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

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Manager
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If x is not equal to 0, is |x| less than 1? (1) x/|x| < x [#permalink]

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23 Jun 2008, 06:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x
Intern
Joined: 20 Jun 2008
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Re: 700 level MGMAT Q [#permalink]

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23 Jun 2008, 06:49
I think answer should be C.
2)gives x negaitive.
if x is negative then 1) gives -1<x<0.

right?

lionheart187 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x
Director
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Re: 700 level MGMAT Q [#permalink]

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23 Jun 2008, 06:50
lionheart187 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x

1: sign(x) < x -> x is either between -1 and 0 or greater than 1 -> insufficient
2: x < 0 -> insufficient

1&2: x is between -1 and 0 -> abs(x) < 1 -> sufficient -> C
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Re: 700 level MGMAT Q [#permalink]

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23 Jun 2008, 06:52
If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x
lets see if we multiple we get x<|x|.x
well since |x| is always positive..if x is negative say a fraction pick x=-1/2 then yeah or if xpositive say x=2 then yes too..insuff

(2) |x| > x
OK so x is negative..
but we still dont know if |x|>1 or not..

together..
sufficient.. cause if x is negative then x<|x|*x cannot be possible for |x|>1...
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Re: 700 level MGMAT Q [#permalink]

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23 Jun 2008, 06:54
Answer is (C):

(1) is possible if and only if -1<x<0 or 1<x: insufficient (try x=-0,5 and x=2)

(2) is possible if and only if x<0: insufficient (try x=-0,5 and x=-2)

(1) and (2) is sufficient since they are possible together if and only if -1<x<0 and then abs(x)<1
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Re: 700 level MGMAT Q [#permalink]

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23 Jun 2008, 10:30
lionheart187 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x| < x

(2) |x| > x

I say C.

Try values of 5 and 1/2 for S1

Try values -5 and -1/2 for S2.

why 1/2? B/c the question is basically asking is -1<x<1 and x =/ 0
Re: 700 level MGMAT Q   [#permalink] 23 Jun 2008, 10:30
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# If x is not equal to 0, is |x| less than 1? (1) x/|x| < x

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