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 this train 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 after that, the fast train passes. $B$ 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.)
I. Solution: Let the sought $BC$ distance be denoted by $x$ and let the speeds of the passenger, freight, and express trains be $c_{1}, c_{2}$, and $c_{3} \mathrm{~km} /$ minute, respectively. According to the student's observation,

\[
\begin{aligned}
& \frac{5}{c_{1}}+\frac{5}{c_{2}}=15 \\
& \frac{5}{c_{2}}+\frac{5}{c_{3}}=11
\end{aligned}
\]

The distance $BC = x$ is covered by the passenger train in $\frac{x}{c_{1}}$ minutes, and during this time, the freight train is on the way from $B$ to $A$ for $\frac{10}{c_{2}}$ minutes, and then the express train is on the way from $A$ to $C$ for $\frac{10+x}{c_{3}}$ minutes, so

\[
\frac{x}{c_{1}}=\frac{10}{c_{2}}+\frac{10+x}{c_{3}}
\]

Thus, we have 3 equations with 4 unknowns, which means the unknowns cannot be fully determined, but in this case, $x$ can be determined, while $c_{1}, c_{2}$, and $c_{3}$ remain undetermined.

Subtracting (2) from (1),

\[
\frac{5}{c_{1}}-\frac{5}{c_{3}}=4, \text { from which } \frac{1}{c_{1}}-\frac{1}{c_{3}}=\frac{4}{5}
\]

From (2),

\[
\frac{1}{c_{2}}+\frac{1}{c_{3}}=\frac{11}{5}
\]

From (3),

\[
\frac{x}{c_{1}}-\frac{x}{c_{3}}=\frac{10}{c_{2}}+\frac{10}{c_{3}}
\]

or

\[
x\left(\frac{1}{c_{1}}-\frac{1}{c_{3}}\right)=10\left(\frac{1}{c_{2}}+\frac{1}{c_{3}}\right)
\]

Considering (4) and (5),

\[
\frac{4}{5} x=10 \cdot \frac{11}{5}
\]

from which

\[
x=\frac{110}{4}=27.5 \mathrm{~km}
\]

II. Solution: According to the student's observation, the passenger train and the freight train cover 5-5 km in succession in a total of 15 minutes, while the freight train and the express train cover 5-5 km in succession in a total of 11 minutes. This implies that the express train covers 5 km in 4 minutes less time than the passenger train. The express train passes the midpoint $F$ of the $AB$ distance 15+11=26 minutes after the passenger train. Since 5 km brings 4 minutes of time advantage, 26 minutes of time advantage is brought after $6.5 \cdot 5 \mathrm{~km}=32.5 \mathrm{~km}$. Therefore, $FC=32.5 \mathrm{~km}$ and thus $BC=FC-FB=32.5-5=27.5 \mathrm{~km}$.

III. Solution: Let's graphically represent the movement of the trains. First, let's plot the kilometers covered by the train departing from $A$ at 0 minutes as a function of $t$ minutes, with an arbitrary speed, i.e., draw the line $AP$ at an arbitrary angle (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_ad9daece9b1239b12c20g-1.jpg?height=558&width=702&top_left_y=2155&top_left_x=695)
After covering 10 km, the freight train departs from $B$ back to $A$ at an arbitrary speed. The freight train reaches the 5 km point 15 minutes after the passenger train, so on our diagram, $PQ=15$ minutes, which implies that the freight train arrives in $A$ 30 minutes after the passenger's departure (which occurred at 0 minutes). (This can be geometrically read from our diagram. Since the speed of the express train must be positive and bounded from above, the speed of the freight train is not entirely arbitrary. If we take 150 km/h as the upper limit for the express train's speed, then the express train needs at least 2 minutes to cover 5 km, and the passenger train needs at least $2+4=6$ minutes, so the distance $5P \geq 6$ minutes. This implies that the distance $PQ=15$ minutes is at most 2.5 times the distance $5P$.)

The express train departs at the 30th minute, but its speed is already determined by the two arbitrarily chosen speeds, as it must reach the 5 km point 11 minutes after the freight train. Thus, $QR=11$ minutes, and the point $R$ is determined, and thus the line representing the movement of the express train is also determined. The intersection point $M(x, y)$ of this line with the line representing the movement of the passenger train. The $y$-coordinate of $M$, which represents kilometers, can be read from our diagram, but it can also be precisely calculated from our diagram:

\[
y:(y-5)=MA:MP=30:26
\]

from which

\[
\begin{gathered}
26 y=30 y-150 \\
y=37.5 \text { and thus } BC=y-10=27.5 \mathrm{~km}
\end{gathered}
\]

Essentially, this is how András Ádám (Hajdúszoboszló, Irinyi János g. IV. o. t.) solved the problem.