10.2. To transport 80 tons of cargo, a truck required 5 more trips than to transport 40 tons of the same cargo. On all trips, except possibly the last one in each transportation, the truck was fully loaded. What values can the truck's carrying capacity take?
Answer: $7 \frac{3}{11} \leq x<8 \frac{8}{9}$.

Solution. Let $x$ be the load capacity of the truck, and $n$ be the number of trips required to transport 40 tons of cargo.

Consider 2 cases:

When transporting 40 tons, in the last trip the truck was loaded to no more than half, i.e., in the $(n-1)$ trips the truck was fully loaded, and in the last trip it was loaded to less than half. Then, for transporting 80 tons, it took $2(n-1)+1=2n-1$ trips. According to the condition, $2n-1-n=5$, from which $n=6$. In this case: $5x<40.5$, $5x \geq 40$, hence $7 \frac{3}{11} \leq x<8$.

When transporting 40 tons, in the last trip the truck was loaded to more than half. Then, for transporting 80 tons, it took $2(n-1)+2=2n$ trips. According to the condition, $2n-n=5$, $n=5$. Then $4.5x<40.5$, $5x \geq 40$, hence $8 \leq x<8 \frac{8}{9}$. Thus, $7 \frac{3}{11} \leq x<8 \frac{8}{9}$.

Comment. Fully considering 1 case - 4 points, 2 case - 3 points.