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

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Manager
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If x and y are positive, is x^3 > y ? (1) x^1/2 > y [#permalink]

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New post 06 Sep 2008, 00:26
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If x and y are positive, is x^3 > y ?

(1) x^1/2 > y
(2) x > y

Kudos [?]: 104 [0], given: 0

SVP
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Re: Zumit DS 010 [#permalink]

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New post 06 Sep 2008, 00:50
dancinggeometry wrote:
If x and y are positive, is x^3 > y ?

(1) x^1/2 > y
(2) x > y


E. if x = 0.49 and y = 0.5, no.
if x = 1 and y = 0.5, yes.
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Re: Zumit DS 010 [#permalink]

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New post 06 Sep 2008, 00:50
dancinggeometry wrote:
If x and y are positive, is x^3 > y ?

(1) x^1/2 > y
(2) x > y



E

scenario 1: x = .49, y = .4, this would satisfy both statements, but x^3 < y

scenario 1: x = 49, y = 1, this would satisfy both statements, but x^3 > y

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VP
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Re: Zumit DS 010 [#permalink]

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New post 06 Sep 2008, 12:07
E is my pick as well.

Both cases by themselves and together hold for integers. Does not hold for fractions.

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Manager
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Re: Zumit DS 010 [#permalink]

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New post 06 Sep 2008, 23:43
OA is E. Bin 2 problem.

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Re: Zumit DS 010 [#permalink]

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New post 07 Sep 2008, 10:44
i think its best to analyse using the no. line.
we know taht when a no. is a fraction , then its cube and square are smaller than the no. itself.

below x and y are such that x>y and there is infinitesimally small gap bet'n the two.

stat 1) x^1/2>y

stat 2) x>y.

_________0___x2_____y_x__x1_1___
on taking the root, x jumps to x1 say, so x1>y , but x^3 shifts x to x2 and here x^3<y.

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Re: Zumit DS 010   [#permalink] 07 Sep 2008, 10:44
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If x and y are positive, is x^3 > y ? (1) x^1/2 > y

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