### All SAT Math Resources

## Example Questions

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

The volume of a cylinder is 36π. If the cylinder’s height is 4, what is the cylinder’s diameter?

**Possible Answers:**

9

12

3

4

6

**Correct answer:**

6

Volume of a cylinder? V = πr^{2}h. Rewritten as a diameter equation, this is:

V = π(d/2)^{2}h = πd^{2}h/4

Sub in h and V: 36p = πd^{2}(4)/4 so 36p = πd^{2}

Thus d = 6

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

A cylinder has a height of 5 inches and a radius of 3 inches. Find the lateral area of the cylinder.

**Possible Answers:**

8π

24π

15π

30π

45π

**Correct answer:**

30π

LA = 2π(r)(h) = 2π(3)(5) = 30π

### Example Question #1 : Cylinders

A cylinder has a volume of 20. If the radius doubles, what is the new volume?

**Possible Answers:**

80

60

40

100

20

**Correct answer:**

80

The equation for the volume of the cylinder is πr^{2}h. When the radius doubles (r becomes 2r) you get π(2r)^{2}h = 4πr^{2}h. So when the radius doubles, the volume quadruples, giving a new volume of 80.

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

A cylinder has a height that is three times as long as its radius. If the lateral surface area of the cylinder is 54π square units, then what is its volume in cubic units?

**Possible Answers:**

81π

27π

9π

54π

243π

**Correct answer:**

81π

Let us call r the radius and h the height of the cylinder. We are told that the height is three times the radius, which we can represent as h = 3r.

We are also told that the lateral surface area is equal to 54π. The lateral surface area is the surface area that does not include the bases. The formula for the lateral surface area is equal to the circumference of the cylinder times its height, or 2πrh. We set this equal to 54π,

2πrh = 54π

Now we substitute 3r in for h.

2πr(3r) = 54π

6πr^{2} = 54π

Divide by 6π

r^{2} = 9.

Take the square root.

r = 3.

h = 3r = 3(3) = 9.

Now that we have the radius and the height of the cylinder, we can find its volume, which is given by πr^{2}h.

V = πr^{2}h

V = π(3)^{2}(9) = 81π

The answer is 81π.

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

What is the volume of a hollow cylinder whose inner radius is 2 cm and outer radius is 4 cm, with a height of 5 cm?

**Possible Answers:**

50π cm^{3}

100π cm^{3}

60π cm^{3}

20π cm^{3}

80π cm^{3}

**Correct answer:**

60π cm^{3}

The volume is found by subtracting the inner cylinder from the outer cylinder as given by V = πr_{out}^{2} h – πr_{in}^{2} h. The area of the cylinder using the outer radius is 80π cm^{3}, and resulting hole is given by the volume from the inner radius, 20π cm^{3}. The difference between the two gives the volume of the resulting hollow cylinder, 60π cm^{3}.

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

What is the volume of a right cylinder with a circumference of 25π in and a height of 41.3 in?

**Possible Answers:**

1032.5π in^{3}

25812.5π in^{3}

3831.34π in^{3}

6453.125π in^{3}

4813.33π in^{3}

**Correct answer:**

6453.125π in^{3}

The formula for the volume of a right cylinder is: V = A * h, where A is the area of the base, or πr^{2}. Therefore, the total formula for the volume of the cylinder is: V = πr^{2}h.

First, we must solve for r by using the formula for a circumference (c = 2πr): 25π = 2πr; r = 12.5.

Based on this, we know that the volume of our cylinder must be: π*12.5^{2}*41.3 = 6453.125π in^{3}

### Example Question #3 : Cylinders

An 8-inch cube has a cylinder drilled out of it. The cylinder has a radius of 2.5 inches. To the nearest hundredth, approximately what is the remaining volume of the cube?

**Possible Answers:**

157.08 in^{3}

203.34 in^{3}

354.92 in^{3}

391.33 in^{3}

462 in^{3}

**Correct answer:**

354.92 in^{3}

We must calculate our two volumes and subtract them. The volume of the cube is very simple: 8 * 8 * 8, or 512 in^{3}.

The volume of the cylinder is calculated by multiplying the area of its base by its height. The height of the cylinder is 8 inches (the height of the cube through which it is being drilled). Therefore, its volume is πr^{2}h = π * 2.5^{2 }* 8 = 50π in^{3}

The volume remaining in the cube after the drilling is: 512 – 50π, or approximately 512 – 157.0795 = 354.9205, or 354.92 in^{3}.

### Example Question #4 : Cylinders

An 12-inch cube of wood has a cylinder drilled out of it. The cylinder has a radius of 3.75 inches. If the density of the wood is 4 g/in^{3}, what is the mass of the remaining wood after the cylinder is drilled out?

**Possible Answers:**

4791.43 g

2594.11 g

3193.33 g

4921.4 g

3813.3 g

**Correct answer:**

4791.43 g

We must calculate our two volumes and subtract them. Following this, we will multiply by the density.

The volume of the cube is very simple: 12 * 12 * 12, or 1728 in^{3}.

The volume of the cylinder is calculated by multiplying the area of its base by its height. The height of the cylinder is 8 inches (the height of the cube through which it is being drilled). Therefore, its volume is πr^{2}h = π * 3.75^{2 }* 12 = 168.75π in^{3}.

The volume remaining in the cube after the drilling is: 1728 – 168.75π, or approximately 1728 – 530.1433125 = 1197.8566875 in^{3}. Now, multiply this by 4 to get the mass: (approx.) 4791.43 g.

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

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?

**Possible Answers:**

103.33 in^{3}

237.17 in^{3}

187.73 in^{3}

203.44 in^{3}

249.73 in^{3}

**Correct answer:**

237.17 in^{3}

The general form of our problem is:

Gel volume = Prism volume – Can volume

The prism volume is simple: 5 * 6 * 10 = 300 in^{3}

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: 2^{2}*π* = 4*π*. The total volume is therefore: 4*π* * 5 = 20*π* in^{3}.

The gel volume is therefore: 300 – 20*π* or (approx.) 237.17 in^{3}.

### Example Question #6 : Cylinders

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 in^{3}, and the gel has a mass of 2.2 g per in^{3}. What is the approximate overall mass of the contents of the prism?

**Possible Answers:**

11.48 kg

973.44 g

15.22 kg

13.95 kg

139.44 g

**Correct answer:**

13.95 kg

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 in^{3}

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: 4.5^{2}*π* = 20.25*π*. The total volume is therefore: 20.25*π* * 10.5 = 212.625*π* in^{3}.

The gel volume is therefore: 6552 – 212.625*π* or (approx.) 5884.02 in^{3}.

The approximate volume for the can is: 667.98 in^{3}

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.

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