 8 = 64$ . That gives 160, choice D. So the answer is B, 96 sq cm.

Check your answer

Each face fits inside an 8 by 6 rectangle, with an area of 48 sq cm. The triangle is half of that, 24 sq cm, and four faces make 96 sq cm. That is less than 160, which would include the base.

More questions like this

The original DOE question

Question 69 as printed on page 83 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 69. Open page 83 of the handbook (PDF)

Question 22 of 25

2018 Test A Math, question 18

Question

$Coordinate grid with the origin O, the x-axis and y-axis each marked 1 and 2, and grid lines at 1 and 2. A straight line starts at O and rises through the grid points (1, 1) and (2, 2). The triangle under that line with corners O, (1, 0) and (1, 1) is shaded.$

What is the area of the shaded region in the graph above?

A0.25 square unit
B0.5 square unit
C1 square unit
D1.5 square units

Show the answer

Correct answer: B

The shaded region is a right triangle. Each leg is 1 unit in length.

So the area is

$A=\frac{1}{2}bh=\frac{1}{2}(1)(1)=\frac{1}{2}$ or 0.5 sq unit

Question analysis

A coordinate grid question: find the area of a shaded right triangle with corners at the origin, (1, 0) and (1, 1), under a diagonal line. Students read the base and height from the grid and apply the triangle area formula. The figure does most of the work; the task is recalling the formula correctly.

NYS standards: 6.G.1 area of polygons
6.G.3 polygons in the coordinate plane
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a triangle's area from a grid is grade 6 work, and this takes one step with whole-unit legs. It is easier than typical grade 7 area problems, which use composite figures or scale drawings.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 2, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a quick formula question with legs of 1. The main risk is the one-half in the triangle formula: leaving it out gives the area of the whole grid square, which is a listed choice. Seeing that the diagonal cuts the unit square exactly in half gives a check that needs no formula at all.

Reading a figure, table or graph minor Understanding the question
The legs of the shaded triangle, from O to (1, 0) and up to (1, 1), are read off the grid lines; the line running on to (2, 2) is not part of the region. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The triangle area formula, one half times base times height, is needed and not given; the shaded triangle is half of one grid square. (Anderson & Krathwohl, 2001)

Strategy: Read the two legs straight off the grid, then apply the triangle area formula, base times height divided by two.

Traps

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

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Judges the shaded triangle by eye as a quarter of the 1-by-1 grid square instead of the half that the diagonal from O to (1, 1) cuts off.
C Wrong formula (King et al., 2004)
Reveals a slip Expected pull: strong
Multiplies base by height, 1 × 1 = 1, and leaves out the one-half, giving the whole unit square instead of the triangle.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 1.5, which is more than the whole grid square around the shaded triangle; it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

Sort out the shaded region
The shaded region is a triangle. Its corners are O, the point (1, 0) and the point (1, 1). O is the origin, the point (0, 0) where the axes meet. The line keeps going up to (2, 2), but only the small triangle is shaded. You need its area in square units.
Read the base and height
The base lies along the x-axis, from O to 1. So the base is 1 unit. The height goes straight up from (1, 0) to (1, 1). So the height is 1 unit too. The base and height meet at a right angle, as the grid lines show.
Use the triangle formula
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} = 0.5$
The area is 0.5 square unit. Watch out: don’t forget the one half. $1 \times 1 = 1$ is the whole grid square, choice C. So the answer is B, 0.5 square unit.

Check your answer

The grid square from 0 to 1 has an area of 1 square unit. The slanted line cuts it corner to corner into two equal triangles. The shaded one is half the square, so 0.5 square unit.

More questions like this

The original DOE question

Question 75 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 75. Open page 84 of the handbook (PDF)

Question 23 of 25

2018 Test B Math, question 35

Question

$Square MNPQ with N at the top left, P at the top right, Q at the bottom right and M at the bottom left; side PQ is labeled 6 cm. S is on the top side NP, T is on the bottom side MQ, and R is on the left side MN. Segments ST, RS and RT are drawn, and the two triangles RNS (top left corner) and TMR (bottom left corner) are shaded.$

R, S, and T are midpoints of the sides of square MNPQ, as shown above. What is the sum of the areas of the shaded triangles?

A9 sq cm
B12 sq cm
C18 sq cm
D36 sq cm

Show the answer

Correct answer: A

The length of one side of the square is 6 cm. Since $R$ , $S$ , and $T$ are midpoints, then $\overline{TM}$ , $\overline{MR}$ , $\overline{RN}$ , and $\overline{NS}$ are all equal to 3 cm. Triangles $TMR$ and $RNS$ are both right triangles, so the area of one of the triangles is

$\frac{1}{2}\times 3\times 3=\frac{9}{2}$

The triangles are congruent, so the sum of the areas is

$\frac{9}{2}+\frac{9}{2}=9\text{ sq cm}$

Question analysis

Points R, S and T are midpoints of the sides of square MNPQ, whose sides are 6 cm long, and students find the total area of the two shaded triangles in its corners. It tests the area of a triangle. The common slips are forgetting the one-half or measuring the wrong region.

NYS standards: 6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Triangle area is grade 6 content, and the corner triangles here are right triangles with easy legs. Typical grade 7 area problems are similar or harder; the only extra step is using midpoints to get the lengths.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 1, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The shaded triangles are right triangles tucked into the corners, so their legs lie along the square's sides and are easy to measure once the midpoints are used. Most lost points come from the one half in the area formula, or from reporting a larger region of the figure, such as the left half or the whole square. Labeling each 3 cm leg on the figure before computing pays off here.

Reading a figure, table or graph moderate Understanding the question
Only one length is labeled, the 6 cm side PQ; the legs NR, NS, MR and MT must each be read as half a side, 3 cm, from the word midpoints. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The area of a triangle, one half times base times height, is not given, and leaving out the one half doubles each 4.5 sq cm triangle. (Anderson & Krathwohl, 2001)

Strategy: Use the midpoints to get each triangle's legs as half the square's side, apply one-half base times height, and remember there are two shaded triangles.

Traps

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

The wrong answers, classified

B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Adds the four 3 cm legs of the two shaded triangles, 3 + 3 + 3 + 3 = 12, treating a total length as an area.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 3 by 3 for each triangle without the one half, 9 + 9 = 18, which also equals the area of the left half, rectangle NSTM.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Squares the only labeled length, 6 × 6 = 36, giving the area of the whole square instead of the two shaded corners.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.

Read the figure
MNPQ is a square, and side PQ is 6 cm. All sides of a square are equal, so every side is 6 cm. A midpoint is the point exactly halfway along a side. R is the midpoint of side NM. S is the midpoint of NP, and T is the midpoint of MQ.
The shaded triangles are RNS at the top and RMT at the bottom.
Find the short sides
Each midpoint cuts a 6 cm side into two equal parts.
$6 \div 2 = 3$
So NR, NS, RM and MT are each 3 cm long.
Find the area of one triangle
Triangle RNS has a square corner at N. A square corner is a right angle. So NS can be the base and NR the height.
$A = \frac{1}{2} \times 3 \times 3 = 4.5$
One triangle is 4.5 sq cm. Triangle RMT has the same sides, so it is also 4.5 sq cm.
Watch out: do not forget the one half. Without it, each triangle is 9 sq cm and the total is 18. That is choice C.
Add the two triangles
There are two shaded triangles, so add their areas to get the total.
$4.5 + 4.5 = 9$
So the answer is A, 9 sq cm.

Check your answer

The left half of the square is rectangle NSTM, 3 cm by 6 cm. Its area is $3 \times 6 = 18$ sq cm. The white triangle RST fills half of it. So the shaded part is the other half, 9 sq cm. Choice D, 36, is the whole square.

More questions like this

The original DOE question

Question 92 as printed on page 159 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 92. Open page 159 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 25

2018 Test B Math, question 39

Question

A wooden box has a square base. The height of this box is 3 times the length of one side of the base. If one side of the base is 3 feet long, what is the volume of this box?

A9 cu ft
B27 cu ft
C36 cu ft
D81 cu ft

Show the answer

Correct answer: D

One side of the square base is 3 ft long. Since the height of the box is 3 times the length, then the height is

3 × 3 = 9 ft.

The volume of a rectangular prism is length × height × width. The volume of the wooden box is

V= 3 × 3 × 9 = 81 cu ft.

Question analysis

A wooden box has a square base, and its height is 3 times the length of a side of the base; students find its volume in cubic feet. It tests the volume of a rectangular prism with dimensions described in words. The work is short, but the height must be built from the side length before multiplying.

NYS standards: 6.G.2 volume with fractional edge lengths
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Volume of a box is grade 5 and 6 content, and this is a two-step version with whole numbers. Grade 7 volume problems usually involve composite or triangular prisms, so this sits below.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 1, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a routine volume question once all three dimensions are written down. Errors come from treating the box as a 3-foot cube, from stopping at the base area, or from mixing perimeter into the product. Listing the dimensions 3, 3 and 9 before multiplying prevents each of them.

Turning words into math minor Understanding the question
"The height of this box is 3 times the length of one side of the base" must become a height of 9 feet before the volume formula can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The volume of a box, length times width times height, is not given, and the square base means the length and width are both 3 feet. (Anderson & Krathwohl, 2001)

Strategy: Write down all three dimensions first, computing the height from the base side, then multiply length, width and height.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Stops at the area of the square base, 3 × 3 = 9, or at the 9-foot height, without bringing in the third dimension.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Uses 3 feet for the height too, as if the box were a cube, 3 × 3 × 3 = 27; multiplying only 3 × 9 also gives 27.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Multiplies the base perimeter, 4 × 3 = 12 feet, by 3, putting a perimeter where the base area belongs.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a box: $V = l \times w \times h$; V is the volume, the space inside the box. l is the length, w is the width and h is the height. l times w is the area of the bottom. Multiplying by h stacks that many layers of cubes.

Sort out what you know
The box has a square base with sides of 3 feet. A square has equal sides, so the length and width are both 3 feet. The height is “3 times the length of one side of the base”. You need the volume, the space inside the box.
Find the height
The height is 3 times the 3-foot side.
$h = 3 \times 3 = 9$
The height is 9 feet.
Watch out: the height is not 3 feet. If you use 3 for the height, you get $3 \times 3 \times 3 = 27$ . That is choice B.
Find the volume
Use the volume formula with all three sizes.
$V = 3 \times 3 \times 9 = 81$
The volume is 81 cubic feet.
Watch out: $3 \times 3 = 9$ is only the area of the bottom. Stopping there gives choice A.
Give the answer
The box is 3 feet long, 3 feet wide and 9 feet tall. So the answer is D, 81 cu ft.

Check your answer

Think in layers. The bottom layer holds $3 \times 3 = 9$ cubes, each 1 foot on a side. The box is 9 feet tall, so it holds 9 layers. Then $9 \times 9 = 81$ cubic feet. The units are cubic feet because the answer is a volume.

More questions like this

The original DOE question

Question 96 as printed on page 160 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 96. Open page 160 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 25 of 25

2018 Test B Math, question 48

Question

$Coordinate axes with origin O. A shaded triangle stands on the x-axis: its base runs from the point marked m to the point marked n, both to the right of O, and its top vertex touches a dashed horizontal line drawn at 4 on the y-axis.$

What is the area of the shaded triangle shown above?

A $m+n$
B $n-m$
C $2\left ( n-m \right )$
D $4\left ( n-m \right )$

Show the answer

Correct answer: C

The height of the triangle is 4 units. The length of the base is $n-m$ . So the area is

$A=\frac{1}{2}(n-m)(4)=2(n-m)$

Question analysis

Students find the area of a shaded triangle drawn on coordinate axes, with its base between the points m and n on the x-axis and its top vertex at a height of 4. It tests the triangle area formula with an algebraic side length. Writing the base as an expression in m and n is the step that needs care.

NYS standards: 6.G.1 area of polygons
6.EE.2a writing algebraic expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Triangle area is grade 6 content, but a base written with two variables is less familiar. With only two steps, the demand matches typical grade 7 work that mixes geometry with algebraic expressions.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Reading a figure, table or graph moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 7, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is trivial, but students must treat a length as a difference of coordinates and keep a formula's one half when the answer is a letter expression. Those who forget the half, or who stop at the base length, find their result among the choices. Trying m = 1 and n = 4 gives a triangle of area 6 that checks the expressions quickly.

Variables instead of numbers moderate Understanding the question
The base runs from m to n on the x-axis, so its length is the expression n − m, and the area stays in letters with no number to check. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph moderate Understanding the question
The height, 4, is read from the dashed line through the apex rather than labeled on the triangle, and the base endpoints appear only as tick labels. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer moderate Choosing the answer
4(n − m) is the right 2(n − m) doubled, written the same way except for the coefficient; only remembering the one half in a triangle's area separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the base as the difference between its endpoint values, read the height from the figure, and apply one-half times base times height.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two labels, m + n, as if the base were the sum of the endpoints, and goes no further with the height.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the base length n − m correctly and gives it as the area, never multiplying by the height or taking one half.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies the base n − m by the height 4 but forgets the one half in the triangle area formula.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.
Length along the x-axis: $\text{length} = \text{right end} - \text{left end}$; The right end and the left end are the numbers where a segment starts and stops on the x-axis. Subtracting the smaller from the larger tells how long the segment is.

Read the figure
The shaded triangle sits on the x-axis, the flat number line. Its base runs from $m$ to $n$ . Its top corner touches the dashed line at height 4. You need the area, written with the letters $m$ and $n$ .
Find the length of the base
The base starts at $m$ and ends at $n$ . Use the length rule: right end minus left end.
$\text{base} = n - m$
Watch out: adding the ends, $m+n$ , does not give a length. That is choice A.
Find the height
The height is the straight-up distance from the base to the top corner. The top corner is on the dashed line, which is at 4 on the y-axis. So the height is 4.
Find the area
Use the triangle area formula.
$A = \frac{1}{2} \times (n - m) \times 4 = 2(n - m)$
Half of 4 is 2, so the area is $2(n-m)$ .
Watch out: forgetting the one half gives $4(n-m)$ , choice D. Stopping at the base gives $n-m$ , choice B.
Give the answer
The area is 2 times the length of the base. So the answer is C, $2(n-m)$ .

Check your answer

Try numbers. Say $m = 1$ and $n = 4$ . The base is 3 and the height is 4, so the area is $\frac{1}{2} \times 3 \times 4 = 6$ . Choice C gives $2(4-1) = 6$ . It matches. Choice D would give 12.

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The original DOE question

Question 105 as printed on page 162 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 105. Open page 162 of the handbook (PDF)

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