6. Let's introduce the following notations: $S$ - the area of the trapezoid, $S_{1}$ - the area of triangle $ABC$, $S_{2}$ - the area of triangle $ACD$, $h$ - the height of the trapezoid, $\gamma=\frac{q}{p}$, $\mu=\frac{n}{m}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8b002ac84843a2e42fafg-09.jpg?height=277&width=760&top_left_y=455&top_left_x=258)

We have $S_{1}: S_{2}=\frac{n}{m}$ because these triangles have the same height. Therefore, $S_{1}=\frac{n}{n+m} \cdot S$. From the similarity of triangles $AND$ and $CNT$, it follows that $\frac{TC}{AD}=\frac{q}{p}$. From this equality, we get: $TC=\frac{q}{p} AD$. Then $TB=TC-BC=\frac{q}{p} AD-\frac{n}{m} AD=\left(\frac{q}{p}-\frac{n}{m}\right) AD$. From the similarity of triangles $AMD$ and $BTM$, it follows that

$$
\mathrm{BM}: \mathrm{AM}=TB: AD=\left(\frac{q}{p}-\frac{n}{m}\right)
$$

Therefore,

$$
\frac{AM}{AB}=\frac{AM}{AM+MB}=\frac{1}{1+BM: AM}=\frac{1}{1+\frac{q}{p}-\frac{n}{m}}
$$

Then

$$
S_{AMN}=\frac{AM}{AB} \cdot \frac{p}{p+q} \cdot S_{1}=\frac{1}{1+\frac{q}{p}-\frac{n}{m}} \cdot \frac{1}{1+\frac{q}{p}} \cdot S_{1}=\frac{1}{1+\gamma-\mu} \cdot \frac{1}{1+\gamma} \cdot S_{1}
$$

From this, we get

$$
\begin{gathered}
S_{MBCN}=S_{1}-S_{AMN}=\left(1-\frac{1}{1+\gamma-\mu} \cdot \frac{1}{1+\gamma}\right) S_{1}= \\
=\left(1-\frac{1}{1+\gamma-\mu} \cdot \frac{1}{1+\gamma}\right) \frac{\mu}{1+\mu} S
\end{gathered}
$$

In our problem, $\gamma=2, \mu=\frac{1}{3}$, so $S_{MBCN}: S=7: 32$.
Answer: $\quad S_{\text {MVCN }}: S=7: 32$.