### All SAT Math Resources

## Example Questions

### Example Question #591 : Problem Solving

An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?

**Possible Answers:**

300π

270π

345π

295π

325π

**Correct answer:**

295π

The height of the submerged part of the cylinder is 27cm. 2πrh + πr^{2 }is equal to 270π + 25π = 295π

### Example Question #33 : Cylinders

A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in^{2}. What is the area of its bases?

**Possible Answers:**

312.5 in^{2}

None of the other answers

78.125π in^{2}

156.25 in^{2}

39.0625π in^{2}

**Correct answer:**

78.125π in^{2}

The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:

A = h * π * d or A = h * π * 2r = 2πrh

Now, substituting in our values, we get:

512.5π = 2 * 41*rπ; 512.5π = 82rπ

Solve for r by dividing both sides by 82π:

6.25 = r

From here, we can calculate the area of a base:

A = 6.25^{2}π = 39.0625π

**NOTE: **The question asks for the area of the **bases**. Therefore, the answer is 2 * 39.0625π or 78.125π in^{2}.

### Example Question #591 : Problem Solving

The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)

**Possible Answers:**

$0.08

$0.29

$0.16

$0.24

$1.18

**Correct answer:**

$0.24

The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.

### Example Question #1 : Cylinders

Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in^{2} to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?

**Possible Answers:**

$0.45

$0.77

$0.57

$0.84

$0.91

**Correct answer:**

$0.84

We have the following categories to consider:

<Aluminum Cost> = (<Area of the top and bottom of the can> + <Lateral area of the can>) * 0.0015

<Label Cost> = (<Area of Label>) * 0.0001

<Attachment cost> = 2 * 0.00125 = $0.0025

The area of ends of the can are each equal to π*5^{2} or 25π. For two ends, that is 50π.

The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.

Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.

The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.

Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.

### Example Question #5 : Cylinders

The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If *r* and *h* represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to *r* in terms of *h*?

**Possible Answers:**

r = 2h/(h – 2)

r = h^{2} + 2h

r = h/(2h – 2)

r = h^{2}/(h + 2)

r = 2h^{2} + 2

**Correct answer:**

r = 2h/(h – 2)

We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.

surface area of cylinder = surface area of bases + lateral surface area

The bases of the cylinder will be two circles with radius *r*. Thus, the area of each will be *πr*^{2}, and their combined surface area will be 2*πr*^{2}.

The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2*πr*, and the height is *h*, so the lateral area is 2*πrh*.

surface area of cylinder = 2*πr*^{2} + 2*πrh*

Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is *πr*^{2}, and the height is *h*, so the volume of the cylinder is *πr*^{2}*h*.

volume = *πr*^{2}*h*

Then, we must set the volume and surface area expressions equal to one another and solve for *r* in terms of *h*.

2*πr*^{2} + 2*πrh* = *πr*^{2}*h*

First, let's factor out 2*πr* from the left side.

2*πr*(*r *+ *h*) = *πr*^{2}*h*

We can divide both sides by π.

2*r*(*r *+ *h*) = *r*^{2}*h*

We can also divide both sides by *r*, because the radius cannot equal zero.

2(*r *+ *h*) = *rh*

Let's now distribute the 2 on the left side.

2*r* + 2*h* = *rh*

Subtract 2*r* from both sides to get all the *r*'s on one side.

2*h* = *rh* – 2*r*

*r*h – 2*r* = 2*h*

Factor out an *r* from the left side.

*r*(*h –* 2) = 2*h*

Divide both sides by *h* – 2

*r* = 2*h*/(*h –* 2)

The answer is *r* = 2*h*/(*h –* 2).

### Example Question #2 : Cylinders

What is the surface area of a cylinder with a radius of 4 and a height of 7?

**Possible Answers:**

96*π*

88*π*

108*π*

164*π*

225*π*

**Correct answer:**

88*π*

Surface area of a cylinder = 2*πr*^{2} + 2*πrh* = 2 * 4^{2 }* *π* + 2 * 4 * 7 * *π* = 32*π* + 56*π* = 88*π*

Certified Tutor