Discussion

Explanation:

Given, $\frac{4{\displaystyle \frac{1}{3}}+3{\displaystyle \frac{1}{3}}\times 1{\displaystyle \frac{4}{5}}÷3{\displaystyle \frac{3}{4}}\times \left(6{\displaystyle \frac{1}{4}}of1{\displaystyle \frac{1}{15}}\right)}{{\displaystyle \frac{2}{3}}÷{\displaystyle \frac{5}{6}}\times {\displaystyle \frac{2}{3}}}$

= $\frac{{\displaystyle \frac{13}{3}}+{\displaystyle \frac{10}{3}}\times {\displaystyle \frac{9}{5}}\times {\displaystyle \frac{4}{15}}\times \left({\displaystyle \frac{25}{4}}\times {\displaystyle \frac{16}{15}}\right)}{{\displaystyle \frac{2}{3}}\times {\displaystyle \frac{6}{5}}\times {\displaystyle \frac{2}{3}}}$

= $\frac{{\displaystyle \frac{13}{3}}+{\displaystyle \frac{8}{5}}\times \left({\displaystyle \frac{20}{3}}\right)}{{\displaystyle \frac{8}{15}}}$

= $\frac{{\displaystyle \frac{13}{3}}+{\displaystyle \frac{32}{3}}}{{\displaystyle \frac{8}{15}}}$

= $15\times \frac{15}{{\displaystyle 8}}$ = $\frac{225}{8}$

= $28\frac{1}{8}$

Hence, option (a).

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