The second problem of the second round of the Rákosi Mátyás competition read as follows: "On a railway line, a passenger train departs from location $A$ to location $C$. When the train passes through $B$, a freight train departs from $B$ towards $A$. When the freight train arrives at $A$, a fast train departs from $A$ towards $C$, and $\mathrm{s}$ it catches up with the previously mentioned passenger train exactly at the $C$ station. Halfway between $A$ and $B$, a student observes that 15 minutes after the passenger train passes, the freight train passes, and 11 minutes later, the fast train. $B$ is 10 km from $A$. The question is, how many kilometers is $C$ from $B$? (We assume that the trains travel at a constant speed)."

The solution to this problem can be found on page 10 of issue No. 10.

In connection with this problem, we raise the following question: What are the limits within which the hourly speeds of the fast, passenger, and freight trains vary, if we limit the maximum speed of the fast train to 150 km/h and stipulate that the speed of the freight train cannot be greater than the speed of the passenger train?
Let's denote the speeds of the passenger, freight, and express trains as $c_{1}$, $c_{2}$, and $c_{3} \mathrm{~km} /$ per minute, respectively.

According to the problem,

$$
\frac{5}{c_{1}}+\frac{5}{c_{2}}=15
$$

where

$$
c_{2} \leqq c_{1}
$$

is our constraint. From (1),

$$
\frac{1}{c_{2}}=3-\frac{1}{c_{1}}
$$

and thus, considering (2),

$$
\frac{1}{c_{1}} \leqq 3-\frac{1}{c_{1}}
$$

from which

$$
c_{1} \geqq \frac{2}{3}
$$

and thus from (1),

$$
c_{2} \leqq \frac{2}{3}
$$

Since according to the problem,

$$
\frac{5}{c_{2}}+\frac{5}{c_{3}}=11
$$

if $c_{2}$ takes its maximum value of $\frac{2}{3}$, then the value of $c_{3}$ is minimal, and this minimal value can be calculated from (3):

$$
\frac{5}{c_{3}}=11-\frac{5}{c_{2}}=11-\frac{15}{2}=\frac{7}{2}
$$

from which

$$
c_{3}=\frac{10}{7}
$$

thus

$$
c_{3} \geqq \frac{10}{7}
$$

The maximum value of $c_{3}$ according to the condition is $\frac{5}{2}$, so $c_{3} \leqq \frac{5}{2}$.

If we substitute the maximum value of $c_{3}$ into (3), we get the minimum value for $c_{2}$:

$$
\frac{5}{c_{2}}=11-\frac{5}{c_{3}}=11-2=9
$$

from which

$$
c_{2}=\frac{5}{9}
$$

thus

$$
c_{2} \geqq \frac{5}{9} \text {. }
$$

Finally, if we substitute the minimum value of $c_{2}$ into (1), we get the maximum value of $c_{1}$:

$$
\frac{5}{c_{1}}=15-9=6
$$

from which

$$
c_{1}=\frac{5}{6}
$$

thus

$$
c_{1} \leqq \frac{5}{6}
$$

Therefore, the limits in $\mathrm{km} /$ per minute are:

$$
\begin{array}{r}
\frac{2}{3} \leqq c_{1} \leqq \frac{5}{6} \\
\frac{2}{3} \geqq c_{2} \geqq \frac{5}{9} \\
\frac{10}{7} \leqq c_{3} \leqq \frac{5}{2}
\end{array}
$$

Converting to $\mathrm{km} /$ per hour:

$$
\begin{gathered}
40 \leqq c_{1} \leqq 50 \\
40 \geqq c_{2} \geqq 33 \frac{1}{3} \\
85 \frac{5}{7} \leqq c_{3} \leqq 150
\end{gathered}
$$

As the speed of the passenger and express trains increases, the speed of the freight train decreases.

Pál Bártfai (Bp. I., Petőfi g. II. o. t.)