At twelve o'clock, four cars set off from $A$ to $B$, which is $210 \mathrm{~km}$ away. The speed of the first car is the geometric mean of the sum of the speeds of the second and third cars and the speed of the fourth car. The speed of the fourth car is half the arithmetic mean of the speeds of the second and third cars. The ratio of the speed of the first car to the speed of the fourth car is the same as the ratio of the speed of the second car to the speed of the third car. Furthermore, we know that the car whose speed is the sum of the speeds of the four cars would reach $B$ from $A$ in 1 hour. When do the individual cars reach $B$?
I. Solution: Let the speeds of the individual cars be $x, y, z, u \mathrm{~km} /$ hour.

According to the problem,

$$
\begin{gathered}
x^{2}=(y+z) u \\
\frac{y+z}{4}=u \\
\frac{x}{u}=\frac{y}{z} \\
x+y+z+u=210
\end{gathered}
$$

From (2), substituting the value of $(y+z)$ into (1),

$$
x^{2}=4 u^{2}, \quad \text { or } \quad \text { (5) } \quad x=2 u
$$

From (5) and (3), it follows that

$$
y=2 z
$$

Considering (5) and (6), (4) becomes

$$
3 u+3 z=210
$$

On the other hand, from (1), using (6) and (5), $y+z=\frac{x^{2}}{u}$, thus

$$
3 z=4 u
$$

Substituting (8) into (7),

$$
7 u=210
$$

from which

$$
\begin{aligned}
& u=\frac{210}{7}=30 \mathrm{~km} / \text { hour. } \\
& z=\frac{4}{3} u=40 \mathrm{~km} / \text { hour. } \\
& y=2 z=80 \mathrm{~km} / \text { hour. } \\
& x=2 u=60 \mathrm{~km} / \text { hour }
\end{aligned}
$$

Thus, the travel times of the cars are

$$
\frac{210}{60}=\frac{7}{2} \text { hours, } \quad \frac{210}{80}=\frac{21}{8} \text { hours, } \quad \frac{210}{40}=\frac{21}{4} \text { hours, } \quad \frac{210}{30}=7 \text { hours. }
$$

and thus their arrival times in $B$ are: 15:30; 14:37.5; 17:15 and 19:00.

Pachert Zsuzsa (Tata, Eötvös J. g. II. o. t.)

II. Solution: From (2), substituting the value of $u$ into (1), we get

$$
x=\frac{y+z}{2}
$$

and thus, considering (2) and (3), $\frac{x}{u}=2=\frac{y}{z}$.

From this,

$$
y=2 z, \quad \text { and thus } \quad x=\frac{3 z}{2}, \quad u=\frac{x}{2}=\frac{3 z}{4}
$$

Substituting these values into (4),

$\frac{3 z}{2}+2 z+z+\frac{3 z}{4}=210$, from which $z=40 \mathrm{~km} /$ hour, etc., similarly as in the first solution.

Meskó Attila (Bp. VII., Madách g. II. o. t.)