81 Given that $A B C D$ is a tetrahedron, $P, Q, R, S$ are internal points on the edges $A B, B D, C D, C A$ respectively, and $\frac{P A}{P B}=\frac{A S}{S C}=1, \frac{B Q}{Q D}=\frac{C R}{R D}=\frac{1}{2}$, the plane $P Q R S$ divides the tetrahedron into two parts with volumes $V_{1}, V_{2}$ respectively $\left(V_{1}<V_{2}\right)$, find $\frac{V_{1}}{V_{2}}$.
81 As shown in Figure 32, since $B P=\frac{1}{2} B A, B Q=\frac{1}{3} B D$, therefore
$$
V_{B-P Q C}=\frac{1}{2} \times \frac{1}{3} V_{A B C D}=\frac{1}{6} V_{A B C D} .
$$

Similarly,
$$
\begin{array}{l}
V_{C-P Q S}=\frac{1}{2} V_{C-P Q A}=\frac{1}{2} \times \frac{1}{2} V_{A-B C Q} \\
=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{3} V_{A B C D}=\frac{1}{12} V_{A B C D}, \\
V_{C-Q S R}=\frac{1}{2} \times \frac{1}{3} V_{C-A A D}=\frac{1}{2} \times \frac{1}{3} \times \frac{2}{3} V_{A B C D}=\frac{1}{9} V_{A B C D} . \\
\end{array}
$$

Therefore,
$$
\begin{array}{c}
V_{1}=\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{9}\right) V_{A B C D}=\frac{13}{36} V_{A B C D} \\
\frac{V_{1}}{V_{2}}=\frac{13}{23} .
\end{array}
$$