Péter sent his son with a message to his brother, Károly, who at the same time sent his son to Péter. The cousins met 720 meters from Péter's house, then after a 2-minute conversation, they continued on. Each spent 10 minutes at the relative's house. On their way back, they met again 400 meters from Károly's house. How far do the two families live from each other? What assumptions can we make to answer this question?
I. solution. The two boys spent the same amount of time both with each other and in conversation at the relative's house, so due to their simultaneous departure, they covered the distances to their first meeting and between the two meetings in the same amount of time, let these times be $t_{1}$ and $t_{2}$. Our data is only sufficient to answer the question assuming that the movements between the mentioned time points were uniform. Let the speed of "little Péter" and "little Károly" up to the first meeting be $v_{1}$ and $v_{2}$, and between the two meetings be $v_{1}^{\prime}$ and $v_{2}^{\prime}$, and let the distance between the houses be $x$. - Thus, before the first meeting, Péter covered a 720-meter distance, and Károly covered a distance of $x-720$ meters, and the time spent moving can be expressed in two ways:

$$
t_{1}=\frac{720}{v_{1}}=\frac{x-720}{v_{2}}
$$

$P$ covered a distance of $(x-720)+400=x-320$ meters between the two meetings, and $K$ covered $720+(x-400)=x+320$ meters, the time elapsed in between:

$$
t_{2}=\frac{x-320}{v_{1}^{\prime}}=\frac{x+320}{v_{2}^{\prime}}
$$

(1) and (2) contain five unknowns, so we can only proceed with a new assumption. This could be that the ratio of the boys' speeds is the same between the two time periods. Thus, from (1) and (2), we get a one-variable equation for $x$:

$$
\frac{v_{1}}{v_{2}}=\frac{v_{1}^{\prime}}{v_{2}^{\prime}}=\frac{720}{x-720}=\frac{x-320}{x+320}
$$

$x>720$ (otherwise, $P$ would have found $K$ at home), so neither denominator is 0, and from $x^{2}-1760 x=0$ we get $x=1760$ meters (the root $x=0$ has no meaning here).

Denes Horvath (Kisújszállás, Móricz Zs. g. 1I. o. t.)

Remark. Most of the solutions assumed that the boys' speed was constant throughout the movement. This is a special case of the assumption stated here.

II. solution. The two boys covered the distance between the houses once up to the first meeting, and three times up to the second meeting, so they "tramped" the distance between the houses twice between the two meetings. Thus, if we again assume that the ratio of the two boys' speeds is constant, then the part of the distance covered together between the two meetings is the same as the part covered up to the first meeting, i.e., $2 \cdot 720=1440$ meters. On the other hand, from the first meeting to the relative's house, they covered the distance between the houses minus $720 \mathrm{~m}$, and on the way back home, they covered another $400 \mathrm{~m}$, so in total, they were $320 \mathrm{~m}$ short of covering the entire distance between the two houses. Therefore, the two families live $1440+320=1760 \mathrm{~m}$ apart.