Simplify :

$\dfrac{7}{15}$ of $\Big(\dfrac{2}{3} + \dfrac{7}{12}\Big)$ ÷ $\Big(\dfrac{5}{6}-\dfrac{3}{5}\Big)$

We have:

$\dfrac{7}{15}$ of $\Big(\dfrac{2}{3} + \dfrac{7}{12}\Big)$ ÷ $\Big(\dfrac{5}{6}-\dfrac{3}{5}\Big)$

According to BODMAS rule, we simplify brackets first

$\begin{array}{ll} = \dfrac{7}{15}\text{ of }\Big(\dfrac{8 + 7}{12}\Big) ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) \\ = \dfrac{7}{15}\text{ of } \Big(\dfrac{15}{12}\Big) ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) \\ = \dfrac{7}{15}\text{ of } \dfrac{5}{12} ÷ \Big(\dfrac{5}{6}-\dfrac{3}{5}\Big) & \text{[First bracket simplified]} \\ = \dfrac{7}{15}\text{ of } \dfrac{15}{12} ÷ \Big(\dfrac{25 - 18}{30}\Big) \\ = \dfrac{7}{15}\text{ of } \dfrac{15}{12} ÷ \dfrac{7}{30} & \text{[Second bracket simplified]} \\ = \dfrac{7}{15} \times \dfrac{15}{12} ÷ \dfrac{7}{30} \\ = \dfrac{7}{12} ÷ \dfrac{7}{30} & \text{[Of simplified]} \\ = \dfrac{7}{12} \times \dfrac{30}{7} & [\text{Reciprocal of } \dfrac{7}{30} \text{ is } \dfrac{30}{7}] \\ = \dfrac{1}{12} \times \dfrac{30}{1} & \text{[Dividing 7 and 7 by 7]} \\ = \dfrac{1}{2} \times \dfrac{5}{1} & \text{[Dividing 30 and 12 by 6]} \\ = \dfrac{5}{2} & \text{[Division simplified]} \\ = 2\dfrac{1}{2} \end{array}$

∴ The answer is $2\dfrac{1}{2}$