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12 Jun 2010, 05:40
Kudos
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If \(a,b,c,d\) are positive numbers, is \(\frac{a}{b}<\frac{(a+c)}{(b+d)}\)?
1. \(\frac{a}{b}<\frac{c}{d}\)
2. \(a+c<b+d\)
The OA is A, but, I have one doubt:
From S1, we can say \(\frac{a}{c}<\frac{b}{d}\)
On adding 1 on both sides, the equation becomes:
\(\frac{(a+c)}{c}<\frac{(b+d)}{d}\)
\(\frac{(a+c)}{(b+d)}<\frac{c}{d}\)
But, we are given \(\frac{a}{b}<\frac{c}{d}\)
So, S1 is insufficient. Please explain where I am missing the point.
1. \(\frac{a}{b}<\frac{c}{d}\)
2. \(a+c<b+d\)
The OA is A, but, I have one doubt:
From S1, we can say \(\frac{a}{c}<\frac{b}{d}\)
On adding 1 on both sides, the equation becomes:
\(\frac{(a+c)}{c}<\frac{(b+d)}{d}\)
\(\frac{(a+c)}{(b+d)}<\frac{c}{d}\)
But, we are given \(\frac{a}{b}<\frac{c}{d}\)
So, S1 is insufficient. Please explain where I am missing the point.
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12 Jun 2010, 08:47
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ykaiim
Given: \(a\), \(b\), \(c\), and \(d\) are positive numbers. Question: is \(\frac{a}{b}<\frac{(a+c)}{(b+d)}\)? --> as all numbers are positive we can safely crossmultiply --> is \(ab+ad<ab+bc\)? --> is \(ad<bc\)? is \(\frac{a}{b}<\frac{c}{d}\)?
(1) Directly gives the answer. Sufficient.
(2) Not sufficient.
Answer: A.
We can get the answer the way you were doing too: \(\frac{a}{b}<\frac{c}{d}\) --> \(\frac{d}{b}<\frac{c}{a}\) --> add 1 to both sides --> \(\frac{d}{b}+1<\frac{c}{a}+1\) --> \(\frac{d+b}{b}<\frac{c+a}{a}\) --> \(\frac{a}{b}<\frac{a+c}{b+d}\).
Hope it helps.
12 Jun 2010, 08:56
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Bunuel,
Thanks for the explainations. But, I am still not clear where in my explaination I am missing. Can you please check?
Thanks for the explainations. But, I am still not clear where in my explaination I am missing. Can you please check?
