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If a#-b, is a-b/b+a>1? [#permalink]
 03 Oct 2009, 21:51
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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
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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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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]
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)} > 0B. a=-3, b=1, so \frac{b}{(b+a)} < 0As 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]
26 Jan 2012, 10:15
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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)} > 0B. a=-3, b=1, so \frac{b}{(b+a)} < 0As you can see, different results. Insufficient. What I am missing? :s 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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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. NEW!!!
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. NEW!!!
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Re: If a#-b, is a-b/b+a > 1? [#permalink]
26 Jan 2012, 10:33
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)} > 0B. a=-3, b=1, so \frac{b}{(b+a)} < 0As you can see, different results. Insufficient. What I am missing? :s 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]
26 Jan 2012, 10:33
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