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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, 20:51
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If a#-b, is (a-b)/(b+a) > 1?

(1) b^2 > a^2
(2) a-b>1
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Joined: 02 Sep 2009
Posts: 50728

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04 Oct 2009, 06:45
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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, 18:43
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, 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
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Re: If a#-b, is a-b/b+a > 1?  [#permalink]

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26 Jan 2012, 09:15
1
1
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
Joined: 04 Oct 2009
Posts: 1194
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
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Re: If a#-b, is a-b/b+a > 1?  [#permalink]

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26 Jan 2012, 09:33
1
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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Joined: 02 Apr 2014
Posts: 471
GMAT 1: 700 Q50 V34
If a#-b, is (a-b)/(b+a) > 1?  [#permalink]

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11 Dec 2017, 12:43
$$(a-b)/(a+b) > 1$$ ?
=> $$((a-b)/(a+b)) - 1 > 0$$?
=> $$(-2b/(a+b)) > 0$$?
=> $$b/(a+b) < 0$$ ?
=> $$1 / (1 + (a/b)) < 0$$ ?, 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

If a#-b, is (a-b)/(b+a) > 1? &nbs [#permalink] 11 Dec 2017, 12:43
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