4) The arithmetic mean of 11 numbers is 4850. If each of the 11 numbers is decreased by 10, their mean becomes:
(A) 4740
(B) 4840
(C) 4830
(D) 4850
(E) the data in the problem is not sufficient to determine it
4. The answer is $\mathbf{( B )}$. Let's call the eleven numbers $a_{1}, a_{2}, \ldots, a_{11}$. We know that

$$
\frac{a_{1}+a_{2}+\cdots+a_{11}}{11}=4850
$$

We can then calculate

$$
\begin{aligned}
\frac{\left(a_{1}-10\right)+\left(a_{2}-10\right)+\cdots+\left(a_{11}-10\right)}{11} & =\frac{a_{1}+a_{2}+\cdots+a_{11}-110}{11} \\
& =\frac{a_{1}+a_{2}+\cdots+a_{11}}{11}-\frac{110}{11} \\
& =4850-10=4840 .
\end{aligned}
$$

[Problem proposed by P. Negrini]