Tag Archive for 'arithmetic'

GRE Math: Two Example Quantitative Comparison Problems and Solutions

Example Quantitative Comparison 1

Example QC #1

If only we had calculators…then this problem would be a piece of cake! Well, for those of you taking the new GRE, you will have a calculator, so this problem would be a breeze. But for those of you who are taking the current GRE, you’ll probably have to find some way to compare them without actually evaluating the radicals.

GRE Math: Two Difficult Example Problems and Solutions

Example Difficult Arithmetic Problem

Example Hard Arithmetic Problem

These problems aren’t difficult because they use particularly complicated mathematical principles — they’re difficult because they’re a little tricky. They require you to think a little beyond what’s given directly to you.

This problem looks a lot like an algebra problem because of the variables, but unfortunately it’s not. With three variables, you would need three equations to solve it, which you don’t have.

So how do you approach this problem? Well, once you realize that it’s not an algebra problem, you should be thinking of how you can arrive at values for x, y, and z without using algebra. You know you need these values because eventually you have to add them together. So let’s look at what you do know from the problem.

3x = 4y = 7z

This means that 3x, 4y, and 7z all equal some number, which we will call n; this also means that

3x = n
4y = n
7z = n

From this, we can see that n must be a multiple of 3, a multiple of 4, and a multiple of 7. What is the smallest number that is a multiple of 3, 4, and 7 (LCM of 3, 4, and 7)? It’s 84! (3 * 4 * 7 = 84)

So if n = 84, then we get the following three equations

3x = 84
4y = 84
7z = 84

Solving for each variable, we get

x = 28
y = 21
z = 12

If we add them together, we get

x + y + z = 28 + 21 + 12 = 61

Thus, the answer is (D).

Example Difficult Geometry Problem

Example Hard Geometry Problem

The goal of this problem is to find the area of the shaded region. There are two ways to do this problem — I’ll go over both.

Method 1: Find the area of the shaded trapezoid
A trapezoid is defined as a quadrilateral with two parallel sides and two non-parallel sides. The shaded region is a trapezoid.

(Brief aside: We know that the shaded region is a trapezoid because both the top line segment (y) and the bottom line segment (x) form right angles with the left line segment (also labeled x). Because both y and x intersect the left line segment at the same angle, they must be parallel.)

So what is the formula for the area of a trapezoid?

Atrapezoid = ½(b1 + b2)h

b1 and b2 are the bases of the trapezoid, which are the two parallel line segments; h is the height of the trapezoid, which is the perpendicular distance from one base to the other base.

What do we already know?

b1 = x
b2 = y

So all we need to know is the height of the trapezoid. Based on the drawing, it sort of looks like the height is simply half of the left segment — but that’s not good enough (and it will lead you to the wrong answer, in this problem)! We should never, ever have to “guess” at a value on the GRE! There must be a way to figure it out.

Notice that both the bottom segment and the left segment have length x. There’s a name for triangles that have two identical sides — isosceles triangles. Isosceles triangles also have two identical angles as well. The angles that are opposite of the identical sides are also identical. Let’s call it a.

We know that the sum of the angles in a triangle is 180º. So the sum of the angles in the whole triangle is

a + a + 90 = 180
2a + 90 = 180
2a = 90
a = 45

Both angles are 45º.

Now let’s examine the smaller triangle (the unshaded portion). The top angle for that triangle is the same — 45º. It also has a right angle, which is 90º, so that means that the remaining angle is also 45º.

Hey, look at that! This triangle is also an isosceles triangle, and the sides that are the same are the bottom (y) and the left segment — which is also y (this concept is called similar triangles).

We know that the entire left segment has length x, and the top part of that segment has length y — this means that the bottom part has length x – y. This bottom part also happens to be the height of our trapezoid!

Let’s plug it all into the area equation now:

Atrapezoid = ½(b1 + b2)h
Atrapezoid = ½(x + y)(x – y)

Note that

(x + y)(x – y) = x2 – y2


Atrapezoid = ½(x2 – y2)

which is answer choice (A) (multiplying by ½ is the same as dividing by 2).

Method 2: Whole Minus Part
The shaded region of the triangle is part of the whole triangle. In order to find the shaded region, we could find the area of the whole triangle and subtract away the area of the smaller triangle (the unshaded region).

The formula for area of a triangle is

Atriangle = ½(bh)

where b is the base of the triangle, and h is the height.

For the large triangle,

Abig_triangle = ½(x*x) = ½x2

For the small triangle, we know the base is y, and from the previous explanation above, we know that the height (the left segment) is also y (because the triangle is isosceles). Thus, the area of the small triangle is

Asmall_triangle = ½(y*y) = ½y2

Subtracting the two, we get

Abig_triangle – Asmall_triangle = ½x2 – ½y2 = ½(x2 – y2)

which is answer choice (A).

Questions? Concerns? Suggestions? Criticisms? Leave a comment below!