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If x is not equal to 0, is |x| less than 1? (1) x/|x| < x

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If x is not equal to 0, is |x| less than 1? (1) x/|x| < x [#permalink] New post 18 Feb 2008, 10:57
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If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x

--

nevermind - I got what I was doing wrong after combining the a,b results
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Re: DS: Abs Value/inequality [#permalink] New post 18 Feb 2008, 13:53
A

X/ abs(X) < X

this tells you that X = -1

since a number divded by itself is 1 but in order for a number divided by itself to be less than 1, one of the terms has to be negative.

since the bottom term can't be negative (abs value) X = -1

SUFF



statement 2


2) |x| > x


this tells me that X is negative but nothing more

INSUFF
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Re: DS: Abs Value/inequality [#permalink] New post 19 Feb 2008, 02:02
terp26 wrote:
A

X/ abs(X) < X

this tells you that X = -1

since a number divded by itself is 1 but in order for a number divided by itself to be less than 1, one of the terms has to be negative.

since the bottom term can't be negative (abs value) X = -1

SUFF



statement 2


2) |x| > x


this tells me that X is negative but nothing more

INSUFF



If X is -1, then you are saying that -1 < -1 as per St1 ...???

St2 says x is -ve. Hence Not suff...

I would mark it E.

Whats the OA ?
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Re: DS: Abs Value/inequality [#permalink] New post 19 Feb 2008, 07:26
oh good point, confused the inequality
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Re: DS: Abs Value/inequality [#permalink] New post 19 Feb 2008, 07:51
OA is C.

By combining A and B , you should get the range of x between -1 and 0 (sufficient for -1<x<1 or |X|<1).
Re: DS: Abs Value/inequality   [#permalink] 19 Feb 2008, 07:51
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