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Welcome back to our fifth and final article on GMAT circles. Last time we explored the possibilities of treating a circle’s of a right triangle. This time we will introduce you to the concept of cylinders and spheres — two 3-dimensional shapes built from circles.

## 1. Cylinder

More than likely, you already know what these things are and could describe them. But let’s try to define them in some interesting ways. A cylinder is a “tall circle” or – to use more proper geometric terminology – a circular prism. A prism is the solid shape that results when you take any polygon and “pull it” upward into something like a pillar. The polygon you started with still exists as the “top and bottom” faces of the prism, and the faces around the sides of the prism are rectangles. (Technically they can be parallelograms, which would produce a “leaning” pillar, but this won’t happen on the GMAT.)

Since a circle doesn’t have sides, a cylinder doesn’t have faces – except for the two circles on its top and bottom. In between, there is one smoothly-curving surface. If you need to find the area of this third surface, you can treat it like a rectangle. The length of this rectangle is the height of the cylinder, and the width of this rectangle is the circumference of the circle. The volume of any prism is the area of its base polygon multiplied by the prism’s height. So for a cylinder, the equation is

V = πr²h

## 2. Sphere

Now for spheres. We all know that a sphere is a perfectly round ball. But think about this: a sphere is like a circle “any way you slice it” – quite literally. If you have some citrus fruits in your kitchen, you can try slicing them in different places at different angles, and the faces of the two resulting pieces will always be circles. Another way to say this is that any cross section taken from a sphere will be a circle. No matter how hard you try, you will never be able to produce an elliptical orange slice. Sorry to disappoint you.

Let’s see how the GMAT employs these shapes in some official problems. Some basic cylinder problems focus on one whole cylinder. More challenging cylinder problems compare one cylinder to another or treat a cylinder as a partially-filled tank.

## 3. A Data Sufficiency Problem Featuring Two Cylinders

It costs \$2,250 to fill right circular cylindrical Tank R with a certain industrial chemical. If the cost to fill any tank with this chemical is directly proportional to the volume of the chemical needed to fill the tank, how much does it cost to fill right circular cylindrical Tank S with the chemical?

1. The diameter of the interior of Tanks R is twice the diameter of the interior of Tank S.
2. The interiors of Tanks R and S have the same height.

(A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
(B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient.

### Solution

Since the cost to fill any tank (including tanks R and S) with this chemical is directly proportional to the volume of the cylindrical tank, the only thing we care about here is the ratio of the two tanks’ volumes. Remember that for the volume of a cylinder, we need to know (or be able to derive) both the area of the circle and the height of the cylinder.

Statement 1 gives us the ratio of the tanks’ diameters: 2:1. This means that the ratio of the areas of the tanks’ bases is 4:1 (if lost here, review on area, circumference, and pi). This is great, but it is still not enough to know the overall ratio of the tanks’ volumes. Statement 1 is insufficient.

Statement 2 tells us that tanks R and S are the same height, specifying “interior” because we are filling up space with a chemical and can’t count whatever volume is taken up by the tank walls. On its own, this information is insufficient.

Combining statements 1 and 2, we have the ratio of the tanks’ diameters (2:1) and the ratio of their heights (1:1). This means that the overall ratio of the tanks’ volumes is fixed. Statements 1 and 2 together are sufficient, and the correct answer is C.

## 4. Partially Filled Cylinder-as-a-tank Problem

The figures show a sealed container that is a right circular cylinder filled with liquid to 12 its capacity. If the container is placed on its base, the depth of the liquid in the container is 10 centimeters and if the container is placed on its side, the depth of the liquid is 20 centimeters. How many cubic centimeters of liquid are in the container.

(A) 4,000 π
(B) 2,000 π
(C) 1,000 π
(D) 400 π
(E) 200 π

### Solution

This problem is less complex than it might first appear. It all comes together when you realize that the 20cm depth in the second orientation of the tank represents the radius of the circle!  Now you can get the area of the circle in cm² using A = r² and then multiply the result by 10 (the depth in centimeters of the liquid in the upright tank) to get the volume of the liquid in cm³. If you can mentally square 20 and then multiply by 10, you should be just seconds away from selecting correct answer choice A.

## 5. Final Cylinder-as-a-tank Problem

Solve carefully before reading on.

A tank is filled with gasoline to a depth of exactly 2 feet. The tank is a cylinder resting horizontally on its side, with its circular end oriented vertically. The inside of the tank is exactly 6 feet long. What is the volume of the gasoline in the tank?

1. The inside of the tank is exactly 4 feet in diameter.
2. The top surface of the gasoline forms a rectangle that has an area of 24 square feet.

(A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
(B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient.

### Solution

#### Statement 1.

Evaluating statement 1 is fairly straightforward. Combined with the information from the question stem that the depth of the gasoline in the tank is 2 feet, the additional information that the inside of the tank has a 4-foot diameter means that the tank is filled halfway with gasoline. If 4 is the diameter, then 2 is the radius, and the gas fills the tank up to its center line. This looks just like the half-filled tank in the previous problem. The question stem also gave us the length of the tank (called a length rather than a height since this tank is a cylinder “lying down”), so the cylinder’s total and fractional volumes are calculable. Statement 1 is sufficient.

#### Statement 2.

Statement 2 performs something like a “double flip.” We are told that the top surface of the gasoline is a 24ft² rectangle. Remembering from the question stem that the tank is 6 feet long, you may realize that 24/6 = 4 and think that this tells you the same thing as statement 1: that the tank has a 4-foot diameter. This would be a mistake. The 24ft² rectangle formed by the surface of the gasoline indeed has a length of 6 and a width of 4, but this width of 4 is not necessarily the diameter of the tank. It could just as easily happen in a larger tank that is less than half (or more than half) full.

Does this make statement 2 insufficient? Well so far, yes. But there’s something we’ve left out that makes it sufficient after all! From the question stem, the depth of the gasoline in the tank is 2 feet. Imagine that the circular end of this tank is transparent. Looking at it this way, the top surface of the gasoline makes a horizontal chord across the circle, and this chord has a length of 4. Simultaneously, this chord is a vertical distance of 2 feet from the bottom of the circle (since the depth of the gasoline in the tank is 2 feet). The only way this can happen is if the 4-foot chord is the diameter of the circle!

Therefore the tank is still half full, and the volume of the gasoline is half of the (calculable) volume of the cylinder. Statement 2 is also sufficient, and the correct answer choice is D.

## 6. Sphere Problem

For the final problem in our circles series, we’ll work with spheres. Spheres are less common on the GMAT than cylinders, and you will never have to memorize any of their formulas. If you need a sphere formula for a problem, it will be supplied with the problem.

For a party, three solid cheese balls with diameters of 2 inches, 4 inches, and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is 433, where r is the radius.)

(A) 12
(B) 16
(C) ∛16
(D) 8
(E) 236

### Solution – Long Way

This sounds like a party you don’t want to miss. I don’t know exactly how to combine three solid cheese balls into one, but I do know how to calculate the diameter.

There are two ways to solve this problem: the long way and the best way. The long way is to calculate the volumes of the three original cheese balls, sum your answers into one volume, and then solve for the radius of the combined cheese ball. First you must divide the given diameters of the original cheese balls by 2, since the volume equation uses radius instead.

V = (4π/3)r³
V = (4π/3)1³ + (4π/3)2³ + (4π/3)3³
V = (4π/3)(1³ + 2³ + 3³)
V = (4π/3)(1 + 8 + 27)
V = (4π/3)(36)
V = 48π
V = (4π/3)r³
48π = (4π/3)r³
48 = (4/3)r³
36 = r³
∛36 = r
2(∛36) = D

And the correct answer choice is E.

### Solution – Short Way

That was the long way. The best way is to think logically and exploit the answer choices. Since we are effectively adding some cheese onto a ball that already has a diameter of 6 inches, the diameter of the combined cheese ball will be greater than 6 inches. This means that answer choices C and D are nonstarters. (C is somewhere between 2 and 3, and D is exactly 6.) Let’s think next about choices A and B, since they are integers and easier to evaluate than choice E.

Can the diameter of the combined cheese ball be as great as 12 (choice A) or even 16 (choice B)? No, it can’t. Picture a “cheese ball snowman” made of the three original cheese balls – a cooler idea for a party than smashing them into one ball, I argue. His height is 12 inches, but this is not the same as having a single ball with a 12-inch diameter. Three spheres whose diameters sum to 12 cannot combine their volumes to produce a single sphere with a diameter of 12. Therefore choices A and B are also out, leaving us with correct choice E. If we approximate the value of E, it is greater than 6 but less than 8, since the cube root of 36 is greater than 3 but less than 4. A combined cheese ball this size makes logical sense.

This concludes our fifth and final article on GMAT circles. Cheers.

Contributor: Elijah Mize (Apex GMAT Instructor)