3-rd 5. From point \(A\) to other points, one can get there in two ways:

1. Exit immediately and walk.

2. Call a car and, after waiting for a certain amount of time, ride in it.

In each case, the mode of transportation that requires the least time is used. It turns out that

\begin{tabular}{|c|c|}
\hline if the final point is & then the travel time will be \\
\hline \(1 \mathrm{km}\) & 10 min \\
\(2 \mathrm{km}\) & 15 min \\
\(3 \mathrm{km}\) & \(17 \frac{1}{2}\) min \\
\hline
\end{tabular}

The speeds of the pedestrian and the car, as well as the waiting time for the car, are assumed to be constant. How much time will be required to reach a point that is 6 km away from \(A\)?
Solution 5. Answer: 25 min.

Let \(v\) be the pedestrian's speed, \(V\) be the car's speed, and \(T\) be the waiting time for the car. (We will measure speed in km/h and time in hours.) If walking, the time required for 1.2 km and 3 km would be \(1 / v\), \(2 / v\), and \(3 / v\) hours, respectively, and if traveling by car, it would be \(T+\frac{1}{V}\), \(T+\frac{2}{V}\), and \(T+\frac{3}{V}\) hours. Suppose that walking 2 km is no slower than traveling by car, i.e., \(T+\frac{2}{V} \geqslant \frac{2}{v}\). Then \(T+\frac{1}{V} \geqslant \frac{T}{2}+\frac{1}{V} \geqslant \frac{1}{v}\), meaning that 1 km can also be walked. However, if 1 km and 2 km can be walked, then the time spent on 2 km would be exactly twice that of 1 km, which is not the case according to the problem. Therefore, \(T+\frac{2}{V} \leqslant \frac{2}{v}\), and thus \(T+\frac{3}{V} \leqslant \frac{3}{2}\left(T+\frac{2}{V}\right) \leqslant \frac{3}{v}\). Consequently, \(1 / v=1 / 6\), \(T+2 / V=1 / 4\), and \(T+3 / V=7 / 24\). Solving this system of equations, we find: \(v=6\), \(T=1 / 6\), and \(V=24\). If walking 6 km, it would take 1 hour, and if traveling by car, it would take \(\frac{1}{6}+\frac{6}{24}=\frac{5}{12}\), i.e., 25 min.

\section*{9th - 10th Grades}