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x, y are positive numbers. Is x>y? 1. x > sqrt(y) 2.

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x, y are positive numbers. Is x>y? 1. x > sqrt(y) 2. [#permalink] New post 16 May 2004, 12:50
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B
C
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x, y are positive numbers. Is x>y?
1. x > sqrt(y)
2. sqrt(x) > y
Senior Manager
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 [#permalink] New post 16 May 2004, 14:58
i am kind of confused here..
i guess it should be B?
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shubhangi

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 [#permalink] New post 16 May 2004, 15:22
The answer should be B.


Is x>y?

1. x > sqrt(y) - INSUFFICIENT
3 > sqrt(4) ----- x < y
2 > sqrt(1) ----- x > y

2. sqrt(x) > y - SUFFICIENT
Square both sides
x > y^2
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 [#permalink] New post 16 May 2004, 15:29
Regarding 2: x > y^2
x = 9, y = 2; x > y
x = .3, y = 1/2; x > y^2, however x < y
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 [#permalink] New post 16 May 2004, 15:49
You're right. I forgot about fractions.

I guess the answer should be C then.

We're looking for cases where x > y^2 and x^2 > y

Take for example x = 2/3 and y= 1/3. In cases where both conditions are satisfied, x is greater than y.
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Re: DS-112 [#permalink] New post 17 May 2004, 08:25
1. x^2 > y
2. x > y^2

combined: x^2 + x > y^2 + y
x^2 - y^2 + x - y > 0
(x-y)(x+y+1) > 0
as x and y are +ve, x+y+1 > 0
Therefore x-y > 0 or x > y.

C it is.
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 [#permalink] New post 17 May 2004, 08:28
Super.

That's an excellent explanation. I need to be able to approach these questions in the same way.
  [#permalink] 17 May 2004, 08:28
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x, y are positive numbers. Is x>y? 1. x > sqrt(y) 2.

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