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# If x,y,n are +ve integers, is (x/y)^n> 1000 ? 1. x=y^3, n

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Intern
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If x,y,n are +ve integers, is (x/y)^n> 1000 ? 1. x=y^3, n [#permalink]

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01 Jul 2006, 05:06
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If x,y,n are +ve integers, is (x/y)^n> 1000 ?

1. x=y^3, n >y
2. x > 5y, n >x.

I dont know OA

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Manager
Joined: 12 Apr 2006
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Location: India

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01 Jul 2006, 06:11
Its B

1. x=y^3, n >y

if y=2, x=8, n=3 eq. in question become (8/2)^3 which is not greater than 1000 but higher number make this true so insufficient.

2. x > 5y, n >x

if y=1, x=6, n=7 eq. in question become (6/1)^7 which is certainly greater than 1000. Sufficient to answer as this is true for all the other +ve integers also.

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Intern
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01 Jul 2006, 11:00
i agree ..the answer looks like B

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Director
Joined: 06 May 2006
Posts: 790

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01 Jul 2006, 13:18
#1. Not sufficient. try with y=1;n=4 and y=2;n=10
#2. Sufficient. (x/y)>5. n>x>=5...
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Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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Senior Manager
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02 Jul 2006, 05:04
Agree 'B' is sufficient, same reasoning with humans.

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Director
Joined: 07 Jun 2006
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02 Jul 2006, 07:54
gmatinjune wrote:
If x,y,n are +ve integers, is (x/y)^n> 1000 ?

1. x=y^3, n >y
2. x > 5y, n >x.

I dont know OA

let's take 'A', using x = y = 1 and n =2 - we can't say. INSUFF.

So that leaves us with BCE

let's take B with lowest integers
y = 1, x = 5 and n = 6 which gives us
5^6 which is > 1000. SUFF.

(B)

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Re: DS Question   [#permalink] 02 Jul 2006, 07:54
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# If x,y,n are +ve integers, is (x/y)^n> 1000 ? 1. x=y^3, n

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