4. 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 12:00, meets the person on the bicycle at 14:00, and meets the person on the motorcycle at 16:00. The person on the motorcycle meets the person on the moped at 17:00 and catches up with the person on the bicycle at 18:00. 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
$$3 x=10\left(v_{3}+v_{4}\right)$$

That is
$$x=\frac{10\left(v_{3}+v_{4}\right)}{3}$$

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

That is
$$t-12=\frac{10}{3}$$

Thus
$$t=15 \frac{1}{3} .$$

Therefore, the bicycle rider meets the moped rider at 15:20.