67 (1001). A tourist left from $A$ to $B$ at a speed of 4 km/h. An hour later, a second tourist, who walked 5 km per hour, set out from $A$ in pursuit. Another hour later, a cyclist set out from $A$, who, overtaking one tourist, overtook the other after 10 minutes. Find the speed of the cyclist.
S o l u t i o n. Let the speed of the cyclist be $v$ kilometers per hour. Then, the first tourist, who walked 8 km in 2 hours, will be caught by the cyclist in $\frac{8}{v-4}$ hours. The second tourist will be caught by the cyclist in $\frac{5}{v-5}$ hours. If the cyclist first overtakes the second tourist, we have the equation

$$
\frac{8}{v-4}-\frac{5}{v-5}=\frac{1}{6}, \text { or } v^{2}-27 v+140=0
$$

Both roots of the equation satisfy the condition of the problem ( $v>5$ ):

$$
v_{1}=20, v_{2}=7
$$

If the cyclist first overtakes the first tourist, we have the equation

$$
\frac{5}{v-5}-\frac{8}{v-4}=\frac{1}{6}, \text { or } v^{2}+9 v-100=0
$$

The condition of the problem ( $v>5$ ) is satisfied by one root

$$
v=\frac{-9+\sqrt{ } 481}{2} \approx 6.5
$$

A n s w e r: $\approx 6.5$ km $/$ h, or 7 km $/ h$, or 20 km $/$ h. 130