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# If x and y are positive, which of the following must be

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Manager
Joined: 19 Aug 2007
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If x and y are positive, which of the following must be [#permalink]  01 Jun 2008, 19:30
If x and y are positive, which of the following must be greater than 1/sqrt(x+y)?

I. sqrt(x+y) / 2x
II. sqrt(x) + sqrt(y) / (x+y)
III. sqrt(x) – sqrt(y) / (x+y)
Senior Manager
Joined: 23 May 2006
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Re: square roots [#permalink]  01 Jun 2008, 20:02
gmat blows wrote:
If x and y are positive, which of the following must be greater than 1/sqrt(x+y)?

I. sqrt(x+y) / 2x
II. sqrt(x) + sqrt(y) / (x+y)
III. sqrt(x) – sqrt(y) / (x+y)

I. sqrt(x+y) / 2x > 1/sqrt(x+y)
(x+y) > 2x
y > x Can this be true - Maybe

II. sqrt(x) + sqrt(y) / (x+y) > 1/sqrt(x+y)
sqrt(x) + sqrt(y) > sqrt(x+y) x sqrt(x+y) / sqrt(x+y)
sqrt(x) + sqrt(y) > sqrt(x+y)
This will always be true: sqrt2 + sqrt2 > sqrt(2+2); 1.4 + 1.4 > 2
This is definitely true for every case

III. sqrt(x) - sqrt(y) / (x+y) > 1/sqrt(x+y)
sqrt(x) - sqrt(y) > sqrt(x+y) x sqrt(x+y) / sqrt(x+y)
sqrt(x) - sqrt(y) > sqrt(x+y)
This will never be true: sqrt2 - sqrt2 > sqrt(2+2); Is 1.4 - 1.4 > 2
This case can never be true.

Hence the answer has to be II.
Manager
Joined: 28 May 2008
Posts: 98
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Re: square roots [#permalink]  01 Jun 2008, 22:46
x97agarwal wrote:
gmat blows wrote:
If x and y are positive, which of the following must be greater than 1/sqrt(x+y)?

I. sqrt(x+y) / 2x
II. sqrt(x) + sqrt(y) / (x+y)
III. sqrt(x) – sqrt(y) / (x+y)

I. sqrt(x+y) / 2x > 1/sqrt(x+y)
(x+y) > 2x
y > x Can this be true - Maybe

II. sqrt(x) + sqrt(y) / (x+y) > 1/sqrt(x+y)
sqrt(x) + sqrt(y) > sqrt(x+y) x sqrt(x+y) / sqrt(x+y)
sqrt(x) + sqrt(y) > sqrt(x+y)
This will always be true: sqrt2 + sqrt2 > sqrt(2+2); 1.4 + 1.4 > 2
This is definitely true for every case

III. sqrt(x) - sqrt(y) / (x+y) > 1/sqrt(x+y)
sqrt(x) - sqrt(y) > sqrt(x+y) x sqrt(x+y) / sqrt(x+y)
sqrt(x) - sqrt(y) > sqrt(x+y)
This will never be true: sqrt2 - sqrt2 > sqrt(2+2); Is 1.4 - 1.4 > 2
This case can never be true.
Hence the answer has to be II.

In my opinion , the best way to solve such questions is by substituting easy values.
Lets say x=5 y=4
If we solve it we'll see that option 2 clearly wins out
Manager
Joined: 19 Aug 2007
Posts: 206
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Re: square roots [#permalink]  02 Jun 2008, 09:33
zeenie wrote:
x97agarwal wrote:
gmat blows wrote:
If x and y are positive, which of the following must be greater than 1/sqrt(x+y)?

I. sqrt(x+y) / 2x
II. sqrt(x) + sqrt(y) / (x+y)
III. sqrt(x) – sqrt(y) / (x+y)

I. sqrt(x+y) / 2x > 1/sqrt(x+y)
(x+y) > 2x
y > x Can this be true - Maybe

II. sqrt(x) + sqrt(y) / (x+y) > 1/sqrt(x+y)
sqrt(x) + sqrt(y) > sqrt(x+y) x sqrt(x+y) / sqrt(x+y)
sqrt(x) + sqrt(y) > sqrt(x+y)
This will always be true: sqrt2 + sqrt2 > sqrt(2+2); 1.4 + 1.4 > 2
This is definitely true for every case

III. sqrt(x) - sqrt(y) / (x+y) > 1/sqrt(x+y)
sqrt(x) - sqrt(y) > sqrt(x+y) x sqrt(x+y) / sqrt(x+y)
sqrt(x) - sqrt(y) > sqrt(x+y)
This will never be true: sqrt2 - sqrt2 > sqrt(2+2); Is 1.4 - 1.4 > 2
This case can never be true.
Hence the answer has to be II.

In my opinion , the best way to solve such questions is by substituting easy values.
Lets say x=5 y=4
If we solve it we'll see that option 2 clearly wins out

thanks.
OA is II.

I just wanted to see what alternative ways there were in solving these kind of problems. I hate plug in chug - I always feel like I may miss something..
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Posts: 1593
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Re: square roots [#permalink]  02 Jun 2008, 09:36
i picked B based on picking numbers...
Re: square roots   [#permalink] 02 Jun 2008, 09:36
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