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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.

Re: If a#-b, is a-b/b+a > 1? [#permalink]
26 Jan 2012, 08:32

tkarthi4u wrote:

Can some help me how to approach such questions?

If a#-b, is a-b/b+a > 1?

1) b^2 > a^2 2) a-b>1

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 _________________

"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."

Re: If a#-b, is a-b/b+a > 1? [#permalink]
26 Jan 2012, 09:15

1

This post received KUDOS

Expert's post

metallicafan wrote:

tkarthi4u wrote:

Can some help me how to approach such questions?

If a#-b, is a-b/b+a > 1?

1) b^2 > a^2 2) a-b>1

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-----

Re: If a#-b, is a-b/b+a > 1? [#permalink]
26 Jan 2012, 09:33

1

This post received KUDOS

Bunuel wrote:

metallicafan wrote:

tkarthi4u wrote:

Can some help me how to approach such questions?

If a#-b, is a-b/b+a > 1?

1) b^2 > a^2 2) a-b>1

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."

Re: If a#-b, is (a-b)/(b+a) > 1? [#permalink]
24 Jul 2014, 04:15

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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