Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB = 3CD$. Let $E$ be then  midpoint of the diagonal $BD$. If $[ABCD] = n \times  [CDE]$, what is the value of $n$? 

(Here $[t]$ denotes the area of the geometrical figure$ t$.)
1. Let $h$ be the altitude of the trapezium, and let $AB = 3CD = 3x$. This implies $CD = x$.
2. The area of the trapezium $ABCD$ can be calculated using the formula for the area of a trapezium:
   \[
   [ABCD] = \frac{1}{2} \times (AB + CD) \times h = \frac{1}{2} \times (3x + x) \times h = \frac{1}{2} \times 4x \times h = 2xh
   \]
3. The area of the triangle $BCD$ can be calculated using the formula for the area of a triangle:
   \[
   [BCD] = \frac{1}{2} \times CD \times h = \frac{1}{2} \times x \times h = \frac{1}{2} xh
   \]
4. Since $E$ is the midpoint of the diagonal $BD$, the area of the triangle $CDE$ is half the area of the triangle $BCD$:
   \[
   [CDE] = \frac{1}{2} [BCD] = \frac{1}{2} \times \frac{1}{2} xh = \frac{1}{4} xh
   \]
5. To find the ratio $\frac{[ABCD]}{[CDE]}$, we substitute the areas we have calculated:
   \[
   \frac{[ABCD]}{[CDE]} = \frac{2xh}{\frac{1}{4} xh} = \frac{2xh \times 4}{xh} = 8
   \]

The final answer is $\boxed{8}$