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15 Feb 2011, 13:09
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
65% (01:54) correct
35%
(01:23)
wrong
based on 191
sessions
History
Date
Time
Result
Not Attempted Yet
15 Feb 2011, 13:39
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banksy
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.
Answer: E.
General Discussion
17 Feb 2011, 23:52
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banksy
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.
Hence, the answer is E.
