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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.

Bunuel, I have a a doubt: In statement (1), we have \(b^2 > a^2\), so this means that \(b>a\) or \(b<a\). Let's pick numbers: In the first scenario \((b>a)\): A. a=2, b=3, so \(\frac{b}{(b+a)} > 0\) B. a=-3, b=1, so \(\frac{b}{(b+a)} < 0\)

As you can see, different results. Insufficient. What I am missing? :s
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"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

Bunuel, I have a a doubt: In statement (1), we have \(b^2 > a^2\), so this means that \(b>a\) or \(b<a\). Let's pick numbers: In the first scenario \((b>a)\): A. a=2, b=3, so \(\frac{b}{(b+a)} > 0\) B. a=-3, b=1, so \(\frac{b}{(b+a)} < 0\)

As you can see, different results. Insufficient. What I am missing? :s

\(b^2 > a^2\) basically means that \(b\) is further from zero than \(a\): \(|b|>|a|\). So your second example is not valid.

Basically we can have following cases: ----------0--a--b-- -------a--0-----b-- ----b--a--0-------- ----b-----0--a-----

Bunuel, I have a a doubt: In statement (1), we have \(b^2 > a^2\), so this means that \(b>a\) or \(b<a\). Let's pick numbers: In the first scenario \((b>a)\): A. a=2, b=3, so \(\frac{b}{(b+a)} > 0\) B. a=-3, b=1, so \(\frac{b}{(b+a)} < 0\)

As you can see, different results. Insufficient. What I am missing? :s

\(b^2 > a^2\) basically means that \(b\) is further from zero than \(a\): \(|b|>|a|\). So your second example is not valid.

Basically we can have following cases: ----------0--a--b-- -------a--0-----b-- ----b--a--0-------- ----b-----0--a-----

For all these cases \(\frac{b}{a+b}>0\).

Hope it's clear.

Thanks buddy!, where did you study? You are a genius!
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

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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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