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:
$\textbf{(A)}\ 225 \qquad \textbf{(B)}\ 210 \qquad \textbf{(C)}\ 200 \qquad \textbf{(D)}\ 180 \qquad \textbf{(E)}\ 175$
The desired percentage of red balls is more than $90$ percent, so write an [inequality](https://artofproblemsolving.com/wiki/index.php/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 $\boxed{\textbf{(B)}}$.