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

Substituting,

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!

12 Responses to “GRE Math: Two Difficult Example Problems and Solutions”


  • it was very helpful

  • Nice problems! Keep publishing challenging exercises.
    My 5 cents..

    The second problem can be solved used triangle proportions.

    As the width and height of the ‘big’ triangle are equivalent to X, therefore the ‘little’ triangle has the same relationship i (width and height are equivalent to Y)
    ==>
    Area = Big /\ – Small /\ = (X*X / 2) – (Y*Y/2)

  • Big thank you for the wonderful explanations!

  • The answer to the second problem can be achieved much more quickly than this explanation lets on. We know that the area of the shaded region equals the area of the entire triangle minus the area of the unshaded region. By the area of a triangle equation we can see that the area of the whole triangle is x^2/2. We can also see from the diagram that the unshaded region is a similar triangle to the shaded region. We know this because the unshaded triangle shares all three angle measurements with the original triangle. By AAA, the triangles are similar. Therefore, we know the formula representing the unshaded region is going to be y^2/2 (because we know that the triangles are isosceles) now to get our final equation, we subtract the unshaded region from the shaded region: x^2/2 – y^2/2 = (x^2-y^2)/2. And there we are.

  • @Kathryn: Yes I did it exactly the same way. In fact the hypotenuse is a 45degree line slopping 1. so for any y (the horizontal parallel line) the vertical counterpart will be exactly equal to y. So answer is 1/2(x^2-y^2). Bingo!!

    please post more exercises.

  • You could use similarity of the smaller triangle and bigger triangle to find the length of the height of the smaller triangle. :)

  • A=0.5(x^2-y^2)

    y/x=h/x
    h = y
    A= 0.5*(x^2-y*h)
    A=0.5*(x^2-y^2)

  • I think this is not correct and the correct answer is forth one.
    Suppose x=2 and y=1
    calculate dashed area
    Its simple whole area of triangle is 2
    and area of upper triangle is 0.5
    so dashed triangle area would be 2-0.5=1.5
    So the correct answer would be the forth one dude!

    • Amin,

      Plugging in x=2 and y=1 does give you 1.5 for answer choice (D), but it also gives you the same result for answer choice (A)! Plugging in numbers can be a useful technique to solving a problem like this. However, the problem is that the numbers you picked give you two possible answers.

      Try plugging in x=3 and y=2. This would give you an area of 4.5 for the large triangle and 2 for the small triangle. The area of the shaded region would then be 2.5. Go through the answer choices and you’ll find that the only answer that results in 2.5 is answer choice (A).

      Hope this helps, dude!

      Let us know if you have any other questions.

  • Thanks a lot! #1 is a great example of when to try a strategy other than a system of equations when the number of unknowns > than the equations!

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