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If x and y are positive integers and y = 9 - x, what is the

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If x and y are positive integers and y = 9 - x, what is the [#permalink] New post 27 Dec 2010, 12:49
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If x and y are positive integers and y=\sqrt{9 - x}, what is the value of y?

(1) x < 8
(2) y > 1
[Reveal] Spoiler: OA

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Re: If x and y are positive integers [#permalink] New post 27 Dec 2010, 13:08
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ajit257 wrote:
If x and y are positive integers and y = \sqrt{9 - x} , what is the value of y?
(1) x < 8
(2) y > 1

Not sure about the ans


Given: x and y are positive integers and y=\sqrt{9-x} --> y^2=9-x --> y^2 is a positive perfect square less than 9: so y^2=4=2^2 if x=5 or y^2=1=1^2 if x=8.

(1) x < 8 --> x=5 and y=2. Sufficient.
(2) y > 1 --> y=2. Sufficient.

Answer: D.
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Re: If x and y are positive integers [#permalink] New post 27 Dec 2010, 13:15
Hi Bunuel,
why is 2nd statement sufficient. why cant y be root 3 ..it is also greater than 1 or for that matter any integer greater than 1
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Re: If x and y are positive integers [#permalink] New post 27 Dec 2010, 13:31
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ajit257 wrote:
Hi Bunuel,
why is 2nd statement sufficient. why cant y be root 3 ..it is also greater than 1 or for that matter any integer greater than 1


You should read the stem carefully: "If x and y are positive integers..."

Also y^2=9-x --> as given that y is a positive integer then y^2 is a positive perfect square less than 9 (perfect square is a square of an integer): so y^2=4=2^2 if x=5 or y^2=1=1^2 if x=8.
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Re: If x and y are positive integers [#permalink] New post 16 Feb 2011, 12:24
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Re: If x and y are positive integers [#permalink] New post 16 Feb 2011, 23:57
Testing numbers can be an easy approach here. I solved it in less than 1 min. Answer is D.
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Re: If x and y are positive integers and y = 9 - x, what is the [#permalink] New post 03 Jul 2013, 00:18
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Re: If x and y are positive integers and y = 9 - x, what is the [#permalink] New post 03 Jul 2013, 09:05
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ajit257 wrote:
If x and y are positive integers and y=\sqrt{9 - x}, what is the value of y?

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


For the given expression we have x,y are positive integers and therefore squaring both sides

y^2= 9-x
From St 1 we have x< 8 -----> so we have possible value of in the range 0<x<8
Only for x=5, we have y=2

So options B,C and E ruled out

St 2 says y>1 so we have y^2=9-x
Now let us say y=2 so x=5
y=3, we get 9=9-x ----> x=0 which is not possible as "x" is a positive integer
hence y=2

Ans should be D

I messed it with choosing A as ans trying to beat the average time and not carefully solving the st2 :oops:
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Re: If x and y are positive integers and y = 9 - x, what is the   [#permalink] 03 Jul 2013, 09:05
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