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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] 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
[Reveal] Spoiler: OA
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Re: Inequalities [#permalink] New post 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.

Answer: A.
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Re: Inequalities [#permalink] 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] 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] New post 26 Jan 2012, 09:15
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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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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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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Re: If a#-b, is a-b/b+a > 1? [#permalink] New post 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!
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Re: If a#-b, is (a-b)/(b+a) > 1? [#permalink] New post 24 Jul 2014, 04:15
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Re: If a#-b, is (a-b)/(b+a) > 1?   [#permalink] 24 Jul 2014, 04:15
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