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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