Section - 24 Question - 13

[camitava]

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]

[kajcha]

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.

[camitava]

Thanks Kacha! But dnt mind - not too much clear to me...

[kajcha]

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.

[camitava]

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?

[Preetics]

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

[camitava]

Hi Preetics! Thanks a lot for ur appraoch and yop, it is very easy to follow for me. Thanks once again ...