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# Is |X| > 1? st1. (1-2x)*(1+x) < 0 st2. (1-x) * (1+2x)

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Manager
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Is |X| > 1? st1. (1-2x)*(1+x) < 0 st2. (1-x) * (1+2x) [#permalink]

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08 Apr 2005, 09:00
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Is |X| > 1?
st1. (1-2x)*(1+x) < 0
st2. (1-x) * (1+2x) < 0
Manager
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08 Apr 2005, 09:02
This question took me a bit to solve.
What would be an efficient way to crack these types of questions?

Also, Do the rules of inequalites apply when square of a variable is involved in the inequality?.

VP
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08 Apr 2005, 11:17
"C"...will explain in detail when I am back home. Ans is YES |X| > 1
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08 Apr 2005, 18:22
try plugin in the values for x....

as 1) is satisfied at 1,-2.....
as 2) is satified at -1,2.......

comparig them together also does not give any fial result.......
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VP
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09 Apr 2005, 02:00
C...

1) (1-2x)*(1+x) < 0 => 1-2x<0 => 1/2<x or 1+x<0 => x<-1 => insuff

2) same reasoning as above => 1<x or x<-1/2 => insuff

1)+2) take the values together so x must be greater than 1 or smaller than -1 that 1) and 2) are true
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14 Apr 2005, 04:44
banerjeea_98 wrote:
"C"...will explain in detail when I am back home. Ans is YES |X| > 1

Still waiting, Bana. And until then, I'll stick by E.
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14 Apr 2005, 06:03
Depapa wrote:
banerjeea_98 wrote:
"C"...will explain in detail when I am back home. Ans is YES |X| > 1

Still waiting, Bana. And until then, I'll stick by E.

oh ! sorry, I didn't post my sol becose it is same as Christoph, can u elaborate how u get "E".
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14 Apr 2005, 06:40
This is how I looked at it:
The question asks: Is x>1 OR x<-1 ALWAYS?
1) (1-2x)(1+x)<0
(1-2x) < 0 --> x<1/2
1+x <0 --> x<-1
INSUFFICIENT
2) (1-x)(1+2x)<0
(1-x) <0 --> x>1
(1+2x)<0 --> x > -1/2
INSUFFICIENT
Combining both statements we get a range from -1/2 to - infinite and 1/2 to +infinite. So answer is E.
Is this method right?
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14 Apr 2005, 07:10
ACT28 wrote:
This is how I looked at it:
The question asks: Is x>1 OR x<-1 ALWAYS?
1) (1-2x)(1+x)<0
(1-2x) < 0 --> x<1/2
1+x <0 --> x<-1
INSUFFICIENT
2) (1-x)(1+2x)<0
(1-x) <0 --> x>1
(1+2x)<0 --> x > -1/2
INSUFFICIENT
Combining both statements we get a range from -1/2 to - infinite and 1/2 to +infinite. So answer is E.
Is this method right?

no...when you divide an inequality by a -ve number you have to change the sign i.e. 1-2x<0 => -2x<-1 (divide by -2) => x>1/2
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Director
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14 Apr 2005, 09:37
christoph wrote:
C...

1) (1-2x)*(1+x) < 0 => 1-2x<0 => 1/2<x or 1+x<0 => x<-1 => insuff

2) same reasoning as above => 1<x or x<-1/2 => insuff

1)+2) take the values together so x must be greater than 1 or smaller than -1 that 1) and 2) are true

christoph, correct me if I am wrong
From 1: x is between -1 & 1/2
From 2: x is between -1/2 & 1

By combining 1 & 2 we get x is between -1 and 1

so shouldn't it be 'E' & not 'C'
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Re: DS: Is |x| > 1? [#permalink]

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14 Apr 2005, 13:22
Is |X| > 1?
Meaning: is x>1 or x<-1?

st1. (1-2x)*(1+x) < 0
1-2x<0
1+x>0
or
1-2x>0
1+x<0
=>
x>1/2 or x<-1
Insufficient, for x could be 3/4 which makes |x|<1 but x could also be 2 which makes |x|>1.

st2. (1-x) * (1+2x) < 0
1-x<0
1+2x>0
or
1-x>0
1+2x<0
=>
x>1 or x<-1/2
Insufficient, same rationale.

Combined:
(1-2x)*(1+x) < 0 and
(1-x)*(1+2x) < 0
=>
x>1/2 or x<-1
x>1 or x<-1/2
=>x>1 or x<-1
Sufficient to make sure |x|>1

(C)
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keep on seeking, and you will find;
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Re: DS: Is |x| > 1? [#permalink]

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14 Apr 2005, 16:02
HongHu wrote:
Is |X| > 1?
Meaning: is x>1 or x<-1?

st1. (1-2x)*(1+x) < 0
1-2x<0
1+x>0
or
1-2x>0
1+x<0
=>
x>1/2 or x<-1
Insufficient, for x could be 3/4 which makes |x|<1 but x could also be 2 which makes |x|>1.
(C)

Honghu, I have a basic question:
why do you say
(1-2x)*(1+x) < 0
is
1 - 2x < 0 ; 1 + x > 0
(or)
1 - 2x > 0; 1 + x < 0

Shouldn't it be
(1-2x) < 0 ; (1+x) < 0

Can you be more brief - I do not follow why you flipped the signs.
VP
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14 Apr 2005, 16:15
If both (1-2x) and (1+x) are -ve wudn't that make their product +ve....so one has to be +ve and other -ve and vice versa so u have to scenarios in order to make the product -ve.
Director
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14 Apr 2005, 18:58
banerjeea_98 wrote:
If both (1-2x) and (1+x) are -ve wudn't that make their product +ve....so one has to be +ve and other -ve and vice versa so u have to scenarios in order to make the product -ve.

Thanks banerjeea. I was thinking in terms of a quad. equation, just realised by mistake.
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14 Apr 2005, 22:22
For st1, x<-1 and x>1/2
For x < -1 , x can be -2,-3,-4,-5 etc all of which |X| > 1
But x > 1/2, x can be 1 which is not >1, and x can be 3/4 which is not >1.
So st1 is not sufficient

For st2, x<-1/2, and x>1
For x>1, |x|>1 since x can take any values above 1
For x<-1/2, x can be -1 for which |x| is not greater 1.

Combining, x>1 and x<-1 so |x|>1.

Ans: C
14 Apr 2005, 22:22
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# Is |X| > 1? st1. (1-2x)*(1+x) < 0 st2. (1-x) * (1+2x)

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