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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] New post 01 Jul 2006, 04:06
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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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 [#permalink] New post 01 Jul 2006, 05:11
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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 [#permalink] New post 01 Jul 2006, 10:00
i agree ..the answer looks like B
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 [#permalink] New post 01 Jul 2006, 12:18
#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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 [#permalink] New post 02 Jul 2006, 04:04
Agree 'B' is sufficient, same reasoning with humans.
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Re: DS Question [#permalink] New post 02 Jul 2006, 06:54
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



Using AD-BCE elimination,

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, 06:54
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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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