Let $x$ and $y$ be two-digit positive integers with mean $60$. What is the maximum value of the ratio $\frac{x}{y}$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac{33}{7} \qquad \textbf{(C)}\ \frac{39}{7} \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \frac{99}{10}$
If $x$ and $y$ have a mean of $60$, then $\frac{x+y}{2}=60$ and $x+y=120$. To maximize $\frac{x}{y}$, we need to maximize $x$ and minimize $y$. Since they are both two-digit positive integers, the maximum of $x$ is $99$ which gives $y=21$. $y$ cannot be decreased because doing so would increase $x$, so this gives the maximum value of $\frac{x}{y}$, which is $\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}$