$5y$ varies inversely as the square of $x$. When $y=16, x=1$. When $x=8, y$ equals: 
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024$
An inverse variation can be expressed in the form $xy = n$, where $n$ is any number (except perhaps zero). Since $5y$ varies inversely with the square of $x$, this particular one will be $5yx^2 = n$. 
We can plug in $16$ for $y$ and $1$ for $x$, which makes $n=80$. The equation is now $5yx^2 = 80$.
When $x = 8$, we can solve for $y$ using the equation $320y = 80$, which makes $y$ $\boxed{\textbf{(D)} \frac{1}{4}}$.