If a#-b, is (a-b)/(b+a) > 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!

\((a-b)/(a+b) > 1\) ?
=> \(((a-b)/(a+b)) - 1 > 0\)?
=> \((-2b/(a+b)) > 0\)?
=> \(b/(a+b) < 0\) ?
=> \(1 / (1 + (a/b)) < 0\) ?, for this term to be less than zero, denominator must be < 0
=> \(1 + (a/b) < 0\) ?
=> \(a/b < -1\) ? This is the simplified form of the question.

Statement 1 : \(b^2 > a^2\)
Since both terms are positive, dividing by \(b^2\) => 1 > \(a^2/b^2\)
=> Taking sq root, \(|a/b| < 1\)
=> \(-1 < a/b < 1\) => Sufficient to answer the question, is \(a/b < - 1\)?

Statement 2: \(a - b > 1\),
best way to attack by plugging in numbers, if a = -4, b = - 6, then a/b > -1
if a = 3, b = -2, then a/b < -1 => Insufficient

Answer (A)

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.