$28 \cdot 21$ There is a set of 50 numbers, the arithmetic mean of these 50 numbers is 38. If the two numbers 45 and 55 are removed from this set, then the arithmetic mean of the remaining numbers in the set is 
(A) 38.5.
(B) 37.5.
(C) 37.
(D) 36.5.
(E) 36.
(14th American High School Mathematics Examination, 1963)
[Solution] From the problem, we know that the sum of the numbers in the set is $38 \cdot 50=1900$.
When removing the two numbers 45 and 55, the sum of the remaining 48 numbers is 1800. Thus, their arithmetic mean is
$$
\frac{1800}{48}=37.5 \text{.}
$$

Therefore, the answer is $(B)$.