alexandrabiorka wrote:What is the average of 101² and 99²?

a) √101 + √99
b) 100
c) 100² - 1
d) 100²
e) 100² + 1

I have doubts about the "GMAT-ness" of this question, since the answer choices seem to force us into a certain approach. For example, the correct answer could also be written as 99² + 200 or 100² - 200, which might force a different approach.

Here's an alternate approach that would favor the other 2 options:

First recognize that the average of 2 values will be HALFWAY between those 2 values.
For example, 10 is the average of 7 and 13, and 10 is HALFWAY between 7 and 13.

So, to determine the average of 101² and 99², let's first determine the DISTANCE between these two values:
The distance between the two values = 101² - 99² [we can factor this difference of squares]
= (101 + 99)(101 - 99)
= (200)(2)
= 400

Since 101² is 400 more than 99², the HALFWAY point between 99² and 101² will be 200 more than 99².
So the average will be 99² + 200

Or we can say that the HALFWAY point between 99² and 101² will be 200 less than 101², which means the average will 101² - 200

Having said all of that, we can still make these two answer choices resemble answer choice E.

99² + 200 = (100 - 1)² + 200
= 100² - 200 + 1 + 200
= 100² + 1
= E
We can use similar algebra to make 101² - 200 resemble answer choice E as well.

Cheers,
Brent

Hi alexandrabiorka,

This question actually has a great "organizational" shortcut to it. It also helps to know that 100^2 = 10,000....

We're asked to figure out (101^2 + 99^2)/2

101^2 = 101(101)
99^2 = 99(99)

Let's "pair up" one 101 and one 99 and add them together....101 + 99 = 200

How many of those pairs do we have in total? 99

So that's 99(200)

What's "left over"? Two 101s (they weren't paired up)

In total, we have 99(200) + 2(101)

99(200) + 202

Let's "break up" the 202 into 200 + 2.

Now we have...
99(200) + 200 + 2
100(200) + 2

20,002

Now we divide by 2...

10,001

There's only one answer that's greater than 10,000...

Final Answer: E

GMAT assassins aren't born, they're made,
Rich