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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?

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 _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

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) _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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.

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.

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.

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.

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