The route from $B$ to $C$ starts with a $3 \mathrm{~km}$ ascent, then proceeds horizontally for $5 \mathrm{~km}$, and finally reaches its destination via a $6 \mathrm{~km}$ descent, which has the same gradient as the first section. A traveler set out from $B$ and, turning back halfway, returned to $B$ after 3 hours and 36 minutes. After a break, they reached $C$ in 3 hours and 27 minutes, and finally completed the return trip in 3 hours and 51 minutes. At what speed did they travel on the flat, uphill, and downhill sections?
I. Solution: In the first trip, the traveler also covered a distance equal to $BC$, that is, $14 \mathrm{~km}$, since they returned to $B$ from halfway. Thus, this trip can be easily compared with the other two. The traveler covered $3-3 \mathrm{~km}$ uphill and downhill, and $8 \mathrm{~km}$ horizontally in the first trip. This only differs from the second trip in that a $3 \mathrm{~km}$ downhill segment is replaced by $3 \mathrm{~km}$ of horizontal travel, and therefore the travel time is 9 minutes longer. Similarly, in the first trip compared to the third, a $3 \mathrm{~km}$ uphill segment is replaced by $3 \mathrm{~km}$ of horizontal travel, and the travel time is 15 minutes shorter.

Accordingly, if we were to replace the remaining uphill and downhill segments of the first trip with equally long horizontal segments, the travel time would be shortened by 15 minutes on one hand and extended by 9 minutes on the other, making it 3 hours 30 minutes = 210 minutes, and the traveler would move at a constant speed throughout. Thus, the traveler needs $210: 14=$ 15 minutes to cover $1 \mathrm{~km}$ horizontally, or in other words: the speed is $60: 15=4 \mathrm{~km}$ per hour.

Now, since according to our comparisons, the travel time for $1 \mathrm{~km}$ of uphill is $15: 3=5$ minutes longer, and for $1 \mathrm{~km}$ of downhill it is $9: 3=3$ minutes shorter than for $1 \mathrm{~km}$ horizontally, it takes $15+5=20$ minutes to cover $1 \mathrm{~km}$ uphill, and $15-3=12$ minutes to cover $1 \mathrm{~km}$ downhill. Therefore, the traveler's speed uphill is $60: 20=3 \mathrm{~km}$ per hour, and downhill it is $60: 12=5 \mathrm{~km}$ per hour.

Porpáczy Erzsébet (Jászberény, Kállai Éva Secondary School II. grade)

II. Solution: Let's denote the speed of the traveler on uphill, horizontal, and downhill segments, measured in km/min, as $x, y$, and $z$ respectively. Then the traveler's travel times for the consecutive $3, 8, 3 \mathrm{~km}$ segments in the first trip are $3 / x, 8 / y, 3 / z$ minutes, and their sum is 3 hours 36 minutes = 216 minutes. Thus, using the other two trips as well, we get the system of equations:

$$
\begin{aligned}
& \frac{3}{x}+\frac{8}{y}+\frac{3}{z}=216 \\
& \frac{3}{x}+\frac{5}{y}+\frac{6}{z}=207 \\
& \frac{6}{x}+\frac{5}{y}+\frac{3}{z}=231
\end{aligned}
$$

This system becomes

$$
3 p+8 q+3 r=216, \quad 3 p+5 q+6 r=207, \quad 6 p+5 q+3 r=231
$$

by introducing new variables $1 / x=p, 1 / y=q, 1 / z=r$. The coefficients of $p$ and $r$ in the last two equations are the same as in the first, so subtracting the sum of these two from three times the first equation, $p$ and $r$ cancel out: $24 q-(5 q+5 q)=3 \cdot 216-(207+231)$, from which $q=15$. Then, from the first and second equations, after simplification,

$$
p+r=32, \quad p+2 r=44
$$

from which $r=12$ and $p=20$. Finally, returning to the original variables, $x=1 / 20, y=1 / 15, z=1 / 12 \mathrm{~km}$ per minute, which only differs in form from the result of the first solution.

Kovács Ferenc (Ózd, József A. Secondary School II. grade)

Note. The clarity and simple relationships of the numerical data, as well as the possibility of mental calculation, made the first solution quick, - however, the method is also applicable with multi-digit data and non-$1: 2$ ratio downhill lengths. It is instructive to compare the "substitutions" of the first solution with the elimination in the second.