average with exponents

[alexandrabiorka]

What is the average of 101^2 and 99^2?

a) square root of 101 + square root of 99
b) 100
c) 100^2 - 1
d) 100^2
e) 100^2 + 1

[GMATinsight]

What is the average of 101^2 and 99^2?

a) square root of 101 + square root of 99
b) 100
c) 100^2 - 1
d) 100^2
e) 100^2 + 1

average of 101^2 and 99^2 = (101^2 + 99^2) / 2

i.e. average of 101^2 and 99^2 = [(100+1)^2 + (100-1)^2] / 2
i.e. average of 101^2 and 99^2 = [(100^2+1^2+2x100x1) + (100^2+1^2-2x100x1)] / 2
i.e. average of 101^2 and 99^2 = [(100^2 + 1+ 2x100x1 + 100^2 + 1 - 2x100x1)] / 2
i.e. average of 101^2 and 99^2 = [(100^2 + 1 + 100^2 + 1)] / 2
i.e. average of 101^2 and 99^2 = 2 x [100^2 + 1] / 2 = 100^2 + 1

Answer: Option E

[GMATGuruNY]

What is the average of 101^2 and 99^2?

a) square root of 101 + square root of 99
b) 100
c) 100^2 - 1
d) 100^2
e) 100^2 + 1

Since most of the answer choices are in terms of 100, rephrase the given expression in terms of 100.
Let x=100.
Then:
101² + 99² = (x+1)² + (x-1)² = (x² + 2x + 1) + (x² - 2x + 1) = 2x² + 2.
Thus:
Average of 101² + 99² = (2x² + 2)/2 = x² + 1 = 100² + 1.

The correct answer is E.

[Brent@GMATPrepNow]

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

[BestGMATEliza]

I found it helpful to go into the GMAT having understood and memorized the following:
(a+b)²+(a-b)² =2a²+2b²
(a+b)²-(a-b)²=4ab
(a-b)²-(a+b)²=-4ab

so that when problems like this came up, I could easily translate them into a more calculable form.

For this problem, you can recognize that 101²+99²=(a+b)²+(a-b)² with a=100 and b=1
so it equals 2a²+2b². However we are looking for the average, so we divide by 2 and get a²+b² or 100²+1²
which equals E

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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