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

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If a b > a + b, where a and b are integers, which of the [#permalink]  30 Jan 2011, 21:18
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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

though the OA is revealed, but I am not conviced with the answer
[Reveal] Spoiler: OA
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Re: AB positive or negitive [#permalink]  30 Jan 2011, 21:28
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

though the OA is revealed, but I am not conviced with the answer

Simplify the inequality:
a – b > a + b
b<0

We know for certain that II is true.

We don't have info about the value of a. Hence, I cannot be derived.

For III ab can be positive or negative depending on the value of a. Since we don't know about a, III cannot be inferred.

Hence, B is the answer
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Re: AB positive or negitive [#permalink]  31 Jan 2011, 00:07
well agreed.....however, if we multiply both side of the equation by "a", we get to know that III is also true.....which follows that I is also true?
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Re: AB positive or negitive [#permalink]  31 Jan 2011, 01:17
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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
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Re: AB positive or negitive [#permalink]  02 Feb 2011, 05:02
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.

Ans: B
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Re: AB positive or negitive [#permalink]  02 Feb 2011, 05:33
Expert's post
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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.

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Re: If a b > a + b, where a and b are integers, which of the [#permalink]  15 Oct 2014, 15:22
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Re: If a b > a + b, where a and b are integers, which of the   [#permalink] 15 Oct 2014, 15:22
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# If a b > a + b, where a and b are integers, which of the

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