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# If x and y are positive, is x^3 > y? (1) Sqrt(x) > y

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Joined: 22 Jul 2008
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If x and y are positive, is x^3 > y? (1) Sqrt(x) > y [#permalink]

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05 Aug 2008, 08:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x and y are positive, is x^3 > y?
(1) Sqrt(x) > y
(2) x > y

Kudos [?]: 16 [0], given: 0

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Joined: 07 Nov 2007
Posts: 1792

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Location: New York

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05 Aug 2008, 09:35
rnemani wrote:
If x and y are positive, is x^3 > y?
(1) Sqrt(x) > y
(2) x > y

(1) Sqrt(x) > y (not sufficient)
x=1/4 y=1/3 Sqrt(1/4) > 1/3
x^3 > y --> (1/2)^3 > (1/3) false

x=16 y=3 Sqrt(16) > 3
x^3 > y --> (16)^3 > (3) true

(2) x > y (sufficient)
x=1/2 y=1/3
x^3 > y --> (1/2)^3 > (1/3) false
x=16 y=3 x^3 > y --> (16)^3 > (3) true

Not sufficient

Combined not sufficient

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05 Aug 2008, 09:38
E.

We have to think of how positive fractions work with these inequalities.

1) $$\sqrt{x} > y$$. This also means that $$x > y^2$$ What if x = 1/4? Then $$sqrt{x} = \frac{1}{2}$$. Y could be 3/8 and so #1 would be true, but we can get multiple answers for the stem with this information, so It's insufficient. (Non fractions are easy to figure out that we get a different answer for the stem).

2) Again, x = 1/2, y = 3/8. x^3 is 1/8 and y = 3/8. Then if we use a positive integer, then the stem is true. 2 different answers = insufficient.

Together. Still don't know because nothing tells us that we cannot use fractions, or that x and Y must be integers. This would clear up the issue with fractions and make the two sufficient. Since this is not happening in the question, the answer is E.

rnemani wrote:
If x and y are positive, is x^3 > y?
(1) Sqrt(x) > y
(2) x > y

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Joined: 17 Jun 2008
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05 Aug 2008, 18:19
rnemani wrote:
If x and y are positive, is x^3 > y?
(1) Sqrt(x) > y
(2) x > y

here given x>0,y>0

(1) sgrt(x)>y => INSUFFI
say x=1/4 y=1/3 => sgrt(x)>y and x^3 <y
say x=16,y=3 => sgrt(x)>y and x^3 >y

x<1 and x>1 have different results.INSUFFI

(2) x>y and x>0 ,y>0 => INSUFFI
sine x<1 and x>1 have different results
say x=1/2 y=1/3 => x^3 <y
x=2,y=1 => x^3>y

combining both :

ay x=1/4 y=1/3 => sgrt(x)>y ,x>y and x^3 <y
say x=16,y=3 => sgrt(x)>y ,x>y and x^3 >y INSUFFI
IMO E
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Re: Inequality problem   [#permalink] 05 Aug 2008, 18:19
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