Cylinders & Spheres In The GMAT
Posted on
19
Apr 2022

Cylinders & Spheres In The GMAT

Welcome back to our fifth and final article on GMAT circles. Last time we explored the possibilities of treating a circle’s radius as the hypotenuse 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 article 1 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)

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Radius as Hypotenuse
Posted on
05
Apr 2022

Radius As Hypotenuse – Problems & Solutions

Welcome back to our fourth article on GMAT circles. Last time we considered inscribed angles and learned that where there is a 90-degree inscribed angle, there is a hypotenuse that is also a diameter of the circle. This time we will explore a class of problems where the radius, rather than the diameter, pulls double duty as a hypotenuse. Let’s dive right in with the following official problem.

1. Radius as Hypotenuse  – GMAT Official Problem

Semicircular archway over a flat street problemThe figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?

A. √2
B. 2
C. 3
D. 4√2
E. 6

Problem SolutionThis problem is a straightforward application of the Pythagorean theorem. Since we are told that the radius of the semicircle is 6 feet, we can draw a 6-foot radius from center O to the point where height h meets the semicircle. Voila – a right triangle.

h = √(62 – 22)
h = √(36 – 4)
h = √32

This is where you should stop and mark answer choice D since we are taking the square root of a number that is not a perfect square. When we simplify this radical, something will get left inside. Therefore answers B, C, and E are out (Answer A is out because √2 =/= √32), and the correct choice is D.

2. Radius as Hypotenuse Problem 1 

Let’s try something a little different:

In the xy-plane, point (r,s) lies on a circle with center at the origin. What is the value of + s²?

1. The circle has radius 2.
2. The point (2,-2) lies on the circle.

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. 

This is the first problem we’ve seen where a circle is placed on the xy-plane. In such problems, it is usually helpful to remember the basic circle principle that every point on the circle (meaning on its edge or perimeter) is equidistant from its center.

Solution

If you’re unfamiliar with these problems, statement 1 may trip you up. Is the radius of the circle sufficient to determine + ? Yes, it is. If you are concerned about the unknown positivity/negativity of the coordinates r and s, recall that the square of any number (except 0) is positive. This means that for any positive/negative combination of r and s, the sum + will have the same value. 

But what you really need here is to see that the expression + matches the famous + from the Pythagorean theorem, and in fact, it functions the exact same way.

Radius as Hypotenuse ProblemIn this setup, the radius is the hypotenuse of the right triangle with legs r and s. Therefore, applying the Pythagorean theorem, the value + represents the square of the radius. So if we know the value of the radius (2), we know the value r² + s², and statement 1 is sufficient.

Statement 2 offers that the point (√2, -√2) lies on the circle. This statement should be “easier” to evaluate than statement 1. Seeing the radicals in the coordinates ought to help you make the connection to the Pythagorean theorem if you didn’t already while evaluating statement 1. But using the principle that every point on a circle is equidistant from its center, we know that this given point (√2, -√2) is the same distance from the center as the point (r, s) in the question. Therefore if we sum the squares of √2 and -√2, the result (4) will also represent the value r² + s² we were asked about.

3. Problem 2

Let’s try one more:

In cross section, a tunnel that carries one lane of one-way traffic is a semicircle with radius 4.2 m. Is the tunnel large enough to accommodate the truck that is approaching the entrance to the tunnel?

1. The maximum width of the truck is 2.4 m.
2. The maximum height of the truck is 4 m.

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. 

This one is a little more complex. Sometimes on GMAT quant problems, it is helpful to ask why certain details were specified. In this case, we are told that the tunnel “carries one lane of one-way traffic.” This is important because if it were not the case, the truck would have to drive on one side or the other, and there’s no way it would be able to get through the tunnel. Since there is only one lane going through the tunnel, the truck can “center up” to give itself the best chance of fitting through.

Solution

This is one of those less-common DS problems where each statement on its own is clearly insufficient. If all we know is that the truck is 2.4m wide at its widest point (statement 1), it may still be too tall to fit through the tunnel. If all we know is that the truck is 4m tall at its tallest point, we don’t know whether the truck is narrow enough to make it through the tunnel while being this tall.

Problem 2 - Solution But if we combine statements 1 and 2, we can use the Pythagorean theorem to calculate the max distance of a point on the “centered up” truck from the point at the “center” of the semicircle.

Now here’s the key step: don’t calculate! Running the Pythagorean theorem with our values here would be a waste of time. As long as the value p [from the graphic] is less than 4.2 (the radius of the tunnel), the truck will fit. But for DS, we don’t have to know whether the truck will fit. All we have to know is whether the value p can be calculated, and in this case, it can be. Statements 1 and 2 together are sufficient, and the correct answer choice is C.

 

This concludes our fourth article on the GMAT’s treatment of circles. Next time we will look at circles in two different 3-dimensional shapes: cylinders and spheres.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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Inscribed Angles & Inscribed Polygons In The GMAT
Posted on
29
Mar 2022

Inscribed Angles & Inscribed Polygons In The GMAT

Welcome back to our third article on GMAT circles. In the second article, we explored central angles, sectors, and arcs. This time we will introduce another kind of angle: the inscribed angle.

An inscribed angle is an angle drawn by using line segments to connect one point on a circle to two other points on the same circle, as in the graphic below:

Inscribed AngleLike a central angle, an inscribed angle creates a “wedge” shape, like a triangle where one side is rounded. The rounded side is an arc of the circle. For a central angle, the measure of the angle corresponds to the measure of the associated arc in a 1:1 relationship. For an inscribed angle, the measure of the angle corresponds to the measure of the associated arc in a 1:2 relationship. A 30 degree inscribed angle creates a 60-degree arc on the other side of the circle. A 60-degree inscribed angle creates a 12- degree arc on the other side of the circle. And, importantly, a 90-degree inscribed angle creates a 180-degree arc (half a circle or a semicircle) on the other side of the circle.

1. Inscribed Angle – GMAT Official Guide Problem

GMAT problems rarely use the term “inscribed angle” or feature an inscribed angle in isolation. Usually, the inscribed angle is part of an inscribed polygon, a polygon drawn inside a circle so that its vertices are points on the circle. Take a look at this official GMAT problem:

Inscribed Angle Official GMAT ProblemIn the figure shown, the triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?

 

 

A. 15π
B. 12π
C. 10π
D. 7π
E. 5π

The problem refers not to an angle inscribed in a circle but to a triangle inscribed in a semicircle. Still, knowing the “1:2” factor of relationship between an inscribed angle and its associated arc is the key to solving this problem. Your logic might go something like this:

  1. This is a semicircle or a 180-degree arc.
  2. The angle at point B “opens up” to the straight edge of the semicircle, which is like the diameter of a circle. Another semicircle or 180-degree arc could be drawn across from this angle so that it makes a whole circle with the existing piece.
  3. Since the measure of an inscribed angle is 1/2 the measure of the arc it “creates” on the other side of the circle, the angle at point B is a 90-degree angle, and the triangle is a right triangle. 

At this point, your attention should return to the given information about the lengths of line segments AB and BC, which we now know to be the legs of a right triangle. These legs have lengths 6 and 8, which have a 3:4 relationship. Therefore we are looking at a 3-4-5 triangle, and the length of the hypotenuse is 10.

Finally, you must recall that this hypotenuse is the diameter of the circle. Therefore the diameter of the whole circle is 10. However, marking answer choice C would be a mistake, since we were asked for the length of arc ABC, where arc ABC is a semicircle (half a circle). So your final step is to divide your diameter of 10π by 2, leading you to the correct answer choice: E.

2. Inscribed Square – GMAT Official Guide Problem

Let’s try another problem, this time with an inscribed square:

Inscribed Square GMAT Official Guide ProblemThe figure shows a drop-leaf. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?
A. 1/2
B. 
√2/2
C. 1
D. √2
E. 2

Notice that the problem doesn’t mention “a square inscribed in a circle,” but that is nonetheless what we have here. Many GMAT quant problems create scenarios that correspond to some mathematical phenomenon without using the math language. In this case, the fact that we are dealing with a square inscribed in a circle is relatively easy to see.

As in the previous problem, we are asked for a value of the circle (this time it is the radius instead of an arc length) but given only information about the inscribed shape: a square. As in the previous problem, the key is realizing that with any 90-degree inscribed angle, the line segments forming the angle are legs of a right triangle whose hypotenuse is also a diameter of the circle.

Using the Pythagorean theorem, the hypotenuse of this triangle (or the diagonal of the inscribed square) is √2. As before, forgetting to divide this value by 2 (since we were asked for the radius, not the diameter) will lead you to an incorrect answer choice. Don’t trip at the finish line. The value you need is √2 /2, answer choice B.

Here is a related problem:

Square inscribed in a circle problemIf rectangle ABCD is inscribed in the circle above, what is the area of the circular region?

A. 36.00
B. 42.25
C. 64.00
D. 84.50
E. 169.00

Again, we are asked for a value of the circle (its total area) but given only information about the inscribed rectangle. For our purposes, this rectangle is just as good as the square in the previous problem. With the square, we only needed the length of one side, because we know that all four sides are the same length. With a rectangle, we need both the length and the width in order to calculate the diagonal – the diameter of the circle – via the Pythagorean theorem. If you know your Pythagorean triples (like 3-4-5), you may realize immediately that the diagonal of this rectangle is 13.

D = √(5² + 12²)
D = √(25 + 144)
D = √169
D = 13

Now that we have the circle’s diameter, we can solve for its area. The radius of the circle is 13/2 or 6.5, and since Area = r², the square of 13/2 or 6.5 will be the coefficient of in the correct answer choice. It would be a waste of time to fully multiply out 6.5 * 6.5. We know that it will be of form __.25, and the only answer choice that matches this is B.

3. Data Sufficiency – GMAT Official Guide Problem

Let’s transition to data sufficiency for one final problem. Using the diagonal/diameter relationship in the previous problems, it would be possible to construct a variety of DS problems. But some DS inscription problems rely on another property of inscribed polygons.

Square ABCD is inscribed in circle O. What is the area of square region ABCD?

  1. The area of circular region O is 64π.
  2. The circumference of circle O  is 16π.

Solution

To answer this problem, all you need to know is that there is only one way to inscribe a square in a circle. The vertices of the square must lie on the circle. The perimeters and areas of the square and the circle will scale together. This means that if we know any value for either shape, we can calculate every value for both of them. Therefore each statement on its own is sufficient, and the answer to this problem is D.

As long as any polygon can be established as regular (having sides of equal length and angles of equal measure), there is only one way to inscribe it in a circle. A square is a regular quadrilateral, so this works for squares every time. But this same problem could have used a regular pentagon, a regular hexagon, or any regular polygon you like, and the correct answer would still be D. The regularity of the polygon is sufficient – and necessary – for this to work. If the regularity of the polygon cannot be established, then there are an infinite number of ways to inscribe it in a circle, each with its own unique area and perimeter.

It is also possible to flip the relationship and inscribe a circle inside a polygon. A related term is circumscription. The shape on the inside is inscribed in the shape on the outside. The shape on the outside is circumscribed around the shape on the inside. GMAT problems where the circle is on the inside usually use a square, so that the diameter of the circle is equal to the length of each side of the square. Such problems tend to be of lower difficulty level.

 

This concludes our third article on the GMAT’s treatment of circles. Next time we will look at what happens when the radius – rather than the diameter – pulls double duty as the hypotenuse of a right triangle.

 

Contributor: Elijah Mize (Apex GMAT Instructor)

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