If x and y are positive integers and y =

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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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by Jay@ManhattanReview » Tue Jul 31, 2018 12:31 am

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BTGmoderatorDC 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 = (9 - x)^(1/2)
Question: what is the value of y?

We have y = (9 - x)^(1/2)

Squaring both the sides, we have y^2 = (9 - x)

You may plug in the possible values of x such that y is a positive integer.

Only two set of values would work: x = 5 and y = 2; x = 8 and y = 1

Let's take each statement one by one.

(1) x < 8

=> x = 5 and y = 2. Sufficient.

(2) y > 1

y = 2. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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