Question : What is the simplified value of $\left(1-\frac{1}{4-\frac{2}{1+\frac{1}{\frac{1}{3}+2}}}\right) \times \frac{15}{16} \div \frac{2}{3}$ of $2 \frac{1}{4}-\frac{3+4}{3^3+4^3}$
Option 1: $\frac{5}{13}$
Option 2: $\frac{4}{13}$
Option 3: $\frac{8}{13}$
Option 4: $\frac{6}{13}$
Correct Answer: $\frac{4}{13}$
Solution : Given: $\left(1-\frac{1}{4-\frac{2}{1+\frac{1}{\frac{1}{3}+2}}}\right) \times \frac{15}{16} \div \frac{2}{3}$ of $2 \frac{1}{4}-\frac{3+4}{3^3+4^3}$ $=(1-\frac{1}{4-\frac{2}{1+\frac{3}{7}}}) \times \frac{15}{16} \div (\frac{2}{3} \times \frac{9}{4})-\frac{3+4}{27+64}$ $=(1-\frac{1}{4-\frac{14}{10}}) \times \frac{15}{16} \div \frac{3}{2} -\frac{7}{91}$ $=(1-\frac{10}{26}) \times \frac{15}{16} \times \frac{2}{3} -\frac{1}{13}$ $=\frac{16}{26} \times \frac{5}{8} -\frac{1}{13}$ $=\frac{10}{26} -\frac{1}{13}$ $=\frac{10-2}{26}$ $=\frac{8}{26}$ $=\frac{4}{13}$ Hence, the correct answer is $\frac{4}{13}$.
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