A) \[\frac{1}{2}\]
B) \[2\]
C) \[1\]
D) \[0\]
Correct Answer: D
Solution :
We have, \[3\frac{1}{12}-\left[ 1\frac{3}{4}+\left\{ 2\frac{1}{2}-\left( 1\frac{1}{2}-\frac{1}{3} \right) \right\} \right]\] \[=\frac{37}{12}-\left[ \frac{7}{4}+\left\{ \frac{5}{2}-\left( \frac{3}{2}-\frac{1}{3} \right) \right\} \right]\] \[=\frac{37}{12}-\left[ \frac{7}{4}+\left\{ \frac{5}{2}-\left( \frac{9-2}{6} \right) \right\} \right]\left[ \begin{align} & \because \text{LCM}\,\,\text{of} \\ & 2,\text{ 3 = 6} \\ \end{align} \right]\] \[=\frac{37}{12}-\left[ \frac{7}{4}+\left\{ \frac{5}{2}-\frac{7}{6} \right\} \right]\] \[=\frac{37}{12}-\left[ \frac{7}{4}+\left\{ \frac{15-7}{6} \right\} \right][\because \text{LCM of 2,6=6 }\!\!]\!\!\text{ }\] \[=\frac{37}{12}-\left[ \frac{7}{4}+\frac{8}{6} \right]\] \[=\frac{37}{12}-\left[ \frac{21+16}{12} \right]\] \[\left[ \begin{align} & \because \text{LCM of} \\ & 4,\text{ 6 = 12} \\ \end{align} \right]\] \[=\frac{37}{12}-\frac{37}{12}=0\]
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