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# m12#05 - DS

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Director
Joined: 25 Aug 2007
Posts: 936
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
m12#05 - DS [#permalink]

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12 Jun 2010, 05:40
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.
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Math Expert
Joined: 02 Sep 2009
Posts: 39593
Re: m12#05 - DS [#permalink]

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12 Jun 2010, 08:47
ykaiim wrote:
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.

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.

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.
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Director
Joined: 25 Aug 2007
Posts: 936
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Re: m12#05 - DS [#permalink]

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12 Jun 2010, 08:56
Bunuel,

Thanks for the explainations. But, I am still not clear where in my explaination I am missing. Can you please check?
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Joined: 02 Sep 2009
Posts: 39593
Re: m12#05 - DS [#permalink]

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12 Jun 2010, 09:21
ykaiim wrote:
Bunuel,

Thanks for the explainations. But, I am still not clear where in my explaination I am missing. Can you please check?

You are not missing anything.

If we are given that $$a$$, $$b$$, $$c$$, and $$d$$ are positive numbers and $$\frac{a}{b}<\frac{c}{d}$$, then $$\frac{(a+c)}{(b+d)}<\frac{c}{d}$$ (the inequality you derived) is true. We are just asked to derive another inequality (the one I derived) not this one.
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Re: m12#05 - DS   [#permalink] 12 Jun 2010, 09:21
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# m12#05 - DS

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