$3 \cdot 42$ A certain project, the number of days required for Team A to complete it alone is $a$ times the number of days required for Teams B and C to complete it together; the number of days required for Team B to complete it alone is $b$ times the number of days required for Teams A and C to complete it together; the number of days required for Team C to complete it alone is $c$ times the number of days required for Teams A and B to complete it together. Then the value of $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1990)
[Solution] Let the time required for teams A, B, and C to complete the project be $x, y, z$ days respectively. According to the problem, we have 
\[
\left\{
\begin{array}{l}
\frac{a}{x}=\frac{1}{y}+\frac{1}{z}, \\
\frac{b}{y}=\frac{1}{x}+\frac{1}{z}, \\
\frac{c}{z}=\frac{1}{x}+\frac{1}{y} .
\end{array}
\quad
\left\{
\begin{array}{l}
a+1=x\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right), \\
b+1=y\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right), \\
c+1=z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) .
\end{array}
\right.
\right.
\]
Let $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=k$, then from the above equations, we have
\[
\begin{aligned}
& \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \\
= & \frac{1}{k x}+\frac{1}{k y}+\frac{1}{k z} \\
= & \frac{1}{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\
= & \frac{1}{k} \cdot k=1 .
\end{aligned}
\]

Therefore, the answer is $(A)$.