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by ShruAj » Sun Aug 30, 2015 10:40 pm
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

The answer is - only (II)

But I could derive III as well by multiplying both the sides by a resulting in :
-ab > ab or 2ab <0 ; ab<0

Could anyone tell me where i went wrong?
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by MartyMurray » Mon Aug 31, 2015 3:49 am
ShruAj 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

The answer is - [spoiler]only (II)[/spoiler]

But I could derive III as well by multiplying both the sides by a resulting in :
-ab > ab or 2ab <0 ; ab<0

Could anyone tell me where i went wrong?
III says that ab < 0.

This means that neither a nor b can equal 0.

It also means that either a > 0 and b < 0 or a < 0 and b > 0.

a > 0 and b < 0 is obviously possible, as we can see by plugging in a = 2 and b = -2

a - b = 4 > 0 = a + b

However, III does not have to be the case. a and b could both be less than 0, and a - b > a + b, would still work.

Let's use a = -2 and b = -4

a - b = 2 > -6 = a + b

Where you went wrong is in multiplying an inequality by a variable without knowing whether the variable represents a positive number or a negative number.

If a is negative, then when you multiply both sides by a, the inequality becomes reversed, as inequalities always do when multiplied by negative numbers.

So in multiplying by a you get two possibilities -ab > ab or -ab < ab.
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by ShruAj » Mon Aug 31, 2015 3:56 am
Thank you, Marty!
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by GMATGuruNY » Mon Aug 31, 2015 3:57 am
ShruAj 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 - b > a + b
-b > b
0 > 2b
0 > b
b < 0.

Since a can be any integer value, only II must be true.
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by Max@Math Revolution » Wed Sep 02, 2015 9:36 pm
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


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-b>a+b, erasing a from both sides gives us -b>b, 0>b+b, 0>2b, 0>b therefore only (II) is the answer..

But I could derive III as well by multiplying both sides by a resulting in :
-ab > ab or 2ab <0 ; ab<0

Could anyone tell me where i went wrong?

==> since we don't know whether a is positive or negative, we can't assure that the inequality sign will stay the same if you multiply a to both sides.



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by [email protected] » Thu Sep 03, 2015 9:16 am
Hi ShruAj,

When dealing with PS questions, it's really important to include the answer choices - without those, you end up limiting how you can approach the question (and you limit the advice that you can receive). In Roman Numeral questions, the answers choices are often designed to offer you a 'shortcut' (and a way to avoid some of the 'work' involved).

Beyond the algebra that you can do, this question can also be solved by TESTing VALUES:

We're told that A-B > A+B and that A and B are INTEGERS. We're asked which of the following MUST be true.

IF...
A = 0
B = -1
0 - (-1) > 0 + (-1)
Eliminate Roman Numerals I and III.

Depending on how the answer choices were designed, we might have enough information right here to answer the question.

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