A student was asked to find the value of $$\frac{\left(2 \frac{1}{3} + 2 \frac{1}{2} - \frac{1}{6} \right) \div 2\frac{1}{3} \times 5 \frac{2}{3} \div 1\frac{2}{3}  of  4\frac{1}{4}}{3 \frac{1}{5} \div 4 \frac{1}{2}  of  5 \frac{1}{3} + 5\frac{1}{3} \times \frac{3}{4} \div 2\frac{2}{3}}$$. His answer was $$\frac{6}{7}$$.what is the difference between correct answer and his answer?

As per the question,

$$\frac{\left(2 \frac{1}{3} + 2 \frac{1}{2} - \frac{1}{6} \right)\div 2\frac{1}{3} \times 5 \frac{2}{3} \div 1\frac{2}{3} of 4\frac{1}{4}}{3 \frac{1}{5} \div 4 \frac{1}{2} of 5 \frac{1}{3} + 5\frac{1}{3} \times \frac{3}{4} \div 2\frac{2}{3}}$$

$$\Rightarrow \frac{\left(2 \frac{1}{3} + 2 \frac{1}{2} - \frac{1}{6} \right)\div2\frac{1}{3} \times 5 \frac{2}{3} \div \frac{7}{3} of \frac{17}{4}}{3\frac{1}{5} \div  \frac{9}{2} of  \frac{16}{3} + 5\frac{1}{3} \times\frac{3}{4} \div 2\frac{2}{3}}$$

$$\Rightarrow \frac{\left(2 \frac{1}{3} + 2 \frac{1}{2} - \frac{1}{6} \right)\div2\frac{1}{3} \times 5 \frac{2}{3} \div \frac{119}{12}}{3\frac{1}{5} \div 24 + 5\frac{1}{3} \times\frac{3}{4} \div 2\frac{2}{3}}$$

$$\Rightarrow \frac{\left(\frac{7}{3} + \frac{5}{2} - \frac{1}{6}\right)\div\frac{7}{3} \times \frac{17}{3} \div \frac{119}{12}}{\frac{16}{5} \div 24 + \frac{16}{3} \times\frac{3}{4} \div \frac{2}{3}}$$

$$\Rightarrow \frac{\left(\frac{28}{6}\right)\div\frac{7}{3} \times \frac{17}{3} \div \frac{119}{12}}{\frac{2}{15}+ \frac{16}{3} \times\frac{9}{8}}$$

$$\Rightarrow \frac{2 \times \frac{17\times 12}{3\times 119}}{\frac{2}{15}+ \frac{16}{3} \times\frac{9}{8}}$$

$$\Rightarrow 2 \times \frac{17\times 12\times 15}{3\times 119\times 92}$$

$$\Rightarrow \frac{ 12\times 5}{ 7\times 41}=\dfrac{60}{287}$$
Hence the required difference $$=\dfrac{6}{7}-\dfrac{60}{287}=\dfrac{246-60}{287}$$

$$=\dfrac{246-60}{287}=\dfrac{186}{287}$$