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:

This I solved by trial and error...with a logic that what RHS has (a - b) and LHS has addition of the same terms. So for this to be true one has to be be zero hence answer will be zero.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

07 Sep 2012, 14:48

1

This post received KUDOS

SergeNew wrote:

If |a+b|=|a-b|, then a*b must be equal to:

A. 1 B. -1 C. 0 D. 2 E. -2

If \(b=0\), the equality obviously holds. \(|a+b|=|a-b|\) means the distance between \(a\) and -\(b\) is the same as the distance between \(a\) and \(b\). For \(b\neq0,\) it means that \(a\) is the average of -\(b\) and \(b\) (or the midpoint between -\(b\) and \(b\)), so necessarily \(a=0.\) Altogether, the product \(ab\) must be \(0.\)

Answer C _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

08 Sep 2012, 03:26

honggil wrote:

Bunuel,

Can you always just square absolute values like you did on this problem? I am wondering if squaring is a valid operation on absolute values.

Absolute value is always non-negative, so if you have an equality between two absolute values, either they are both 0 or they are equal to the same positive number. Squared, they still remain equal. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

09 Sep 2012, 00:59

1

This post received KUDOS

SergeNew wrote:

If |a+b|=|a-b|, then a*b must be equal to:

A. 1 B. -1 C. 0 D. 2 E. -2

I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities:

From there, I concluded that a*b must be zero since the only way a = -a is if a = 0 and the only way b = -b is if b = 0. Did I do this problem incorrectly?

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

09 Sep 2012, 06:56

HImba88 wrote:

SergeNew wrote:

If |a+b|=|a-b|, then a*b must be equal to:

A. 1 B. -1 C. 0 D. 2 E. -2

I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities:

From there, I concluded that a*b must be zero since the only way a = -a is if a = 0 and the only way b = -b is if b = 0. Did I do this problem incorrectly?

It is absolutely correct. Good job! _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

09 Sep 2012, 14:14

EvaJager wrote:

HImba88 wrote:

SergeNew wrote:

If |a+b|=|a-b|, then a*b must be equal to:

A. 1 B. -1 C. 0 D. 2 E. -2

I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities:

From there, I concluded that a*b must be zero since the only way a = -a is if a = 0 and the only way b = -b is if b = 0. Did I do this problem incorrectly?

It is absolutely correct. Good job!

Thanks Eva. I didn't even think initially to square both sides. That way seems much more efficient than the way I approached the problem. Guess my brain is wired differently

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

09 Sep 2012, 15:33

I found the right answer. But not sure if the procedure is right |a+b| = |a-b|

|a| + |b| = |a| -|b|

|b| = -|b| this is possible only with 0. So it is possible in only 2 cases. 1 and 0. But if it is 1, a+b neq to a-b. so b=0 and a*b = 0. Please let me know if this is a right approach.

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

09 Sep 2012, 16:03

ravipprasad wrote:

I found the right answer. But not sure if the procedure is right |a+b| = |a-b|

|a| + |b| = |a| -|b|

|b| = -|b| this is possible only with 0. So it is possible in only 2 cases. 1 and 0. But if it is 1, a+b neq to a-b. so b=0 and a*b = 0. Please let me know if this is a right approach.

Not sure if you can split the absolute value that way. For example:

a = -5 b = 3

|a+b| gets you 2 while |a| + |b| gets you 8.

That approach does get you the correct answer though so I may be incorrect

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

07 Dec 2012, 23:59

Can we plug in nos here? when we plug in random nos we realize that this can be equal only when a*b=0 _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

04 Jul 2013, 03:28

1

This post received KUDOS

SergeNew wrote:

If |a+b|=|a-b|, then a*b must be equal to:

A. 1 B. -1 C. 0 D. 2 E. -2

Think the equation without "a". We know that -b and +b are equal in magnitude. We are adding "a" to each. That still doesn't change the equality of the magnitude. That is possible only when 0 is added or "a" is 0. We can say the reverse also and say that to a , we add b and -b and the equality still holds. In this case b is 0. Either a or b is 0. So a*b must be equal to 0. _________________

Re: If |a+b|=|a-b|, then a*b must be equal to: [#permalink]

Show Tags

09 Jul 2013, 17:02

If |a+b|=|a-b|, then a*b must be equal to:

|a+b|=|a-b| (a+b)*(a+b) = (a-b)*(a-b) a^2+2ab+b^2 = a^2-2ab+b^2 4ab=0 In other words, a or b must = 0, therefore, the product of a*b is 0 regardless of what a or b are...one of them is 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...