From point $A$, a passenger train travels towards $B$ at a speed of $30 \mathrm{~km}$ per hour. A fast train, also departing from $A$ but later, travels at a speed of $60 \mathrm{~km}$ per hour and catches up with the passenger train in $B$ under normal conditions. The passenger train, after covering ${ }^{2} / 3$ of its journey, is forced to continue at half its original speed due to a technical failure, which results in the fast train catching up $27^{1} / 9 \mathrm{~km}$ before $B$. What is the distance between $A$ and $B$?
I. Solution: Since the fast one, traveling at twice the speed, will catch up with the person at the end of the path, the person is at the halfway point of the path when the fast one starts and when the person has covered $2 / 3$ of the path, the fast one has covered $1 / 3$ of the path. From this point on, the speed of the fast one is four times the speed of the person. Considering the distance $A B$ as a unit and denoting the distance traveled by the person, which is covered due to a technical error until the moment of catching up, by $x$:

$$
\frac{2}{3}+x=\frac{1}{3}+4 x
$$

which means

$$
3 x=\frac{1}{3}
$$

from which

$$
x=\frac{1}{9}
$$

Since at the moment the technical error occurred, the person had only $\frac{1}{3}$ of the path left to travel, by the time of catching up, only $\frac{1}{3}-\frac{1}{9}=\frac{2}{9}$ of the path remains to $B$, and according to the problem, this is $271 / 9=\frac{244}{9} \text{ km}$. Therefore, $A B=\frac{244}{9} \cdot \frac{9}{2}=122 \text{ km}$.

Ferenc Beleznay (Bp. V., Piarist g. I. o. t.)

Note: We see, therefore, that for the solution, we do not need the actual speeds, knowing the ratio of the speeds is sufficient.

II. Solution: Let the distance $A B$ be $x \text{ km}$. Using the formula time $=\frac{\text{distance}}{\text{speed}}$, we set up the equation. The time for the fast one from the start until the catch-up, according to the problem, is $\frac{x-\frac{244}{9}}{60}$ hours. During the same time, the person, who had already traveled $\frac{x}{2} \text{ km}$ when the fast one started, travels $\frac{2 x}{3}-\frac{x}{2} \text{ km}$ at a speed of $30 \text{ km/h}$ and $\frac{x}{3}-\frac{244}{9} \text{ km}$ at a speed of $15 \text{ km/h}$, and thus the time is:

$$
\frac{\frac{2 x}{3}-\frac{x}{2}}{30}+\frac{\frac{x}{3}-\frac{244}{9}}{15} \text{ hours. }
$$

Therefore,

$$
\frac{x-\frac{244}{9}}{60}=\frac{\frac{2 x}{3}-\frac{x}{2}}{30}+\frac{\frac{x}{3}-\frac{244}{9}}{15}
$$

Removing the first-order fractions:

$$
x-\frac{244}{9}=\frac{4 x}{3} \quad x+\frac{4 x}{3}-4 \frac{244}{9}
$$

which means

$$
\begin{gathered}
\frac{8 x}{3}-2 x=3 \cdot \frac{244}{9} \\
\frac{2 x}{3}=\frac{244}{3}
\end{gathered}
$$

from which

$$
x=122
$$

Györgyi Lackner (Bp. XIII., 1. sz. textile techn. I. o. t.)