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

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Example Questions

Example Question #21 : Volume Of A Three Dimensional Figure

Give the volume of a cube with surface area 3 square meters.

Possible Answers:

$250,000\sqrt{2} \textup{ cm}^{3}$ $\displaystyle 250,000\sqrt{2} \textup{ cm}^{3}$

$250,000 \textup{ cm}^{3}$ $\displaystyle 250,000 \textup{ cm}^{3}$

$125,000 \textup{ cm}^{3}$ $\displaystyle 125,000 \textup{ cm}^{3}$

$125,000\sqrt{2} \textup{ cm}^{3}$ $\displaystyle 125,000\sqrt{2} \textup{ cm}^{3}$

$125,000\sqrt{3} \textup{ cm}^{3}$ $\displaystyle 125,000\sqrt{3} \textup{ cm}^{3}$

Correct answer:

$250,000\sqrt{2} \textup{ cm}^{3}$ $\displaystyle 250,000\sqrt{2} \textup{ cm}^{3}$

Explanation:

Let $\displaystyle s$ be the length of one edge of the cube. Since its surface area is 3 square meters, one face has one-sixth of this area, or $\frac{3}{6} = \frac{1}{2}$ $\displaystyle \frac{3}{6} = \frac{1}{2}$ square meters. Therefore, $s^{2} = \frac{1}{2}$ $\displaystyle s^{2} = \frac{1}{2}$, and $s = \sqrt{ \frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}}=\frac{1 \cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}= \frac{\sqrt{2}}{2}$ $\displaystyle s = \sqrt{ \frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}}=\frac{1 \cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}= \frac{\sqrt{2}}{2}$ meters.

The choices are in centimeters, so multiply this by 100 - the sidelength is

$\frac{\sqrt{2}}{2} \cdot 100 = 50 \sqrt{2}$ $\displaystyle \frac{\sqrt{2}}{2} \cdot 100 = 50 \sqrt{2}$ centimeters.

The volume is the cube of this, or $V= \left ( 50 \sqrt{2} \right )^{3}= 50^{3} \cdot \left ( \sqrt{2} \right )^{3}= 125,000 \cdot 2 \sqrt{2}= 250,000\sqrt{2}$ $\displaystyle V= \left ( 50 \sqrt{2} \right )^{3}= 50^{3} \cdot \left ( \sqrt{2} \right )^{3}= 125,000 \cdot 2 \sqrt{2}= 250,000\sqrt{2}$ cubic centimeters.

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Example Question #7 : How To Find The Volume Of A Cube

The length of a diagonal of a cube is $15\sqrt{2}$ $\displaystyle 15\sqrt{2}$. Give the volume of the cube.

Possible Answers:

$750 \sqrt{2}$ $\displaystyle 750 \sqrt{2}$

$750 \sqrt{6}$ $\displaystyle 750 \sqrt{6}$

$1,500 \sqrt{2}$ $\displaystyle 1,500 \sqrt{2}$

$750 \sqrt{3}$ $\displaystyle 750 \sqrt{3}$

$1,500 \sqrt{3}$ $\displaystyle 1,500 \sqrt{3}$

Correct answer:

$750 \sqrt{6}$ $\displaystyle 750 \sqrt{6}$

Explanation:

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

$s^{2}+s^{2}+s^{2}= (15 \sqrt{2})^{2}$ $\displaystyle s^{2}+s^{2}+s^{2}= (15 \sqrt{2})^{2}$

$3s^{2}=15 ^{2}\cdot \left ( \sqrt{2} \right )^{2}$ $\displaystyle 3s^{2}=15 ^{2}\cdot \left ( \sqrt{2} \right )^{2}$

$3s^{2}=225 \cdot 2 = 450$ $\displaystyle 3s^{2}=225 \cdot 2 = 450$

$s^{2} = 150$ $\displaystyle s^{2} = 150$

$s = \sqrt{150 } = \sqrt{25} \cdot \sqrt{6} = 5\sqrt{6}$ $\displaystyle s = \sqrt{150 } = \sqrt{25} \cdot \sqrt{6} = 5\sqrt{6}$

Cube the sidelength to get the volume:

$s ^{3 }= \left ( 5\sqrt{6} \right )^{3 }$ $\displaystyle s ^{3 }= \left ( 5\sqrt{6} \right )^{3 }$

$s ^{3 }= 5^{3 } \left (\sqrt{6} \right )^{3 } = 125 \cdot 6 \sqrt{6} = 750 \sqrt{6}$ $\displaystyle s ^{3 }= 5^{3 } \left (\sqrt{6} \right )^{3 } = 125 \cdot 6 \sqrt{6} = 750 \sqrt{6}$

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Example Question #1 : How To Find The Volume Of A Cube

The length of a diagonal of one face of a cube is $16\sqrt{2}$ $\displaystyle 16\sqrt{2}$. Give the volume of the cube.

Possible Answers:

4,096 $\displaystyle 4,096$

32,768 $\displaystyle 32,768$

$8,192\sqrt{2}$ $\displaystyle 8,192\sqrt{2}$

$65,536\sqrt{2}$ $\displaystyle 65,536\sqrt{2}$

The correct answer is not among the other responses.

Correct answer:

4,096 $\displaystyle 4,096$

Explanation:

A diagonal of a square has length $\sqrt{2}$ $\displaystyle \sqrt{2}$ times that of a side, so each side of each square face of the cube has length $16 \sqrt{2} \div \sqrt{2}= 16$ $\displaystyle 16 \sqrt{2} \div \sqrt{2}= 16$. Cube this to get the volume:

$16^{3} = 4,096$ $\displaystyle 16^{3} = 4,096$

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Example Question #11 : How To Find The Volume Of A Cube

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

Possible Answers:

$64 \sqrt{3} \textup{ in}^{3}$ $\displaystyle 64 \sqrt{3} \textup{ in}^{3}$

$192 \sqrt{3} \textup{ in}^{3}$ $\displaystyle 192 \sqrt{3} \textup{ in}^{3}$

$216 \textup{ in}^{3}$ $\displaystyle 216 \textup{ in}^{3}$

$266 \sqrt{2} \textup{ in}^{3}$ $\displaystyle 266 \sqrt{2} \textup{ in}^{3}$

$288 \textup{ in}^{3}$ $\displaystyle 288 \textup{ in}^{3}$

Correct answer:

$192 \sqrt{3} \textup{ in}^{3}$ $\displaystyle 192 \sqrt{3} \textup{ in}^{3}$

Explanation:

Since we are looking at inches, we will look at one foot as twelve inches.

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

$s^{2}+s^{2}+s^{2}=12^{2}$ $\displaystyle s^{2}+s^{2}+s^{2}=12^{2}$

$3s^{2}=144$ $\displaystyle 3s^{2}=144$

$s^{2}=48$ $\displaystyle s^{2}=48$

$s= \sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}$ $\displaystyle s= \sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}$ inches.

Cube this sidelength to get the volume:

$s^{3}= \left ( 4\sqrt{3} \right )^{3} = 4^{3} \cdot \left ( \sqrt{3} \right ) ^{3} = 64 \cdot 3 \sqrt{3}= 192 \sqrt{3}$ $\displaystyle s^{3}= \left ( 4\sqrt{3} \right )^{3} = 4^{3} \cdot \left ( \sqrt{3} \right ) ^{3} = 64 \cdot 3 \sqrt{3}= 192 \sqrt{3}$ cubic inches.

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Example Question #12 : How To Find The Volume Of A Cube

The distance from one vertex of a perfectly cubic aquarium to its opposite vertex is 4.5 $\displaystyle 4.5$ meters. Give the volume of the aquarium in liters.

$\displaystyle 1$ cubic meter = 1,000 $\displaystyle 1,000$ liters.

Possible Answers:

$6,750 \sqrt{2} \textup{ L}$ $\displaystyle 6,750 \sqrt{2} \textup{ L}$

$10,125\sqrt{3}\textup{ L}$ $\displaystyle 10,125\sqrt{3}\textup{ L}$

$6,750 \textup{ L}$ $\displaystyle 6,750 \textup{ L}$

$10,125 \textup{ L}$ $\displaystyle 10,125 \textup{ L}$

$91,125 \textrm{ L}$ $\displaystyle 91,125 \textrm{ L}$

Correct answer:

$10,125\sqrt{3}\textup{ L}$ $\displaystyle 10,125\sqrt{3}\textup{ L}$

Explanation:

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

$s^{2}+s^{2}+s^{2}=4.5^{2}$ $\displaystyle s^{2}+s^{2}+s^{2}=4.5^{2}$

$3s^{2}=20.25$ $\displaystyle 3s^{2}=20.25$

$s^{2} =6.75 = \frac{27}{4}$ $\displaystyle s^{2} =6.75 = \frac{27}{4}$

$s = \sqrt{\frac{27}{4}} = \frac{\sqrt{27}}{\sqrt{4}}} = \frac{ \sqrt{9} \cdot \sqrt{3}}{2}= \frac{3\sqrt{3}}{2}$ $\displaystyle s = \sqrt{\frac{27}{4}} = \frac{\sqrt{27}}{\sqrt{4}}} = \frac{ \sqrt{9} \cdot \sqrt{3}}{2}= \frac{3\sqrt{3}}{2}$ meters.

Cube this sidelength to get the volume:

$s ^{3 } =\left ( \frac{3\sqrt{3}}{2} \right )^{3 }= \frac{3^{3 } \left (\sqrt{3}\right )^{3 }}{2^{3 } } = \frac{ 27 \cdot 3\sqrt{3} }{8 } = \frac{ 81 \sqrt{3} }{8 }$ $\displaystyle s ^{3 } =\left ( \frac{3\sqrt{3}}{2} \right )^{3 }= \frac{3^{3 } \left (\sqrt{3}\right )^{3 }}{2^{3 } } = \frac{ 27 \cdot 3\sqrt{3} }{8 } = \frac{ 81 \sqrt{3} }{8 }$ cubic meters.

To convert this to liters, multiply by 1,000:

$\frac{ 81\sqrt{3} }{8 } \cdot 1,000 = \frac{81\sqrt{3} }{8 } \cdot \frac{1,000}{1} = \frac{ 81\sqrt{3} }{1 } \cdot \frac{125}{1}= 10,125\sqrt{3}$ $\displaystyle \frac{ 81\sqrt{3} }{8 } \cdot 1,000 = \frac{81\sqrt{3} }{8 } \cdot \frac{1,000}{1} = \frac{ 81\sqrt{3} }{1 } \cdot \frac{125}{1}= 10,125\sqrt{3}$ liters.

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Example Question #861 : Ssat Upper Level Quantitative (Math)

Find the volume of a cube with a side length of $\displaystyle 20$.

Possible Answers:

$8\textup{,}000$ $\displaystyle 8\textup{,}000$

$6\textup{,}000$ $\displaystyle 6\textup{,}000$

400 $\displaystyle 400$

800 $\displaystyle 800$

$4\textup{,}000$ $\displaystyle 4\textup{,}000$

Correct answer:

$8\textup{,}000$ $\displaystyle 8\textup{,}000$

Explanation:

Write the formula for the volume of a cube.

$V=a^3=(20)^3=8\textup{,}000$ $\displaystyle V=a^3=(20)^3=8\textup{,}000$

The correct answer is $8\textup{,}000$ $\displaystyle 8\textup{,}000$.

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Example Question #862 : Ssat Upper Level Quantitative (Math)

Find the volume of a square pyramid that has a height of $\displaystyle 9$ and a side length of $\displaystyle 4$.

Possible Answers:

$\displaystyle 48$

$\displaystyle 60$

$\displaystyle 54$

$\displaystyle 42$

Correct answer:

$\displaystyle 48$

Explanation:

The formula to find the volume of a square pyramid is

$\text{Volume}=(side)^2\frac{height}{3}$ $\displaystyle \text{Volume}=(side)^2\frac{height}{3}$

So plugging in the information given from the question,

$\text{Volume}=4^2\frac{9}{3}=16\times3=48$ $\displaystyle \text{Volume}=4^2\frac{9}{3}=16\times3=48$

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Example Question #863 : Ssat Upper Level Quantitative (Math)

Find the volume of a square pyramid with a height of $\displaystyle 12$ and a length of a side of its square base of $\displaystyle 3$.

Possible Answers:

$\displaystyle 24$

$\displaystyle 36$

$\displaystyle 60$

$\displaystyle 48$

Correct answer:

$\displaystyle 36$

Explanation:

The formula to find the volume of a square pyramid is

$\text{Volume}=(side)^2\frac{height}{3}$ $\displaystyle \text{Volume}=(side)^2\frac{height}{3}$

So plugging in the information given from the question,

$\text{Volume}=(3^2)\frac{12}{3}=9\times 4=36$ $\displaystyle \text{Volume}=(3^2)\frac{12}{3}=9\times 4=36$

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Example Question #864 : Ssat Upper Level Quantitative (Math)

Find the volume of a regular hexagonal prism that has a height of $\displaystyle 10$. The side length of the hexagon base is $\displaystyle 12$.

Possible Answers:

$4320\sqrt3$ $\displaystyle 4320\sqrt3$

$2160\sqrt3$ $\displaystyle 2160\sqrt3$

$1440\sqrt3$ $\displaystyle 1440\sqrt3$

$660\sqrt3$ $\displaystyle 660\sqrt3$

Correct answer:

$2160\sqrt3$ $\displaystyle 2160\sqrt3$

Explanation:

The formula to find the volume of a hexagonal prism is

$\text{Volume}=\frac{3\sqrt3}{2}(side)^2(height)$ $\displaystyle \text{Volume}=\frac{3\sqrt3}{2}(side)^2(height)$

Plugging in the values given by the question will give

$\text{Volume}=\frac{3\sqrt3}{2}(12^2)(10)=2160\sqrt3$ $\displaystyle \text{Volume}=\frac{3\sqrt3}{2}(12^2)(10)=2160\sqrt3$

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Example Question #25 : Volume Of A Three Dimensional Figure

In terms of $\displaystyle x$, find the volume of a regular hexagonal prism that has a height of $\displaystyle 5$. The hexagon base has side lengths of $\displaystyle 4x$.

Possible Answers:

$240x\sqrt3$ $\displaystyle 240x\sqrt3$

$120x\sqrt3$ $\displaystyle 120x\sqrt3$

$120x^2\sqrt3$ $\displaystyle 120x^2\sqrt3$

$360x^2\sqrt3$ $\displaystyle 360x^2\sqrt3$

Correct answer:

$120x^2\sqrt3$ $\displaystyle 120x^2\sqrt3$

Explanation:

The formula to find the volume of a hexagonal prism is

$\text{Volume}=\frac{3\sqrt3}{2}(side)^2(height)$ $\displaystyle \text{Volume}=\frac{3\sqrt3}{2}(side)^2(height)$

Plugging in the values given by the question will give

$\text{Volume}=\frac{3\sqrt3}{2}(4x)^2(5)=120x^2\sqrt3$ $\displaystyle \text{Volume}=\frac{3\sqrt3}{2}(4x)^2(5)=120x^2\sqrt3$

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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 #21 : Volume Of A Three Dimensional Figure

Example Question #7 : How To Find The Volume Of A Cube

Example Question #1 : How To Find The Volume Of A Cube

Example Question #11 : How To Find The Volume Of A Cube

Example Question #12 : How To Find The Volume Of A Cube

Example Question #861 : Ssat Upper Level Quantitative (Math)

Example Question #862 : Ssat Upper Level Quantitative (Math)

Example Question #863 : Ssat Upper Level Quantitative (Math)

Example Question #864 : Ssat Upper Level Quantitative (Math)

Example Question #25 : Volume Of A Three Dimensional Figure

All SSAT Upper Level Math Resources

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