If $\log_8{3}=p$ and $\log_3{5}=q$, then, in terms of $p$ and $q$, $\log_{10}{5}$ equals
$\mathrm{(A)\ } pq \qquad \mathrm{(B) \ }\frac{3p+q}{5} \qquad \mathrm{(C) \  } \frac{1+3pq}{p+q} \qquad \mathrm{(D) \  } \frac{3pq}{1+3pq} \qquad \mathrm{(E) \  }p^2+q^2$
Notice that $\frac{1}{p}=\log_{3}{8}$, so it would probably be easier to work in base $3$. From change of base, $\log_{10}{5}=\frac{\log_{3}{5}}{\log_{3}{10}}$. We're given that $\log_{3}{5}=q$, so now we just need to find $\log_{3}{10}$. 
We have $\frac{1}{p}=\log_{3}{8}=3\log_{3}{2}$, so $\log_{3}{2}=\frac{1}{3p}$. Also, $\log_{3}{10}=\log_{3}{5}+\log{3}{2}=q+\frac{1}{3p}$. Therefore, $\log_{10}{5}=\frac{q}{q+\frac{1}{3p}}=\frac{3pq}{3pq+1}, \boxed{\text{D}}$.