Find the number which bears the same ratio to $\dfrac{7}{33}$ that $\dfrac{8}{21}$ does to $\dfrac{4}{9}$.

Let the number be x.

According to question,

$\Rightarrow \dfrac{x}{\dfrac{7}{33}} = \dfrac{\dfrac{8}{21}}{\dfrac{4}{9}} \\[1em] \Rightarrow \dfrac{33x}{7} = \dfrac{8 \times 9}{21 \times 4} \\[1em] \Rightarrow \dfrac{33x}{7} = \dfrac{72}{84} \\[1em] \Rightarrow x = \dfrac{72 \times 7}{84 \times 33} \\[1em] \Rightarrow x = \dfrac{504}{2772} \\[1em] \Rightarrow x = \dfrac{2}{11}.$

Hence, number = $\dfrac{2}{11}$.