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 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

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)

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

The 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

This 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.

In 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.

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)

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:

Like 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:

In 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:

The 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:

If 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)

Posted on
22
Mar 2022

## Pieces of Pi: Sectors, Arcs, and Central Angles

Welcome back to our series on GMAT circles. In the first article, we introduced the properties of radius/diameter, circumference, and area, discussing the relationships between all of these. This time we will introduce something called a central angle, which creates portions of a circle’s area and circumference called sectors and arcs, respectively.

The best way to define these things is probably with a simple visual.

A central angle is an angle created by using line segments to connect a circle’s center to two points on its edge. A sector is the part of a circle’s area bounded by this central angle, and an arc is the part of a circle’s circumference between the two points used to draw the angle. A 90-degree central angle creates both a 90-degree sector and a 90-degree arc. An important note is that the lines used to form the central angle are radii of the circle.

As a further illustration, think about a pizza (something I do regularly). The pizza is a circle, the pieces are sectors separated by central angles, the crust is the circle’s circumference, and each piece’s section of crust is an arc. From this, you can see that any central angle creates both a sector and an arc that correspond to one another. When you pull a piece from a pizza or cut out a piece from a pie, you use a central angle to create a sector with an arc on its rounded edge.

To represent these things mathematically, we consider a circle to be like a 360-degree central angle. In this setup, the fractional relationship of a central angle to 360 corresponds to two things:

1. The fractional relationship of the resulting sector to the circle’s total area
2. The fractional relationship of the resulting arc to the circle’s total circumference

Since 90 is ¼ of 360, the area of a 90-degree sector is ¼ of its circle’s total area, and the length of a 90-degree arc is ¼ of its circle’s circumference.

To show all of this algebraically, let’s use the variable x for the degree measure of a central angle:

x / 360 = arc length / circumference
x / 360 = sector area / circle area

Most pizzas are divided into 8 slices. This means that each slice has a central angle of 360/8 = 45° and that each slice is ⅛ of the area of the entire pizza.

### Examples:

1. What is the central angle for three slices of pizza?
The central angle formed by 3 slices of pizza is 3 * 360 / 8 = 135 degrees.

2. What’s the area of a slice if the diameter is 20cm and there are six slices?
The area of a slice of pizza is 1/6 of its pizza’s total area. So, the area of a pizza can be found by using this formula A= π*r2 = 3.14*102 = 314cm
The area of a slice of pizza is  314/ 6 = 52.33cm

Keep in mind that you may have to consider this relationship in either direction. You may be given some info about the whole circle and then tasked with concluding something about a sector or an arc. Or you may be given some info about a sector or an arc and then tasked with concluding something about the whole circle. You may even be given info about both the whole circle (its area or circumference) and a sector or arc and then tasked with calculating the central angle. Each of these represents a perspective shift, and when doing a problem form, you can rewrite the problem from each of these perspectives to make sure you can fully navigate problems of this sort.

## Pieces of Pi: Official GMAT Problems

Now for some official GMAT problems. Let’s start with two straightforward sector problems, one problem solving and one data sufficiency.

### Problem-Solving Problem

The annual budget of a certain college is to be shown on a circle graph. If the size of each sector of the graph is to be proportional to the amount of the budget it represents, how many degrees of the circle should be used to represent an item that is 15 percent of the budget?

A. 15°
B. 36°
C. 54°
D. 90°
E. 150°

From the question, we can tell that the “circle graph” mentioned here is what we usually call a “pie chart,” a handy way to show the breakdown of a whole (like a budget) into its various parts. If we want to represent 15% of the budget, we need a sector with a central angle using 15% of the (360) available degrees in the circle. 0.15 * 360 = 54, so the correct answer is C. Piece of cake. Or piece of pie.

### Now for a DS pie chart problem

TOTAL EXPENSES FOR THE FIVE DIVERSIONS OF COMPANY H

The figure represents a circle graph of Company’s H total expenses broken down by the expenses for each of its five diversions. If O is the center of the circle and if Company H’s total expenses are \$5,400,000, what are the expenses for Division R?

1. x = 94
2. The total expenses for Division S and T are twice as much as the expenses for Division R.

Once again, this pie chart (which the GMAT apparently prefers to call a “circle graph”) is being used to represent a budget breakdown. Here we are told that the value represented by the whole circle is \$5,400,000. We can think of this value as the area of the circle. We are asked for the expenses for division R, or in circle terms, the area of sector R.

#### Statement 1: x = 94

This is the measure of the central angle bounding the sector whose area we need to know (sector R). Since we already know the area of the whole circle, the measure of this central angle is the final piece of the puzzle. (Area of sector R = 94/360 * \$5,400,000) Statement 1 is sufficient.

#### Statement 2: The total expenses for Divisions S and T are twice as much as the expenses for Division R.

This statement relates the total of two unknown sectors to another unknown sector. Given this statement alone, we don’t know the relationship of any of these sectors to the whole circle, so we can’t solve for any of their areas. Statement 2 is insufficient.

Statement 1 is sufficient.
Statement 2 is insufficient.

## Pieces of Pi: More Difficult Problems

Let’s ratchet up the difficulty a bit with another sector problem that involves more smoke and mirrors.

The figure consists of three identical circles that are tangent to each other. If the area of the shaded region is 64√3 – 32*π, what’s the radius of each circle?

A. 4
B. 8
C. 16
D. 24
E. 32

(tangent means just touching and not overlapping)

The problem mentions only three circles and a shaded region, but the graphic includes something more: an equilateral triangle drawn by connecting the centers of the three circles. You can solve this problem without knowing anything about the formula for the area of an equilateral triangle (although you should know this formula).  It should occur to you that the area of the shaded region could be expressed as the area of the triangle minus the combined area of those three sectors, which matches the given expression 64√3 – 32*π. So the (irrelevant) area of the triangle is 64√3, and the area of the three sectors is 32*π.

You might start by trying to get the area of a single sector by dividing 32*π by 3. But 32 won’t divide nicely by 3, which should signal you to try something else. If you can’t go from the combined area of the three sectors to the area of one sector, maybe you can go from the combined area of the three sectors to the area of one circle. You might use your spatial reasoning and conclude that the three sectors together look like they make up half a circle. Or you might recall that each interior angle of an equilateral triangle measures 60 degrees. Therefore each of these sectors is ⅙ (60/360) of a whole circle, and the three of them together do indeed make up half a circle (3 * ⅙ = ½).

Now, if the sectors’ combined area is 32*π, and this is half a circle, then the area of the whole circle is 2 * 32*π = 64. Having found the area of a circle, we can now solve for the radius.

A = 64 = π*r2
64 = r2
r = 8

And the correct answer choice is B.

### Arc Length Problem

Let’s try one more problem, this time focusing on arc length.

The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is 4*π/3 what is the length of line segment RU?

A. 4/3
B. 8/3
C. 3
D. 4
E.  6

A note about points that “lie on a circle.” This always means that the points are on the edge or perimeter of the circle.

It may be helpful to visualize or even draw out what has been described here.

This is a good opportunity to introduce some terminology. We see that line segment RU connects two points on the circle. Such a line segment is called a chord. If the line continues on to either side of the circle so that the circle is “skewered,” the line is called a secant (the GMAT does not expect you to know this term). When a chord or secant passes through the circle’s center, it creates a diameter. A line outside a circle that just touches the circle at one point is called a tangent.

If you aren’t sure how to calculate the length of a chord like RU, start with what you know. We are given the length of arc RTU (4*π/3) and the radius of the circle. A good step is to calculate the circumference of the circle so that we can see how it relates to arc RTU.

C = 2*π*r
C = 2*π*4
C = 8*π

The circumference is 8*π, and arc RTU is 4*π/3. 8*π/6 = 4*π/3. Therefore arc RTU represents ⅙ of the circumference of the circle, and its corresponding central angle is 60 degrees (360/6). Drawing out this information helps us to see its relevance.

The central angle and chord RU form an equilateral triangle. Since the radius of the circle is 4, chord RU also has length 4, and the correct answer is D.

This concludes our second article on the GMAT’s treatment of circles. Next time we will look at another kind of angle inside a circle: an inscribed angle, and at the related topic of inscribed polygons.

Contributor: Elijah Mize (Apex GMAT Instructor)

Posted on
15
Mar 2022

## Circles On The GMAT 101 – Area, Circumference, and Pi

Circles on the GMAT function like any other GMAT quant topic: the list of “knowledge bits” you need is short, but the questions creatively combine and/or disguise these few “knowledge bits” to create complex problems.

In this first article, we will discuss the most basic properties of circles and the formulas that relate to these properties. The properties in question are area, circumference, and radius/diameter.

## Circle Properties: Radius & Diameter

I treat radius (r) and diameter (D) together because they essentially express the same thing, and because the relationship between them is so simple.

Diameter tells you how “wide” a circle is at its widest point. If you draw a straight line all the way across a circle, hitting the center on the way, that line is a diameter, and the length of that line is the diameter of the circle. The radius of the circle is half of this line – or any straight line drawn from the center of the circle to a point on its edge.

This is the most basic property of a circle. The two next most basic properties involve bringing in a specific number, one of the most famous numbers in mathematics: pi.

## Circle Properties: Pi

Pi is the name of a Greek letter that looks (in its lowercase form) like this: π.

The value represented by this character is irrational (the numbers after the decimal never stop), but for most purposes and for the GMAT, 3.14 is enough. If you happen to be working with fractions or if you prefer fractions to decimals, 22/7 is a rather precise way to express pi. Rounded to the thousandths place, pi is 3.142, and 22/7 is 3.143. That’s close enough for jazz and certainly close enough for GMAT quant.

Interestingly – and for reasons relating to math beyond what is required for the GMAT – this same value can be combined with the radius/diameter to calculate both the circumference (C)  and area (A) of the circle. Circumference is the distance around the circle (its perimeter), and area is the space inside the circle.

C = π*D     or     C = 2*π*r
A = π*r²

The two options for the circumference equation are, in fact, equivalent, since D = 2r. You may be wondering why mathematicians split the 2 and the r around the π, instead of just saying π2r. One reason is simply the conventions that have developed for algebraic notation; it “looks wrong” to have the 2 in the middle after the π. But another reason is a (non-GMAT math) connection between the circumference and area equations. Both equations have a π, an r, and a 2. In the area equation, the 2 functions as an exponent on the r. In the circumference equations, it functions as a coefficient multiplying the expression. Area is pi times the square of the radius. Circumference is pi times the radius doubled.

Examples:

1. If a radius is 3, what’s the area of the circle?

A = π*r²= π*3² ≈ 28.27433

2. If the area of a circle is 25π, what is the diameter of the circle?

A = π*
r² = 25*π÷π
r = √25 = 5

3. If a radius is 4, what’s the circumference of the circle?

C = 2*r*π
C = 2*4*π
C = 8*π = 25.13274

## Official GMAT Problem

In all likelihood, none of this looks new. But like we said at the beginning, GMAT quant can get creative with common mathematical knowledge. Take a look at this official GMAT problem, and try to answer it before reading on:

In the figure shown, if the area of the shaded region is 3 times the area of the smaller circular region, then the circumference of the larger circle is how many times the circumference of the smaller circle?

A. 4
B. 3
C. 2
D. √3
E. √2

### Understanding the Problem

The only properties at play here are area and circumference, perhaps with radius/diameter as a “bridge” between the two, but the answer to the problem may not be immediately obvious. Part of this is due to a common GMAT technique – the removal of numbers to make the problem more abstract.

This problem disguises the relationship between the two circles by referring not to the area of each circle, but to the area of the shaded region. A helpful preliminary deduction is that if, as the problem says, the area of the shaded region is 3 times the area of the smaller circular region, then the area of the larger circle is 4 times the area of the smaller circle (comprising the 3 “parts” in the shaded region and the 1 “part” in the smaller circle).

So we have the factor relating the circles’ areas: 4. But we were asked about circumference, not area. How do we get from area to circumference? Well, both the area and circumference equations have the radius in them, so one option is to pick two values for area, with one being 4 times the other, solve for the radius of each circle, and then plug each of these radius values into the circumference equation.

### Solving the Problem

Let’s say the area of the larger circle is 16, and the area of the smaller circle is 4. Since π is common to all the circle equations, in this case it is irrelevant, and we should just remove it instead of keeping it as “dead weight” to move around algebraically.

Large Circle                             Small Circle
A = π*r²                                     A = π*r²
16 = π*r²                                   4 = π*r²
r = 4                                         r = 2

So when the areas of two circles are related by a factor of 4, their radii (plural for radius) are related by a factor of 2: that is, the square root of 4. This makes sense, since radius is a linear value and area is a square value. This is just like how when you double the length of the side of a square, you quadruple its area (since 2² = 4). When you triple the length of the side of a square, you make its area nine times what it was before (since 3² = 9). The factor of increase for area is the square of the factor of increase for the linear measure.

C = 2*π*r
C = 2*π*4
C = 2*π*2

Since we are only concerned about the factor relating the circumferences of the circles, we can ignore whatever is common to both (2*π), leaving us with 4 and 2. Since 4 is twice as much as 2, the circumference of the larger circle is twice the circumference of the smaller circle, and the correct answer choice is C.

### The Final Step

That last step (and any real calculation) is avoided with two insights:

1. There is a square relationship between area and radius. Since the area of the larger circle is 4 times that of the smaller circle, the radius of the larger circle is the square root of 4 (2) times the radius of the smaller circle.
2. There is a linear relationship between circumference and radius. Since the radius of the larger circle is twice that of the smaller circle, the circumference of the larger circle is also twice that of the smaller circle.

When you move beyond rote memorization and understand the circle equations, these insights happen naturally and lead you to the correct answer choice quickly, with virtually no calculation.

This concludes the first article in our series on the GMAT’s treatment of circles. Next, we will dive into what happens when you use something called a central angle to mark off only a part of the area (a sector) and circumference (an arc) of a circle.

Contributor: Elijah Mize (Apex GMAT Instructor)

Posted on
01
Feb 2022

## Executive Assessment Exam (EA) – Quant Section

We know what you’re thinking: math is a scary subject and not everyone can excel at it. When attempting math on the Executive Assessment – better known as the EA – the stakes may seem much higher. Especially since there is a whole section dedicated to math that you need to prepare for. There is good news though, the EA is not actually testing your math skills, but rather your creative problem-solving skills through math questions. Furthermore, the EA Quant Section only requires that you have sound knowledge of high-school-level mathematics. So, you just need to practice your fundamentals and learn how to use them to solve specific EA problems and find solution paths that work to your advantage.

The EA Quant Section contains a total of 14 questions, and you are given 30 minutes to complete all of them. This gives you about 2 minutes to solve each question, so in most cases, the regular way of solving math equations that you were taught in high school will not cut it. To succeed on the EA you must find the optimal problem-solving strategy for each question type. This can seem a daunting start, so our expert instructors at Apex GMAT recommend that you start your quant section prep with a review. Look over the types of EA questions asked in the test and review the math fundamentals which you may not have been using in your day-to-day life.

## What types of questions will you find in the EA Quant Section?

There are 2 main types of questions you should look out for when preparing to take the GMAT exam:

### Data Sufficiency Questions

For this type of EA question you don’t generally need to do calculations. However, you will have to determine whether the information that is provided to you is sufficient enough to answer the question. These questions aim to evaluate your critical thinking skills.

They generally contain a question, 2 statements, and 5 answer choices that are the same in all EA data sufficiency questions.

Here is an example of a Data Sufficiency Question:

(1) 9x-1 = 3
(2) 3x-3 = 19

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.

### Problem Solving Questions

This question type is pretty self-explanatory: you’ll have to solve the question and come up with a solution. However, you’ll be given 5 answer choices to choose from. Generally, the majority of questions in the quant section of the EA will be problem-solving questions as they clearly show your abilities to use mathematical concepts to solve problems.

Here is an example of an EA problem-solving question from the official GMAC itself:

In a certain town of 4,000 residents, 40 percent of the residents are registered voters and 25 percent of the registered voters voted in the mayoral election. How many of the town’s registered voters voted in the mayoral election?

A) 400
B) 600
C) 1,000
D) 1,600
E) 2,600

## The main concepts you should focus on

The one thing that you need to keep in mind when starting your EA prep is the level of math you need to know before going in for the Quant section. All you’ll need to master is high-school-level math. That being said, once you have revised and mastered these math fundamentals, your final step is learning how to apply this knowledge to actual EA problems and you should be good to go. This is the more challenging side of things but doing this helps you tackle all the other problem areas you may be facing such as time management, confidence levels, and test anxiety

Here are the 3 main groups of questions on the quant section of the EA and the concepts that you should focus on for each:

### Algebra

• Algebraic expressions
• Equations
• Functions
• Polynomials
• Permutations and combinations
• Inequalities
• Exponents
• Coordinate Planes

• Profit
• Sets
• Rate
• Interest
• Percentage
• Ratio
• Mixtures

### Arithmetic

• Number theory
• Percentages
• Basic statistics
• Power and root
• Integer properties
• Decimals
• Fractions
• Probability
• Real numbers

## 5 tips to improve your  EA Quant skills

1. Master the fundamentals! This is your first step towards acing this section of the EA. As this section only contains math that you have already studied thoroughly in high school, you’ll only need to revise what you have already learned and you’ll be ready to start practicing some real EA problems.

2. Practice time management! This is a crucial step as every single question is timed and you won’t get more than 2 minutes to spend on each question. That is why you should start timing yourself early on in your EA prep, so you get used to the time pressure.

3. Know the EA question types! This is something that you will learn once you get enough practice with some actual EA questions. That way, you’ll be able to easily recognize different question types and you’ll be able to use your preferred solution path without losing time.

4. Memorize the answer choices for the data sufficiency questions! These answers are always the same and their order never changes. Memorizing them will help you save precious time that you can spend elsewhere. To help you better memorize them, we are sharing an easier and less wordy way to think of them:

A) Only statement 1
B) Only statement 2
C) Both statements together
D) Either statement
E) Neither statement

5. Make use of your scrap paper! There is a reason why you’re provided with scrap paper, so make sure to take advantage of it. You will definitely need it to take notes and make calculations, especially for the problem-solving questions that you will come across in this EA section.

## Final Thoughts

It’s true that math might seem like a scary subject and that’s why many people fear struggling with the EA quant section. Yet, it can be easily conquered with the right strategy and prep process. You just need to get acquainted with the question types, assess your skill level related to them and work, work, work until you become confident enough to crack that EA Quant section.

## Solutions:

EA Data Sufficiency Questions: The answer is D. EACH statement ALONE is sufficient.
EA Problem Solving Question: The answer is A) 400.

Contributor: Bilhen Sali

Posted on
07
Sep 2021

## How-to GMAT: No Calculator? Use These Mental Math Tips Instead

The GMAT is an exam largely focused on numbers and numerical data. And while doing math on the GMAT should be avoided sometimes it is inevitable. True, the test-taker is given a calculator for the duration of the Integrated Reasoning section but the same cannot be said for the Quantitative Reasoning Section.

###### The Basics

Before explaining any methods for dividing and multiplying with ease, let’s make sure we have revised a few simple rules:

• Numbers with an even last digit are divisible by 2 – 576 is and 943 is not;
• Numbers with a sum of digits divisible by 3 are also divisible by 3 – 3,465 for example (3+4+6+5=18);
• If the last 2 digits of a number a divisible by 4, the number itself is divisible by 4 – 5,624 for example (because 24/4=6);
• Numbers with last digit 0 or 5 are divisible by 5;
• Numbers that can be divided by both 2 and 3 can be divided by 6;
• Similar to numbers divisible by 3, numbers divisible by 9 must have a sum of digits divisible by 9 – 6,453 for example;
• If the last digit of a number is 0 it is divisible by 10;

With that out of the way, we can move onto some more advanced mental math techniques.

###### Avoid division at all costs

Don’t divide unless there is no other option. And that is especially true with long division. The reason why long division is so perilous is that it is very easy to make a careless mistake as there are usually several steps included in the calculation, it takes too much time, and to be honest, few people are comfortable doing it.

Fortunately, the GMAT doesn’t test the candidates’ human-calculator skills but rather their capacity to think outside the box and show creativity in their solution paths, especially when under pressure – exactly what business schools look for.

However, sometimes you cannot avoid division, and when that is the case remember: Factoring is your best friend. Always simplify fractions especially if you’ll need to turn them into decimals. For example, if you have 234/26 don’t start immediately trying to calculate the result. Instead, factor them little by little until you receive something like 18/2 which is a lot easier to calculate.

A tip for factoring is to always start with smaller numbers as they are easier to use (2 is easier to use compared to 4, 6, or 8) and also look for nearby round numbers.

If you have to calculate 256/4 it would be far less tedious and time-consuming to represent 256 as 240+16 and calculate 240/4+16/4=60+4=64. Another example is 441/3. If we express it like 450-9 it is far easier to calculate 450/3-9/3=150-3=147.

###### Dividing and Multiplying by 5

Sometimes when you have to divide and multiply by 5 (you’ll have to do it a lot) it would be easier to substitute the number with 10/2. It might not always be suitable for your situation but more often than not it can be utilized in order to save some time.

###### Using 9s

With most problems, you could safely substitute 9 with 10-1. For example, if you have to calculate 46(9) you can express it as 46(10 – 1) which is a lot more straightforward to compute as 46(10) – 46(1) = 460 – 46 = 414

You can also use the same method for other numbers such as 11, 8, 15, 100, etc:

18(11) = 18(10 + 1) = 180 + 18 = 198

28(8) = 28(10 – 2) = 280 – 56 = 224

22(15) = 22(10 + 5) = 220 + 110 = 330

26(99) = 26(100 – 1) = 2600 – 26 = 2574

###### Dividing by 7

The easiest way to check if a number is divisible by 7 is to find the nearest number you know is divisible by 7. For instance, if you want to check if you can divide 98 by 7 you should look for the nearest multiple of 7. In this instance either 70, 77, or 84. Start adding 7 until you reach the number: 70 + 7 = 77 + 7 = 84 + 7 = 91 + 7 = 98. The answer is yes, 98 is divisible by 7 and it equals 14

###### Squaring

When you have to find the square of a double-digit number it might be easier to break the number into components. For example, 22^2 would be calculated like this:

22^2
= (20 + 2)(20 + 2)
= 400 + 40 + 40 + 4
= 484

Similarly, if you have to find the square of 39 instead of calculating (30 + 9)(30 +9) you could express it like this:

39^2
= (40 – 1)(40 – 1)
= 1600 – 40 – 40 + 1
= 1521

You can use the same approach when multiplying almost any double-digit numbers, not only squaring. For example 37 times 73:

(40 – 3)(70 + 3)
= 2800 + 120 – 210 – 9
= 2701

###### Conclusion

This ends the list of mental math tips and tricks you can utilize to make the Quant section a bit less laborious. Keep in mind that no strategy or shortcut would be able to compensate for the lack of proper prep so it all comes down not only to practicing but doing so the right way.

For more information regarding the GMAT Calculator, GMAT Calculator & Mental Math – All You Need To Know, is a very insightful article to read.

Posted on
01
Sep 2021

## Additional Voters – GMAT Quant Problem

Hey guys, today we’re going to look at a particularly challenging GMAT Quant problem that just about everyone resorts to an algebraic solution path on, but there’s a very elegant part solution path. When we take a look at this problem we observe immediately that the difficulty is that we have no baseline for the number of voters that we start with. That’s the confusing part here and this is one of the ways that the GMAT modulates difficulty; when they give us a problem without fixed numbers, and where we’re not free to run a scenario because there are add-on numbers that change the relative values.

Here they’re adding the 500 and the 600 which means there exist fixed values at the beginning, but we don’t know what they are. What we want to do here is remove ourselves a bit from the problem and let the ratios that they give us guide our way.

We start out with three parts Republicans, five parts Democrats. These eight parts constitute everything, but we don’t know how many voters are in each part – it could be one voter in each, or a hundred, or a thousand, and we can’t speculate yet. So, what we need to do is not worry about it, and this is where a lot of people get really uncomfortable. Let it go for a second, and notice that, after we add all the new voters, we end up with an extra part on the Republican side and the same number of parts on the Democrat side.

What does this mean? Well, the parts are obviously getting bigger from the before to the after. But because we have an overall equivalence between the number of parts we can actually reverse engineer the solution out of this.

###### Reverse Engineering the Solution

We’re adding 500 Democrats and we’re maintaining five parts from the before to the after. This means that each part is getting an extra 100 voters for the total of plus 500. On the Republican side, we’re adding 600 voters. We already know, from the Democratic side, that each part needs to increase by 100 to keep pace with all the other parts. So, 300 voters are used in the three republican parts, leaving 300 extra voters to constitute the entirety of the fourth part.

Now we know that each part after we add the voters equals 300 and therefore each part before we added the voters was 200. From there we get our answer choice. I forget what they were asking us at this point, and this is actually a really great moment because it’s very common on these complex problems to get so caught up, even if you’re doing it mentally, with a more conducive solution path, to forget what’s being asked. When you’re doing math on paper, which is something we really don’t recommend, it’s even easier to do so because you get so involved processing the numbers in front of you that you lose conceptual track of what the problem is about.

So, they’re asking for the difference between the Democratic and Republican voters after the voters are added. Now we know there’s one part difference and we know that after voters are added a part equals 300 voters so the answer choice is B, 300.

###### Something to Keep in Mind

This one is not easy to get your head around, but it’s easier than dealing with the mess of algebra that you’d otherwise have to do.
Review this one again. This is a GMAT Quant problem you may have to review several days in a row. It’s one where you might attain an understanding, and then when you revisit it four hours later or the next day, you lose it and you have to fight for it again. It’s in this process of dense contact and fighting that same fight over and over again that you will slowly internalize this way of looking at it, which is one that is unpracticed. The challenge in this problem isn’t that it’s so difficult. It’s that it utilizes solution pads and way of thinking that we weren’t taught in school and that is entirely unpracticed. So, much of what you see as less difficult on the GMAT is less difficult only because you’ve been practicing it in one form or another since you were eight years old. So, don’t worry if you have to review this again and I hope this was helpful.

Check out this link for another super challenging GMAT Quant problem.

Posted on
18
Aug 2021

## GMAT Prime Factors Problem – GMAT Quant

Hey guys, check out this problem. This is an example of a problem that requires daisy-chaining together or linking together several key algebraic insights in order to answer it.

###### GMAT Prime Factors Problem – Applied Math Solution Path

Notice there’s an applied math solution path. We want prime factors of 3⁸ - 2⁸, and it’s just reasonable enough that we can do the math here. And the GMAT will do this a lot, they’ll give us math that’s time-consuming, but not unreasonably time-consuming in order to just draw us into an applied math solution path. We’ll take a look at this really quickly.

3⁸ is the same as 9⁴.
3⁸ = 3²*⁴= (3²)= 9
9 = (9 * 9)² = 81²
81 * 81 = 6,561

9 * 9 is 81² – about 6,400 or if we want to get exact, which we do need to do here because we’re dealing with factors, 81 * 81 is 6,561. Don’t expect you to know that, it can be done in 20 seconds on a piece of paper or mentally. And then 2⁸, that one you should know, is 256. And then, 6,561 – 256 = 6,305.

So now we need to break down 6,305 into prime factors. You know how to do that using a factor tree, so I’m going to zoom us right into a better solution path because I don’t want to give away the answer.

###### GMAT Prime Factors Problem – Another Solution Path

Notice that 3⁸ and 2⁸ are both perfect squares so we have the opportunity to factor this into (3– 2) * (3 + 2). Once again, the first term is a difference of two squares, the second term we can’t do anything with. So we break down that term, and lo and behold, (3² – 2²) * (3² + 2²) * (3 + 2), and once again we can factor that first term out into (3 + 2), (3 – 2), and so on. We work these out mathematically, and they’re much easier and more accessible mathematically, and we get 3 – 2 = 1 which obviously is a factor of everything. 3 + 2 = 5, 3² + 2² = 9 + 4 = 13, and then 3 + 2⁴ = 81 + 16 = 97.

So now we’ve eliminated everything, except B and C, 65 and 35. This is where the other piece of knowledge comes in. Since we have factors of 5 and 13. 65 must also be a factor because it’s comprised of a 5 and a 13. 35 requires a 7. We don’t have a 7 anywhere, so the correct answer choice is C, 35.

###### GMAT Prime Factors Problem – Takeaways

So the big takeaways here are, that, when provided with some sort of algebraic expression like this, look for a factoring pattern. And, when it comes to prime factorization, remember, that if you break it down into the basic prime factor building blocks, anything that is a product of those building blocks also exists as a factor.

Hope this helped and good luck!

Found it helpful? Try your hand at this GMAT problem, GMAT Prime Factorization (Anatomy of a Problem).

Posted on
17
Aug 2021

## GMAT Arithmetic 101- All You Need To Know

By: Apex GMAT
Date: 17 August 2021

While studying and preparing for the GMAT quant section, you might have come across some different types of GMAT arithmetic questions. These are actually quite common in the quantitative reasoning section and can be often intertwined with other GMAT algebra and GMAT geometry questions.

These usually come in 2 different formats: data sufficiency problems and problem-solving. The former have a very particular structure where you will have to determine whether the 2 statements are enough to come up with a solution. The latter type of problem requires you to actually solve the problem and derive a proper solution.

#### Arithmetic Concepts you need to revise

These are the arithmetic concepts you’ll need to know before you start practicing. Make sure to revise these fundamentals:

#### How to solve a GMAT arithmetic problem?

The number 1 thing you need to keep in mind when dealing with GMAT arithmetic problems is that the concepts that you’ll come across are fairly simple. You can easily revise these concepts because they are all things we study in high-school-level math. But here’s the kicker: the way these concepts are incorporated into the GMAT problems makes them more challenging, especially when the GMAT arithmetic problems are intertwined with GMAT algebra problems or even GMAT geometry problems. That is where things get tricky, as you need to apply your knowledge in a much more complicated setting that incorporates more than one concept. However, it all comes down to knowing the basics of arithmetics, which we can also refer to as the mechanics of the problem.

In order to help you better understand how to go about a GMAT arithmetic question, we will discuss an arithmetic problem and its solution and solution paths. In this GMAT problem, we are going to see how even the simplest mathematical concepts can become more challenging given the way the problem is formulated and structured.

#### Problem (GMAT Official Guide 2018)

When positive integer x is divided by positive integer y, the remainder is 9.
If x/y = 96.12, what is the value of y?

(A) 96
(B) 75
(C) 48
(D) 25
(E) 12

Solution:
In this case, you will have to revise the properties of numbers in order to properly find a solution to the problem.
When x is divided by y, the remainder is 9. So x=yq + 9 for a random positive integer q.
After dividing both sides by y, we get: x/y = q + 9/y.
But, x/y= 96.12 = 96 + 0.12.
Equating the two expressions for x/y gives q + 9/y= 96+0.12.
Thus:

q=96 and 9/y= 0.12
9=0.12y
y=9/0.12
y=75