SSAT Upper Level Quantitative Quiz: 3 D Volume

5 questions · exam conditions

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3 D VolumeQuestion 1 of 5

A rectangular swimming pool is 20 feet long, 15 feet wide, and has a uniform depth of 8 feet. The pool is currently filled to 75% of its capacity. How many cubic feet of water must be added to fill the pool to 90% of its capacity?

180

cubic feet

300

cubic feet

450

cubic feet

360

cubic feet

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SSAT Upper Level Quantitative Quiz

SSAT Upper Level Quantitative Quiz: 3 D Volume

Practice 3 D Volume in SSAT Upper Level Quantitative with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 3 D Volume, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Upper Level Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$180$ cubic feet
$300$ cubic feet
$450$ cubic feet
$360$ cubic feet (correct answer)

Explanation: When you encounter volume problems involving percentages, you need to find the total capacity first, then calculate the difference between two percentage levels. Start by finding the pool's total volume:

20 \times 15 \times 8 = 2,400

cubic feet. Now you can work with the percentages. Currently, the pool is 75% full, which means it contains

2,400 \times 0.75 = 1,800

cubic feet of water. You want to fill it to 90% capacity, which would be

2,400 \times 0.90 = 2,160

cubic feet of water. The amount you need to add is the difference:

2,160 - 1,800 = 360

cubic feet. Looking at the wrong answers: Choice A (180) represents half the correct amount—you might get this if you miscalculated the percentage difference as 7.5% instead of 15%. Choice B (300) could result from incorrectly using 12.5% of the total volume instead of 15%. Choice C (450) is what you'd get if you calculated 18.75% of the total volume, perhaps by adding percentages incorrectly. The key insight is that going from 75% to 90% means adding 15% of the total capacity, and

2,400 \times 0.15 = 360

cubic feet. Strategy tip: In percentage volume problems, always calculate the total capacity first, then find what each percentage represents in actual units. The difference between two percentages of the same whole gives you your answer directly.

Question 2

A manufacturer produces metal cubes and then drills a cylindrical hole completely through each cube from one face to the opposite face. If the cube has side length 10 cm and the cylindrical hole has radius 3 cm, what is the volume of metal remaining in each cube?

$1000 - 90\pi$ cubic centimeters (correct answer)
$1000 - 180\pi$ cubic centimeters
$1000 - 270\pi$ cubic centimeters
$1000 - 45\pi$ cubic centimeters

Explanation: The cube volume is

10^3 = 1000

cubic cm. The cylindrical hole goes completely through the cube, so its height equals the cube's side length (10 cm). The volume of the cylindrical hole is

V = \pi r^2 h = \pi(3)^2(10) = 90\pi

cubic cm. The remaining volume is

1000 - 90\pi

cubic cm. Choice B incorrectly doubles the hole volume. Choice C triples the hole volume. Choice D halves the hole volume.

Question 3

A water tank in the shape of a right circular cone (vertex pointing down) has a radius of 12 feet at the top and a height of 16 feet. When the tank is filled to a depth of 8 feet from the vertex, what is the volume of water in the tank?

$96\pi$ cubic feet (correct answer)
$192\pi$ cubic feet
$384\pi$ cubic feet
$144\pi$ cubic feet

Explanation: When a cone is partially filled, the water forms a smaller similar cone. The full cone has radius 12 ft and height 16 ft. The water reaches height 8 ft from the vertex. By similar triangles, if the water height is 8 ft, the radius at water surface is

r = \frac{12 \times 8}{16} = 6

ft. The volume of water is

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6)^2(8) = \frac{1}{3}\pi(36)(8) = \frac{288\pi}{3} = 96\pi

cubic feet. Choice B doubles the correct answer. Choice C uses the wrong radius calculation. Choice D uses incorrect height proportion.

Question 4

A cone and a cylinder have the same base radius $r$ and the same height $h$ . If the volume of the cylinder is 216 cubic units, what is the volume of the cone?

$108$ cubic units
$72$ cubic units (correct answer)
$144$ cubic units
$648$ cubic units

Explanation: The volume of a cylinder is

V_{cylinder} = \pi r^2 h = 216

. The volume of a cone is

V_{cone} = \frac{1}{3}\pi r^2 h

. Since both shapes have the same base radius and height,

V_{cone} = \frac{1}{3} \times V_{cylinder} = \frac{1}{3} \times 216 = 72

cubic units. Choice A incorrectly uses

\frac{1}{2}

instead of

\frac{1}{3}

. Choice C incorrectly uses

\frac{2}{3}

of the cylinder volume. Choice D incorrectly multiplies by 3 instead of dividing.

Question 5

A spherical balloon has a radius of 9 inches. If the radius is decreased by $\frac{1}{3}$ , by what factor does the volume decrease?

$\frac{1}{3}$
$\frac{8}{27}$
$\frac{27}{8}$ (correct answer)
$\frac{2}{3}$

Explanation: Original radius is 9 inches. Decreased by

\frac{1}{3}

means the new radius is

9 - \frac{1}{3}(9) = 9 - 3 = 6

inches. Original volume:

V_1 = \frac{4}{3}\pi(9)^3 = \frac{4}{3}\pi(729) = 972\pi

. New volume:

V_2 = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi(216) = 288\pi

. The ratio of original to new volume is

\frac{V_1}{V_2} = \frac{972\pi}{288\pi} = \frac{972}{288} = \frac{27}{8}

. So the volume decreases by a factor of

\frac{27}{8}

. Choice A uses linear scaling. Choice B gives the reciprocal. Choice D incorrectly uses

\frac{2}{3}