\section*{Problem 4 - 041024}

The path from place A to place B is 11.5 km long and first goes uphill, then runs at the same level, and finally goes downhill. A pedestrian who walked from A to B covered this distance in 2 hours 54 minutes.

For the return trip on the same route, he needed \(3 \mathrm{~h} 6 \mathrm{~min}\). He walked uphill at a speed of \(3 \frac{\mathrm{km}}{\mathrm{h}}\), on the middle section at a speed of \(4 \frac{\mathrm{km}}{\mathrm{h}}\), and downhill at \(5 \frac{\mathrm{km}}{\mathrm{h}}\).

How long are the individual sections, assuming that on each section the respective speed was constant?
}

Let the measures of the three sections be \(a, b\), and \(a+x\).

Then we have:

\[
\begin{aligned}
2 a+b+x & =11.5 \\
\frac{a}{3}+\frac{b}{4}+\frac{a+x}{5} & =2.9 \\
\frac{a+x}{3}+\frac{b}{4}+\frac{a}{5} & =3.1
\end{aligned}
\]

From (2) and (3), we obtain \(x=1.5\), and from (1) \(2a+b=10\) and from (2) \(32a+15b=156\).

From these equations, we calculate \(b=4\) and \(c=3\).

The path leads \(3 \text{ km}\) uphill, then \(4 \text{ km}\) on a level, and finally \(4.5 \text{ km}\) downhill when going from A to B. The verification confirms the result.

Solutions of the second round 1964 adopted from [5]

\subsection*{7.6.3 Third Round 1964, Class 10}