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Section - 24 Question - 13

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by camitava » Sat Sep 22, 2007 6:49 am
Guys I am having a problem to solve. I am not getting the approach to solve the problem. Do we do the calculation manually or we have some other technique to solve this kind of problem -

[spoiler]
OA is A[/spoiler]
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Amitava
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by kajcha » Sat Sep 22, 2007 6:01 pm
For any fraction x/y, greater the denominator smaller the fraction.

Now, evaluate all options.

In choice A you would notice that denominators are smaller than options B, C and E. So you can quickly eliminate them.

Now evaluate choice D. This can be written as (1/2)+(1/12) and option A can be written as (1/2)+(1/root2)+(1/root3)+(1/root5). Just by looking you can say option A is greater than E.
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by camitava » Mon Sep 24, 2007 12:37 am
Thanks Kacha! But dnt mind - not too much clear to me... :cry:
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by kajcha » Mon Sep 24, 2007 9:07 am
Choice A = 1/root(2)+1/root(3)+1/root(4)+1/root(5)

Choice B = (1/4)+(1/9)+(1/16)+(1/25)

4>root(2) so (1/4)<(1/root(2)) --------- (1)

Similarly,
(1/9)<(1/root(3)) ---------(2)
(1/16)<(1/root(4)) --------(3)
(1/25)<(1/root(5)) --------(4)

Sum up (1), (2), (3) and (4)

LHS < RHS => Choice A>Choice B

Similarly, you can check other options. Let me know if you want me to explain this more.
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by camitava » Tue Sep 25, 2007 12:59 am
Yah! Kajcha, I got ur point. But now look at the option D, It is not summation. It is like 1 - 1/2 + 1/3 - 1/4. Both summation and substraction appear together. If so, even sticking to ur point I can say D is less than A. Got my point, Kajcha?
Correct me If I am wrong


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by Preetics » Sun Oct 14, 2007 11:31 pm
Hi camitava,

Might be late with this response and u might have figured out how to do this by now. But my approach was this.
A can also be written as root2/2 +root3/3 +root4/4 + root5/5

we also know that root of any no goes 2 times with its whole no. (for eg root4 =2 and root4/4=1/2 ) so rt3/3=1/2 and same for other root. therefore sum of 4 halves=2 which is greater than all other options

Hope this helps and hope i dont sound confusing
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by camitava » Mon Oct 15, 2007 2:14 am
Hi Preetics! Thanks a lot for ur appraoch and yop, it is very easy to follow for me. Thanks once again ... :wink: :D
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