Is |n| < 1?

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Is |n| < 1?

by VJesus12 » Mon Apr 08, 2019 5:14 am

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Is |n| < 1?

(1) n^x - n < 0
(2) x^(-1) = -2

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

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by Jay@ManhattanReview » Mon Apr 08, 2019 10:05 pm

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VJesus12 wrote:Is |n| < 1?

(1) n^x - n < 0
(2) x^(-1) = -2

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT
We have to determine whether |n| < 1. This means that we have to determine whether - 1 < n < 1.

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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