$3 \cdot 40 A$ the time required to complete a job is $m$ times the time required for $B$ and $C$ to complete the job together; the time required for $B$ to complete the job is $n$ times the time required for $A$ and $C$ to complete the job together, and the time required for $C$ to complete the job is $x$ times the time required for $A$ and $B$ to complete the job together, then $x$ is
(A) $\frac{2 m n}{m+n}$.
(B) $\frac{1}{2(m+n)}$.
(C) $\frac{1}{m+n-m n}$.
(D) $\frac{1-m n}{m+n+2 m n}$.
(E) $\frac{m+n+2}{m n-1}$.
(16th American High School Mathematics Examination, 1965)
［Solution］Let $A, B, C$ complete the work alone in times $a, b, c$ respectively. From the given conditions, we have
$$
m \cdot \frac{1}{a}=\frac{1}{b}+\frac{1}{c}, \quad n \cdot \frac{1}{b}=\frac{1}{a}+\frac{1}{c}, \quad x \cdot \frac{1}{c}=\frac{1}{a}+\frac{1}{b} \text {. }
$$

From the first two equations, we have $\frac{m}{a}-\frac{n}{b}=\frac{1}{b}-\frac{1}{a}$, hence $\frac{a}{b}=\frac{m+1}{n+1}$.
At the same time, we have

which means
$$
\begin{array}{c}
\frac{m}{a}+\frac{n}{b}=\frac{1}{b}+\frac{1}{a}+\frac{2}{c}=\frac{1}{b}+\frac{1}{a}+\frac{2}{x}\left(-\frac{1}{a}+\frac{1}{b}\right), \\
\frac{1}{a}\left(m-1-\frac{2}{x}\right)=\frac{1}{b}\left(1-n+\frac{2}{x}\right),
\end{array}
$$

thus
$$
\frac{a}{b}=\frac{m-1-\frac{2}{x}}{1-n+\frac{2}{x}}=\frac{m+1}{n+1},
$$

leading to
$$
x=\frac{m+n+2}{m n-1} \text {. }
$$

Therefore, the answer is $(E)$.