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:

Unless you know whether a is positive or not you cannot determine how the inequality will look like after you multiply it by a. You cannot still prove III to be true.

Lets consider the two cases here: Suppose a is positive (3, say) 3-b>3+b Multiply the inequality by 3 9-3b>9+3b 6b<0 b<0

hence, ab<0

Now, suppose a=-3 plug in the values:

-3-b>-3+b Multiply by a(-3) 9+3b<9-3b 6b<0 b<0

hence, ab > 0

Thus the sign of ab will vary depending on the sign of a. Hope this clears up the confusion

Q: If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

Through inequality; I just derived that b is -ve a – b > a + b -b > b --> subtracting 'a' from both sides -2b > 0 --> subtracting 'b' from both sides -b > 0 --> dividing both sides by 2 b < 0 --> multiplying -ve sign on both sides

The rest two; I ruled out using numbers

Let b, as we know is -ve, to be equal to -1

Case I a=+ve, say 100 a - b = 100 – (-1) = 101 a + b = 100 + (-1) = 99

a-b > a+b

Case II a=-ve, say -100 a - b = -100 – (-1) = -100 + 1 = -99 a + b = -100 + (-1) = -100 - 1 = -101

So, a - b > a + b

Thus, a - b > a + b is true for both +ve and -ve 'a'

We just proved that a-b>a+b is true for both +ve and -ve values of a.

We can't conclusively say that a < 0. Statement I is ruled out.

for a=+ve; ab = +ve * -ve = -ve and for a=+ve; ab = -ve * -ve = +ve

So, we can't conclusively say ab < 0. Statement III is ruled out.

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A I only B II only C I and II D I and III E II and III

You don't need number plugging at all.

Given: \(a-b>a+b\) --> \(a\) cancels out (which means that from given info we can say nothing about it: it can be positive, negative or zero, so it can be ANY number) --> \(2b<0\) --> \(b<0\).

As we know nothing about \(a\) and know that \(b<0\), so only II must be true.

Re: If a b > a + b, where a and b are integers, which of the [#permalink]

Show Tags

15 Oct 2014, 15:22

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.
_________________

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A. I only B. II only C. I and II only D. I and III only E. II and III only

Merging topics.

Please refer to the discussion above.

can we not sqaure both sides of a – b > a + b ?

1. No. 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). Check for more here: inequalities-tips-and-hints-175001.html

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A. I only B. II only C. I and II only D. I and III only E. II and III only

Merging topics.

Please refer to the discussion above.

can we not sqaure both sides of a – b > a + b ?

You are only complicating the simple expression given by squaring it.

Also, as a general rule of inequalities, squaring variables in the inequalities is not a good idea until you know for sure the sign of a-b as multiplication by a negative number reverses the inequality.
_________________

If a b > a + b, where a and b are integers, which of the [#permalink]

Show Tags

19 Feb 2016, 21:40

Bunuel wrote:

girishkakkar wrote:

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A I only B II only C I and II D I and III E II and III

You don't need number plugging at all.

Given: \(a-b>a+b\) --> \(a\) cancels out (which means that from given info we can say nothing about it: it can be positive, negative or zero, so it can be ANY number) --> \(2b<0\) --> \(b<0\).

As we know nothing about \(a\) and know that \(b<0\), so only II must be true.

Answer: B.

Hi Bunuel,

I cannot understand, how could we cancel 'a' here when we do not know about its sign? Please explain this.

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A I only B II only C I and II D I and III E II and III

You don't need number plugging at all.

Given: \(a-b>a+b\) --> \(a\) cancels out (which means that from given info we can say nothing about it: it can be positive, negative or zero, so it can be ANY number) --> \(2b<0\) --> \(b<0\).

As we know nothing about \(a\) and know that \(b<0\), so only II must be true.

Answer: B.

Hi Bunuel,

I cannot understand, how could we cancel 'a' here when we do not know about its sign? Please explain this.

Thanks.

There is a difference between reducing (dividing/multiplying) an inequality by a variable and subtracting/adding a variable to both parts of an inequality. The first one we cannot do unless we know the sign of the variable but subtracting/adding is always applicable.

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

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...