SAT Math - Cylinders | Practice Hub

1

Claire's cylindrical water bottle is 9 inches tall and has a diameter of 6 inches. How many cubic inches of water will her bottle hold?

$81\pi$

$324\pi$

$54\pi$

$18\pi$

$45\pi$

Explanation

The volume is the area of the base times the height. The area of the base is $\pi r^{2}$ , and the radius here is 3. $\pi\cdot 3^{2}\cdot 9=81\pi$

2

Claire's cylindrical water bottle is 9 inches tall and has a diameter of 6 inches. How many cubic inches of water will her bottle hold?

$81\pi$

$324\pi$

$54\pi$

$18\pi$

$45\pi$

Explanation

The volume is the area of the base times the height. The area of the base is $\pi r^{2}$ , and the radius here is 3. $\pi\cdot 3^{2}\cdot 9=81\pi$

3

What is the volume of a cylinder with a diameter of 13 inches and a height of 27.5 inches?

$381.44\pi \ in^{3}$

$2583.1\pi \ in^{3}$

$357.5\pi \ in^{3}$

$4647.5\ in^{3}$

$1161.88\pi \ in^{3}$

Explanation

The equation for the volume of a cylinder is V = Ah, where A is the area of the base and h is the height.

Thus, the volume can also be expressed as V = πr2h.

The diameter is 13 inches, so the radius is 13/2 = 6.5 inches.

Now we can easily calculate the volume:

V = 6.52π * 27.5 = 1161.88π in3

4

What is the volume of a cylinder with a diameter of 13 inches and a height of 27.5 inches?

$381.44\pi \ in^{3}$

$2583.1\pi \ in^{3}$

$357.5\pi \ in^{3}$

$4647.5\ in^{3}$

$1161.88\pi \ in^{3}$

Explanation

The equation for the volume of a cylinder is V = Ah, where A is the area of the base and h is the height.

Thus, the volume can also be expressed as V = πr2h.

The diameter is 13 inches, so the radius is 13/2 = 6.5 inches.

Now we can easily calculate the volume:

V = 6.52π * 27.5 = 1161.88π in3

5

A hollow prism has a base 12 in x 13 in and a height of 42 in. A closed, cylindrical can is placed in the prism. The remainder of the prism is then filled with gel, surrounding the can. The thickness of the can is negligible. Its diameter is 9 in and its height is one-fourth that of the prism. The can has a mass of 1.5 g per in3, and the gel has a mass of 2.2 g per in3. What is the approximate overall mass of the contents of the prism?

15.22 kg

139.44 g

973.44 g

11.48 kg

13.95 kg

Explanation

We must find both the can volume and the gel volume. The formula for the gel volume is:

Gel volume = Prism volume – Can volume

The prism volume is simple: 12 * 13 * 42 = 6552 in3

The volume of the can is found by multiplying the area of the circular base by the height of the can. The height is one-fourth the prism height, or 42/4 = 10.5 in. The area of the base is equal to πr_2. Note that the prompt has given the diameter. Therefore, the radius is 4.5, not 9. The base's area is: 4.52_π = 20.25_π_. The total volume is therefore: 20.25_π_ * 10.5 = 212.625_π_ in3.

The gel volume is therefore: 6552 – 212.625_π_ or (approx.) 5884.02 in3.

The approximate volume for the can is: 667.98 in3

From this, we can calculate the approximate mass of the contents:

Gel Mass = Gel Volume * 2.2 = 5884.02 * 2.2 = 12944.844 g

Can Mass = Can Volume * 1.5 = 667.98 * 1.5 = 1001.97 g

The total mass is therefore 12944.844 + 1001.97 = 13946.814 g, or approximately 13.95 kg.

6

A hollow prism has a base 12 in x 13 in and a height of 42 in. A closed, cylindrical can is placed in the prism. The remainder of the prism is then filled with gel, surrounding the can. The thickness of the can is negligible. Its diameter is 9 in and its height is one-fourth that of the prism. The can has a mass of 1.5 g per in3, and the gel has a mass of 2.2 g per in3. What is the approximate overall mass of the contents of the prism?

15.22 kg

139.44 g

973.44 g

11.48 kg

13.95 kg

Explanation

We must find both the can volume and the gel volume. The formula for the gel volume is:

Gel volume = Prism volume – Can volume

The prism volume is simple: 12 * 13 * 42 = 6552 in3

The volume of the can is found by multiplying the area of the circular base by the height of the can. The height is one-fourth the prism height, or 42/4 = 10.5 in. The area of the base is equal to πr_2. Note that the prompt has given the diameter. Therefore, the radius is 4.5, not 9. The base's area is: 4.52_π = 20.25_π_. The total volume is therefore: 20.25_π_ * 10.5 = 212.625_π_ in3.

The gel volume is therefore: 6552 – 212.625_π_ or (approx.) 5884.02 in3.

The approximate volume for the can is: 667.98 in3

From this, we can calculate the approximate mass of the contents:

Gel Mass = Gel Volume * 2.2 = 5884.02 * 2.2 = 12944.844 g

Can Mass = Can Volume * 1.5 = 667.98 * 1.5 = 1001.97 g

The total mass is therefore 12944.844 + 1001.97 = 13946.814 g, or approximately 13.95 kg.

7

A metal cylindrical brick has a height of $6\ in$ . The area of the top is $25\pi\ in^2$ . A circular hole with a radius of $2\ in$ is centered and drilled half-way down the brick. What is the volume of the resulting shape?

$175\pi\ in^3$

$150\pi\ in^3$

$125\pi\ in^3$

$100\pi\ in^3$

$138\pi\ in^3$

Explanation

To find the final volume, we will need to subtract the volume of the hole from the total initial volume of the cylinder.

The volume of a cylinder is given by the product of the base area times the height: $V=A_{base}h$ .

Find the initial volume using the given base area and height.

$V_i=(25\pi)(6)=150\pi\ in^3$

Next, find the volume of the hole that was drilled. The base area of this cylinder can be calculated from the radius of the hole. Remember that the height of the hole is only half the height of the block.

$V_{hole}=(\pi r^2)h$

$V_{hole}=(\pi(2)^2)(3)=(4\pi)(3)=12\pi\ in^3$

Finally, subtract the volume of the hole from the total initial volume.

$150\pi-12\pi=138\pi\ in^3$

8

A metal cylindrical brick has a height of $6\ in$ . The area of the top is $25\pi\ in^2$ . A circular hole with a radius of $2\ in$ is centered and drilled half-way down the brick. What is the volume of the resulting shape?

$175\pi\ in^3$

$150\pi\ in^3$

$125\pi\ in^3$

$100\pi\ in^3$

$138\pi\ in^3$

Explanation

To find the final volume, we will need to subtract the volume of the hole from the total initial volume of the cylinder.

The volume of a cylinder is given by the product of the base area times the height: $V=A_{base}h$ .

Find the initial volume using the given base area and height.

$V_i=(25\pi)(6)=150\pi\ in^3$

Next, find the volume of the hole that was drilled. The base area of this cylinder can be calculated from the radius of the hole. Remember that the height of the hole is only half the height of the block.

$V_{hole}=(\pi r^2)h$

$V_{hole}=(\pi(2)^2)(3)=(4\pi)(3)=12\pi\ in^3$

Finally, subtract the volume of the hole from the total initial volume.

$150\pi-12\pi=138\pi\ in^3$

9

A hollow prism has a base 5 in x 6 in and a height of 10 in. A closed, cylindrical can is placed in the prism. The remainder of the prism is then filled with gel around the cylinder. The thickness of the can is negligible. Its diameter is 4 in and its height is half that of the prism. What is the approximate volume of gel needed to fill the prism?

103.33 in3

187.73 in3

203.44 in3

249.73 in3

237.17 in3

Explanation

The general form of our problem is:

Gel volume = Prism volume – Can volume

The prism volume is simple: 5 * 6 * 10 = 300 in3

The volume of the can is found by multiplying the area of the circular base by the height of the can. The height is half the prism height, or 10/2 = 5 in. The area of the base is equal to πr_2. Note that the prompt has given the diameter. Therefore, the radius is 2, not 4. The base's area is: 22_π = 4_π_. The total volume is therefore: 4_π_ * 5 = 20_π_ in3.

The gel volume is therefore: 300 – 20_π_ or (approx.) 237.17 in3.

10

A hollow prism has a base 5 in x 6 in and a height of 10 in. A closed, cylindrical can is placed in the prism. The remainder of the prism is then filled with gel around the cylinder. The thickness of the can is negligible. Its diameter is 4 in and its height is half that of the prism. What is the approximate volume of gel needed to fill the prism?

103.33 in3

187.73 in3

203.44 in3

249.73 in3

237.17 in3

Explanation

The general form of our problem is:

Gel volume = Prism volume – Can volume

The prism volume is simple: 5 * 6 * 10 = 300 in3

The volume of the can is found by multiplying the area of the circular base by the height of the can. The height is half the prism height, or 10/2 = 5 in. The area of the base is equal to πr_2. Note that the prompt has given the diameter. Therefore, the radius is 2, not 4. The base's area is: 22_π = 4_π_. The total volume is therefore: 4_π_ * 5 = 20_π_ in3.

The gel volume is therefore: 300 – 20_π_ or (approx.) 237.17 in3.

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