7.5. Several seventh-graders were solving problems. The teacher does not remember how many children there were and how many problems each of them solved. However, he remembers that, on the one hand, each solved more than a fifth of what the others solved, and on the other hand, each solved less than a third of what the others solved. How many seventh-graders could there be? Find all the options and prove that there are no others.
Answer: 5 seventh-graders.

Solution: Let one seventh-grader solve $a$ problems, and the rest solve $S - a$. Then

$$
\begin{gathered}
a < (S-a) / 3 \\
3a < S - a \\
4a < S \\
a < S / 4
\end{gathered}
$$

Similarly,

$$
\begin{gathered}
(S-a) / 5 < a \\
S - a < 5a \\
S < 6a \\
S / 6 < a
\end{gathered}
$$

Thus, if there are 4 or fewer students, and each solved less than a quarter of all problems, then together they solved fewer than all the problems, which is impossible. Similarly, if there are 6 or more students, and each solved more than a fifth of all problems, then together they solved more than all the problems, which is also impossible. Therefore, there must be 5 students in total.

## Criteria:

Answer - 0 points.

Solved under the assumption that each solved more than a fifth of all problems and less than a quarter of all problems - 3 points.