Tricky Number Prop

[anticipation]

If x and y are positive integers and y = √(9 -x) , what is the value of y?
(1) x < 8
(2) y > 1

[shovan85]

If x and y are positive integers and y = √(9 -x) , what is the value of y?
(1) x < 8
(2) y > 1

IMO D

(1) x<8

x and y both positive integers

any integer below 8 and above 0 Only x = 5 satisfies the same. Thus y = 2. (Sufficient)

(2) y>1

x and y both positive integers

Now y^2 where y > 1 can be 4, 9, 16
but y = √(9 -x) => y^2 = 9-x

So only y^2 = 4 will satisfy for a positive integer x and thus y=2. (Sufficient)

[avigyan710]

If x and y are positive integers and y = √(9 -x) , what is the value of y?
(1) x < 8
(2) y > 1

Problem Stem=> Y =sqrt(9-x).

X and Y are positive integers. It means the value of X and Y cannot be -ve and fraction.

0 is neither positive nor negative integer.

sqrt of 0 is 0.

Stem1=> Only one value of X satisfies the condition. X=5. Y=3. SUFF.
Stem2=> Since Y is greater than 1 and it is not negative, X can only be 5. SUFF.

Answer is D.

What is the OA?