GMAT Math : Calculating the volume of a cube

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Example Questions

Example Question #1 : Calculating The Volume Of A Cube

What is the volume of a cube with a side length of ?

Possible Answers:

Correct answer:

Explanation:

V=s^3

V=2^3

V=8

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Example Question #2 : Calculating The Volume Of A Cube

The length, width, and height of a rectangular prism, in inches, are three different prime numbers. All three dimensions are between six feet and seven feet. What is the volume of the prism?

Possible Answers:

It is impossible to tell from the information given.

$444,059\textrm{ in}^{3}$

$478,661 \textrm{ in}^{3}$

$504,889\textrm{ in}^{3}$

$490,779\textrm{ in}^{3}$

Correct answer:

$478,661 \textrm{ in}^{3}$

Explanation:

Six feet and seven feet are equal to, respectively, 72 inches and 84 inches. There are three different prime numbers between 72 and 84 - 73, 79, and 83 - so these are the three dimensions of the prism in inches. The volume of the prism is

$73 \times 79 \times 83 = 478,661$ cubic inches.

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Example Question #3 : Calculating The Volume Of A Cube

The length of a diagonal of one face of a cube is $5\sqrt{2}$ . Give the volume of the cube.

Possible Answers:

250

$125 \sqrt{2}$

125

$\frac{125 \sqrt{2}}{2}$

The correct answer is not among the other responses.

Correct answer:

125

Explanation:

A diagonal of a square has length $\sqrt{2}$ times that of a side, so each side of each square face of the cube has length $5 \sqrt{2} \div \sqrt{2}= 5$ . Cube this to get the volume:

$V= 5^{3} = 125$

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Example Question #4 : Calculating The Volume Of A Cube

The length of a diagonal of a cube is $\sqrt{17}$ . Give the volume of the cube.

Possible Answers:

$\frac{ 17 \sqrt{ 51} }{ 9} \right )$

$\frac{ 17 \sqrt{ 51} }{ 3} \right )$

289

$\frac{ 17 \sqrt{ 34} }{ 2} \right )$

$\frac{ 17 \sqrt{ 34} }{ 6} \right )$

Correct answer:

$\frac{ 17 \sqrt{ 51} }{ 9} \right )$

Explanation:

Let be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem,

$s^{2}+s^{2}+s^{2}= ( \sqrt{17})^{2}$

$3s^{2}= 17$

$s^{2}= \frac{17}{3}$

$s =\sqrt{ \frac{17}{3}}$

Cube the sidelength to get the volume:

$s^{3} =\left (\sqrt{ \frac{17}{3}} \right )^{3}$

$=\frac{ \left (\sqrt{ 17} \right )^{3}}{\left (\sqrt{3} \right )^{3}} \right )$

$=\frac{ 17 \sqrt{ 17} }{ 3 \sqrt{3} } \right )$

$=\frac{ 17 \sqrt{ 17} \cdot \sqrt{3}}{ 3 \sqrt{3} \cdot \sqrt{3}} \right )$

$=\frac{ 17 \sqrt{ 51} }{ 9} \right )$

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Example Question #5 : Calculating The Volume Of A Cube

A sphere with surface area $40 \pi$ is inscribed inside a cube. Give the volume of the cube.

Possible Answers:

8 0

$10 \sqrt{10}$

$40 \sqrt{10}$

$80 \sqrt{10}$

Correct answer:

$80 \sqrt{10}$

Explanation:

The sidelength of the cube is the diameter of the inscribed sphere, which is twice that sphere's radius. The sphere has surface area $40 \pi$ , so the radius is calculated as follows:

$4 \pi r^{2} = SA$

$4 \pi r^{2} = 40 \pi$

$r^{2} = 40 \pi \div 4 \pi = 10$

$r = \sqrt{10}$

The diameter of the sphere - and the sidelength of the cube - is twice this, or $2 \sqrt{10}$ .

Cube this sidelength to get the volume of the cube:

$V= \left ( 2 \sqrt{10} \right )^{3} = 2^{3} \cdot\left ( \sqrt{10 } \right )^{3} = 8 \cdot 10 \sqrt{10} = 80 \sqrt{10}$

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Example Question #3 : Calculating The Volume Of A Cube

A cube is inscribed inside a sphere with volume $36 \pi$ . Give the volume of the cube.

Possible Answers:

The correct answer is not given among the other responses.

$12 \sqrt{6}$

$36 \sqrt{2}$

$24 \sqrt{3}$

Correct answer:

$24 \sqrt{3}$

Explanation:

The diameter of the circle - twice its radius - coincides with the length of a diagonal of the inscribed cube. The sphere has volume $36 \pi$ , so the radius is calculated as follows:

$\frac{4}{3} \pi r^{3} = V$

$\frac{4}{3} \pi r^{3} = 36 \pi$

$r^{3} = 36\pi \div \left (\frac{4}{3} \pi \right )$

$r^{3} = 27$

$r = \sqrt[3]{27} = 3$

The diameter of the sphere - and the length of a diagonal of the cube - is twice this, or 6.

Now, let be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem,

$s^{2}+s^{2}+s^{2}=6^{2}$

$3s^{2}=36$

$s^{2}=12$

$s = \sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2 \sqrt{3}$

The volume of the cube is the cube of this, or

$V=\left (2 \sqrt{3} \right )^{3} =2^{3} \left ( \sqrt{3} \right )^{3} = 8 \cdot 3 \sqrt{3} = 24 \sqrt{3}$

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Example Question #31 : Cubes

A sphere with volume $36 \pi$ is inscribed inside a cube. Give the volume of the cube.

Possible Answers:

$27 \sqrt{3}$

$36 \sqrt{6}$

216

$72\sqrt{3}$

Correct answer:

216

Explanation:

The sidelength of the cube is the diameter of the inscribed sphere, which is twice that sphere's radius. The sphere has volume $36 \pi$ , so the radius is calculated as follows:

$\frac{4}{3} \pi r^{3} = V$

$\frac{4}{3} \pi r^{3} = 36 \pi$

$r^{3} = 36\pi \div \left (\frac{4}{3} \pi \right )$

$r^{3} = 27$

$r = \sqrt[3]{27} = 3$

The diameter of the sphere - and the sidelength of the cube - is twice this, or 6. Cube this sidelength to get the volume of the cube:

$V = 6^{3} = 216$

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Example Question #591 : Geometry

The distance from one vertex of a cube to its opposite vertex is twelve feet. Give the volume of the cube.

Possible Answers:

$\frac{64 }{9} \textup{ yd}^{3}$

$32\sqrt{3} \textup{ yd}^{3}$

$32\sqrt{2} \textup{ yd}^{3}$

$\frac{64 \sqrt{3}}{9} \textup{ yd}^{3}$

$\frac{64 \sqrt{2}}{6} \textup{ yd}^{3}$

Correct answer:

$\frac{64 \sqrt{3}}{9} \textup{ yd}^{3}$

Explanation:

Since we are looking at yards, we will look at twelve feet as four yards.

Let be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem,

$s^{2}+s^{2}+s^{2}=4^{2}$

$3s^{2}=16$

$s^{2}= \frac{16}{3}$

$s^{2}= \sqrt{\frac{16}{3}} = \frac{\sqrt{16}}{\sqrt{3}} = \frac{4}{\sqrt{3}}$ yards.

Cube this sidelength to get the volume:

$s^{3}= \left ( \frac{4}{\sqrt{3}} \right )^{3} = \frac{4^{3}}{(\sqrt{3})^{3}} = \frac{64 }{3\sqrt{3} }= \frac{64 \cdot \sqrt{3}}{3\sqrt{3}\cdot \sqrt{3} } = \frac{64 \sqrt{3}}{9}$ cubic yards.

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Example Question #32 : Cubes

An aquarium is shaped like a perfect cube; the area of each glass face is 6.25 square meters. If it is filled to the recommended $80\%$ capacity, then how much water will it contain?

Note: cubic meter = 1,000 liters.

Possible Answers:

The correct answer is not given among the other choices.

$10,000\textup{ L}$

$8,000\textup{ L}$

$12,500\textup{ L}$

$15,625\textup{ L}$

Correct answer:

$12,500\textup{ L}$

Explanation:

A perfect cube has square faces; if a face has area 6.25 square meters, then each side of each face measures the square root of this, or 2.5 meters. The volume of the tank is the cube of this, or

$2.5^{3} = 15.625$ cubic meters.

Its capacity in liters is $15.625 \times 1,000 = 15,625$ liters.

80% of this is

$15,625 \times 0.8 = 12,500$ liters.

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Example Question #112 : Rectangular Solids & Cylinders

Which choice comes closest to the volume of a cube with surface area 135,000 square centimeters?

Possible Answers:

$4.5 \textup{ m}^{3}$

$3 \textup{ m}^{3}$

$5 \textup{ m}^{3}$

$4 \textup{ m}^{3}$

$3.5 \textup{ m}^{3}$

Correct answer:

$3.5 \textup{ m}^{3}$

Explanation:

The suface area of a cube is six times the square of the length of one side, so solve for in the following:

$6s^{2} =135,000$

$s^{2} = 22,500$

s = 150

This is the sidelength in centimeters; since we are looking at meters, divide this by 100 to convert to $150 \div 100 = 1.5$ meters.

Cube this to get volume

$V =1.5^{3}= 3.375$ cubic meters.

Of the given choices, 3.5 cubic meters comes closest.

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GMAT Math : Calculating the volume of a cube

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Calculating The Volume Of A Cube

Example Question #2 : Calculating The Volume Of A Cube

Example Question #3 : Calculating The Volume Of A Cube

Example Question #4 : Calculating The Volume Of A Cube

Example Question #5 : Calculating The Volume Of A Cube

Example Question #3 : Calculating The Volume Of A Cube

Example Question #31 : Cubes

Example Question #591 : Geometry

Example Question #32 : Cubes

Example Question #112 : Rectangular Solids & Cylinders

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