The sum of three numbers is $98$. The ratio of the first to the second is $\frac {2}{3}$, 
and the ratio of the second to the third is $\frac {5}{8}$. The second number is: 
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$
Let the $3$ numbers be $a,$ $b,$ and $c$. We see that 
\[a+b+c = 98\]
and 
\[\frac{a}{b} = \frac{2}{3} \Rrightarrow 3a = 2b\]
\[\frac{b}{c} = \frac{5}{8} \Rrightarrow 8b = 5c\]
Writing $a$ and $c$ in terms of $b$ we have
$a = \frac{2}{3} b$ and $c = \frac{8}{5} b$. 
Substituting in the sum, we have
\[\frac{2}{3} b + b + \frac{8}{5} b = 98\]
\[\frac{49}{15} b = 98\]
\[b = 98 \cdot \frac{15}{49} \Rrightarrow b = 30\]
$\boxed{C}$
~JustinLee2017