GMATPrep Q

[smehmood]

What is the solution to the attached problems?
Thanks
M

[sudhir3127]

What is the solution to the attached problems?
Thanks
M

its C

plug in the values and it will be cake walk...

Assume X=2 and Y =2

1. sqrt ( x+y)/2x
sqrt (4)/4 = 2/4 = 1/2........................................1

2. (rt x+ rt y)/x+y
(rt2 + rt 2)/4
2rt2/4 = rt2/2 > 1/2...................................................2


3. (sqrtx - rt y)/x+y = 0....................................3

hence C

.Hope it helps..

[kv_ajay]

Sudhir,

You got the correct answer but i think there are various opportunities to plug in numbers and too much work.

Explanation from my point would be
Root (X) + Root (Y) => Root (X+Y)
e.g. X= 4, Y = 16 then 2+4 > 4.47

so Root (X) + Root (Y)
---------------------
Root (X+Y) Root (X+Y)

So 1/Root(X+Y) is multiplied by integer > 1 so it will always be greater than 1/(Root X+Y)

[smehmood]

So, should we plug-in numbers in these type of questions?
I plug in numbers and I was getting both I and II greater then the number in question.

Ajay, you did not explain how you will compare x+y (denominator is II) with 2x (denominator in I)?

Thanks
M

[somail]

Plugging numbers in is okay, but you need to be carefull. The question says "must be greater than" meaning in all situations.

Lets say we pick x=4 and y=12

for I we get:

sqrt(16)/8 = 1/2 which is greater

but if we were to use x=12 and y=4 we get:

sqrt(16)/24 = 1/6 which is less.

So in this situation you can pick numbers, but just be extra careful to think about all possibilities.