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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

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."