In trapezoid $A B C D$, $A D / / B C, \frac{A D}{B C}=\frac{1}{2}$, point $M$ is on side $A B$, such that $\frac{A M}{M B}=\frac{3}{2}$, point $N$ is on side $C D$: such that line segment $M N$ divides the trapezoid into two parts with an area ratio of $3: 1$. Find $\frac{C N}{N D}$.
$$
\begin{array}{c}
=, \because \frac{S_{\triangle M B C}}{S_{\triangle A B C}}=\frac{2}{5}, \\
\frac{S_{\triangle A B C}}{S_{\text {trapezoid }}}=\frac{2}{3}, \\
\therefore \frac{S_{\triangle M B C}}{S_{\text {trapezoid }}}=\frac{4}{15}>\frac{1}{4},
\end{array}
$$

Therefore, it can only be that
$S_{\text {quadrilateral MBCN }}: S_{\text {quadrilateral AMND }}=3: 1$.

Thus, $\frac{C N}{N D}=\frac{S_{\triangle M C N}}{S_{\triangle M D N}}=\frac{\frac{3}{4} S_{\text {trapezoid }}-S_{\triangle M B C}}{\frac{1}{4} S_{\text {trapezoid }}-S_{\triangle M A D}}=\frac{29}{3}$.