BTGmoderatorDC wrote:If p, q, and r are different positive integers such that p + q + r = 6, what is the value of x ?

(1) The average of x^p and x^q is x^r.

(2) The average of x^p and x^r is not x^q.

OA E

Source: Manhattan Prep

Given that p, q, and r are different positive integers such that p + q + r = 6, we can deduce that p, q and r are 1, 2, and 3 (not respectively!). Any of them can have any value.

Let's take each statement one by one.

(1) The average of x^p and x^q is x^r.

Case 1: Say p = 1, q = 2 and r = 3

Thus,

(x + x^2)/2 = x^3
=> x + x^2 = 2x^3

1 + x = 2x^2

2x^2 - x - 1 = 0

2x^2 - 2x + x -1 =0

2x(x - 1) + 1(x - 1) = 0

(x - 1)(2x + 1) = 0

x = 1 or -1/2

No unique value of x. Insufficient.

There is no need to consider other cases now.

(2) The average of x^p and x^r is not x^q.

This information is not of any significance. It can certainly help us dealing with Statement 1.

(1) and (2) together

Since the case discussed in Statement 1 does not invalidate Statement 2, we can conclude that the unique value of x cannot be determined.

The correct answer: E

Hope this helps!

-Jay
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