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

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If a#-b, is (a-b)/(b+a) > 1? [#permalink]

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03 Oct 2009, 21:51
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If a#-b, is (a-b)/(b+a) > 1?

(1) b^2 > a^2
(2) a-b>1
[Reveal] Spoiler: OA
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04 Oct 2009, 07:45
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Expert's post
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.

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04 Oct 2009, 19:43
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

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.
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Re: If a#-b, is a-b/b+a > 1? [#permalink]

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26 Jan 2012, 09:32
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
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Re: If a#-b, is a-b/b+a > 1? [#permalink]

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26 Jan 2012, 10:15
1
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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-----

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

Hope it's clear.
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Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
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Re: If a#-b, is a-b/b+a > 1? [#permalink]

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26 Jan 2012, 10:33
1
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!
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Re: If a#-b, is (a-b)/(b+a) > 1? [#permalink]

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24 Jul 2014, 05:15
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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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Re: If a#-b, is (a-b)/(b+a) > 1?   [#permalink] 24 Jul 2014, 05:15
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