$3 \cdot 32$ Three cars $A, B, C$ each travel at a certain speed from location A to location B. $B$ starts 5 minutes later than $C$, and catches up with $C$ 20 minutes after starting. If $A$ starts 10 minutes later than $B$, and catches up with $C$ 50 minutes after starting. Then the time it takes for $A$ to catch up with $B$ after starting is
(A) 100 min.
(B) 200 min.
(C) 250 min.
(D) 50 min.
(2nd "Jinyun Cup" Junior High School Mathematics Invitational, 1985)
［Solution］Let the speeds of $A, B, C$ be $x, y, z$ respectively. According to the problem, we have 
\[
\left\{
\begin{array}{l}
\frac{5 z}{y-z}=20, \\ 
\frac{(10+5) z}{x-z}=50 .
\end{array}
\right.
\]
which simplifies to 
\[
\left\{
\begin{array}{l}
4 y=5 z, \\ 
10 x=13 z .
\end{array}
\right.
\]

Thus,
\[
\left\{
\begin{array}{l}
y=\frac{5}{4} z \\
x=\frac{13}{10} z
\end{array}
\right.
\]

The time required for $A$ to catch up with $B$ is
\[
\frac{10 y}{x-y}=\frac{10 \cdot \frac{5}{4} z}{\frac{13}{10} z-\frac{5}{4} z}=250 \text { (minutes). }
\]

Therefore, the answer is $(C)$.