Dear Ian\Stuart,
Please suggest the best and the quickest way to solve these type of questions.
Rgds,
Sumit
GMAT Prep - Quant 2
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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.
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.
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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
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
"Great works are performed not by strength but by perseverance."