A pedestrian walked along a bus route where buses depart from the terminals every 5 minutes. He counted that 11 buses passed him and 13 came towards him. For how long did he walk at least and at most, if the speed of the bus is 8 times that of the pedestrian?
First, let's estimate the walking time of the pedestrian, given that 13 buses passed by. Calculate how long the pedestrian walked, if they encountered the 1st bus right at the start and finished walking when the 13th bus passed by. Considering the distance covered by a bus in 1 minute as 1 unit, the distance between the pedestrian and the 13th bus at the start is $12 \cdot 5 = 60$ units. If the pedestrian walked for $x$ minutes, then they covered $\frac{x}{8}$ units, and the bus covered $x$ units:

$$
\frac{x}{8} + x = 60, \quad x = 53 \frac{1}{3} \text{ minutes. }
$$

If the pedestrian started walking when the first bus was 5 minutes away and continued walking until just before encountering the 14th bus, then by similar reasoning we get:

$$
\frac{y}{8} + y = 70, \quad y = 62 \frac{2}{9} \text{ minutes. }
$$

In the first case, therefore, we have the estimate for the pedestrian's walking time $t$ as:

$$
53 \frac{1}{3} \leq t \leq 62 \frac{2}{9}
$$

Now, let's repeat the above reasoning, taking into account that 11 buses passed by.

The shortest distance is obtained if the pedestrian starts walking with the 1st bus and stops when the 11th bus catches up.

In this case, they covered $10 \cdot 5 = 50$ units, and if we denote the time by $x$:

$$
x - \frac{x}{8} = 50 \quad x = 57 \frac{1}{7} \text{ minutes }
$$

For the longest walking time, we get it if the pedestrian starts walking when the 1st bus is 5 minutes behind and continues walking until the 12th bus just overtakes them. In this case:

$$
y - \frac{y}{8} = 60 \quad y = 68 \frac{4}{7} \text{ minutes }
$$

In the second case, therefore, we have the estimate for the pedestrian's walking time $t$ as:

$$
57 \frac{1}{7} \leq t \leq 68 \frac{4}{7}
$$

Considering both conditions simultaneously, we can conclude that the pedestrian walked for at least $57 \frac{1}{7}$ and at most $62 \frac{2}{9}$ minutes.

Simon Károly (Kazincbarcika, Ságvári E. Gymn., I. o. t.)