# 1963 AHSME Problems/Problem 19

## Problem

In counting $n$ colored balls, some red and some black, it was found that $49$ of the first $50$ counted were red. Thereafter, $7$ out of every $8$ counted were red. If, in all, $90$ % or more of the balls counted were red, the maximum value of $n$ is:

$\textbf{(A)}\ 225 \qquad \textbf{(B)}\ 210 \qquad \textbf{(C)}\ 200 \qquad \textbf{(D)}\ 180 \qquad \textbf{(E)}\ 175$

## Solution

The desired percentage of red balls is more than $90$ percent, so write an inequality.

$$\frac{49+7x}{50+8x} \ge 0.9$$

Since $x >0$, the sign does not need to be swapped after multiplying both sides by $50+8x$.

$$49+7x \ge 45+7.2x$$ $$4 \ge 0.2x$$ $$20 \ge x$$

Thus, up to $20$ batches of balls can be used, so a total of $20 \cdot 8 + 50 = 210$ balls can be counted while satisfying the requirements. The answer is $\boxed{\textbf{(B)}}$.

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