A3. For natural numbers $m$ and $n$, it holds that $19 \leq m \leq 49, 51 \leq n \leq 101$. What is the maximum value that the expression $\frac{n+m}{n-m}$ can take?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60
A3. Since $\frac{n+m}{n-m}=\frac{n-m+2 m}{n-m}=1+2 \frac{m}{n-m}$, the value will be greatest when $n-m=2$ and $m=49$. The value will then be 50.