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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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New post 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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Re: Inequalities  [#permalink]

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New post 04 Oct 2009, 06:45
2
2
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.

Answer: A.
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Re: Inequalities  [#permalink]

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New post 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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New post 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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New post 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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Retired Moderator
User avatar
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
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WE 2: Banking
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Re: If a#-b, is a-b/b+a > 1?  [#permalink]

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New post 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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New post 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

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