Problem 3. A clock shows exactly one o'clock. A mosquito and a fly are sitting at the same distance from the center on the hour and minute hands, respectively. When the hands coincide, the insects swap places. How many times greater is the distance that the mosquito has traveled in half a day compared to the distance the fly has traveled in the same time?
Answer: $83 / 73$.

Solution. The mosquito and the fly move in a circle. In the first hour, the mosquito will cover $11 / 12$ of this circle (at the beginning, it was on the hour hand, pointing to 1, at the end - on the minute hand, pointing to 12). In the second hour, the mosquito will cover $3 / 12$ of the circle (it was on the minute hand, pointing to 12, and ended up on the hour hand, pointing to 3). Thus, the mosquito covered $14 / 12$ of the circle in the first 2 hours. In the next 2 hours, he will also cover $14 / 12$ of the circle, and so on. We get that in 10 hours, he will have traveled $5 \cdot 14 / 12=70 / 12$ of the circle. The eleventh hour begins with the mosquito on the hour hand, pointing to 11, and the minute hand pointing to 12. The mosquito will cover $1 / 12$ of the circle in this hour and will end up on the minute hand. In the last, twelfth hour, the minute and hour hands will not meet, and the mosquito will travel one circle. Thus, in half a day, the mosquito will cover a distance of $A=70 / 12+1 / 12+1=83 / 12$. Similarly, for the fly, we get the distance $B=5 \cdot(2 / 12+10 / 12)+1+1 / 12=73 / 12$. Therefore, $A / B=83 / 73$.

Answer to variant 2: $72 / 71$.