Two positive integers $x$ and $y$ have $x y=24$ and $x-y=5$. The value of $x+y$ is
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
The positive integer divisors of 24 are $1,2,3,4,6,8,12,24$.

The pairs of divisors that give a product of 24 are $24 \times 1,12 \times 2,8 \times 3$, and $6 \times 4$.

We want to find two positive integers $x$ and $y$ whose product is 24 and whose difference is 5 .

Since $8 \times 3=24$ and $8-3=5$, then $x=8$ and $y=3$ are the required integers.

Here, $x+y=8+3=11$.

ANswer: (B)