$4 \cdot 210$ On the same route, there are four people: the first person is in a car, the second person is on a motorcycle, the third person is on a moped, and the fourth person is on a bicycle. The speeds of the vehicles are constant. The person in the car catches up with the person on the moped at 12 o'clock, meets the person on the bicycle at 14 o'clock, and meets the person on the motorcycle at 16 o'clock. The person on the motorcycle meets the person on the moped at 17 o'clock and catches up with the person on the bicycle at 18 o'clock. When did the person on the bicycle meet the person on the moped?
[Solution] Let the speeds of the car, motorcycle, moped, and bicycle be $v_{1}, v_{2}, v_{3}, v_{4}$, respectively, and let $x, y$ represent the distances between the bicycle rider and the car passenger, and the bicycle rider and the motorcycle driver at 12 o'clock. According to the problem, we have
$$
\begin{array}{l}
x=2\left(v_{1}+v_{4}\right), \\
x+y=4\left(v_{1}+v_{2}\right), \\
x+y=5\left(v_{2}+v_{3}\right), \\
y=6\left(v_{2}-v_{4}\right) .
\end{array}
$$
((1) + (3) $) \times 2-$ (2) + (4) ), we get
$$
3 x=10\left(v_{3}+v_{4}\right) \text {, }
$$

which means
$$
x=\frac{10\left(v_{3}+v_{4}\right)}{3} .
$$

Suppose the bicycle rider meets the moped rider at time $t$, then
$$
x=(t-12)\left(v_{3}+v_{4}\right),
$$

which means
$$
t-12=\frac{10}{3} \text {, }
$$

thus $\quad t=15 \frac{1}{3}$.
Therefore, the bicycle rider meets the moped rider at 15:20.