GED Math : GED Math

Study concepts, example questions & explanations for GED Math

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Example Questions

Example Question #14 : 3 Dimensional Geometry

Find the volume of a sphere with a radius of \(\displaystyle \frac{3}{4}\pi\).

Possible Answers:

\(\displaystyle \frac{9}{4} \pi ^3\)

\(\displaystyle \pi ^3\)

\(\displaystyle \pi^4\)

\(\displaystyle \frac{9}{16}\pi\)

\(\displaystyle \frac{9}{16} \pi ^4\)

Correct answer:

\(\displaystyle \frac{9}{16} \pi ^4\)

Explanation:

Write the formula for the volume of a sphere.

\(\displaystyle V = \frac{4}{3}\pi r^3\)

Substitute the radius into the equation.

\(\displaystyle V = \frac{4}{3}\pi (\frac{3}{4}\pi)^3= \frac{4}{3}\pi (\frac{3}{4}\pi)(\frac{3}{4}\pi)(\frac{3}{4}\pi)\)

The answer is:  \(\displaystyle \frac{9}{16} \pi ^4\)

Example Question #1531 : Ged Math

Determine the volume of a sphere with a radius of 6.

Possible Answers:

\(\displaystyle 288\pi\)

\(\displaystyle 72\pi\)

\(\displaystyle 48\pi\)

\(\displaystyle 324\pi\)

\(\displaystyle 144\pi\)

Correct answer:

\(\displaystyle 288\pi\)

Explanation:

Write the formula for the volume of a sphere.

\(\displaystyle V= \frac{4}{3}\pi r^3\)

Substitute the radius.

\(\displaystyle V= \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (6) (6) (6) = 288\pi\)

The answer is:  \(\displaystyle 288\pi\)

Example Question #1532 : Ged Math

Find the volume of a sphere with a radius of 6cm.

Possible Answers:

\(\displaystyle 72\pi \text{ cm}^3\)

\(\displaystyle 864\pi \text{ cm}^3\)

\(\displaystyle 648\pi \text{ cm}^3\)

\(\displaystyle 288\pi \text{ cm}^3\)

\(\displaystyle 216\pi \text{ cm}^3\)

Correct answer:

\(\displaystyle 288\pi \text{ cm}^3\)

Explanation:

To find the volume of a sphere, we will use the following formula:

\(\displaystyle V = \frac{4}{3}\pi r^3\)

where r is the radius of the sphere. 

Now, we know the radius of the sphere is 6cm. So, we can substitute. We get

\(\displaystyle V = \frac{4}{3} \cdot \pi \cdot (6\text{cm})^3\)

\(\displaystyle V = \frac{4}{3} \cdot \pi \cdot 216\text{cm}^3\)

Now, the 3 and the 216 can both be divided by 3. So, we get

\(\displaystyle V= \frac{4}{1} \cdot \pi \cdot 72\text{cm}^3\)

\(\displaystyle V= 4 \cdot \pi \cdot 72\text{cm}^3\)

\(\displaystyle V = 4 \cdot 72\text{cm}^3 \cdot \pi\)

\(\displaystyle V = 288\text{cm}^3 \cdot \pi\)

\(\displaystyle V = 288\pi \text{ cm}^3\)

 

Example Question #12 : 3 Dimensional Geometry

Find the volume of a sphere with a surface area of \(\displaystyle 4\pi\).

Possible Answers:

\(\displaystyle \frac{256}{3}\pi\)

\(\displaystyle \frac{16}{3}\pi\)

\(\displaystyle \frac{1}{3}\pi\)

\(\displaystyle \pi\)

\(\displaystyle \frac{4}{3} \pi\)

Correct answer:

\(\displaystyle \frac{4}{3} \pi\)

Explanation:

Write the formulas for the surface area and volume of a sphere.

\(\displaystyle A = 4\pi r^2\)

\(\displaystyle V=\frac{4}{3} \pi r^3\)

Substitute the surface area into the first formula and solve for the radius.

\(\displaystyle 4\pi = 4\pi r^2\)

Divide by \(\displaystyle 4\pi\) on both sides.

\(\displaystyle \frac{4\pi }{4\pi}= \frac{4\pi r^2}{4\pi}\)

\(\displaystyle 1=r^2\)

Square root both sides.

\(\displaystyle r=1\)

Substitute the radius into the volume formula.

\(\displaystyle V=\frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi\)

The answer is:  \(\displaystyle \frac{4}{3} \pi\)

Example Question #11 : Volume Of A Sphere

Find the volume of a sphere with a radius of \(\displaystyle \pi ^6\).

Possible Answers:

\(\displaystyle \frac{4}{3}\pi ^{19}\)

\(\displaystyle \frac{4}{3}\pi ^{10}\)

\(\displaystyle 4\pi^{10}\)

\(\displaystyle \frac{16}{3}\pi ^{10}\)

\(\displaystyle \frac{16}{3}\pi ^{19}\)

Correct answer:

\(\displaystyle \frac{4}{3}\pi ^{19}\)

Explanation:

Write the formula for the volume of a sphere.

\(\displaystyle V=\frac{4}{3}\pi r^3\)

Substitute the radius.

\(\displaystyle V=\frac{4}{3}\pi (\pi ^6)^3= \frac{4}{3}\pi (\pi ^6) (\pi ^6) (\pi ^6)\)

The answer is:  \(\displaystyle \frac{4}{3}\pi ^{19}\)

Example Question #12 : 3 Dimensional Geometry

You are 3D printing a ball which will be an integral piece to the project you are creating. If you need the ball to have a radius of 12cm, what volume of material will be required?

Possible Answers:

\(\displaystyle 144\pi cm^3\)

\(\displaystyle 2304 cm^3\)

\(\displaystyle 7234.56 cm^3\)

\(\displaystyle 1728 cm^3\)

Correct answer:

\(\displaystyle 7234.56 cm^3\)

Explanation:

You are 3D printing a ball which will be an integral piece to the project you are creating. If you need the ball to have a radius of 12cm, what volume of material will be required?

We are asked to find the volume of a sphere, to do so, we should use the formula for volume of a sphere.

\(\displaystyle V_{sphere}=\frac{4}{3} \pi r^3\)

So, let's plug in our radius and solve:

\(\displaystyle V_{sphere}=\frac{4}{3} \pi (12cm)^3=\frac{4}{3} \pi 1728cm^3=7234.56 cm^3\)

So our answer is:

\(\displaystyle 7234.56 cm^3\)

Example Question #1533 : Ged Math

You are given a spherical floating speaker for your birthday. If the speaker has a radius of 5 inches, what is its volume?

Possible Answers:

\(\displaystyle 166.67in^3\)

\(\displaystyle 523.56in^3\)

\(\displaystyle 500in^3\)

\(\displaystyle 125in^3\)

Correct answer:

\(\displaystyle 523.56in^3\)

Explanation:

You are given a spherical floating speaker for your birthday. If the speaker has a radius of 5 inches, what is its volume?

We need to find the volume of a sphere. To do so, we need the following formula:

\(\displaystyle V=\frac{4}{3}\pi r^3\)

We know r is 5, so just plug and simplify

\(\displaystyle V=\frac{4}{3}\pi (5in)^3\approx 523.56in^3\)

Example Question #21 : 3 Dimensional Geometry

A sphere has surface area \(\displaystyle 20 \pi\). Give its volume.

Possible Answers:

\(\displaystyle \frac{20 \pi }{3}\)

\(\displaystyle \frac{20 \pi \sqrt{5}}{3}\)

\(\displaystyle \frac{40 \pi}{3}\)

\(\displaystyle \frac{20 \pi \sqrt{3}}{3}\)

Correct answer:

\(\displaystyle \frac{20 \pi \sqrt{5}}{3}\)

Explanation:

The formula for the surface area \(\displaystyle A\) of a sphere, given its radius \(\displaystyle r\), is

\(\displaystyle A= 4 \pi r^{2}\).

Set \(\displaystyle A = 20 \pi\) and solve for \(\displaystyle r\) to get the radius.

\(\displaystyle 4 \pi r^{2} = 20 \pi\)

Divide both sides by \(\displaystyle 4 \pi\):

\(\displaystyle \frac{4 \pi r^{2}}{4 \pi} = \frac{20 \pi}{4 \pi}\)

\(\displaystyle r^{2} = 5\)

Take the positive square root of both sides to obtain the radius:

\(\displaystyle r = \sqrt{5}\)

The formula for the volume \(\displaystyle V\) of a sphere, given its radius \(\displaystyle r\), is

\(\displaystyle V = \frac{4 \pi r^{3}}{3}\)

Substitute \(\displaystyle \sqrt{5}\) for \(\displaystyle r\) in this formula and evaluate the expression:

\(\displaystyle V = \frac{4 \pi (\sqrt{5})^{3}}{3}\)

\(\displaystyle = \frac{4 \pi \cdot \sqrt{5}\cdot \sqrt{5} \cdot \sqrt{5}}{3}\)

\(\displaystyle = \frac{4 \pi \cdot5 \cdot \sqrt{5}}{3}\)

\(\displaystyle = \frac{20 \pi \sqrt{5}}{3}\)

Example Question #1534 : Ged Math

A ball in the shape of a sphere has a radius of \(\displaystyle 8\) inches. What volume of air, in cubic inches, is needed to fully inflate the ball?

Possible Answers:

\(\displaystyle 268.08\)

\(\displaystyle 2811.24\)

\(\displaystyle 1905.21\)

\(\displaystyle 2144.66\)

Correct answer:

\(\displaystyle 2144.66\)

Explanation:

Recall how to find the volume of a sphere:

\(\displaystyle \text{Volume}=\frac{4}{3}\pi r^3\)

Plug in the given radius to solve for the volume.

\(\displaystyle \text{Volume}=\frac{4}{3}\pi (8^3)=\frac{2048\pi}{3}=2144.66\)

Make sure to round to two places after the decimal point.

Example Question #1535 : Ged Math

A sphere has a radius of \(\displaystyle 3\). Find its volume. 

Possible Answers:

\(\displaystyle 50\Pi\)

\(\displaystyle 18\Pi\)

\(\displaystyle 60\Pi\)

\(\displaystyle 45\Pi\)

\(\displaystyle 36\Pi\)

Correct answer:

\(\displaystyle 36\Pi\)

Explanation:

The formula for the volume of a sphere is \(\displaystyle (4/3)\Pi r^3\), with \(\displaystyle r\) standing for radius. Since you know that the radius of the sphere is \(\displaystyle 3\), all you have to do is plug in \(\displaystyle 3\) for \(\displaystyle r\) and solve. Once you do that you learn that the volume is \(\displaystyle 36\Pi\).

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