Is |n| < 1?
(1) n^x - n < 0
(2) x^(-1) = -2
[spoiler]OA=C[/spoiler]
Source: Manhattan GMAT
Is |n| < 1?
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We have to determine whether |n| < 1. This means that we have to determine whether - 1 < n < 1.VJesus12 wrote:Is |n| < 1?
(1) n^x - n < 0
(2) x^(-1) = -2
[spoiler]OA=C[/spoiler]
Source: Manhattan GMAT
Let's take each statement one by one.
(1) n^x - n < 0
Case 1: Say n = 1/2, and x = 2. We see that n^x - n < 0 => (1/2)^2 - (1/2) < 0 => 1/4 - 1/2 < 0 => -1/2 < 0. We see that n = 1/2 lies within - 1 < n < 1. The answer is Yes.
Case 2: Say n = 4, and x = 1/2. We see that n^x - n < 0 => √4 - 4 < 0 => 2 - 4 < 0 => -2 < 0. We see that n = 4 does not lie within - 1 < n < 1. The answer is No.
Insufficient.
(2) x^(-1) = -2
No information about n. Clearly insufficient.
(1) and (2) together
From (2) x^(-1) = -2, we have x = -1/2
From (1) n^x - n < 0, we have n^(-1/2) < n => 1/√n < n => 1 < n*√n; we can multiply both sides by √n without caring for the change of sign of the inequality since √n is always a positive number.
So, we have 1 < n*√n
We see that n cannot be negative since √n would then be an unreal number. n can only be greater than 1. Thus, the answer is no, |n| is not less than 1.
The correct answer: C
Hope this helps!
-Jay
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