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

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If x and y are positive integers and y = square root (9-x) , [#permalink]

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11 Feb 2005, 15:05
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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
Joined: 01 Jan 2005
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11 Feb 2005, 15:21
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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11 Feb 2005, 15:22
HongHu , can you explain how you got D?...isn't 0 considered a positive integer?
Director
Joined: 07 Jun 2004
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11 Feb 2005, 15:26
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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11 Feb 2005, 15:29
Vijo 0 is not a Positve number its only a integer with no + / - sign with it
SVP
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11 Feb 2005, 16:43
0 is not positive, nor negative.

X<8 means it cannot be equal to 8.
Y>1 means it cannot be equal to 1.
Current Student
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11 Feb 2005, 20:35
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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11 Feb 2005, 23:37
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.
Manager
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15 Feb 2005, 19:35
one more for D

bytw, I think we just need to know that y is not 1 to find the solution
15 Feb 2005, 19:35
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