Rózsa, Ibolya, and Viola decided that they would solve all the problems in their problem collection. Rózsa works on $a$, Ibolya on $b$, and Viola on $c$ problems daily. (Only one of them works on a given problem.) If Rózsa solved 11 times, Ibolya 7 times, and Viola 9 times more problems daily, they would finish in 5 days; whereas if Rózsa solved 4 times, Ibolya 2 times, and Viola 3 times as many problems, they would finish in 16 days. In how many days will they complete the solutions?
I. solution. In the first case, Rózsa solves $11 a$, Ibolya $7 b$, and Viola $9 c$ problems daily; thus, in 5 days, they solve $5(11 a + 7 b + 9 c)$ problems. In the second case, the number of problems solved is $16(4 a + 2 b + 3 c)$. Since in both cases all problems are solved, we have

$$
5(11 a + 7 b + 9 c) = 16(4 a + 2 b + 3 c)
$$

After rearranging, we get $b = 3 a + c$.

If it takes $x$ days to complete the solutions, we can write the following equation:

$$
5(11 a + 7 b + 9 c) = x(a + b + c)
$$

Substituting the previously obtained equality for $b$ and rearranging, we get the equation: $160 a + 80 c = 4 x a + 2 x c$. Dividing the equation by $4 a + 2 c \neq 0$, we arrive at $x = 40$. Therefore, it will take 40 days to solve all the problems.

This is possible, for example, if $a = c = 1, b = 4$, and the number of problems is: $5(11 + 7 \cdot 4 + 9) = 240$.

II. solution. Let the number of problems to be solved be $n$. Then

$$
5(11 a + 7 b + 9 c) = n \quad \text{and} \quad 16(4 a + 2 b + 3 c) = n
$$

In the first case, $\frac{n}{5}$ problems were solved per day, and in the second case, $\frac{n}{16}$ problems were solved per day.

$$
11 a + 7 b + 9 c = \frac{n}{5}, \quad 4 a + 2 b + 3 c = \frac{n}{16}
$$

Subtracting twice the second equation from the first, we get

$$
3(a + b + c) = \frac{n}{5} - \frac{n}{8} = \frac{3 n}{40}
$$

Thus, $a + b + c = \frac{n}{40}$ is the number of problems solved per day. Therefore, it will take 40 days to solve all the problems.