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If is an integer, is mod(x) > 1 ? 1. (1-2x)(1+x) < 0

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If is an integer, is mod(x) > 1 ? 1. (1-2x)(1+x) < 0 [#permalink] New post 24 May 2010, 17:20
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33% (01:32) correct 66% (01:13) wrong based on 3 sessions
If is an integer, is mod(x) > 1 ?

1. (1-2x)(1+x) < 0
2.(1-x)(1+2x) < 0

[Reveal] Spoiler:
oa : c


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Re: integer. [#permalink] New post 24 May 2010, 20:46
sushma0805 wrote:
If is an integer, is mod(x) > 1 ?

1. (1-2x)(1+x) < 0
2.(1-x)(1+2x) < 0

[Reveal] Spoiler:
oa : c


src : gmat club test.


mod(x) > 1 --> we need to find out if x is a fraction or not. In other words x<-1 or x >1 or we can disprove -1<x<1

1) Two cases
i) 1-2x > 0 and 1+x <0
1>2x and x<-1
1/2>x and x<-1
So x <-1
ii) 1-2x < 0 and 1+x >0
1/2<x and x > -1
So 1/2 < x
Thus insufficient because

2) i) 1-x > 0 and 1 + 2x <0
1>x and x < -1/2
Thus x <-1/2
ii) 1-x < 0 and 1 + 2x >0
1 < x and x > -1/2
Thus x > 1
Thus insufficient

C) Thus from 1) and 2) we know that x < -1 or x>1. Thus x won't be a fraction. Sufficient.

My answer: C
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Re: integer. [#permalink] New post 07 Jul 2010, 07:40
HI
ANS IS D ..taking the range u can find out the range of X in each condition ..
!.. -1<x<1/2 and in
2nd condition -1/2 <x<1 so clearly it is d
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Re: integer. [#permalink] New post 07 Jul 2010, 16:51
If is an integer, is mod(x) > 1 ?

1. (1-2x)(1+x) < 0
2.(1-x)(1+2x) < 0

From statement 1 we can get

x > 0.5
x < -1

From statement 2 we can get

x > 1
x < -0.5

So by combining above 2 statements we get

x < -1 and
x > 1

hence sufficient

So Answer = C
Re: integer.   [#permalink] 07 Jul 2010, 16:51
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If is an integer, is mod(x) > 1 ? 1. (1-2x)(1+x) < 0

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