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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]

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New post 27 Dec 2010, 12:49
1
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A
B
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D
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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
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Re: If x and y are positive integers  [#permalink]

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New post 27 Dec 2010, 13:08
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3
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]

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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]

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New post 27 Dec 2010, 13:31
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 and y = 9 - x, what is the  [#permalink]

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Re: If x and y are positive integers and y = 9 - x, what is the  [#permalink]

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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]

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New post 11 Sep 2017, 22:20
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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



\(y=\sqrt{9 - x}\)

x and y are positive integers.
If x could be 0, y = 3
If x = 9, y = 0 (not allowed)

So y can be either 1 (x = 8) or 2 (x = 5)

(1) x < 8
So x cannot be 8. It must be 5 and y must be 2.
Sufficient alone.

(2) y > 1
So y cannot be 1. It must be 2.
Sufficient alone.

Answer (D)
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Re: If x and y are positive integers and y = 9 - x, what is the  [#permalink]

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New post 04 Jan 2019, 10:28
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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