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M14 #13

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M14 #13 [#permalink] New post 11 Nov 2008, 20:40
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Hello. I did not see this question under the search, although I'm not certain I searched correctly.
Please explain:

If x is an integer, is |x| \gt 1 ?

1. (1 - 2x)(1 + x) \lt 0
2. (1 - x)(1 + 2x) \lt 0

[Reveal] Spoiler: OA
C

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Answer is C because combined we come conclude that neither 0,-1 and 1 can be X and therefore [x]>1

My question: What is the most effiecient way of answering this question? It it best to just plug in numbers and by doing so, conclude that between S1 and S2 X is neither 0, 1, or -1?

My route was:

1) Simplifying S1 to.... 2X>1 so X>1/2 OR X<-1
2) Simplifying S2 to....X>1 OR 2X<-1 so X>-1/2

Is that incorrect? And if so, how would you combine S1 and S2 and exclude numbers?
Thank you.
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Re: M14 #13 [#permalink] New post 11 Nov 2008, 23:00
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I will try possible options.

From stmt1:
either (1-2x) > 0 and (1+x) < 0
or, (1-2x) < 0 and (1+x) > 0
Hence, x > 1/2 or x < -1...insufficient.

Similarly, from stmt2:
x > 1 or x < -1/2....insufficient.

Combining stmt1 and stmt2:
x > 1 or, x < -1....sufficient.
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Re: M14 #13 [#permalink] New post 12 Nov 2008, 10:12
jallenmorris... I was using the wrong keys on my computer, but yes, I meant the absolute value of X.

scthakur... thank you for the clarification. You can see that I messed up by decreasing S2 to X>-1/2....You only change signs when you divide by a negative but not when you divide into a negative. Thank you. Now, it makes sense. :-D
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Re: M14 #13 [#permalink] New post 25 Jan 2010, 05:48
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lets rephrase the question |x|>1 is x> 1 or x<-1
1.(1-2x)(1+x) <0
so x<-1 or x>1/2
statement 1 statisfies x<-1 but it says x>1/2 not x>1 so insufficient
2.(1-x)(1+2x)<0
so x>1 or x<-1/2
statement 2 statisfies x>1 but it says x<1-1/2 not x<-1 so insufficient

but combining both we will know that x<-1 and x>1

so ans is C
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Re: M14 #13 [#permalink] New post 28 Jan 2011, 06:55
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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.
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Re: M14 #13 [#permalink] New post 28 Jan 2011, 09:46
Bunuel wrote:
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.




nice explanation bunuel.
i followed the same procedure.
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Re: M14 #13 [#permalink] New post 29 Jan 2011, 07:17
I suppose it could be (D).
We are told that x is an integer.
So, from (1) we have x<-1 and x>1/2, but if x is an int, so second part turns to be x>1.
The same logic for (2)
x<-1/2 and x>1 turns into x<-1 and x>1
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Re: M14 #13 [#permalink] New post 29 Jan 2011, 07:31
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ulm wrote:
I suppose it could be (D).
We are told that x is an integer.
So, from (1) we have x<-1 and x>1/2, but if x is an int, so second part turns to be x>1.
The same logic for (2)
x<-1/2 and x>1 turns into x<-1 and x>1


OA for this question is C, not D.

x>1/2 and x=integer means that x can be any integer more than 1/2: 1, 2, 3, ... so x could still equal to 1 for statement (1), so this statement is not sufficient. Similarly for statement (2) x could still equal to -1, so it's also not sufficient.

Hope it's clear.
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Re: M14 #13 [#permalink] New post 23 May 2011, 14:08
sonnco wrote:
silasaaa2 wrote:
lets rephrase the question |x|>1 is x> 1 or x<-1
1.(1-2x)(1+x) <0
so x<-1 or x>1/2
statement 1 statisfies x<-1 but it says x>1/2 not x>1 so insufficient
2.(1-x)(1+2x)<0
so x>1 or x<-1/2
statement 2 statisfies x>1 but it says x<1-1/2 not x<-1 so insufficient

but combining both we will know that x<-1 and x>1

so ans is C


That was a very good setup. Thank you!


Thanks a lot.

However, for statement one you have to consider that either (1-2x) is < 0 and (1+x) > 0 OR (1-2x) is > 0 and (1 + x) < 0. Considering all four possible values will make you arrive at X < -1 or X > 1/2.

You have to do the same for statement 2.

However, the problem with this approach is that its too time consuming. Will take around 3 minutes I believe. Is there any short cut?
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 05:44
Why isn't E instead of C ? when you combine the two statements, don't you still have the possibility that x could be = +/- 1 ?
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 06:23
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imadkho wrote:
Why isn't E instead of C ? when you combine the two statements, don't you still have the possibility that x could be = +/- 1 ?


Hi, and welcome to GMAT Club.

Below is an answer to your doubt:

The ranges we get from (1)+(2) are not inclusive, so x cannot be -1 or 1 (all inequalities have either strict < or strict >).

Hope it's clear.
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 06:34
thanks, could you please simplify more?
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 10:41
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imadkho wrote:
thanks, could you please simplify more?


Please refer to my solution above. The question asks: if x is an integer, is |x| > 1? So, basically the question asks whether: x is an integer more than 1: 2, 3, 4, 5, ... or an integer less than -1: -2, -3, -4, -5, ...

For (1)+(2) we get that x>1 or x<-1, which is exactly what we wanted to know.

Hope it's clear.
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 14:20
In statement 1) x <-1 or x>1/2 ;
In statement 2) x <-1/2 or x>1 ;
so my question is simply the following: why would we, upon combining the two statements, consider only x<-1 and x>1 while not considering x>1/2 and x<-1/2, and thus choose E instead of C (the same logic used with each of the statements alone)?
thanks for your help.
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 17:37
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imadkho wrote:
In statement 1) x <-1 or x>1/2 ;
In statement 2) x <-1/2 or x>1 ;
so my question is simply the following: why would we, upon combining the two statements, consider only x<-1 and x>1 while not considering x>1/2 and x<-1/2, and thus choose E instead of C (the same logic used with each of the statements alone)?
thanks for your help.


When statements are taken together we should consider intersection of the ranges, so the ranges which are true for both statements.
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 17:55
I was wondering how you go about combing the statements? Do you just know that we must satisfy the greater of statements? E.g. X>1 must be used because everything can be satisfied in this where x>1/2 cannot. And same as other. Is this right?

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Re: M14 #13 [#permalink] New post 01 Feb 2012, 18:00
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geno5 wrote:
I was wondering how you go about combing the statements? Do you just know that we must satisfy the greater of statements? E.g. X>1 must be used because everything can be satisfied in this where x>1/2 cannot. And same as other. Is this right?

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Consider this:
If x is an integer, what is the value of x?

(1) 2<x<5
(2) 0<x<4

What is the answer?
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Re: M14 #13 [#permalink] New post 01 Feb 2012, 19:07
C or 3. So because x>1 it cannot be x>1/2 and x<-1 it cannot be x<-1/2.
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Re: M14 #13 [#permalink] New post 27 Feb 2012, 19:07
Hey Bunuel or anyone,

In statement (1) I found the domain of x by finding the roots and plugging different values within the sections to find if it satified the inequility. But how do we find the domain of x through this other approach???


either {(1-2x) > 0 and (1+x) < 0 } OR { (1-2x) < 0 and (1+x) > 0 }

Hence, x > 1/2 or x < -1 ???


What I arrived at using this approach was:

{ 1/2 > x and x < -1 } OR { 1/2 < x and x>-1} ------->>> { 1/2 > x and x < -1 } OR { 1/2 < x }

but how do we combine this OR statement to arrive ---->>>> x > 1/2 or x < -1 ????

We have two contradictory expressions 1/2 > x or 1/2 < x


Please Help!!!
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Re: M14 #13 [#permalink] New post 27 Feb 2012, 23:39
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Hey Bunuel or anyone,

In statement (1) I found the domain of x by finding the roots and plugging different values within the sections to find if it satified the inequility. But how do we find the domain of x through this other approach???


either {(1-2x) > 0 and (1+x) < 0 } OR { (1-2x) < 0 and (1+x) > 0 }

Hence, x > 1/2 or x < -1 ???


What I arrived at using this approach was:

{ 1/2 > x and x < -1 } OR { 1/2 < x and x>-1} ------->>> { 1/2 > x and x < -1 } OR { 1/2 < x }

but how do we combine this OR statement to arrive ---->>>> x > 1/2 or x < -1 ????

We have two contradictory expressions 1/2 > x or 1/2 < x


Please Help!!!


(1-2x)(1+x)<0 --> two multiples have the opposite signs:

(1-2x)>0 and (1 + x)<0 --> x<\frac{1}{2} and x<-1 --> x<-1 (you take the intersection of the ranges since both must be true simultaneously);

(1-2x)<0 and (1 + x)>0 --> x>\frac{1}{2} and x>-1 --> x>\frac{1}{2} (you take the intersection of the ranges since both must be true simultaneously);

So the the ranges for which (1-2x)(1+x)<0 holds true are: x<-1 and x>\frac{1}{2}.

Hope it's clear.
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Re: M14 #13   [#permalink] 27 Feb 2012, 23:39
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5 M14 #27 dczuchta 19 13 Nov 2008, 16:55
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M14 #13

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