$13 \cdot 40$ has $n$ colored balls, among which there are red and black ones. In the first 50 balls, 49 are red, and thereafter, every 8 balls contain 7 red ones. If at least $90\%$ of all the balls are red, then the maximum value of $n$ is
(A) 225.
(B) 210.
(C) 200.
(D) 180.
(E) 175.
(14th American High School Mathematics Examination, 1963)
［Solution］From the given, we have the inequality
$$
\frac{49+\frac{7}{8}(n-50)}{n} \geqslant \frac{9}{10} \text {, }
$$

Solving this, we get $n \leqslant 210$.
Therefore, the correct choice is $(B)$.