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 500,000 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:

1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Please explain more why second condition is sufficient. I am not able to understand it

Bunuel wrote:

SMAbbas wrote:

1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Please explain more why second condition is sufficient. I am not able to understand it

Bunuel wrote:

SMAbbas wrote:

1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Answer D.

Hope it's clear.

You should specify what didn't you understand there. Thank you. _________________

Please explain more why second condition is sufficient. I am not able to understand it

Bunuel wrote:

SMAbbas wrote:

1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Answer D.

Hope it's clear.

Dear piyushmnit,

the question asks whether -1<x<1 is true, so any range of x, that would fall into that one would be sufficient.

The second statement tells that 0<x<1, which fall into the the aforementioned range.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

It would be helpful if someone explains why we don't open the mod under the second option. As well, what's wrong with my algebra here?

\(|x| < \frac{1}{x}\) if \(x<0\), then \(-x*x<1\) and therefore, \(x\) could take all -ve number.

However, picking any negative number surely invalidates that claim. Obviously, having trouble reconciling my algebra with picking numbers.

Thanks for your help,

Question 1:

We did not open the mod because the LHS of the inequality (=|x|) is a positive quantity for ALL x ---> If 1/x is > a positive quantity ---> x > 0 . This is the reason the mod was 'not opened'.

\(1/x > a '+' quantity\) ---> x > 0

Question 2:

If \(x<0\) --> \(|x| = -x\) ---> \(-x<\frac{1}{x}\) ---> \(-x^2>1\) (reversed the sign of inequality as x<0 and multiplying an inequality by a negative number reverses the sign of inequality!)

---> This is not possible as \(x^2 = +\), \(-x^2 < 0\) and thus \(-x^2\) can never be > 1.

Thus , \(x<0\) is not a valid substitution. You did not reverse the sign of inequality when you multiplied the inequality by x as x<0 _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...