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If x and y are positive integers and y = square root (9-x) ,

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Intern
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If x and y are positive integers and y = square root (9-x) , [#permalink] New post 11 Feb 2005, 14:05
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If x and y are positive integers and y = square root (9-x) , what is the value of y?

(1) x < 8
(2) y > 1
Manager
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 [#permalink] New post 11 Feb 2005, 14:21
A.


I) The only poss values are x=5 which makes y=+2 and -2 but since x is positive y=2. Sufficient.

II) The poss values are

x=5, y=2
x=8, y=1
Thus, Insufficient.
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 [#permalink] New post 11 Feb 2005, 14:22
HongHu , can you explain how you got D?...isn't 0 considered a positive integer?
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 [#permalink] New post 11 Feb 2005, 14:26
I have to go with D on this one as

in statement 1 y can be 9-5 = 4 only y = 2 possible suff

II) is suff as Y can be again only 2 as 9-9 = 0 and thats not positive integer so x has to be less than 9 but > 0
Director
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 [#permalink] New post 11 Feb 2005, 14:29
Vijo 0 is not a Positve number its only a integer with no + / - sign with it
SVP
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 [#permalink] New post 11 Feb 2005, 15:43
0 is not positive, nor negative.

X<8 means it cannot be equal to 8.
Y>1 means it cannot be equal to 1.
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 [#permalink] New post 11 Feb 2005, 19:35
D as well

since we have positive integers only....y = sqrt(perfect square)...since 0 is not a positive integer....only option is y=2

both statements are sufficient!
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Re: DS: sqrt(9-x) [#permalink] New post 11 Feb 2005, 22:37
maheshw wrote:
If x and y are positive integers and y = square root (9-x) , what is the value of y?

(1) x < 8
(2) y > 1


one more good question.
go with D.
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 [#permalink] New post 15 Feb 2005, 18:35
one more for D

bytw, I think we just need to know that y is not 1 to find the solution
  [#permalink] 15 Feb 2005, 18:35
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If x and y are positive integers and y = square root (9-x) ,

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