Data Sufficiency

Dear Ian\Stuart,

Please suggest the best and the quickest way to solve these type of questions.

Rgds,
Sumit

Lets pick numbers, since both are positive. Let x=16 and y=9 (i pick perfect squares to its easy to compute values)

you need the answer which is greater than 1/sqrt(x+y). i.e 1/sqrt(16+9)=1/5=>0.2

Now plug the above value of x and y and you will find:

1.sqrt(x+y)/2x=sqrt(16+9)/2*16=>5/32=>0.15 Not correct.

2.sqrt(16)+sqrt(y)/(x+y)=>sqrt(16)+sqrt(9)/(16+9)=>9/25=>0.36 Correct.

3.sqrt(16)-sqrt(y)/(x+y)=>sqrt(16)-sqrt(9)/(16+9) =>1/25 => 0.04 Not correct

Hence answer II only.

If you don't want to choose numbers then the best method is to make denominators of the fractions same and compare the numerators.
The fractions reduce down to following comparisons

Option1: Is 2x(sqrt(x+y)) > (x+y)(sqrt(x+y)) => is 2x>(x+y)
True only when x>y - can't say

Option2: Is (sqrt(x) + sqrt(y)) > (sqrt(x+y))
squaring both sides ( x+y+2sqrt(x)sqrt(y)) > (x+y) => yes true

Option3: Is (sqrt(x) - sqrt(y)) > (sqrt(x+y)) ; again squaring both sides
we get ( x+y-2sqrt(x)sqrt(y)) > (x+y) => No

My 2 cents.....pick numbers appropriate. Its the fastest way to solve these kind of questions. You don't want to be bogged down by time in the real exam.