13.235. Two cars and a motorcycle participated in a race over the same distance. The second car took 1 minute longer to complete the entire distance than the first car. The first car moved 4 times faster than the motorcycle. What part of the distance did the second car cover in one minute, if it covered $1 / 6$ of the distance more per minute than the motorcycle, and the motorcycle took less than 10 minutes to cover the distance?
Solution.

Let $l$ be the length of the distance, $v_{1}, v_{2}, v_{x}$ be the speeds of the first, second car, and the motorcycle, respectively. According to the problem, we have the system

$\left\{\begin{array}{l}\frac{l}{v_{2}}-\frac{l}{v_{1}}=\frac{1}{60}, \\ v_{1}=4 v_{\mathrm{M}}, \\ \frac{v_{2}}{60}-\frac{v_{\mathrm{M}}}{60}=\frac{l}{6},\end{array}\right.$

from which $v_{1}=120 l, v_{2}=40 l, v_{\mathrm{M}}=30 l$ or $v_{1}=20 l, v_{2}=15 l, v_{\mathrm{M}}=5 l$.

Since $\frac{l}{v_{\mathrm{M}}}<\frac{10}{60}$, the value $v_{\mathrm{M}}=5 l$ does not fit, so $v_{2}=40 l$. Then the second car covers $\frac{40 l}{60 l}=\frac{2}{3}$ of the distance in 1 minute.

Answer: $\frac{2}{3}$.