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If x and y are positive

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Pathik
 Post subject: If x and y are positive
 Post Posted: Thu May 08, 2008 12:34 am 
If x and y are positive, which of the following must be greater than 1/root(x+y)

I. root(x+y)/2x

II [root(x) + root(y)] / (x+y)

III [ root(x) - root(y) ] / (x+y)

A. none
B. I only
C. II only
D. I and III
E II and III

What is the fastest way to solve this.
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Captain
 Post subject:
 Post Posted: Thu May 08, 2008 11:01 am 
Quote:
What is the fastest way to solve this.

I dont know a fastest way. But this is how I thought -
First I note that 1/root(x+y) is symmetric wrt x,y
Code:
I. root(x+y)/2x

This is not symmetric with respect to x,y. So this cant be correct. What if I take a very small value of y; 1/root(x+y) goes to 1/root(x). but this equation goes to 1/(2root(x)). if x = 4 then 1/root(x+y) = 1/2, but I. becomes 1/4. Thus not always true.
Code:
III [ root(x) - root(y) ] / (x+y)

This one is symmetric. so I will use x = y; III. becomes 0, and 1/root(x+y) becomes 1/root(2x); III will be less atleast in one case.

Now
Code:
II [root(x) + root(y)] / (x+y)

This one is also symmetric. And putting x=y we will get II = 1/root(x) and the other as 1/ {root(2)*root(x)}, Thus II is greater.
At this point I declared II as the answer. To be sure I wrote both the expressions and proved II>1/root(x+y), for generic cases.
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RonPurewal
 Post subject:
 Post Posted: Sun May 11, 2008 1:55 am 
Offline
ManhattanGMAT Staff


Posts: 7146
http://www.manhattangmat.com/forums/post4838.html
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Guest
 Post subject:
 Post Posted: Mon Nov 17, 2008 5:33 pm 
Ron ,

I still feel lost with this Q..Stacey's explanation on the link you provided helped, but I don' get how to solve the problem. Please explain..
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RonPurewal
 Post subject:
 Post Posted: Sat Nov 29, 2008 7:57 am 
Offline
ManhattanGMAT Staff


Posts: 7146
Anonymous wrote:
Ron ,

I still feel lost with this Q..Stacey's explanation on the link you provided helped, but I don' get how to solve the problem. Please explain..


try this.
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