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If [(a-b)/c] < 0, is a > b ? (1) c < 0 (2) a+b < 0

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If [(a-b)/c] < 0, is a > b ? (1) c < 0 (2) a+b < 0 [#permalink] New post 31 May 2010, 09:59
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Question Stats:

33% (01:24) correct 67% (00:52) wrong based on 2 sessions
Here is the question

If [(a-b)/c] < 0, is a > b ?

(1) c < 0
(2) a+b < 0

Correct answer is a.

I am having trouble with why isn't (2) also sufficient?
If a+b < 0, then couldn't we just solve it further so a < -b ??
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Re: Pos and Neg Question [#permalink] New post 31 May 2010, 10:29
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St. 2 is insufficient.

Plug-in numbers to check :

St. 2 : a + b < 0

Case 1 : a > b

a=1 b= -2, 1+ (-2) = -1 < 0

Case 2 : a < b

a=-2 b= 1, (-2) + 1 = -1 < 0

Hence, insufficient. Hope this helps !
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Re: Pos and Neg Question [#permalink] New post 31 May 2010, 10:40
Sorry I didn't follow. Do we really have to plug in numbers in a question like this. It seems to be a pos/neg inequality.
If a+b < 0, then naturally solving further would yield [ a < -b ], no? I might be really confused but I didn't follow how plugging in numbers helped.

Thank you.
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Re: Pos and Neg Question [#permalink] New post 31 May 2010, 10:50
It is not mentioned in the question whether a and b are +ve or -ve numbers.

For a < -b (or -a > b) , it is possible that a can be greater than b or less than b based on the 2 cases in my previous post.
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Re: Pos and Neg Question   [#permalink] 31 May 2010, 10:50
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