Problem 3. A team of workers was laying linoleum in a store's warehouse and in the cash hall, with the warehouse area being 3 times larger than the cash hall area. In the part of the team working in the warehouse, there were 5 more workers than in the part working in the cash hall. When the work in the warehouse was completed, the part of the team working in the cash hall was still working. What is the maximum number of workers that could have been in the team?
Answer: 9.

Solution. Let the number of workers in the cash hall be denoted as $n$, then the number of workers in the warehouse is $n+5$, and the total number of people in the team is $2n+5$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivity of each part of the team is $an$ and $a(n+5)$. If the area of the cash hall is $S$, then the area of the warehouse is $3S$. The time spent on completing the entire work by each part of the team is $\frac{S}{an}$ and $\frac{3S}{a(n+5)}$, respectively. According to the problem, $\frac{S}{an} > \frac{3S}{a(n+5)}$. Due to the positivity of all variables, this inequality is equivalent to $n+5 > 3n$, or $2n < 5$. Therefore, $n \leqslant 2$, and $2n+5 \leqslant 9$. The equality situation is clearly possible: just take any positive $S$ and $a$.