Two cities $A$ and $B$, $a$ km apart, lie along a navigable river, next to which a railway runs. A train departs from $A$ towards $B$ at a speed of $n \mathrm{~km}$ per hour, while at the same time, two boats depart from $B$ in opposite directions but at the same speed. The train meets the oncoming boat in half the time it takes to reach the other boat. What is the speed of the boats?
I. solution: Let the speed of the boats be $x \mathrm{~km} /$ hour. The time until the train meets the oncoming boat is $\frac{a}{n+x}$ hours, because the train and the boat together cover $n+x \mathrm{~km}-\mathrm{t}$ per hour.

According to the problem, the train catches up with the other boat $\frac{2 a}{n+x}$ hours after. During this time, the difference in the distances covered by the train and the other boat is $a$, that is,

$$
\frac{2 a n}{n+x}-\frac{2 a x}{n+x}=a
$$

Dividing by $2 a$ ($a \neq 0$)

$$
\frac{n-x}{n+x}=\frac{1}{2}, \quad \text { from which } \quad x=\frac{n}{3}
$$

Thus, the speed of the boats is one-third of the train's speed.

Benkö György (Sárospatak, Rákóczi g. I. o. t.)

II. solution: Keeping the meaning of $x$, let $t$ be the number of hours until the first meeting, then according to the problem $t n+t x=a$, and $2 t n-2 t x=a$, that is,

$$
t n+t x=2 t n-2 t x
$$

Dividing by $t$ ($t \neq 0$) and rearranging

$$
3 x=n, \quad \text { from which } \quad x=\frac{n}{3}
$$

Gergó Éva (Bp. XIV. Gorkij school VII. = I. g. o. t.)

III. solution: We can also solve our problem without equations, purely by reasoning.

The distance of the first meeting point from the second is composed of the distance covered by the first boat in the first half-time and the distance covered by the second boat in two half-times, that is, this distance - which the train covers in the second half-time - is three times the distance covered by the boat in the first half-time. Therefore, the speed of the boats is one-third of the train's speed.

Bencze Ernő (Székesfehérvár, József A. g. I. o. t.)