John $(J)$ and Louis $(L)$, who live far from each other, agree over the phone to meet in Magyarfan $(M)$, which lies between their villages, and both set out resolutely. $J$ walks at a speed of $5 \mathrm{~km}$ per hour, and $L$ at $4 \frac{1}{2} \mathrm{~km}$. $L$ plans to have time to shop in $M$, as $J$'s journey is one and a half times longer than his; - all the more so because he covers $3 / 5$ of his journey by a car traveling at $30 \mathrm{~km} / \mathrm{h}$ to $M$. But as soon as he gets off in $M$, $J$ arrives 4 minutes later. $J$ was also lucky; he covered $3 / 5$ of his journey in the same car as $L$. How far apart do $J$ and $L$ live from each other?
If $L$'s journey is $x(\mathrm{~km})$, then $J$'s journey is $3 x / 2$. Thus, $L$ travels $3 x / 5$ on foot and $2 x / 5$ by car, during which $3 x / 5 \cdot 4.5$ and $2 x /(5 \cdot 30)$ hours elapse, so the travel time is

$$
\frac{3 x}{5 \cdot 4.5}+\frac{2 x}{5 \cdot 30}=\frac{2 x}{15}+\frac{x}{75}=\frac{11 x}{75} \text { hours. }
$$

$J$ travels $(3 / 5) \cdot(3 x / 2)=9 x / 10$ by car and $(2 / 5) \cdot(3 x / 2)=3 x / 5$ on foot, so the travel time is

$$
\frac{3 x}{25}+\frac{9 x}{300}=\frac{3 x}{20} \text { hours, }
$$

and this is 4 minutes, $1 / 15$ hours longer than (1). Thus

$$
\frac{11 x}{75}+\frac{1}{15}=\frac{3 x}{20} \quad x=20 \mathrm{~km}
$$

Our travelers covered 20 and $30 \mathrm{~km}$ respectively, so they live $50 \mathrm{~km}$ apart.

János Köves (Budapest, Kossuth L. Mechanical Engineering High School I. Class)

Remarks. 1. If we had immediately taken the unknown distance $L J=y$, then from the $1: 1.5=2: 3$ ratio, $L$ and $J$'s journeys would be $2 y /(2+3)=2 y / 5$ and $3 y / 5$. Further, with the above calculations, we would have immediately obtained $y=50 \mathrm{~km}$.

2. Several considered two "different" solutions if they once calculated with $9 / 2, 3 / 2, 3 / 5, 2 / 5$ and another time with 4.5, 1.5, 0.6, 0.4. However, this does not give us a different solution if we use different notations for the numerical values involved, just as it is not a different solution if we calculate the time once in hours and another time in minutes.