Practice Test

Wrong formula

Question 1 of 35

2026 Practice Test B Math, question 28

Question

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

Use $\pi$ to express your answer.

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

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

Show the answer

Correct answer: 25π

Solution:

$25\pi$

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

$A = \pi r^2$

$r = 5$

$A = \pi(5)^2$

$A = 25\pi$

Question analysis

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

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the area of a circle from its radius is the grade 7 circle standard applied directly, in one routine calculation. Typing the answer in terms of π, with no choices to compare, adds a little risk.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 2 of 35

2026 Practice Test C Math, question 6

Question

$A parallelogram that leans to the right. The angle at its upper left vertex is labeled 3x°, and the angle at its lower left vertex is labeled 2x°.$

In the parallelogram above, what is the value of x?

Show the answer

Correct answer: A

In a parallelogram, any two adjacent angles are supplementary, so the sum of the two given angles is equal to 180°.

Write an equation showing that the sum of $3x^{\circ}$ and $2x^{\circ}$ is equal to 180°:

$3x + 2x = 180$

Solve for x:

$5x = 180$

$\frac{5x}{5} = \frac{180}{5}$

$x = 36$

Question analysis

A parallelogram is drawn with the two angles at the ends of its left side labeled 3x° and 2x°, and students find the value of x. It tests the angle facts of a parallelogram and solving a simple equation. Knowing what the two angles must add up to is the key fact, since the figure gives no other number.

NYS standards: 7.G.5 angle relationships and equations
8.G.5 angle sums, transversals, angle-angle similarity
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 The angle fact needed, about consecutive angles of a parallelogram, rests on grade 8 parallel-line reasoning. Once it is known, the work is one short equation, like a typical grade 7 problem about angle sums.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the angle fact is in place, 5x = 180 is a one-line equation, so the question tests recall more than computation. Students who use 90° for the pair land on half the right value, and the two smallest choices come only from combining the labels' 3 and 2. A sense check helps: x = 36 makes the angles 108° and 72°, one obtuse and one acute, just as the drawing shows.

Something you must already know moderate Working it out
The figure gives no number of degrees, only 3x° and 2x°, so everything rests on recalling that the two angles at the ends of one side of a parallelogram add to 180°. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The marked angles sit at the upper and lower ends of the left side, consecutive rather than opposite, which is what makes them supplementary instead of equal. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide what two angles that share a side of a parallelogram add up to, set 3x + 2x equal to that total, solve for x, and check that the two angles fit the figure.

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

The wrong answers, classified

B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses 90° as the total of the two angles, the complementary total, so 5x = 90 gives 18, though consecutive angles of a parallelogram add to 180°.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the two coefficients from the labels, 3 × 2 = 6, without using any angle fact or writing an equation.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Combines 3x + 2x into 5x and gives the coefficient 5, stopping before dividing the 180° total by it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Neighboring angles of a parallelogram: $a + b = 180^{\circ}$; a and b are two angles of a parallelogram that share a side, like the top left and bottom left angles. They always add to 180°. Here is why. The four angles add to 360°, and opposite angles are equal. So each pair of neighbors is half of 360°.

Sort out the question
The figure is a parallelogram. That is a four-sided shape whose opposite sides are parallel, so they never meet. Two angles are marked: 3x° at the upper left and 2x° at the lower left. You need the value of x.
Write an equation for the angles
The two marked angles are at the two ends of the left side. So they are neighbors, not opposite angles. Use the rule that neighbors add to 180°:
$3x + 2x = 180$
Watch out: the total is 180°, not 90°. Using 90 gives 5x = 90 and x = 18, choice B.
Solve for x
Three x’s plus two x’s make five x’s:
$5x = 180$
Divide both sides by 5 to get x alone:
$x = \frac{180}{5} = 36$
Watch out: stopping at 5, the number in front of x, gives choice D. Multiplying 3 × 2 = 6 gives choice C, but that uses no angle fact.
Choose the answer
The value of x is 36. It is a number, not the size of an angle. The angles themselves are 3 × 36 and 2 × 36 degrees.
So the answer is A, 36.

Check your answer

Find the two angles. The top one is $3 \times 36 = 108$ degrees. The bottom one is $2 \times 36 = 72$ degrees. They add to 180°. In the figure, the top angle is wide, more than 90°. The bottom one is narrow, less than 90°. That matches.

More questions like this

The original DOE question

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

Question 3 of 35

2026 Practice Test C Math, question 20

Question

Two circles, with centers at A and C, are shown. Line segment AB is congruent to line segment DE.

$Two circles. The larger circle has center A, and a radius is drawn from A to point B on the circle, labeled 5. The smaller circle has center C, and a segment is drawn from point D on the circle, through C, to point E on the circle.$

Determine the area, A, and the circumference, C, of each circle to complete the table.

Select one answer in each row.

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

Show the answer

Correct answer: A = 25π: Circle A; C = 5π: Circle C

$A = 25\pi$ : Circle A

$C = 5\pi$ : Circle C

As shown in the illustration, AB (the radius of Circle A) is 5 units in length. Since DE (the diameter of Circle C) is congruent to AB, we know it is also 5 units in length.

Circle A:

$A = \pi r^{2}$

$A = 25\pi$

$r = 5$

We know the radius of Circle A is 5 so the answer to row 1 is A.

Circle C:

$C = \pi \cdot d$

$C = 5\pi$

We have to look for the circle with a diameter of 5. That is circle C.

Question analysis

Two circles centered at A and C are shown: radius AB of Circle A is labeled 5, and segment DE, drawn through C, is congruent to AB. Students decide whether an area of 25π and a circumference of 5π each belong to Circle A, Circle C or neither. Keeping radius and diameter apart, and area and circumference, is the challenge.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying the area and circumference formulas is the grade 7 circle standard. This gives a radius for one circle and, for the other, a segment known only through congruence, and both rows of the table must be right.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know moderate
Much to keep track of minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Circle A is direct: its labeled radius of 5 gives an area of 25π. Circle C is where points are lost, since its size comes from a sentence rather than a label, and the segment through its center is a diameter, so its circumference is 5π while its area is only 6.25π. The row labels favor guessers, because the A and C in A = 25π and C = 5π match the circles' names, though they stand for area and circumference.

Reading a figure, table or graph moderate Understanding the question
Circle C's size is shown only by segment DE, drawn through center C with no label, and only the stem's sentence that DE is congruent to AB makes it a diameter of 5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know moderate Working it out
Neither A = πr² nor C = πd is given, and one circle is measured by a radius and the other by a diameter, so each formula needs the right length. (Anderson & Krathwohl, 2001)
Much to keep track of minor Working it out
Four values are in play, areas of 25π and 6.25π and circumferences of 10π and 5π, and both rows must be matched correctly for the item to count. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write down the radius and the diameter of each circle, compute both circles' areas and circumferences in terms of π, then match each value in the table.

Traps

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

The wrong answers, classified

1:B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Squares the diameter instead of the radius, so Circle C's area seems to be π × 5² = 25π while Circle A's becomes 100π.
1:C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses 2πr for the area, so Circle A seems to have an area of 10π and Circle C one of 5π, and no circle seems to have 25π.
2:A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses π times the radius for the circumference, so Circle A's radius of 5 seems to give a circumference of 5π.
2:C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Overlooks that DE is congruent to AB, so Circle C has no known size, and with Circle A's circumference at 10π picks Neither.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle: $A = \pi r^{2}$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. Square the radius, then multiply by π.
Circumference of a circle: $C = \pi d$; C is the circumference, the distance around the circle. d is the diameter, the distance across the circle through its center. A diameter is twice as long as a radius. Multiply the diameter by π.

Sort out the question
Circle A has its center at A. Its radius, AB, is 5. Circle C has its center at C. Segment DE goes through C, so it is a diameter. Congruent means the same length, so DE is 5 too.
Careful: in the table, A means area and C means circumference. Circle A and Circle C are just names.
Find each radius and diameter
Circle A: the radius is 5, so the diameter is $2 \times 5 = 10$ .
Circle C: the diameter is 5, so the radius is $5 \div 2 = 2.5$ .
Watch out: without the sentence about DE, Circle C seems to have no size. Then you might pick Neither for the second row.
Check the area row
Use the area formula for each circle. For Circle A:
$A = \pi \times 5^{2} = 25\pi$
For Circle C:
$A = \pi \times 2.5^{2} = 6.25\pi$
Only Circle A has an area of 25π.
Watch out: squaring Circle C’s diameter instead of its radius gives 25π by mistake.
Check the circumference row
Use the circumference formula for each circle. For Circle A:
$C = \pi \times 10 = 10\pi$
For Circle C:
$C = \pi \times 5 = 5\pi$
Only Circle C has a circumference of 5π.
Watch out: π times the radius is not the circumference. It would give Circle A 5π.
Fill in the table
Choose one button in each row. In the row for A = 25π, choose the Circle A column. In the row for C = 5π, choose the Circle C column. Neither stays empty.
So the answers are A = 25π: Circle A and C = 5π: Circle C.

Check your answer

Compare sizes. Circle A’s radius, 5, is twice Circle C’s radius, 2.5. So the distance around Circle C should be half of Circle A’s 10π. Half of 10π is 5π, which matches. And the bigger circle has the bigger area, 25π.

More questions like this

The original DOE question

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

Question 4 of 35

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 5 of 35

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 6 of 35

2026 Practice Test C Math, question 49

Question

The box plot illustrates the number of hours eighth-grade students spend on screens during a typical week. What is the difference between the range and the interquartile range (IQR) of the screen time data?

$A box plot titled NUMBER OF SCREEN HOURS FOR 8th GRADE STUDENTS above a number line labeled Hours, from 5 to 20 with a labeled tick at every 1. The left whisker starts at 5, the box runs from 8 to 15 with the median line at 12, and the right whisker ends at 20.$

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

The range of screen time hours is hours.

The IQR is hours.

The difference between the range and the IQR is hours.

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

Show the answer

Correct answer: blank 1: 15; blank 2: 7; blank 3: 8

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

Maximum value $= 20\text{ hr}$ , Minimum value $= 5\text{ hr}$

$20\text{ hr} - 5\text{ hr} = 15\text{ hr}$

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

Upper Quartile Value $= 15\text{ hr}$ , Lower Quartile Value $= 8\text{ hr}$

$15\text{ hr} - 8\text{ hr} = 7\text{ hr}$

Determine the difference between the range (15 hr) and the interquartile range (7 hr):

$15\text{ hr} - 7\text{ hr} = 8\text{ hr}$

Question analysis

A box plot shows how many hours a week eighth-grade students spend on screens, on a number line marked every hour. Students move number tiles to complete three sentences: the range, the interquartile range, and the difference between them. Each tile can be used only once, and keeping the range and the interquartile range apart is the tricky part.

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Reading a box plot and finding its range and interquartile range are grade 7 statistics skills. The values sit on labeled whole-number ticks and the three sentences lead through the steps, a typical grade 7 task.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 69 words.
Corrected from the DOE handbook: A missing comma in the explanation was restored.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The plot is drawn on whole-hour ticks, and the three sentences lay out the steps, so this is a direct check of two definitions. The main risk is taking a value off the plot, such as the lower quartile of 8, for a spread between two values. Because the four tiles allow only one use each, a student who confuses the measures can still fill every box, which hides the error.

Something you must already know minor Working it out
The sentences name the range and the IQR but never define them; students must know the range spans the whiskers and the IQR spans only the box. (Anderson & Krathwohl, 2001)
Many steps minor Working it out
Three boxes must all be right, the third depends on the first two, and each tile can be used once, so one mix-up spreads to a second box. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Read the five key values off the plot, subtract the least from the greatest for the range, subtract the lower quartile from the upper for the IQR, then subtract those results.

Traps

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

The wrong answers, classified

DCB Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the lower quartile, 8, as the interquartile range, reading a value off the box instead of its width, so the difference becomes 15 − 8 = 7.
BDC Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Swaps the two measures, giving the box’s width, 7, as the range and the whiskers’ spread, 15, as the IQR, while the difference stays 8.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation runs two values together: “Maximum value = 20 hr Minimum value = 5 hr”. Its next line is “Upper Quartile Value = 15 hr, Lower Quartile Value = 8 hr”, and Question 32’s explanation has “Maximum value = 84° F, Minimum value = 39° F”. The site adds the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The box plot shows the hours of screen time for eighth graders in a week. You need three numbers: the range, the IQR, and the difference between them.
There are four tiles: 5, 7, 8 and 15. Each tile is used at most once, and one tile is left over.
Read the five values
The number line has a tick at every hour, and every value sits on a tick:
- The left whisker starts at 5. This is the minimum.
- The box starts at 8. This is the lower quartile, the middle of the lower half of the data.
- The line inside the box is at 12. This is the median, the middle value.
- The box ends at 15. This is the upper quartile, the middle of the upper half.
- The right whisker ends at 20. This is the maximum.
Find the range
Use the range formula with the two ends of the whiskers:
$\text{range} = 20 - 5 = 15$
The screen times spread over 15 hours, from the least to the most.
Find the IQR
Use the IQR formula with the two edges of the box:
$\text{IQR} = 15 - 8 = 7$
The middle half of the students spend between 8 and 15 hours, a spread of 7 hours.
Watch out: the IQR is a width, not a point on the plot. The tile 8 is the lower quartile. Putting it in the IQR box also makes the last box 7, which is wrong.
Find the difference
Subtract the IQR from the range:
$15 - 7 = 8$
Watch out: do not swap the two measures. The range uses the whiskers’ ends, and the IQR uses only the box.
Fill in the three boxes
On the test, you drag a tile into each box. On this site, you choose each tile from its box’s menu. The range box gets 15, the IQR box gets 7, and the difference box gets 8. The tile 5 is not used.
So the boxes get 15, 7 and 8.

Check your answer

The difference is just the two whiskers added together. The left whisker is $8 - 5 = 3$ hours long. The right whisker is $20 - 15 = 5$ hours long. $3 + 5 = 8$ , which matches the third box.

More questions like this

The original DOE question

Question 49 as shown on the DOE’s digital test, page 122 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 122, question 49. This is the DOE’s version, before the corrections listed in the analysis.

Question 7 of 35

2026 Practice Test C Math, question 50

Question

Terry wants to pour cement around the edge of the circular patio in her backyard. The patio has a radius of 5 feet. What is the distance, in feet, around the edge of the patio? Use 3.14 for $\pi$ .

A15.7
B31.4
C49.3
D78.5

Show the answer

Correct answer: B

To find the distance around a circle if the radius is known, use the formula for circumference, $C = 2\pi r$ .

$C = 2(3.14)(5) = 31.4$

Question analysis

Terry wants to pour cement around the edge of a circular patio with a radius of 5 feet. Students find the distance around the edge, using 3.14 for π. It tests the circumference of a circle, and because the formulas for a circle's area and circumference both use the radius, they are easy to mix up.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying a circle formula is a grade 7 standard, and this is a one-formula problem with a given value for π. The only demand beyond routine is recalling the right formula, which the test does not print.
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
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One formula and one multiplication finish the problem once the right formula is chosen. The cement in the story pulls toward area, and the area, 78.5, is listed, as is half the circumference, from multiplying 3.14 by the radius alone. Remembering that the circumference is π times the diameter, here 10 feet, and that the answer is in feet, not square feet, settles it.

Something you must already know moderate Working it out
The circumference formula is not printed, and the area formula uses the same radius and π, so recalling the wrong one gives a number that looks just as finished. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
Pouring cement suggests covering a surface, but the question asks for “the distance, in feet, around the edge”, which is the circumference, not the area. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Decide which circle measurement the distance around the edge is, recall its formula, and substitute the radius and 3.14, doubling the radius wherever the formula calls for the diameter.

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.
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 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies π by the radius, 3.14 × 5 = 15.7, using the radius where C = πd needs the diameter, so it finds half the distance around.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Starts from the area formula and squares π instead of the radius, 3.14 × 3.14 × 5 ≈ 49.3, mixing the wrong formula with a slip in squaring.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Finds the patio’s area, 3.14 × 5² = 78.5 square feet, using the area formula instead of the circumference for the distance around.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \pi r$; C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, about 3.14. 2r is the diameter, the distance all the way across. The distance around is always about 3.14 times that.
Area of a circle: $A = \pi r^2$; A is the area, the space inside the circle, in square units. r² means r times r. It is here so you can tell the two formulas apart. This question does not use it.

Sort out the question
The patio is a circle with a radius of 5 feet. The radius is the distance from the center to the edge. Terry wants cement around the edge.
You need the distance around the edge, in feet. The distance around a circle is called its circumference. Use 3.14 for π.
Choose the right formula
The distance around is a length, so use the circumference formula, $C = 2\pi r$ .
Watch out: pouring cement may make you think of area. The area, $3.14 \times 5 \times 5 = 78.5$ square feet, is choice D. That is the space inside the patio, not the distance around it.
Put in the numbers
Put 3.14 in for π and 5 in for r. Multiply 2 by 5 first, because that is easy:
$C = 2 \times 3.14 \times 5 = 10 \times 3.14 = 31.4$
The distance around the patio is 31.4 feet.
Watch out: $3.14 \times 5 = 15.7$ , choice A, forgets the 2. It goes only halfway around.
Give the answer
The distance around the edge of the patio is 31.4 feet.
So the answer is B, 31.4.

Check your answer

Estimate. The diameter is 10 feet. The distance around a circle is a little more than 3 times its diameter. $3 \times 10 = 30$ , so about 31 feet makes sense. Choice D, 78.5, is far too big, and it is in square feet.

More questions like this

The original DOE question

Question 50 as shown on the DOE’s digital test, page 123 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 123, question 50.

Question 8 of 35

2026 Practice Test D Math, question 8

Question

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

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 5.86

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

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

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

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

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

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

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

$5.86\text{ sq in.}$

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

Question analysis

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

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a circle's area is a grade 7 standard, and subtracting a hole from a shape is common grade 7 practice. The radius must come from the diameter and the result must be rounded, but the steps are routine.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 49 words.
Corrected from the DOE handbook: A wrong symbol in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 9 of 35

2024 Test A Math, question 2

Question

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

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

Show the answer

Correct answer: 150

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

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

$r + m = 180$

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

$r = 5m$

$5m + m = 180$

$6m = 180$

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

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

Therefore, the measure of angle R is 150°.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 10 of 35

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 11 of 35

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 12 of 35

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 13 of 35

2024 Test B Math, question 5

Question

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

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

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

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

Show the answer

Correct answer: 32

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

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

The height of the triangular prism is 2 inches.

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 35

2024 Test B Math, question 18

Question

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

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

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

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

Show the answer

Correct answer: D

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

Question analysis

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

NYS standards: 7.SP.1 box plots and interquartile range
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Reading box plots, finding the interquartile range and testing for outliers are all grade 7 standards in New York. This item gathers several of them in one question, but each check is short, typical of grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Something you must already know moderate
Every choice must be checked moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 63 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 15 of 35

2024 Test B Math, question 32

Question

Two circular plates have the dimensions shown.

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

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

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

Show the answer

Correct answer: C

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

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

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

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

Question analysis

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

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a circle's area is a grade 7 standard, and this asks for two areas and their difference. It matches typical classwork: the formula must be recalled and each diameter halved first, with no further twist.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 16 of 35

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 17 of 35

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 18 of 35

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 19 of 35

2023 Test A Math, question 15

Question

A swim instructor used the box plot below to display the distribution of the ages of students who signed up for swim lessons.

$Box plot titled STUDENTS’ AGES above a number line from 5 to 13 with a tick and a label at every whole number and arrows at both ends. The left whisker starts at 7, the box runs from 8 to 10 with the median line at 9, and the right whisker ends at 12.$

Which statement about the distribution of ages is true?

AThe data contain an outlier.
BApproximately half the students are exactly 9 years old.
CApproximately one-fourth of the students are at least 10 years old.
DThe interquartile range is 3 greater than the range of the data.

Show the answer

Correct answer: C

The data points in the box plot are as follows:

minimum: 7
first quartile: 8
median: 9
third quartile: 10
maximum: 12

Since the value of the third quartile is 10, approximately one-fourth of the students are 10–12 years old.

Question analysis

A swim instructor's box plot shows the ages of students who signed up for lessons. Students decide which of four statements about the distribution is true, using the minimum, quartiles, median and maximum. The statements test different box plot ideas, including outliers, the median, the quartiles and the range, so each needs its own check.

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 New York places box plots, the interquartile range and outliers in grade 7. Here four statements test those ideas one at a time; each check is short, so the item matches typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Something you must already know moderate
Every choice must be checked moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 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

Reading the five values off the plot is quick; the work is knowing what a box plot can and cannot say. It shows where quarters of the data begin and end, not how many students share any one age, and a long whisker is not an outlier unless it passes the 1.5 × IQR cutoff, here 13. The true statement uses the third quartile: about a quarter of the students are 10 or older.

Something you must already know moderate Working it out
The statements rely on box plot facts the question does not give: each section holds about a quarter of the data, the middle line is the median, and outliers lie beyond 1.5 IQRs. (Anderson & Krathwohl, 2001)
Every choice must be checked moderate Choosing the answer
Four statements test four separate ideas, an outlier, the share at one age, a quarter of the data and two spreads, so ruling one out says nothing about the rest. (Embretson & Wetzel, 1987)
Reading a figure, table or graph minor Understanding the question
The five values, 7, 8, 9, 10 and 12, are read off a 5-to-13 scale, where the longer right whisker draws the eye more than the box does. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the five key values off the plot, then test each statement: compute the range, the interquartile range and the outlier cutoffs, and recall what share of the data each part holds.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the long right whisker, reaching 12, as a sign of an outlier, but 12 is below the cutoff 10 + 1.5 × 2 = 13.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reads the median line at 9 as the age of half the students, though a median only splits the ages into a lower and an upper half.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Swaps the two measures, taking the box's 2 years as the range and the whiskers' 5 years as the interquartile range, so the IQR seems 3 greater.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range: $\text{range} = \text{maximum} - \text{minimum}$; The maximum is the largest value, the right end of a box plot. The minimum is the smallest, the left end. The range is how far apart they are. It shows how spread out all the data are.
Interquartile range: $\text{IQR} = Q_3 - Q_1$; IQR is short for interquartile range. Q1, the first quartile, is the left edge of the box. Q3, the third quartile, is the right edge. The IQR is the width of the box. It is the spread of the middle half of the data. The four parts of a box plot each hold about one-fourth of the data.
Outlier test: $\text{below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}$; An outlier is a value far from the rest of the data. A common rule says a value is an outlier if it lies more than 1.5 IQRs below Q1 or above Q3.

Read the five values off the plot
A box plot shows five numbers. The median is the middle value when the ages are in order.
Part Age
Minimum, the left whisker’s end 7
First quartile, Q1, the box’s left edge 8
Median, the line in the box 9
Third quartile, Q3, the box’s right edge 10
Maximum, the right whisker’s end 12
Check choice A, an outlier
$\text{IQR} = 10 - 8 = 2$
$Q_3 + 1.5 \times 2 = 13 \text{ and } Q_1 - 1.5 \times 2 = 5$
An outlier would be above 13 or below 5. The ages only run from 7 to 12. So there is no outlier. Watch out: a long whisker is not an outlier. It just means the top ages are spread out. Choice A is false.
Check choice B, the median
The median, 9, is the middle age. About half the students are 9 or younger. About half are 9 or older. It does not tell how many are exactly 9. Watch out: the median line does not mean half the students are 9. Choice B is false.
Check choice C, the top fourth
The box plot splits the students into four groups of about the same size. Each group holds about one-fourth of them. The last group runs from Q3, 10, to the maximum, 12. At least 10 means 10 or more. So about one-fourth of the students are at least 10 years old. Choice C is true.
Check choice D, the two spreads
$\text{range} = 12 - 7 = 5$
$\text{IQR} = 2$
The IQR is 3 less than the range, not 3 greater. The box always fits inside the whiskers, so the IQR can never be bigger. Watch out: swapping the two measures gives choice D. So the answer is C: about one-fourth of the students are at least 10 years old.

Part	Age
Minimum, the left whisker’s end	7
First quartile, Q1, the box’s left edge	8
Median, the line in the box	9
Third quartile, Q3, the box’s right edge	10
Maximum, the right whisker’s end	12

Check your answer

Picture 20 students. About 5 would be in each fourth of the plot. The top 5 would be ages 10 to 12. That is about one-fourth, just as choice C says. Each other choice fails a fact you can read straight off the plot.

More questions like this

The original DOE question

Question 72 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 72. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 20 of 35

2023 Test A Math, question 48

Question

The manager of a shipping company recorded the weights, in pounds, of the last 9 packages the company shipped. The manager displayed the distribution of the weights in a box plot with the five-number summary shown below:

minimum: 29
first quartile: 31
median: 42
third quartile: 73
maximum: 98

Which statement about the distribution of weights is supported by the box plot the manager created?

AThe minimum, 29, is an outlier.
BThe maximum, 98, is an outlier.
CThe interquartile range of the data is 11.
DThe interquartile range and the median are equal.

Show the answer

Correct answer: D

First, identify the first quartile and the third quartile:

first quartile: 31
third quartile: 73

Then, find the difference between the first quartile and the third quartile:

$73 - 31 = 42$

The median is given: 42.

So, the interquartile range and the median are equal.

Question analysis

This statistics question gives the five-number summary of nine package weights shown in a box plot and asks which statement about the distribution is supported. Students compute the interquartile range and decide whether the minimum or the maximum is an outlier. Judging outliers by eye, or subtracting the wrong pair of values for the IQR, makes wrong statements look true.

NYS standards: 7.SP.1 box plots and interquartile range
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the interquartile range and testing for outliers are grade 7 standards in New York. There is no plot to read here, only five numbers, but every statement has to be checked by calculation, which makes it a careful grade 7 task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Every choice must be checked moderate
Solution: 3 steps
Reading level: Question and choices: about grade 7, 84 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The numbers are given outright, so the work is knowing what each statement claims and testing it. The outlier statements sound right because 98 looks far from 73 and 29 is the smallest value, but the 1.5 × IQR rule clears both. The true statement needs only one subtraction, 73 − 31, and a look at the median, which happens to have the same value.

Something you must already know moderate Working it out
Neither the interquartile range, the third quartile minus the first, nor the outlier rule, more than 1.5 × IQR beyond a quartile, is defined on the page. (Anderson & Krathwohl, 2001)
Every choice must be checked moderate Choosing the answer
The four statements test different ideas, two possible outliers, a value for the IQR and a comparison of the IQR with the median, so ruling out one says nothing about the next. (Embretson & Wetzel, 1987)

Strategy: Compute the interquartile range from the two quartiles, use it to set the outlier cutoffs 1.5 IQR beyond each quartile, and test every statement against those numbers.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Calls 29 an outlier just because it is the minimum, though it sits only 2 below the first quartile and far inside the cutoff 31 − 63 = −32.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Judges 98 an outlier because it lies 25 above the third quartile, without testing it against the cutoff 73 + 1.5 × 42 = 136.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Subtracts the first quartile from the median, 42 − 31 = 11, instead of from the third quartile, measuring only the lower half of the box.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range: $\text{IQR} = Q_3 - Q_1$; Q1, the first quartile, is the middle of the lower half of the data. Q3, the third quartile, is the middle of the upper half. The interquartile range, or IQR, is the distance between them. It is the width of the box in a box plot. It shows how spread out the middle half of the data is.
Outlier rule: $\text{outlier: below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}$; An outlier is a value far away from the rest of the data. Q1 and Q3 are the quartiles, and IQR is the interquartile range. A value is an outlier if it lies more than 1.5 times the IQR below Q1 or above Q3.

Sort out the question
The manager’s box plot is drawn from the five numbers in the list, so you can work from the list. The weights, in pounds, are: minimum 29, first quartile 31, median 42, third quartile 73, maximum 98. You need the one statement that is true. So check each choice.
Find the interquartile range
$\text{IQR} = Q_3 - Q_1$
$\text{IQR} = 73 - 31 = 42$
The interquartile range is 42 pounds. That makes choice C, which says 11, false.
Compare the IQR with the median
The median is 42 pounds. The interquartile range is also 42 pounds. They are equal, so statement D is true.
Watch out: 42 − 31 = 11 subtracts Q1 from the median, not from Q3. That mistake makes choice C look right.
Test the ends for outliers
$\text{outlier: below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}$
$1.5 \times 42 = 63$
$31 - 63 = -32 \qquad 73 + 63 = 136$
An outlier would be below −32 or above 136. The minimum, 29, is not below −32. The maximum, 98, is not above 136. So A and B are false. Watch out: 98 looks far away, but not far enough to be an outlier.
Choose the answer
Only statement D is supported by the five numbers. So the answer is D, the interquartile range and the median are equal.

Check your answer

Check the subtraction by adding back: $31 + 42 = 73$ , which is Q3. The median, 42, sits inside the box, between 31 and 73, as it must. The two measures mean different things, but here they happen to be the same number.

More questions like this

The original DOE question

Question 105 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 88, question 105. Open page 88 of the handbook (PDF)

Question 21 of 35

2023 Test A Math, question 49

Question

$A shaded rectangle inside a larger, unshaded outer rectangle. The bottom side of the outer rectangle is labeled 4 ft. The left side of the shaded rectangle is labeled 2 ft. The shaded rectangle is wider than it is tall.$

In the figure above, the shaded rectangle is similar to the outer rectangle. The length of the outer rectangle is 4 feet, and the perimeter of the outer rectangle is 14 feet. If the width of the shaded rectangle is 2 feet, what is the area of the shaded rectangle?

A $5\frac{1}{3}$ sq ft
B6 sq ft
C $9\frac{1}{3}$ sq ft
D12 sq ft

Show the answer

Correct answer: A

For the outer rectangle, the length is 4 feet and the perimeter is 14 feet. Write and solve an equation to find the width:

$2(4) + 2w = 14$	Apply the additive inverse property; subtract 8 from both sides of the equation.
$2w = 6$	Apply the multiplicative inverse property; divide both sides of the equation by 2.
$w = 3$

Use that information in a proportion to find the length of the shaded rectangle:

$\frac{x}{4} = \frac{2}{3}$	Cross multiply.
$3x = 4(2)$	Apply the multiplicative inverse property; divide both sides of the equation by 3.
$x = \frac{8}{3}$

Now find the area of the shaded rectangle:

$2 \times \frac{8}{3} = \frac{16}{3} = 5\frac{1}{3}$

Question analysis

This geometry question shows a shaded rectangle similar to an outer rectangle that is 4 feet long with a perimeter of 14 feet; the shaded rectangle is 2 feet wide. Students find the shaded area. They must find a missing side, match corresponding sides, and scale by a factor rather than a fixed amount, which makes it a multistep similar-figures problem.

NYS standards: 8.G.4 similarity and dilations
7.G.1 scale drawings
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Similar figures are grade 8 geometry, though grade 7 scale drawings use the same proportions. Students must first find a missing side from the perimeter, match corresponding sides, solve a proportion with a fractional result, and then find an area.
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
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is familiar, but the three together, perimeter to width, width to scale factor, and sides to area, demand a plan. The strongest pull is scaling by subtraction, shrinking each side by 1 foot because the width drops from 3 to 2, which gives a neat whole-number area. Others stop at the outer rectangle or add the sides instead of multiplying. Checking that 8/3 to 2 matches 4 to 3 confirms the shape is similar.

Combines ideas moderate Working it out
Perimeter, similarity and area are chained: 14 feet gives the outer width of 3 feet, the 2-foot side sets a scale factor of 2/3, and only then can the area be found. (Embretson, 1998; Daniel & Embretson, 2010)
Reading a figure, table or graph minor Understanding the question
The figure marks the shaded rectangle's 2-foot side running up and down and the outer 4-foot length along the bottom, so the 2 feet must be paired with the outer 3-foot side, which is not labeled. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
The scale factor 2/3 makes the shaded length 8/3 feet, and the area 2 × 8/3 = 16/3 has to be written as the mixed number 5 1/3. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Use the perimeter to find the outer rectangle's width, match the shaded width of 2 feet to it, scale the 4-foot length by the same factor, then multiply length by width.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Shrinks each side by the same amount instead of the same factor: the width drops by 1 foot, from 3 to 2, so the length drops from 4 to 3, giving 6.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Finds the shaded sides, 2 and 8/3 feet, correctly but computes the perimeter, 2 × (2 + 8/3) = 28/3 = 9 1/3, instead of the area.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the outer width of 3 feet and gives the outer rectangle's area, 4 × 3 = 12, instead of the smaller shaded one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a rectangle: $P = 2l + 2w$; P is the perimeter, the distance all the way around. l is the length and w is the width. A rectangle has two lengths and two widths, so you add each one twice.
Scale factor of similar shapes: $\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}$; Similar shapes have the same shape but different sizes. Each side of the small shape equals its matching side on the large shape times one number, the scale factor. So one pair of matching sides tells you the scale factor.
Area of a rectangle: $A = l \times w$; A is the area, the space inside. l is the length and w is the width. You multiply because the rectangle is made of rows of unit squares: w rows with l squares in each.

Sort out the question
The outer rectangle is 4 feet long, and its perimeter is 14 feet. The shaded rectangle is similar to it, so it has the same shape. The shaded rectangle’s width, its left side, is 2 feet. You need the shaded area, so you also need the shaded length.
Find the outer rectangle’s width
$P = 2l + 2w$
$14 = 2 \times 4 + 2w$
$14 = 8 + 2w$
$2w = 6, \quad w = 3$
The outer rectangle is 3 feet wide. That is its up-and-down side.
Find the scale factor
The shaded width, 2 feet, matches the outer width, 3 feet.
$\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}$
$\text{scale factor} = \frac{2}{3}$
Every shaded side is $\frac{2}{3}$ of the matching outer side. Watch out: similar shapes shrink by the same factor, not by the same number of feet. Taking 1 foot off each side gives 2 by 3, an area of 6, choice B.
Find the shaded length
Multiply the outer length, 4 feet, by the scale factor.
$l = \frac{2}{3} \times 4 = \frac{8}{3} = 2\frac{2}{3}$
The shaded rectangle is $2\frac{2}{3}$ feet long.
Find the shaded area
$A = l \times w$
$A = \frac{8}{3} \times 2 = \frac{16}{3} = 5\frac{1}{3}$
Watch out: 4 × 3 = 12 is the outer area, choice D. So the answer is A, $5\frac{1}{3}$ sq ft.

Check your answer

Check the shape. For the outer rectangle, length ÷ width is $4 \div 3 = \frac{4}{3}$ . For the shaded one, it is $\frac{8}{3} \div 2 = \frac{4}{3}$ . They match, so the shapes are similar. Also, $5\frac{1}{3}$ square feet is less than the outer 12, as it should be.

More questions like this

The original DOE question

Question 106 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 88, question 106. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 22 of 35

2023 Test A Math, question 51

Question

At a carnival, visitors can win a prize if an employee cannot correctly guess their age. The employee records the first 15 ages that he guesses one Saturday. The box plot shown below displays data about the ages he recorded.

$A box plot above a number line from 7 to 18 with a tick at every integer. The left whisker starts at 7, the box runs from 8 to 17 with the median line at 10, and the right whisker ends at 18.$

What is the interquartile range of the box plot?

Show the answer

Correct answer: B

Quartile 1 is the median of the lower half of the data, 8.

Quartile 3 is the median of the upper half of the data, 17.

$\text{IQR} = \text{Q3} - \text{Q1} = 17 - 8 = 9$

The interquartile range is 9.

Question analysis

This statistics question shows a box plot of 15 ages that a carnival employee guessed and asks for its interquartile range. Students read the quartiles from the edges of the box and subtract. Confusing the interquartile range with the full range, or using the median line in place of a quartile, gives the wrong choices.

NYS standards: 7.SP.1 box plots and interquartile range
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Reading a box plot and finding its interquartile range is a grade 7 standard in New York. The values sit on labeled whole-number ticks and one subtraction is needed, a routine grade 7 task once the box's edges are known as the quartiles.
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
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is one subtraction, so the question tests whether students know which parts of a box plot to use. The whiskers' ends give the range, the line inside the box gives the median, and both are easy to grab in place of the quartiles. The story about the carnival and the 15 ages plays no part in the answer.

Something you must already know minor Working it out
The plot shows five values with no labels, so students must know that the box's ends mark the quartiles and that the interquartile range is the box's width, not the whiskers'. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The values 7, 8, 10, 17 and 18 are read off a scale ticked at every whole number, with the box's ends at 8 and 17 and the median line at 10. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the first and third quartiles on the plot by recalling which parts of a box plot mark them, then subtract the smaller quartile from the larger.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
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.
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: weak
Subtracts the median from the third quartile, 17 − 10 = 7, measuring only the part of the box right of the middle line.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gives the median, 10, read off the line inside the box, taking a single value on the plot for the spread between the quartiles.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Subtracts the minimum from the maximum, 18 − 7 = 11, which is the range shown by the whiskers, not the width of the box.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range: $\text{IQR} = Q_3 - Q_1$; Q1, the first quartile, is the middle of the lower half of the data. Q3, the third quartile, is the middle of the upper half. In a box plot, Q1 is the left end of the box and Q3 is the right end. So the IQR is the width of the box.

Sort out the question
A box plot shows five numbers. The whiskers, the lines sticking out, end at the smallest and largest values. The box covers the middle half of the data. The line inside the box is the median, the middle value. You need the interquartile range of the ages.
Read the five numbers
Read each one off the number line below the plot.
- Smallest age, the left whisker: 7
- Q1, the left end of the box: 8
- Median, the line inside the box: 10
- Q3, the right end of the box: 17
- Largest age, the right whisker: 18
Find the interquartile range
$\text{IQR} = Q_3 - Q_1$
$\text{IQR} = 17 - 8 = 9$
The interquartile range is 9 years. Watch out: 18 − 7 = 11 is the range, the whole spread from whisker to whisker. That is choice D. The median, 10, is choice C.
So the answer is B, 9.

Check your answer

Count the spaces across the box. From 8 to 17 there are 9 spaces of 1 year each, so the box is 9 years wide. The answer is also less than the whole range, 11, as it must be.

More questions like this

The original DOE question

Question 108 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 108. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 23 of 35

2023 Test B Math, question 16

Question

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

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

A30°
B24°
C15°
D12°

Show the answer

Correct answer: B

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

Question analysis

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

NYS standards: 7.G.5 angle relationships and equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding an angle by dividing a full turn into equal parts is early angle measurement. It takes one division, and the only fact needed is the number of degrees in a circle, so it sits below grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with 15 spokes gives 30°, which is 360 ÷ 12, the angle between hours on a clock, so it mostly draws guesses at a familiar angle.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the number of spokes, 15, as the angle in degrees, matching a number in the question without any calculation.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 180 by 15, using a straight angle’s 180° as the full turn around the hub instead of 360°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 73 as printed on page 195 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 73. Open page 195 of the handbook (PDF)

Question 24 of 35

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 25 of 35

2023 Test B Math, question 22

Question

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

4.6	6.0	3.8	3.5	4.5
5.4	3.9	6.5	4.8	6.2

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

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

Show the answer

Correct answer: C

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

3.5	3.8	3.9	4.5	4.6
4.8	5.4	6.0	6.2	6.5

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

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

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

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

3.5

3.8

3.9

4.5

4.6

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

4.8

5.4

6.0

6.2

6.5

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

Question analysis

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

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Building a box plot from data is grade 7 content. This version needs all five summary values from ten unsorted decimals, including a median between two values, then a close comparison of plots, longer and finer than typical.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
A very close wrong answer moderate
Reading a figure, table or graph moderate
Slow to do carefully minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 79 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 198, question 79. Open page 198 of the handbook (PDF)

Question 26 of 35

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 27 of 35

2023 Test B Math, question 26

Question

The box plot below represents a set of data.

$Box plot above a number line with arrows at both ends, a tick every 3 units from 0 to 30, and labels at 0, 6, 12, 18, 24 and 30. The left whisker starts at 3, the box runs from 6 to the unlabeled tick for 27 with the median line at 15, and the right whisker ends at 30.$

What is the interquartile range of the set of data?

Show the answer

Correct answer: C

First, determine the values of the upper and lower quartiles of the set of data:

$The same box plot, with an arrow labeled lower quartile pointing down to the left edge of the box, at 6, and an arrow labeled upper quartile pointing down to the right edge of the box, at the unlabeled tick for 27.$

The upper quartile value is indicated by the vertical line segment at the far right edge of the box plot above the hash mark representing 27.
The lower quartile value is indicated by the vertical line segment at the far left edge of the box plot above the hash mark representing 6.

To determine the interquartile range, subtract the lower quartile value (6) from the upper quartile value (27):

$27 - 6 = 21$

Question analysis

A box plot on a number line marked every 3 units, with labels every 6, represents a set of data. Students find the interquartile range. They have to read the two ends of the box, one of which falls on an unlabeled tick, and remember exactly which part of the plot the interquartile range measures.

NYS standards: 7.SP.1 box plots and interquartile range
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the interquartile range from a box plot is a grade 7 skill, and this is a typical example. Read two values and subtract; the only extra care is a box edge that falls on a tick with no label.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the two edges of the box are read, one subtraction finishes the question, so the work is reading the scale and knowing which part of the plot is meant. The upper quartile falls on an unlabeled tick, which costs care. The wrong answers measure the whole spread or only half of the box, all common mix-ups of what the interquartile range covers.

Reading a figure, table or graph moderate Understanding the question
The axis is labeled only every 6 units but ticked every 3, and the box’s right edge sits on the unlabeled tick for 27, so it has to be counted from 24. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Nothing defines the interquartile range; it is the width of the box, the upper quartile minus the lower, not the span of the whiskers. (Anderson & Krathwohl, 2001)

Strategy: Find the values at the left and right edges of the box, counting tick marks in steps of 3 where there is no label, and subtract the smaller from the larger.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Gives the median minus the lower quartile, 15 − 6 = 9, as the interquartile range, measuring only the left part of the box.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Gives the upper quartile minus the median, 27 − 15 = 12, as the interquartile range, measuring only the right part of the box.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Subtracts the ends of the whiskers, 30 − 3 = 27, giving the range of the data instead of the interquartile range.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range: $\text{IQR} = Q_3 - Q_1$; IQR stands for interquartile range. Q3, the third quartile, is the right edge of a box plot’s box. Q1, the first quartile, is the left edge. Subtracting gives the width of the box, where the middle half of the data lies.

Sort out the question
You see a box plot and need the interquartile range, or IQR. That is the width of the box: the upper quartile minus the lower quartile. The whiskers don’t matter here.
Read the number line
The labels are 0, 6, 12, 18, 24 and 30. But there is a tick mark halfway between each pair of labels. So the ticks go up by 3: 0, 3, 6, 9 and so on. The tick between 24 and 30 stands for 27.
Find the edges of the box
The left edge of the box is at 6. That is the lower quartile, Q1. The right edge is at the unlabeled tick after 24, which is 27. That is the upper quartile, Q3.
Subtract to find the IQR
$\text{IQR} = Q_3 - Q_1$
$\text{IQR} = 27 - 6 = 21$
Watch out: $30 - 3 = 27$ uses the ends of the whiskers. That is the range, choice D. Also, the median line at 15 splits the box. $15 - 6 = 9$ and $27 - 15 = 12$ are each only part of the box. So the answer is C, 21.

Check your answer

Count the tick spaces across the box, from 6 to 27. There are 7 spaces, and each is worth 3. So the box is $7 \times 3 = 21$ units wide. That matches.

More questions like this

The original DOE question

Question 83 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 83. Open page 199 of the handbook (PDF)

Question 28 of 35

2023 Test B Math, question 40

Question

Angle A and angle B are complementary. If the measure of angle B is three times as great as the measure of angle A, what is the measure of angle B?

A45°
B67.5°
C75°
D135°

Show the answer

Correct answer: B

Let A represent the measure of angle A, and let B represent the measure of angle B, in degrees. Since the angles are complementary, $A+B=90$ . The question states that $B=3A$ , or $A=\frac{B}{3}$ . Substitute that expression for A in the first equation and solve for B:

$A+B=90$

$\frac{B}{3}+B=90$

$\frac{4B}{3}=90$

$4B=270$

$B=67.5$

Question analysis

A geometry question about two complementary angles, where angle B is three times as large as angle A. Students find the measure of angle B. They must recall what complementary means, since the question does not say, and set up an equation that splits the total in the right proportion.

NYS standards: 7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation from complementary angles is a grade 7 standard. This is the familiar version without a figure: one equation from the definition, then a division and a multiplication, typical of grade 7 classwork.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once complementary is read as a 90° total, the problem is one equation with a decimal answer, 4A = 90. The 22.5° for angle A is not listed, so stopping at the smaller angle is not rewarded here, but using 180° is: it gives 45° and 135°, and both appear among the choices. Checking that the chosen angle and one third of it add to 90° settles it.

Something you must already know moderate Working it out
The stem never says what complementary means; the 90° total has to be recalled, and a student who confuses it with the supplementary 180° doubles both angles. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The clause “the measure of angle B is three times as great as the measure of angle A” has to become B = 3A, so the 90° splits into four equal parts, not three. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Recall the total that complementary angles share, think of angle A as one part and angle B as three parts, find one part, then multiply for angle B.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Splits the 90° evenly, 90 ÷ 2 = 45, using the complementary total but ignoring that angle B is three times angle A.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 75°, which would need angle A to be 25° and the pair to add to 100°; it mostly draws guesses.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses 180°, the supplementary total, so 4A = 180 gives angle A = 45° and angle B = 135°, a pair that is not complementary.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Complementary angles: $A + B = 90$; A and B are the measures of the two angles, in degrees. Complementary angles are two angles that add up to 90 degrees, the size of a square corner. Use this whenever a problem says two angles are complementary.

Sort out the question
Complementary angles are two angles that add up to 90 degrees. Angle B is three times as big as angle A. You need the measure of angle B, in degrees.
Write the facts as equations
Let A and B stand for the two measures. The angles are complementary, so use the formula.
$A + B = 90$
Angle B is three times angle A, so:
$B = 3A$
Put 3A in place of B
Replace B in the sum with 3A. Then there is only one letter to find.
$A + 3A = 90$
$4A = 90$
Think of it as 4 equal parts. Angle A is 1 part, and angle B is 3 parts.
Find angle A
Divide both sides by 4.
$A = 90 \div 4 = 22.5$
Angle A measures 22.5 degrees. That is not the answer yet, because the question asks for angle B.
Find angle B
Angle B is three times angle A.
$B = 3 \times 22.5 = 67.5$
Angle B measures 67.5 degrees.
Watch out: Angles that add to 180 degrees are called supplementary. Using 180 by mistake gives 45 and 135, choice D. Splitting 90 in half gives 45, choice A. That ignores the fact that B is three times A.
So the answer is B, 67.5°.

Check your answer

Add the two angles: 22.5 + 67.5 = 90. So they are complementary. Now divide: 67.5 ÷ 22.5 = 3. So angle B is three times angle A. Both facts check out.

More questions like this

The original DOE question

Question 97 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 97. Open page 202 of the handbook (PDF)

Question 29 of 35

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 30 of 35

2023 Test B Math, question 52

Question

$An equilateral triangle drawn with dashed sides. On each side a semicircle bulges outward, with that side as its diameter. The three curved arcs together form the outline of the garden.$

The garden design shown above is created in the shape of three semicircles whose diameters lie on the sides of an equilateral triangle. If each diameter is 2 meters long, what is the perimeter of the garden, in meters?

A6
B12
C $3\pi$
D $6\pi$

Show the answer

Correct answer: C

The perimeter (or circumference) of one of the three semicircles is $\frac{1}{2}(\text{diameter})\pi=\frac{1}{2}(2\pi)=\pi$ . Since there are 3 semicircles that make up the perimeter, the perimeter is $3\pi$ .

Question analysis

A garden is outlined by three semicircles whose diameters, each 2 meters long, are the sides of an equilateral triangle. Students find the perimeter of the garden in meters. It tests the circumference of a circle applied to half circles, and the figure mixes straight dashed sides with curved ones, so deciding what counts as the outline matters.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying the circumference formula is a grade 7 standard. Here it is used for a semicircle and repeated three times, a short extension of routine classwork, with a dashed triangle inside the figure to look past.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The perimeter is just three half-circles, and each is half of π × 2. The figure’s dashed triangle invites measuring straight sides instead, and remembering the whole circle’s circumference without halving it gives a choice twice too large. Tracing the garden’s outer edge with a finger before computing shows that only the curves count.

Reading a figure, table or graph minor Understanding the question
The dashed triangle lies inside the garden, so its three 2-meter sides are not part of the perimeter; only the three curved arcs bound the garden. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The circumference formula is not given, and each arc is half of a circle with a 2-meter diameter, so students must take half of π × 2. (Anderson & Krathwohl, 2001)

Strategy: Find the circumference of a full circle with a 2-meter diameter, take half of it for one semicircle, and multiply by the number of semicircles in the outline.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Adds the three dashed sides of the triangle, 3 × 2 = 6, measuring the triangle inside the garden instead of its curved outline.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12, which is twice the triangle’s perimeter and matches no arc length, so it mostly draws guesses.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses a whole circumference, π × 2, for each of the three arcs, forgetting that each one is only half a circle, which doubles the perimeter.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = \pi \times d$; C is the circumference, the distance around a circle. d is the diameter, the straight line across the circle through its center. π is a number a little more than 3.14. The distance around any circle is π times as long as the distance across it.
Arc of a semicircle: $\text{arc} = \frac{1}{2} \times \pi \times d$; A semicircle is half of a circle. Its arc is the curved part. d is the diameter, the straight side. The arc is half of the full circle’s circumference, so take half of π times d.

Sort out the figure
The garden’s edge is made of three semicircles, or half circles. Each one sits on a side of the dashed triangle. Each side is a diameter, 2 meters long. The perimeter is the distance around the outside. Here that is only the three arcs, the curved parts. The dashed triangle is inside, so it does not count.
Find a full circle’s circumference
Picture the whole circle that one semicircle comes from. Its diameter is 2 meters. Use the circumference formula.
$C = \pi \times 2 = 2\pi$
The full circle would measure $2\pi$ meters around.
Find the length of one arc
Each semicircle’s arc is half of that circle. Use the arc formula.
$\text{arc} = \frac{1}{2} \times \pi \times 2 = \pi$
One arc is $\pi$ meters long.
Watch out: Using the full circle’s $2\pi$ for each arc doubles the answer. That gives $6\pi$ , choice D.
Add the three arcs
The garden’s edge is three equal arcs.
$\pi + \pi + \pi = 3\pi$
The perimeter is $3\pi$ meters.
Watch out: The dashed sides add to 2 + 2 + 2 = 6 meters, choice A. That is the triangle, not the curved edge.
So the answer is C, $3\pi$ meters.

Check your answer

Estimate. $3\pi$ is about 3 × 3.14 = 9.42 meters. A curved path is longer than the straight line across it. So the edge should be longer than the triangle’s 6 meters, and 9.42 is.

More questions like this

The original DOE question

Question 109 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 206, question 109. Open page 206 of the handbook (PDF)

Question 31 of 35

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 32 of 35

2018 Test B Math, question 17

Question

A large circular dinner plate has a radius of 20 centimeters. A smaller circular plate with a circumference of $20\pi$ centimeters is placed in the center of the larger dinner plate. What is the area of the part of the larger dinner plate that is not covered by the smaller plate?

A $20\pi$ sq cm
B $100\pi$ sq cm
C $200\pi$ sq cm
D $300\pi$ sq cm

Show the answer

Correct answer: D

The radius of the large plate is 20 cm. Use that to find the area of the large plate:

$A=\pi r^2=\pi(20^2)=400\pi\text{ sq cm}$

The circumference of the smaller plate is 20π cm. Use that to find the radius, and then the area, of the smaller plate:

$C=2\pi r$

$20\pi=2\pi r$

$r=10$

$A=\pi r^2=\pi(10^2)=100\pi\text{ sq cm}$

Subtract the area of the small plate from the area of the large plate:

400π − 100π = 300π sq cm

Question analysis

This circle question places a smaller plate, described by its circumference, on a larger dinner plate described by its radius, and asks for the area of the large plate left uncovered. Students recall the circumference and area formulas, find the small radius and subtract areas. Because the two plates are described in different ways, circumference and area have to be kept apart.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying circle area and circumference formulas is the grade 7 standard, and subtracting one area from another is a common grade 7 exercise. Working backward from a circumference adds one step but no new content.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Something you must already know moderate
Solution: 4 steps
Reading level: Question and choices: about grade 6, 62 words.
Students, last 30 days: 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 describes the two plates in different measures, one by radius and one by circumference, so students must convert before they can compare. The trap is a halving argument: the small circumference is half the large one, but the area it covers is only a quarter. Students who keep both areas in terms of π and subtract finish quickly; those who reason from the circumferences alone go wrong.

Many steps moderate Working it out
The chain runs through four stages: the large area 400π, the small radius 10 from the circumference 20π, the small area 100π, then the uncovered difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know moderate Working it out
Neither formula is given: students need C = 2πr to recover the small plate's radius and A = πr² for both areas. (Anderson & Krathwohl, 2001)

Strategy: Use C = 2πr to get the small plate's radius, find both areas with A = πr², keep everything in terms of π, and subtract.

Traps

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

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Works with circumferences instead of areas, subtracting the small plate's 20π from the large plate's 40π.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the area of the smaller plate, the part that is covered, instead of the part of the large plate left uncovered.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reasons that a circumference half as long covers half the area, so takes half of 400π, not seeing that area scales with the square.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \times \pi \times r$; C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π, said “pie,” is a number a little more than 3. The distance around is always about 6.28 times the radius.
Area of a circle: $A = \pi \times r^2$; A is the area, the space inside the circle. r is the radius. r² means r times r. π is about 3.14. The radius is squared because area spreads out in two directions.

Write down what you know
The big plate has a radius of 20 cm. The radius is the distance from the center to the edge.
The small plate has a circumference of 20π cm. Circumference is the distance around a circle.
You need the area of the big plate that the small plate does not cover. Area is the space inside a shape.
Find the big plate's area
$A = \pi \times r^2$
$A = \pi \times 20^2 = \pi \times 400 = 400\pi$
The big plate is 400π sq cm. Keep π as a symbol, because the choices use it.
Find the small plate's radius
You know the small plate’s circumference, not its radius. Use the circumference formula to work back.
$C = 2 \times \pi \times r$
$20\pi = 2 \times \pi \times r$
Divide both sides by 2π. That leaves $r = 10$ . The small radius is 10 cm.
Find the small plate's area
Use the area formula again, with the small radius.
$A = \pi \times 10^2 = 100\pi$
The small plate covers 100π sq cm.
Watch out: 100π is the covered part. The question asks for the part not covered, so do not stop at choice B.
Subtract to find the uncovered part
$400\pi - 100\pi = 300\pi$
The uncovered area is 300π sq cm.
Watch out: Do not subtract the distances around the plates. That gives 40π − 20π = 20π, choice A.
So the answer is D, 300π sq cm.

Check your answer

The small radius is half the big radius. So the small area is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the big area. That leaves three quarters uncovered: $\frac{3}{4} \times 400\pi = 300\pi$ . It matches.

More questions like this

The original DOE question

Question 74 as printed on page 156 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 74. Open page 156 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 33 of 35

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 34 of 35

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 35 of 35

2018 Test B Math, question 48

Question

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

What is the area of the shaded triangle shown above?

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

Show the answer

Correct answer: C

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

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