A) >
B) <
C) =
D) Can't be determined
Correct Answer: A
Solution :
We have, \[\left[ \frac{7}{3}-\frac{2}{6} \right]-\frac{1}{2}+\left[ \frac{4}{5}\div \frac{6}{8} \right]+\frac{3}{4}\] \[=\left( \frac{14-2}{6} \right)-\frac{1}{2}+\left[ \frac{4}{5}\times \frac{8}{6} \right]+\frac{3}{4}\] \[=\left( \frac{12}{6} \right)-\frac{1}{2}+\frac{16}{15}+\frac{3}{4}=2+\frac{16}{15}+\frac{3}{4}-\frac{1}{2}\] \[=\frac{120+64+45-30}{60}=\frac{199}{60}\] Also, \[\frac{4}{5}\div \left( \frac{6}{8}+\frac{3}{4} \right)-\left( \frac{3}{4}+\frac{5}{6} \right)+3\frac{1}{3}\] \[=\frac{4}{5}\div \left( \frac{6+6}{8} \right)-\left( \frac{9+10}{12} \right)+\frac{10}{3}\] \[=\frac{4}{5}\div \frac{3}{2}-\frac{19}{12}+\frac{10}{3}=\frac{4}{5}\times \frac{2}{3}-\frac{19}{12}+\frac{10}{3}\] \[=\frac{8}{15}-\frac{19}{12}+\frac{10}{3}=\frac{32-95+200}{60}=\frac{137}{60}\] \[\therefore \] \[\frac{199}{60}\,\,\,\,\frac{137}{60}\] \[\therefore \] \[\left[ \frac{7}{3}-\frac{2}{6} \right]-\frac{1}{2}+\left[ \frac{4}{5}\div \frac{6}{8} \right]+\frac{3}{4}\,\frac{4}{5}\] \[+\left[ \frac{6}{8}+\frac{3}{4} \right]-\left[ \frac{3}{4}+\frac{5}{6} \right]+3\frac{1}{3}\]
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