6.5. The arithmetic mean of four numbers is 10. If one of these numbers is erased, the arithmetic mean of the remaining three increases by 1; if instead another number is erased, the arithmetic mean of the remaining numbers increases by 2; and if only the third number is erased, the arithmetic mean of the remaining increases by 3. By how much will the arithmetic mean of the remaining three numbers change if the fourth number is erased?
Answer: will decrease by 6.

Solution. From the fact that the arithmetic mean of four numbers is 10, it follows that the sum of these numbers is 40. Similarly, the sum of three numbers (excluding the first) is 33, the sum of three numbers (excluding the second) is 36, and the sum of three numbers (excluding the third) is 39. From these conditions, we get: the first number is 7, the second is 4, the third is 1. Thus, the arithmetic mean of the first three numbers is 4, which is 6 less than 10.

Similar reasoning can be conducted algebraically. Denoting the four numbers by $a, b, c$, and $d$ respectively, we get four equations: $(a+b+c+d): 4=10, (b+c+d): 3=11, (a+c+d): 3=12$, $(a+b+d): 3=13$. The solution to this system is the set of numbers: $(7,4,1,28)$.

## Evaluation Criteria:

+ fully justified solution

$\pm$ correct line of reasoning (the system of equations is correctly set up), but an arithmetic error is made

the answer is not obtained, but it is proven that the arithmetic mean of the first three numbers is less than 10

$\mp$ the correct answer is obtained by selecting a set of four numbers and verifying that it satisfies the condition (but it is not proven that the found set of four numbers is the only possible one)

- a set of four numbers is given, but it is not even verified
- an error is made in the definition of the arithmetic mean
- the problem is not solved or is solved incorrectly