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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

Study concepts, example questions & explanations for SSAT Upper Level Math

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All SSAT Upper Level Math Resources

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Example Questions

Example Question #163 : Geometry

A right cone has a volume of $8\pi$ , a height of and a radius of the circular base of . Find .

Possible Answers:

$\sqrt{2}$

$2\sqrt[3]{2}$

$\sqrt[3]{3}$

$\sqrt[3]{2}$

Correct answer:

$\sqrt[3]{2}$

Explanation:

The volume of a cone is given by:

$Volume=\frac{1}{3}\pi r^2h$

where is the radius of the circular base, and is the height; the perpendicular distance from the base to the vertex. Substitute the known values in the formula:

$Volume=\frac{1}{3}\pi r^2h\Rightarrow 8\pi=\frac{1}{3}\pi\times (2t)^2\times 3t$

$\Rightarrow 8\pi=4\pi t^3\Rightarrow t^3=2\Rightarrow t=\sqrt[3]{2}$

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Example Question #4 : How To Find The Volume Of A Cone

A cone has a diameter of $12\:m$ and a height of $4\:m$ . In cubic meters, what is the volume of this cone?

Possible Answers:

$54\pi\:m^3$

$36\pi\:m^3$

$48\pi\:m^3$

$144\pi\:m^3$

Correct answer:

$48\pi\:m^3$

Explanation:

First, divide the diameter in half to find the radius.

$\frac{d}{2}=r$

$\frac{12}{2}=6\:m$

Now, use the formula to find the volume of the cone.

$\text{Volume}=\frac{1}{3}\pi r^2 h$

$\text{Volume}=\frac{1}{3}\pi \times 6^2\times4=\frac{1}{3}\pi \times 36\times4=\frac{144}{3}\pi=48\pi\:m^3$

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Example Question #11 : Volume Of A Three Dimensional Figure

A cone has a radius of inches and a height of inches. Find the volume of the cone.

Possible Answers:

$12\pi\:in^3$

$24\pi\:in^3$

$36\pi\:in^3$

$108\pi\:in^3$

Correct answer:

$36\pi\:in^3$

Explanation:

The volume of a cone is given by the formula:

$\text{Volume}=\frac{1}{3}\pi r^2h$

Now, plug in the values of the radius and height to find the volume of the given cone.

$\text{Volume}=\frac{1}{3}\pi\times3^2\times12=\frac{1}{3}\pi\times9\times12=\frac{108}{3}\pi=36\pi\:in^3$

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Example Question #851 : Ssat Upper Level Quantitative (Math)

A cube has six square faces, each with area 64 square inches. Using the conversion factor 1 inch = 2.5 centimeters, give the volume of this cube in cubic centimeters, rounding to the nearest whole number.

Possible Answers:

2,000 cubic centimeters

400 cubic centimeters

800 cubic centimeters

8,000 cubic centimeters

4,000 cubic centimeters

Correct answer:

8,000 cubic centimeters

Explanation:

The volume of a cube is the cube of its sidelength, which is also the sidelength of each square face. This sidelength is the square root of the area 64:

$\sqrt{64}=8$ inches.

Multiply this by 2.5 to get the sidelength in centimeters:

$\times 2.5$ = 20 centimeters.

The cube of this is

$20^{3}$ =8000 cubic centimeters

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Example Question #852 : Ssat Upper Level Quantitative (Math)

A cube has a side length of 5 inches. Give the volume and surface area of the cube.

Possible Answers:

25 in^3 ,75 in^2

125 in^3 ,150 in^2

25 in^3 ,150 in^2

125 in^3 ,75 in^2

250in^3, 150 in^2

Correct answer:

125 in^3 ,150 in^2

Explanation:

A cube has all edges the same length. The volume of a cube is found by multiplying the length of any edge by itself twice. As a formula:

Volume=s^3 where is the length of any edge of the cube.

The Surface Area of a cube can be calculated as 6s^2 .

So we get:

Volume =s^3=5^3=125 in^3

Surface area $=6s^2=6\times 5^2=6\times 25=150 in^2$

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Example Question #2 : How To Find The Volume Of A Figure

A cheese seller has a 2 foot x 2 foot x 2 foot block of gouda and she wants to cut it into smaller gouda cubes that are 1.5 inches on a side. How many cubes can she cut?

Possible Answers:

4096

2048

512

256

1024

Correct answer:

4096

Explanation:

First we need to determine how many of the small cubes of gouda would fit along one dimension of the large cheese block. One edge of the large block is 24 inches, so 16 smaller cubes $(24\div 1.5)$ would fit along the edge. Now we simply cube this one dimension to see how many cubes fit within the whole cube. $16^{3}=4096$ .

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Example Question #4 : How To Find The Volume Of A Cube

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

Possible Answers:

$\frac{N^{3} \sqrt{2} }{4}$

$\frac{N^{3} }{3 }$

$\frac{N^{3} }{2 }$

$3N\sqrt{N}$

$\frac{N^{3} \sqrt{3} }{9 }$

Correct answer:

$\frac{N^{3} \sqrt{3} }{9 }$

Explanation:

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

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

$3s^{2}=N^{2}$

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

$s =\sqrt{\frac{N^{2}}{3}} = \frac{N}{\sqrt{3}} = \frac{N\sqrt{3}}{3}$

Cube this sidelength to get the volume:

$s^{3} = \left ( \frac{N\sqrt{3}}{3} \right )^{3} = \frac{N^{3} \left ( \sqrt{3}\right ) ^{3} }{3^{3} } = \frac{N^{3} \left (3 \sqrt{3}\right ) }{27 }= \frac{N^{3} \sqrt{3} }{9 }$

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Example Question #853 : Ssat Upper Level Quantitative (Math)

Give the volume of a cube with surface area .

Possible Answers:

$6\sqrt[3]{N^{2}}$

$6N\sqrt{ 6N}$

$\frac{N^{3}}{36}$

$\frac{N^{3}}{216}$

$\frac{N\sqrt{ 6N} }{36}$

Correct answer:

$\frac{N\sqrt{ 6N} }{36}$

Explanation:

Let be the length of one edge of the cube. Since its surface area is , one face has one-sixth of this area, or $\frac{N}{6}$ . Therefore,

$s^{2} = \frac{N}{6}$ ,

and

$s =\sqrt{ \frac{N}{6}}= \frac{ \sqrt{ N}}{\sqrt{ 6}}= \frac{ \sqrt{ 6N}}{6}$

Cube this sidelength to get the volume:

$s ^{3}= \left ( \frac{ \sqrt{ 6N}}{6} \right )^{3} = \frac{\left ( \sqrt{ 6N} \right )^{3}}{6^{3}}= \frac{ 6N\sqrt{ 6N} }{216}= \frac{N\sqrt{ 6N} }{36}$

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Example Question #5 : How To Find The Volume Of A Cube

The length of a diagonal of one face of a cube is 10. Give the volume of the cube.

Possible Answers:

$\frac{1,000\sqrt{3} }{9}$

$500\sqrt{2}$

$\frac{1,000 \sqrt{3} }{3}$

$250\sqrt{2}$

$125\sqrt{2}$

Correct answer:

$250\sqrt{2}$

Explanation:

Since a diagonal of a square face of the cube is 10, each side of each square has length $\frac{\sqrt{2}}{2}$ times this, or $10 \cdot \frac{\sqrt{2}}{2} = 5\sqrt{2}$ .

Cube this to get the volume of the cube:

$\left (5\sqrt{2} \right )^{3}= 5^{3} \left ( \sqrt{2} \right )^{3} = 125 \left (2 \sqrt{2} \right ) = 250 \sqrt{2}$ .

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Example Question #3 : How To Find The Volume Of A Figure

An aquarium is shaped like a perfect cube; the area of each glass face is 1.44 square meters. If it is filled to the recommended 90% capacity, then, to the nearest hundred liters, how much water will it contain?

Note: 1 cubic meter = 1,000 liters.

Possible Answers:

$1,400\textup{ L}$

$1,700\textup{ L}$

$1,200\textup{ L}$

$1,600 \textup{ L}$

$1,300\textup{ L}$

Correct answer:

$1,600 \textup{ L}$

Explanation:

A perfect cube has square faces; if a face has area 1.44 square meters, then each side of each face measures the square root of this, or 1.2 meters. The volume of the tank is the cube of this, or

$1.2^{3} = 1.728$ cubic meters.

Its capacity in liters is $1.728 \times 1,000 = 1,728$ liters.

90% of this is

$1,728 \times 0.9 = 1,555.2$ liters.

This rounds to 1,600 liters, the correct response.

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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #163 : Geometry

Example Question #4 : How To Find The Volume Of A Cone

Example Question #11 : Volume Of A Three Dimensional Figure

Example Question #851 : Ssat Upper Level Quantitative (Math)

Example Question #852 : Ssat Upper Level Quantitative (Math)

Example Question #2 : How To Find The Volume Of A Figure

Example Question #4 : How To Find The Volume Of A Cube

Example Question #853 : Ssat Upper Level Quantitative (Math)

Example Question #5 : How To Find The Volume Of A Cube

Example Question #3 : How To Find The Volume Of A Figure

All SSAT Upper Level Math Resources

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