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If a#b, is (ab)/(b+a) > 1?
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03 Oct 2009, 20:51
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If a#b, is (ab)/(b+a) > 1? (1) b^2 > a^2 (2) ab>1
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Re: Inequalities
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04 Oct 2009, 06:45



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Re: Inequalities
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04 Oct 2009, 18:43
tkarthi4u wrote: Can some help me how to approach such questions?
If a#b, is ab/b+a > 1? 1) b^2 > a^2 2) ab>1 1) b^2 > a^2 i.e. lbl > a however b could be +ve or ve so does a. In either case, ab/b+a is always ve. So Suff.. 2) ab>1 or a > b. In this case: If a and b both are +ve, (ab)/(b+a) > 1. If a and b both are ve, (ab)/(b+a) < 1. NSF... So A is it.
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Re: If a#b, is ab/b+a > 1?
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26 Jan 2012, 08:32
tkarthi4u wrote: Can some help me how to approach such questions?
If a#b, is ab/b+a > 1?
1) b^2 > a^2 2) ab>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 ab/b+a > 1?
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26 Jan 2012, 09:15
metallicafan wrote: tkarthi4u wrote: Can some help me how to approach such questions?
If a#b, is ab/b+a > 1?
1) b^2 > a^2 2) ab>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: 0ab a0b ba0 b0a For all these cases \(\frac{b}{a+b}>0\). Hope it's clear.
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Re: If a#b, is ab/b+a > 1?
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26 Jan 2012, 09:33
Bunuel wrote: metallicafan wrote: tkarthi4u wrote: Can some help me how to approach such questions?
If a#b, is ab/b+a > 1?
1) b^2 > a^2 2) ab>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: 0ab a0b ba0 b0a 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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If a#b, is (ab)/(b+a) > 1?
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11 Dec 2017, 12:43
\((ab)/(a+b) > 1\) ? => \(((ab)/(a+b))  1 > 0\)? => \((2b/(a+b)) > 0\)? => \(b/(a+b) < 0\) ? => \(1 / (1 + (a/b)) < 0\) ?, 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)




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






