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:

so EITHER 1/x > 0 and \(x^2 - 1 > 0\) OR 1/x<0 and \(x^2 - 1 < 0\) [ since \((x^2-1)/x > 0\) ...either both are positive or both are negative]

Case 1:

1/x is greater than 0 then x > 0 and for \(x^2 - 1 > 0\) ... x >1 or x>-1

so finally x>1

Cae 2:

1/x < 0 then x is less than 0 and for \(x^2 - 1 < 0\) ..... x<-1 and x <1

so finally x < -1

PLEASE CORRECT MY APPROACH

apoorvasrivastva, your approach is good for case 1, however for case 2 it will be -1 < x < 0 as nplaneta posted earlier.

The reason being... (x^2 - 1) / x > 0 When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 now as we are considering x < 0 then, when we take square root of both side: sqrt(x^2) < sqrt(1) x < 1 [but x is a negative number so we need to multiply both sides by -1] -x > -1 x > -1 [left side became positive due to negative x and -1] and since the case is for x < 0 , therefore, the solution is -1 < x < 0

The solution for x > 1/x being: x>1 (when x>0) and -1 < x < 0 (when x<0)

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum [sorry I can't post the direct link, as the forum isn't allowing me cos I am new member] This exact problem is explained in this post as well.

~Cheers
_________________

The three most significant times in your life are: 1. When you fall in love 2. The birth of your first child 3. When you prepare for your GMAT

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum

Actually i got this problem from the very link you mentioned of HongHu's post, as it is not explained in detail there , i came up posting it in a separate thread...

scarish wrote:

case 2 it will be -1 < x < 0 as nplaneta posted earlier. When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 .......... x < 1 [but x is a negative number so we need to multiply both sides by -1]

I couldn't understand why you have to multiply x, you said to flip the inequality, but it is not flipped i think. Then why have to multiply by -1. sorry i don't get you. May be i am not comprehending it.. Could you explain clearly, the second case. i was also getting the same answer as apoorvasrivastva

1/x < 0 then x is less than 0 and for \(x^2 - 1 < 0\) ..... x<-1 and x <1

so finally x < -1

PLEASE CORRECT MY APPROACH

apoorvasrivastva, your approach is good for case 1, however for case 2 it will be -1 < x < 0 as nplaneta posted earlier.

The reason being... (x^2 - 1) / x > 0 When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 now as we are considering x < 0 then, when we take square root of both side: sqrt(x^2) < sqrt(1) x < 1 [but x is a negative number so we need to multiply both sides by -1] -x > -1 x > -1 [left side became positive due to negative x and -1] and since the case is for x < 0 , therefore, the solution is -1 < x < 0

The solution for x > 1/x being: x>1 (when x>0) and -1 < x < 0 (when x<0)

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum [sorry I can't post the direct link, as the forum isn't allowing me cos I am new member] This exact problem is explained in this post as well.

~Cheers[/quote]

Thanks i realsed my mistake!!!Kudos given to u !!:)

case 2 it will be -1 < x < 0 as nplaneta posted earlier. When x < 0 then: (x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)] x^2 < 1 .......... x < 1 [but x is a negative number so we need to multiply both sides by -1]

crejoc wrote:

I couldn't understand why you have to multiply x, you said to flip the inequality, but it is not flipped i think. Then why have to multiply by -1. sorry i don't get you. May be i am not comprehending it.. Could you explain clearly, the second case. i was also getting the same answer as apoorvasrivastva

Right, our equation was (x^2-1)/x > 0, so to get rid of x from the denominator you have to multiply both sides of the inequality by x, that is (x^2-1)*x/x > 0*x => (x^2-1)/x < 0 and the sign is flipped because x < 0. Multiplying both sides by x is what you do when we simply "cross-multiply" but we can only safely cross-multiply when it is equality but not in inequality.

Secondly, upon solving our equation we arrived at: x < 1

Now x in this equation is -ve, since we are considering a case where x < 0. So if we leave it like x < 1 then the x in this case is not comparable with x in our solution for the case where x > 0. So in order to compare apples with apples, we need to get x to mean the same thing for both cases. Therefore, we multiply both sides by -1 (which flips the sign again) to arrive at: x > -1

Also, try picking few numbers in and outside the boundaries in the original equation, which may also help.

crejoc wrote:

Actually i got this problem from the very link you mentioned of HongHu's post, as it is not explained in detail there , i came up posting it in a separate thread...

On a separate note, I think if you have doubts on questions in a particular post then its better to post your doubts on that post itself, so as to keep all responses on the issue together, instead of starting the new post. I don't know if this is the policy on this forum or not, may be moderator can clarify?
_________________

The three most significant times in your life are: 1. When you fall in love 2. The birth of your first child 3. When you prepare for your GMAT

i think best approach in those 2 mins in GMAT will be to plug values.....just keep in mind that whenever it's about fraction....intervals should be 1<x; 0<x<1; -1<x<0; -1>x

we find only applicable are 1<x and -1<x<0

i hope this helps
_________________

Bhushan S. If you like my post....Consider it for Kudos

i think best approach in those 2 mins in GMAT will be to plug values.....just keep in mind that whenever it's about fraction....intervals should be 1<x; 0<x<1; -1<x<0; -1>x

we find only applicable are 1<x and -1<x<0

i hope this helps

Yes the faster the better, thanks for your approach...

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...