# GRE Math: Two Example Math Problems

Example Problem #1

Example Geometry Problem

This is a fun problem.  The problem tells us that ∆ABC and ∆DEF have the same area.  It also tells us that AD > CF.  The problem is asking us about altitude, which is height.

From this second piece of information, we know that the base of ∆ABC is longer than the base of ∆DEF.  How do we know this?  From the picture, we see that the base of ∆ABC is composed of two segments: AD + DC.  The base of ∆DEF is also composed of two segments: DC + CF.  Both triangles share DC — this is the same for both triangles.  But we know that AD is bigger than CF, which means that AD + DC > DC + CF.  Thus, the base of triangle ∆ABC is larger.

Now, since we know that both triangles have the same area, but the base of ∆ABC is bigger, this means that the height of ∆DEF has to be bigger in order to compensate.  The answer should be B, but let’s prove it.

In math terms:

Area∆ABC = Area∆DEF
½b1h1 = ½b2h2

We know that b1 (base of ∆ABC) is bigger than b2 (base of ∆DEF), so let’s just choose some arbitrary numbers to make it a bit easier to see the relationship.  Let’s let b1 = 6 and b2 = 4.

½(6)(h1) = ½(4)(h2)
3h1 = 2h2
h1 = ⅔h2

Therefore, we know that the height of ∆ABC is smaller than the height of ∆DEF.  Thus, the answer is B.

Example Problem #2

Example Algebra Problem

The definition of average is:

Average = (a1 + a2 + a3 + … + an) / n

where a1 through an are elements of a list and n represents the number of elements in the list.  This equation means “the sum of all the elements divided by the number of elements.”  Basically, you add everything up and then divide it by the number of things you added together.

So we know from the problem that

n = (6 + 9 + k) / 3

and the problem asks us what the value of k is in terms of n (a different n from the equation above — this one is just a generic variable).  “In terms of n” means that our answer will have an “n” in it.  So basically, the problem says “solve for k, and your answer will have an n in it.”  We have an equation, given above, and we want to solve for a variable in it — this is a basic algebra problem.

(6 + 9 + k) / 3 = n  [copied equation from above]
(15 + k) / 3 = n  [added 6 and 9]
15 + k = 3n  [multiplied 3 to both sides of equation]
k = 3n – 15  [subtracted 15 from both sides of equation]

This entry was posted in Example Problems and tagged , , , , , , . Bookmark the permalink.

### 3 Responses to GRE Math: Two Example Math Problems

1. Quinn Clark says:

Why can’t you do this?…

(6 + 9 + k)/3 = n
(15 + k)/3 = n
(5 + k)/1 = n (written this way, I think I see that k does not have a factor of 3, but it look right written as a fraction)
5 + k = n
k = 5 – n (which is answer B).

I trust you, but I think I’d learn it better if I know where I went wrong.
Thanks

• Jason says:

Hi Quinn,
You identified the problem: k is missing a factor. Think about it this way: if you divided 15 by 3 to get 5, similarly, you have to divide k by 3 to get “k/3.” When you solve for n, the line should read “5 + k/3 = n”. Then, you can subtract 5 from both sides to get “k/3 = n – 5” and then solve for k by multiplying both sides by 3: “k = 3n – 15” which is answer choice (A).

If this is still not convincing, plug in numbers. Let’s say k = 12. If so, the average n would be (6 + 9 + 12)/3 = (27)/3 = 9. Now plug the values of n=9 and k=12 into the answer choices. You’ll see that (A) works: 3n – 15 = 3*9 – 15 = 27 – 15 = 12 = k. However, (B) yields 5 – n = 5 – 9 = -4, which does not equal k.

Hope that helps!

• haris rajput says:

wow what a math i like it