What is the remainder when the average (arithmetic mean) of the intege

The GMAT won't ask remainder questions without first specifying that all numbers are positive.
I sugdest you skip this one.

We need to find What is the remainder when the average (arithmetic mean) of the integers –x^4, -x^3, x and x^2 is divided by 4

Average = Sum of Terms / Total Number of Terms = \(\frac{–x^4 + (-x^3) + x + x^2}{4}\) = \(\frac{–x^4 -x^3 + x + x^2}{4}\) = \(\frac{-x^3(x + 1) + x (1+x)}{4}\) = \(\frac{(x + 1) (-x^3 + x)}{4}\) = \(\frac{(x + 1) x (1 - x^2)}{4}\) = \(\frac{x (x + 1)(1-x)(1+x)}{4}\) = \(\frac{(1-x)*x*(x + 1)^2}{4}\) = -\(\frac{(x-1)*x*(x + 1)^2}{4}\)

STATEMENT 1: x – 3 is even

x - Odd = Even
=> x = Even + Odd = Odd
Since x is odd so x-1 and x+1 will become even
Either x-1 is a multiple of 4 and x+1 is a multiple of 2 or x-1 is a multiple of 2 and x+1 is a multiple of 4
in both the cases (x-1)*\((x+1)^2\) will become a multiple of 4*4 = 16
=> -\(\frac{(x-1)*x*(x + 1)^2}{4}\) will be a multiple of \(\frac{16}{4}\) = 4
=> Remainder when the average (arithmetic mean) of the integers –x^4, -x^3, x and x^2 is divided by 4 = 0
=> SUFFICIENT


STATEMENT 2: The average (arithmetic mean) of –x^3, x and x^2 is 11

\(\frac{–x^3 + x + x^2}{3}\) = 11
=> \(x + x^2 - x^3\) = 33
=> x is odd
Same as Statement 1
=> SUFFICIENT

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Remainders

target is (arithmetic mean) of the integers –x^4, -x^3, x and x^2 is divided by 4

#1
x-3 is even
x has to be odd
so value of 4 odd integer will give even value only which will be 2^2 at least
plug in x= 1,5,
sufficient divisible by 4
#2
The average (arithmetic mean) of –x^3, x and x^2 is 11
x*(x+1-x^2) =33
x= -3
in target we get yes
sufficient
option D