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# If x is an integer, is |x|>1?

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Manager
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If x is an integer, is |x|>1? [#permalink]

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24 May 2010, 17:20
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95% (hard)

Question Stats:

43% (02:55) correct 57% (01:55) wrong based on 35 sessions

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If x is an integer, is |x| > 1?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-is-an-integer-is-x-131452.html
[Reveal] Spoiler: OA
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Re: If x is an integer, is |x|>1? [#permalink]

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24 May 2010, 20:46
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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.

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Re: If x is an integer, is |x|>1? [#permalink]

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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: If x is an integer, is |x|>1? [#permalink]

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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

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Re: If x is an integer, is |x|>1? [#permalink]

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04 Mar 2015, 09:56
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Re: If x is an integer, is |x|>1? [#permalink]

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04 Mar 2015, 10:04
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If x is an integer, is |x| > 1?

First of all: is $$|x| > 1$$ means is $$x<-1$$ (-2, -3, -4, ...) or $$x>1$$ (2, 3, 4, ...), so for YES answer $$x$$ can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as $$(2x-1)(x+1)>0$$ (so that the coefficient of x^2 to be positive after expanding): roots are $$x=-1$$ and $$x=\frac{1}{2}$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-1$$ and $$x>\frac{1}{2}$$. $$x$$ could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as $$(x-1)(2x+1)>0$$: roots are $$x=-\frac{1}{2}$$ and $$x=1$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-\frac{1}{2}$$ and $$x>1$$. $$x$$ could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is $$x<-1$$ and $$x>1$$. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-is-an-integer-is-x-131452.html
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Re: If x is an integer, is |x|>1?   [#permalink] 04 Mar 2015, 10:04
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