Question : What is the value of $\frac{\frac{5}{12} \text { of } \frac{6}{25}-\frac{5}{6} \times \frac{12}{33}+\frac{5}{11} \div \frac{25}{33}}{\frac{1}{6} \div \frac{1}{2}+\frac{1}{6} \times \frac{1}{2}-\frac{1}{2} \text { of } \frac{1}{6}}$?
Option 1: $\frac{131}{110}$
Option 2: $\frac{125}{117}$
Option 3: $\frac{121}{112}$
Option 4: $\frac{133}{115}$
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Correct Answer: $\frac{131}{110}$
Solution : Given: $\frac{\frac{5}{12} \text { of } \frac{6}{25}-\frac{5}{6} \times \frac{12}{33}+\frac{5}{11} \div \frac{25}{33}}{\frac{1}{6} \div \frac{1}{2}+\frac{1}{6} \times \frac{1}{2}-\frac{1}{2} \text { of } \frac{1}{6}}$ = $\frac{\frac{1}{10} -\frac{5}{6} \times \frac{12}{33}+\frac{5}{11} \times \frac{33}{25}}{\frac{1}{6} \times \frac{2}{1}+\frac{1}{6} \times \frac{1}{2}-\frac{1}{2} \times \frac{1}{6}}$ = $\frac{\frac{1}{10}-\frac{10}{33} +\frac{3}{5} }{\frac{1}{3}+ \frac{1}{12}-\frac{1}{12} }$ = $\frac{\frac{33-100+198}{330}}{\frac{1}{3}}$ = $\frac{131}{330}\times\frac{3}{1}$ = $\frac{131}{110}$ Hence, the correct answer is $\frac{131}{110}$.
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