Simplify :

$6\dfrac{1}{3}$ ÷ $\Big(2\dfrac{1}{5} + 3\dfrac{1}{2}\Big)$ of $3\dfrac{1}{3}$

We have:

$6\dfrac{1}{3}$ ÷ $\Big(2\dfrac{1}{5} + 3\dfrac{1}{2}\Big)$ of $3\dfrac{1}{3}$

= $\dfrac{19}{3}$ ÷ $\Big(\dfrac{11}{5} + \dfrac{7}{2}\Big)$ of $\dfrac{10}{3}$ [Converting mixed to improper fraction]

According to BODMAS rule, we simplify brackets first

$\begin{array}{ll} = \dfrac{19}{3} ÷ \Big(\dfrac{22 + 35}{10}\Big)\text{ of }\dfrac{10}{3} \\ = \dfrac{19}{3} ÷ \dfrac{57}{10}\text{ of }\dfrac{10}{3} & \text{[Bracket simplified]} \\ = \dfrac{19}{3} ÷ \dfrac{57}{10} \times \dfrac{10}{3} \\ = \dfrac{19}{3} ÷ \dfrac{570}{30} \\ = \dfrac{19}{3} ÷ \dfrac{19}{1} & \text{[Of simplified]} \\ = \dfrac{19}{3} \times \dfrac{1}{19} & [\text{Reciprocal of } \dfrac{19}{1} \text{ is } \dfrac{1}{19}] \\ = \dfrac{1}{3} & \text{[Division simplified]} \end{array}$

∴ The answer is $\dfrac{1}{3}$