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:

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

Put that value in the equation a+b >ab then (-b) +b > (-b) (b) i.e. 0> -b^2 As b^2 is always +ve this equaliton will hold true. Hence statement 1 is sufficient.

As above explanations say statement B is also sufficient.

Hi Yogesh .. can you please share the access codes for these tests. It will be very nice of you. My email id is sandeepuc@gmail.com. Thanks in advance.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

Thanks for the solution. But I still couldnt understand why are we taking a=0 and b=0 in the 1 st statement. It states a=-b, is it ok to write 0=-0..... It would be of great help if you can explain. It will help in clearing my doubts.

First of all as \(|a|=b\) (\(b\) equals to absolute value of some number) then \(b\geq{0}\), as absolute value is always non-negative.

(1) \(a=-b\) (\(b=-a\)) --> so \(a\leq{0}\) and \(LHS=a+b=0\). But \(RHS=ab\leq{0}\) thus we can not say for sure that \(a+b>ab\), because if \(a=b=0\) then \(a+b=0=ab\) (so in case \(a=b=0\), \(a+b\) is not more than \(ab\) it equals to \(ab\)). Not sufficient.

Thanks for the solution. But I still couldnt understand why are we taking a=0 and b=0 in the 1 st statement. It states a=-b, is it ok to write 0=-0..... It would be of great help if you can explain. It will help in clearing my doubts.

Re: If |a| = b is a + b > ab ? [#permalink]
25 Jun 2013, 15:33

1

This post was BOOKMARKED

If |a| = b is a + b > ab

(1) a = -b (2) a = -3

|a| = b so b MUST be ≥ 0

1.) a = -b b=-a

So what do we know?

|a|=|b| b is positive a is the negative value of b

So, a + b > ab a+b = 0 HOWEVER we are not sure what values a and b are. For example, a and b could be -3 and 3 or a and b could be 0 and 0. INSUFFICIENT

(2) a = -3 We know that |a|=b, so if a = -3 then b must = 3 a + b > ab -3+3 > (-3)(3) 0>-9 TRUE

Re: If |a| = b is a + b > ab ? [#permalink]
26 Sep 2014, 05:52

Hello from the GMAT Club BumpBot!

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. _________________

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...