Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.