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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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27 Dec 2010, 13:49
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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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Ajit

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Re: If x and y are positive integers [#permalink]

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27 Dec 2010, 14:08
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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

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.

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Re: If x and y are positive integers [#permalink]

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27 Dec 2010, 14:15
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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Ajit

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Re: If x and y are positive integers [#permalink]

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27 Dec 2010, 14:31
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 [#permalink]

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16 Feb 2011, 13:24
good question. thanks.
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Re: If x and y are positive integers [#permalink]

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17 Feb 2011, 00:57
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]

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03 Jul 2013, 01:18
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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03 Jul 2013, 10:05
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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
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Re: If x and y are positive integers and y = 9 - x, what is the [#permalink]

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21 Apr 2015, 03:30
Hello from the GMAT Club BumpBot!

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

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11 Sep 2017, 21:52
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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Re: If x and y are positive integers and y = 9 - x, what is the [#permalink]

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11 Sep 2017, 23:20
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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.

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Re: If x and y are positive integers and y = 9 - x, what is the   [#permalink] 11 Sep 2017, 23:20
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