13.430 A passenger can travel from Moscow to city $N$ by train. In this case, he will be on the way for 20 hours. If, however, he waits for the departure of the plane (and he will have to wait more than 5 hours after the train departs), the passenger will reach city $N$ in 10 hours, including the waiting time. How many times faster is the speed of the plane compared to the speed of the train, if it is known that the plane will be over this train 8/9 hours after departure from the airport and will have flown as many kilometers by this moment as the train will have traveled?
Solution. Let $x$ and $y$ be the speeds (km/h) of the train and the airplane, respectively, and $z$ be the waiting time for the airplane to depart. Then $\frac{8}{9} y$ is the distance the airplane will fly until it meets the train. According to the problem, $\frac{8}{9} y = x \left(z + \frac{8}{9}\right)$. Additionally, $20 x = (10 - z) y' - $ is the distance to city $N$.

$$
\begin{gathered}
\text { If now } t = \frac{y}{x}, \text { then } \\
\left\{\begin{array}{l}
t = \frac{9}{8} \left(z + \frac{8}{9}\right), \\
t = \frac{20}{10 - z}, \\
z > 5.
\end{array} \Rightarrow \left\{\begin{array}{l}
z = \frac{8 t - 8}{9}, \\
z = \frac{10 t - 20}{t}, \\
\frac{10 t - 20}{t} > 5
\end{array} \Rightarrow \left\{\begin{array}{l}
4 t^{2} - 49 t + 9 = 0, \\
t > 4
\end{array} \Leftrightarrow t = 10 .\right.\right.\right.
\end{gathered}
$$

Answer: 10 times.