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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
(1) x < 8
(2) y > 1
Given: x and y are positive integers and y = (9 - x)^(1/2)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
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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