• # question_answer What is the value of $3\frac{1}{12}+\left[ 1\frac{3}{4}+\left\{ 2\frac{1}{2}-\left( 1\frac{1}{2}-\frac{1}{3} \right) \right\} \right]$? A) $6\frac{1}{2}$               B) $2\frac{1}{6}$C) $6\frac{1}{6}$                 D) $0$

Correct Answer: C

Solution :

$3\frac{1}{12}+\left[ 1\frac{3}{4}+\left\{ 2\frac{1}{2}-\left( 1\frac{1}{2}-\frac{1}{3} \right) \right\} \right]$ $=\frac{37}{12}+\left[ \frac{7}{4}+\left\{ \frac{5}{2}-\left( \frac{3}{2}-\frac{1}{3} \right) \right\} \right]$ $=\frac{37}{12}+\left[ \frac{7}{4}+\left\{ \frac{5}{2}-\left( \frac{9-2}{6} \right) \right\} \right]$ $=\frac{37}{12}+\left[ \frac{7}{4}+\left\{ \frac{5}{2}-\frac{7}{6} \right\} \right]$ $=\frac{37}{12}+\left[ \frac{7}{4}+\left\{ \frac{15-7}{6} \right\} \right]$ $=\frac{37}{12}+\frac{7}{4}+\frac{8}{6}=\frac{37}{12}+\frac{7}{4}+\frac{4}{3}$ $=\frac{34+21+16}{12}=\frac{74}{12}=\frac{37}{6}=6\frac{1}{6}$

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