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

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

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New post 27 Apr 2012, 16:14
5
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A
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C
D
E

Difficulty:

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Question Stats:

59% (02:13) correct 41% (02:09) wrong based on 129 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

Can somebody please explain this question?

Thanks
Vikram
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Posts: 58335
Re: If x is an integer, is |x|>1.  [#permalink]

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New post 27 Apr 2012, 22:22
3
1
vdadwal wrote:
If x is an integer, is |x|>1.

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

Can somebody please explain this question?

Thanks
Vikram


This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

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.

Answer: C.

This question is also discussed here: m14-72785.html

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: If x is an integer, is |x|>1.  [#permalink]

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New post 10 May 2012, 06:48
roots are x=-1 and x=1/2 and --> ">" sign means that the given inequality holds true for: x<-1 and x>1/2 ... can you please help me with this concept and what will happen if sign was "<"..further, will it be right in stating that when there is a positive sign, x is greater than the positive root and x is less than the negative root?
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Re: If x is an integer, is |x|>1.  [#permalink]

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New post 10 May 2012, 06:49
pavanpuneet wrote:
roots are x=-1 and x=1/2 and --> ">" sign means that the given inequality holds true for: x<-1 and x>1/2 ... can you please help me with this concept and what will happen if sign was "<"..further, will it be right in stating that when there is a positive sign, x is greater than the positive root and x is less than the negative root?


Explained here:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: If x is an integer, is |x|>1.  [#permalink]

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New post 19 Jul 2012, 19:23
Bunuel wrote:
vdadwal wrote:
If x is an integer, is |x|>1.

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

Can somebody please explain this question?

Thanks
Vikram


This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

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.

Answer: C.

This question is also discussed here: m14-72785.html

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.



I have a question regarding the above solution let's say in statement 1, when you solve the inequality why do you say that x<-1 AND x >1/2

why is this an AND condition ....why not OR? If this were a quadratic equation x (can be) = 1/2 OR -1 OR both
For inequality why is the same thing an AND as opposed to OR?
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Solving compounded inequalities - any efficient approach?  [#permalink]

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New post 23 Oct 2013, 19:58
If x is an integer, is |x|>1?

(1) (1−2x)(1+x)<0

(2) (1−x)(1+2x)<0

Hi all - I tried this problem on a GMAT club test and I didn't really understand the method. Any quick approaches to finding the solution to each inequality? Is there any quick method to figure out the range of values of x for which statement 1 and 2 will be accurate?

Help appreciated! I'd like to know the quickest, most efficient way to approach such problems!

Cheers
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Re: Solving compounded inequalities - any efficient approach?  [#permalink]

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New post 24 Oct 2013, 01:05
sidvish wrote:
If x is an integer, is |x|>1?

(1) (1−2x)(1+x)<0

(2) (1−x)(1+2x)<0

Hi all - I tried this problem on a GMAT club test and I didn't really understand the method. Any quick approaches to finding the solution to each inequality? Is there any quick method to figure out the range of values of x for which statement 1 and 2 will be accurate?

Help appreciated! I'd like to know the quickest, most efficient way to approach such problems!

Cheers


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

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Re: If x is an integer, is |x|>1.   [#permalink] 23 Sep 2018, 09:57
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