11.2. Fishermen caught several carp and pike. Each caught as many carp as all the others caught pike. How many fishermen were there if the total number of carp caught was 10 times the number of pike? Justify your answer.
# Solution:

Method 1. Each fisherman caught as many carp and pike together as the total number of pike caught. Summing the catches of all fishermen, we get that the total catch of all fishermen (in terms of the number of fish) is equal to the total number of pike caught, multiplied by the number of fishermen. On the other hand, there are 10 times more carp than pike, so the total catch in terms of the number of fish is 11 times the number of pike. Therefore, there are 11 fishermen in total.

Method 2. Let the total number of fishermen be $n$, and the $i$-th fisherman caught $a_{i}$ pike and $b_{i}$ carp $(i=\overline{1, n})$. Then $b_{k}=\left(\sum_{i=1}^{n} a_{i}\right)-a_{k}$ for all $k$. Summing these equalities over $k$, we get
$\sum_{i=1}^{n} b_{i}=n \cdot \sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} a_{i}=(n-1) \cdot \sum_{i=1}^{n} a_{i}$. But $\sum_{i=1}^{n} b_{i}=10 \cdot \sum_{i=1}^{n} a_{i}$, from which $n-1=10$ and $n=11$.

Answer: 11 fishermen.

Recommendations for checking:

| present in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| A correct approach but an incorrect answer due to  arithmetic or algebraic errors | 5 points |
| Correctly formulated but unsolved system of equations,  fully describing the problem's conditions | 3 points |
| Only specific cases are considered, for example, when  all fishermen caught the same number of pike | 1 point |
| Correct answer without justification or illustrated  by a finite number of examples | 0 points |