3. Schoolchildren Chukov and Gekov are skating at constant speeds around a closed circular running track of an ice stadium. If Chukov skates clockwise and Gekov skates counterclockwise, their meetings occur four times more frequently than when they overtake each other while skating in the same direction. The speed of one of the schoolchildren is 6 m/s. What is the speed of the other?
Answer: Either $10 \mathrm{m} /$ sec, or $3.6 \mathrm{~m} /$ sec.

Instructions. When moving towards each other, the time between meetings is $t_{1}=\frac{L}{V_{1}+V_{2}}$, when moving in the same direction, the time between overtakes: $t_{2}=\frac{L}{V_{1}-V_{2}}$ (here $L-$ is the length of one lap, $V_{1}, V_{2}$ - are the speeds). According to the condition $\frac{t_{2}}{t_{1}}=4$, therefore $\frac{V_{1}+V_{2}}{V_{1}-V_{2}}=4$, from which $\frac{V_{2}}{V_{1}}=\frac{3}{5}$.

The condition does not specify whose speed is given. Therefore, either $\frac{6}{V}=\frac{3}{5}$ (then $V=10$ m/sec), or $\frac{V}{6}=\frac{3}{5}$ (then $V=3.6$ m $/$ sec).