455*. From point A to point B, which is 40 km away from A, two tourists set off simultaneously: the first on foot at a speed of 6 km/h, and the second on a bicycle. When the second tourist overtook the first by 5 km, the first tourist got into a passing car traveling at a speed of 24 km/h. Two hours after leaving A, the first tourist caught up with the second and arrived in B before him. Find the speed of the tourist who was riding the bicycle.
$\triangle$ Let's make a drawing (Fig. 2). $\qquad$
Fig. 2

Let $\mathrm{A}_{1}$ and $\mathrm{B}_{1}$ be the points where the first and second tourists are, respectively, at the moment when the second tourist overtakes the first by $5 \mathrm{km}$, and let $\mathrm{K}$ be the point where the first tourist catches up with the second.

Denote the time the first tourist takes to travel $\mathrm{AA}_{1}$ (and simultaneously the second tourist to travel $\mathrm{AB}_{1}$) by $t_{1}$ hours, and the time the first tourist takes to travel $\mathrm{A}_{1} \mathrm{~K}$ (and the second to travel $\mathrm{B}_{1} \mathrm{~K}$) by $t_{2}$ hours. Let the speed of the cyclist be $v$ km/h.

Since each tourist traveled the distance $\mathrm{AK}$ in 2 hours, we have $t_{1}+t_{2}=2$.

Since in time $t_{1}$ the cyclist traveled 5 km more than the first tourist, we have $v t_{1}-6 t_{1}=5$.

Finally, express the distance $\mathrm{AK}$ in two ways - using quantities related to the first tourist and quantities related to the second, and equate the resulting expressions:

$$
6 t_{1}+24 t_{2}=v\left(t_{1}+t_{2}\right)
$$

We have the system of equations:

$$
\left\{\begin{array}{l}
t_{1}+t_{2}=2 \\
v t_{1}-6 t_{1}=5 \\
6 t_{1}+24 t_{2}=v\left(t_{1}+t_{2}\right)
\end{array}\right.
$$

From the first equation, $t_{2}=2-t_{1}$, and from the second, $-v=\frac{6 t_{1}+5}{t_{1}}$. Substitute these expressions for $t_{2}$ and $v$ into the third equation and solve the new equation with the unknown $t_{1}$:

$6 t_{1}+24\left(2-t_{1}\right)=\frac{2\left(6 t_{1}+5\right)}{t_{1}}, \quad 6 t_{1}^{2}+48 t_{1}-24 t_{1}^{2}=12 t_{1}+10$,

$18 t_{1}^{2}-36 t_{1}+10=0, \quad 9 t_{1}^{2}-18 t_{1}+5=0 ; \quad t_{1}=\frac{9 \pm 6}{9}$.

Thus, $t_{1}=\frac{5}{3}$ or $t_{1}=\frac{1}{3}$. Accordingly, $v=9$ or $v=21$.

Do both these values of $v$ fit? It turns out, no: if $v=21$ km/h, then the cyclist will travel a distance of 42 km in 2 hours, and this distance is greater than $A B=40$ km.

Answer: 9 km/h.