Problem 4. On a circular path 1650 m long, two motorcyclists are moving at a constant speed. If the motorcyclists move in opposite directions, they will meet every minute, and if they move in the same direction, the motorcyclist moving at the higher speed will overtake the other every 11 minutes. Determine the speeds of the motorcyclists.
Solution. Let $v_{1}$ be the speed at which the faster one is moving, and $v_{2}$ be the speed of the slower motorcyclist. From the first condition (time is 1 minute or $1 / 60$ hour), we have $v_{1} \frac{1}{60}+v_{2} \frac{1}{60}=\frac{1650}{1000}$, and from the second condition (time is 11 minutes, or $11 / 60$ hour) $v_{1} \frac{11}{60}-v_{2} \frac{11}{60}=\frac{1650}{1000}$. We obtain a system of two linear equations with two unknowns: $v_{1}+v_{2}=99, v_{1}-v_{2}=9$, whose solution is $v_{1}=54 \mathrm{~km} / \mathrm{h}, v_{2}=45 \mathrm{~km} / \mathrm{h}$.