$153 S-A B C D$ is a regular quadrilateral pyramid, $P, Q$ are the midpoints of $S B, S D$ respectively, the plane through points $A, P, Q$ divides $S-A B C D$ into two parts with volumes $V_{1}, V_{2}\left(V_{1}<V_{2}\right)$, then $\frac{V_{2}}{V_{1}}$ equals ( )
A. 3
B. 4
C. 5
D. 6
153 C. As shown in Figure 6(a), let the midpoint of $PQ$ be $O$, then $O$ lies on the height of the pyramid, and thus within $\triangle SAC$. Let $AO$ intersect $SC$ at $M$, then the section is $APMQ$. From Figure 6(b), it is easy to see that $\frac{MQ}{QC} = \frac{OQ}{AC - OQ} = \frac{1}{3}$. Therefore, $\frac{SM}{SC} = \frac{1}{3}$.

Figure 6
$$
\begin{array}{c}
\frac{V_{S-APQ}}{V_{S-ABD}} = \frac{SP \times SQ}{SB \times SD} = \frac{1}{4}. \\
\frac{V_{S-MPQ}}{V_{S-CBD}} = \frac{SP \times SQ \times SM}{SB \times SD \times SC} = \frac{1}{12}
\end{array}
$$

Therefore,
$$
\begin{array}{c}
V_{1} = V_{S-APMQ} = V_{S-APQ} + V_{S-\triangle PQ} = \frac{V_{S} \cdot ABCD}{2} \times \left(\frac{1}{4} + \frac{1}{12}\right) \\
= \frac{1}{6} V_{S \cdot ABCD} = \frac{1}{6} (V_{1} + V_{2}). \\
\frac{V_{2}}{V_{1}} = 5.
\end{array}
$$