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

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Senior Manager
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If x ≠ 0, is x^2/|x|< 1? (1) x < 1 (2) x > −1 [#permalink] New post 13 Oct 2008, 20:30
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A
B
C
D
E

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

Could some please explain as to how to simply the question. Thanks!
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Re: DS [#permalink] New post 13 Oct 2008, 20:54
vksunder wrote:
If x ≠ 0, is x^2/|x|< 1?
(1) x < 1
(2) x > −1

Could some please explain as to how to simply the question. Thanks!


"x^2 / |x| < 1?" can be re-arranged as "x^2 < |x| ?"

then from
1: if x < 1, x could be +ve fraction, -ve fraction or -ve integer/number.. nsf.
2: if x > -1, x could be a -ve fraction, +ve fraction or +ve integer/number.. nsf.

from 1 and 2: -1 < x < 1. so x is either a +ve or -ve fraction. in that case, lxl is always > x^2.

so it is C.
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Re: DS [#permalink] New post 13 Oct 2008, 21:09
vksunder wrote:
If x ≠ 0, is x^2/|x|< 1?
(1) x < 1
(2) x > −1

Could some please explain as to how to simply the question. Thanks!



x^2 can be written as |x| * |x| and hence x^2/|x| can reduce to |x|.

|x| will be smaller than 1 only when -1<x<1 and hence C should be the answer.
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Re: DS [#permalink] New post 14 Oct 2008, 12:07
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vksunder wrote:
If x ≠ 0, is x^2/|x|< 1?
(1) x < 1
(2) x > −1

Could some please explain as to how to simply the question. Thanks!


REWRITE

X^4 / X^2 <1

X^2<1

ie: is /x/< 1

obviously C
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Re: DS [#permalink] New post 14 Oct 2008, 14:26
Thanks for the explanation guys! OA - C.
Re: DS   [#permalink] 14 Oct 2008, 14:26
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