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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. [#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:
C
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Joined: 02 Sep 2009
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Kudos [?]: 9577 [1] , given: 826

Re: DS : absolute value and inequalities [#permalink]  03 Jan 2010, 07:51
1
KUDOS
msand wrote:
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:
C

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.

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