If x 0, is x^2 / |x| < 1? (1) x < 1 (2) x > -1 A.

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If x 0, is x^2 / |x| < 1? (1) x < 1 (2) x > -1 A. [#permalink]

03 Jan 2010, 07:14

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If x ≠ 0, is x^2 / |x| < 1?
(1) x < 1
(2) x > -1

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Can someone please explain the process of solving this ?
OA :

[Reveal] Spoiler:

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Re: DS : absolute value and inequalities [#permalink]

03 Jan 2010, 07:51

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msand wrote:

[Reveal] Spoiler:

We can safely reduce LHS of inequality \frac{x^2}{|x|}<1 by |x|. We'll get: is |x|<1? Basically question asks whether x is in the range -1<x<1.

Statements 1 and 2 are not sufficient, but together they are defining the range for x as -1<x<1.

Answer: C.
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Re: DS : absolute value and inequalities [#permalink] 03 Jan 2010, 07:51

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If x 0, is x^2 / |x| < 1? (1) x < 1 (2) x > -1 A.

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If x 0, is x^2 / |x| < 1? (1) x < 1 (2) x > -1 A.

If x 0, is x^2 / |x| < 1? (1) x < 1 (2) x > -1 A.

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