A group of hikers sets out on a journey at a constant speed of $5 \mathrm{~km}$ per hour. 4 hours and 24 minutes later, a cyclist messenger sets out after them. After $8 \mathrm{~km}$, the cyclist's inner tube bursts. It takes 10 minutes to change the tire. Then, traveling at a speed $2 \mathrm{~km}$ per hour faster, the cyclist catches up to the group at the same point as if they had continued at the initial speed without stopping. What is the cyclist's initial speed?
I. Solution: Let the initial speed of the cyclist be $x \mathrm{~km} /$ hour and assume that $t$ hours after departure, they meet the group of hikers. If the cyclist had traveled at the initial speed without stopping, they would have covered $t x \mathrm{~km}-\mathrm{t}$ by the time of the meeting. However, in reality, they covered $8 \mathrm{~km}-\mathrm{t}$ in $\frac{8}{x}$ hours, then after a $\frac{1}{6}$ hour break, they traveled at $x+2 \mathrm{~km} /$ hour for $t-\frac{8}{x}-\frac{1}{6}$ hours. The group of hikers would have covered $5(t+4.4)=5 t+22 \mathrm{~km}-\mathrm{t}$ by the time of the meeting.

According to the problem,

$$
t x=5 t+22
$$

and

$$
8+\left(t-\frac{8}{x}-\frac{1}{6}\right)\left(x^{\prime}+2\right)=5 t+22
$$

From (1),

$$
t=\frac{22}{x-5}
$$

Substituting this value into (2),

$$
8+\left(\frac{22}{x-5}-\frac{8}{x}-\frac{1}{6}\right)(x+2)=\frac{110}{x-5}+22
$$

which simplifies to

$$
\frac{22 x-66}{x-5}-\frac{x}{6}-\frac{16}{x}-\frac{67}{3}=0
$$

Multiplying both sides by $6 x(x-5)$ (where $x \neq 0, x-5 \neq 0$),

$$
6 x(22 x-66)-x^{2}(x-5)-96(x-5)-134 x(x-5)=0
$$

or, after rearranging,

$$
x^{3}-3 x^{2}-178 x-480=0
$$

Among the divisors of 480, we find the root $x=-3$ of this polynomial equation. Dividing the polynomial by the factor $x+3$, we arrive at the quadratic equation

$$
x^{2}-6 x-160=0
$$

whose roots are 16 and -10.

Thus, among the three roots of the equation, only $x=16 \mathrm{~km} /$ hour satisfies the problem.

$$
\text { Györgyi Lackner (Bp., V., Fonóip. techn. III. o. t.) }
$$

II. Solution: Let the initial speed of the cyclist be $x \mathrm{~km} /$ hour and the meeting point be $8+s$ km from the starting point.

The group of hikers walked for $\frac{8+s}{5}$ hours. The cyclist would cover the distance in $\frac{8+s}{x}$ hours. According to the problem,

$$
\frac{8+s}{5}=4.4+\frac{8+s}{x}
$$

When the tire burst, the cyclist was $s$ km from the finish line. They would have covered this distance in $\frac{s}{x}$ hours, but in reality, $\frac{1}{6}+\frac{s}{x+2}$ hours passed until the meeting, so according to the problem,

$$
\frac{s}{x}=\frac{1}{6}+\frac{s}{x+2}
$$

From (3),

$$
s=\frac{14 x+60}{x-5}
$$

Substituting this value into (4),

$$
\frac{14 x+40}{x(x-5)}=\frac{1}{6}+\frac{14 x+40}{(x+2)(x-5)}
$$

which leads to

$$
x^{3}-3 x^{2}-178 x-480=0
$$

and so on, as in the first solution.