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03 Oct 2009, 21:51
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
43% (02:32) correct
57%
(02:36)
wrong
based on 146
sessions
History
Date
Time
Result
Not Attempted Yet
04 Oct 2009, 07:45
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If \(a\neq{-b}\), is \(\frac{a-b}{b+a}>1\)?
Is \(\frac{a-b}{b+a}>1\)? --> is \(\frac{-2b}{a+b}>0\) --> is \(\frac{b}{a+b}<0\)?
(1) b^2>a^2 --> \((b-a)(b+a)>0\), 2 cases:
A. \(b-a>0\) and \(b+a>0\) --> sum these two: \(b>0\). So \(b>0\) and \(b+a>0\) --> \(\frac{b}{a+b}>0\) --> answer to the question is NO;
B. \(b-a<0\) and \(b+a<0\) --> sum these two: \(b<0\). So \(b<0\) and \(b+a<0\) --> \(\frac{b}{a+b}>0\) --> answer to the question is NO;
So, in both cases answer to the question " is \(\frac{b}{a+b}<0\)?" is NO.
Sufficient.
(2) a-b>1 --> a>b. But b/(a+b) can be < or > 0 (plugging numbers a=3, b=1 and a=3 b=-1) so, not sufficient.
Answer: A.
Is \(\frac{a-b}{b+a}>1\)? --> is \(\frac{-2b}{a+b}>0\) --> is \(\frac{b}{a+b}<0\)?
(1) b^2>a^2 --> \((b-a)(b+a)>0\), 2 cases:
A. \(b-a>0\) and \(b+a>0\) --> sum these two: \(b>0\). So \(b>0\) and \(b+a>0\) --> \(\frac{b}{a+b}>0\) --> answer to the question is NO;
B. \(b-a<0\) and \(b+a<0\) --> sum these two: \(b<0\). So \(b<0\) and \(b+a<0\) --> \(\frac{b}{a+b}>0\) --> answer to the question is NO;
So, in both cases answer to the question " is \(\frac{b}{a+b}<0\)?" is NO.
Sufficient.
(2) a-b>1 --> a>b. But b/(a+b) can be < or > 0 (plugging numbers a=3, b=1 and a=3 b=-1) so, not sufficient.
Answer: A.
04 Oct 2009, 19:43
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tkarthi4u
1) b^2 > a^2
i.e. lbl > a however b could be +ve or -ve so does a.
In either case, a-b/b+a is always -ve. So Suff..
2) a-b>1 or a > b. In this case:
If a and b both are +ve, (a-b)/(b+a) > 1.
If a and b both are -ve, (a-b)/(b+a) < 1. NSF...
So A is it.
