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# Is ab > 0?

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Manager
Joined: 10 Feb 2011
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15 Feb 2011, 13:09
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45% (medium)

Question Stats:

57% (02:05) correct 43% (00:41) wrong based on 58 sessions

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Is ab > 0?

(1) a – b > 0
(2) a + b < 0
[Reveal] Spoiler: OA
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Re: Is ab > 0? (1) a – b > 0. (2) a + b <0. [#permalink]

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15 Feb 2011, 13:39
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banksy wrote:
173. Is ab > 0?
(1) a – b > 0.
(2) a + b <0.

Note that:
You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Back to the original question:

Is ab > 0?

Question basically asks whether $$a$$ and $$b$$ have the same sign.

(1) a – b > 0 --> $$a>b$$. Not sufficient to say whether $$a$$ and $$b$$ have the same sign.

(2) a + b <0 --> $$a<-b$$. Again not sufficient to say whether $$a$$ and $$b$$ have the same sign.

(1)+(2) subtract (2) from (1): $$(a-b)-(a+b)>0$$ --> $$b<0$$ --> but $$a$$ could still be positive or negative (or even zero), for example: $$a=-1$$ and $$b=-2$$ or $$a=1$$ and $$b=-2$$. Not sufficient.

Or: as from (1) $$b<a$$ and from (2) $$a<-b$$ then $$b<a<-b$$ --> $$a<|b|$$ --(b)-----0-----(-b)-- --> $$a$$ is somewhere between $$b$$ and $${-b}$$ so it can be positive, negative or zero). Not sufficient.

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Re: Is ab > 0? (1) a – b > 0. (2) a + b <0. [#permalink]

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16 Feb 2011, 13:14
same trick as the other one you posted. good luck. thanks for posting.
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Re: Is ab > 0? (1) a – b > 0. (2) a + b <0. [#permalink]

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17 Feb 2011, 23:52
banksy wrote:
173. Is ab > 0?
(1) a – b > 0.
(2) a + b <0.

Try picking numbers and applying them considering different possible scenarios.

(1) a-b > 0
Case 1. a= -2 b= -5
a-b=-2-(-5)= 3. Condition satified.
So is ab>0? Yes.

Case 2. A= 5, B=(-3)
a-b= 5-(-3)= 8. Condition satisfied.
So is ab>0? No.

2 different answers. Hence this statement is insufficient.

(2) a+b<0

Case 1. a=- -10 , b= -5.
a+b=(-10-5)=-15. Condition Satisfied.
So is ab>0? Yes.

Case 2. a= -10 , b= 5
a+b= (-10+5)=-5. Condition satisfied.
So is ab>0?. No.

Again, 2 different answers. Hence this statement is also not sufficient.

1&2 Combined: Not sufficient as we can pick any combination of positive/ negative numbers.

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Re: Is ab > 0? [#permalink]

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23 Feb 2014, 05:11
Bumping for review and further discussion.
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Re: Is ab > 0? [#permalink]

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13 Mar 2016, 07:44
The basic rule => If any linear algebraic equation contains one algebraic symbol => we can alter that
so there is no way we convert the + sign or the - sign to the * sign
hence E
Also B must be always negative (from combining the inequalities)
and A>0 or A<0 so AB>0 or AB<O
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Re: Is ab > 0? [#permalink]

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22 Apr 2017, 09:28
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).

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Re: Is ab > 0?   [#permalink] 22 Apr 2017, 09:28
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