# Solve the following$9 \frac{3}{4} \div\left[2 \frac{1}{6}+\left\{4 \frac{1}{3}-\left(1 \frac{1}{2}+1 \frac{3}{4}\right)\right\}\right]$

Solution:

$9 \frac{3}{4} \div\left[2 \frac{1}{6}+\left\{4 \frac{1}{3}-\left(1 \frac{1}{2}+1 \frac{3}{4}\right)\right\}\right]$

$=\frac{9 \times 4+3}{4} \div\left[\frac{2 \times 6+1}{6}+\left\{\frac{4 \times 3+1}{3}-\left(\frac{1 \times 2+1}{2}+\frac{4 \times 1+3}{4}\right)\right\}\right]$

$=\frac{39}{4} \div\left[\frac{13}{6}+\left\{\frac{13}{3}-\left(\frac{3}{2}+\frac{7}{4}\right)\right\}\right]$

LCM of 4 and 2 is 2

=$\frac{39}{4} \div\left[\frac{13}{6}+\left\{\frac{13}{3}-\left(\frac{6+7}{4}\right)\right\}\right]$

=$\frac{39}{4} \div\left[\frac{13}{6}+\left\{\frac{13}{3}-\left(\frac{13}{4}\right)\right\}\right]$

=$\frac{39}{4} \div\left[\frac{13}{6}+\left\{\frac{13}{3}-\frac{13}{4}\right\}\right]$

LCM of 3 and 4 is 12

=$\frac{39}{4} \div\left[\frac{13}{6}+\left\{\frac{13\times4-13\times3}{12}\right\}\right]$

=$\frac{39}{4} \div\left[\frac{13}{6}+\left\{\frac{52-39}{12}\right\}\right]$

=$\frac{39}{4} \div\left[\frac{13}{6}+\frac{13}{12}\right]$

LCM of 6 and 12 is 12

=$\frac{39}{4} \div\left[\frac{13\times2+13}{12}\right]$

=$\frac{39}{4} \div\left[\frac{26+13}{12}\right]$

=$\frac{39}{4} \div\left[\frac{39}{12}\right]$

=$\frac{39}{4} \div \frac{39}{12}$

=$\frac{39}{4} \times \frac{12}{39}$

=$\frac{39}{4} \times \frac{12}{39}$ $(39\times1 =39 and 4\times3 =12)$

=$3$

Therefore the answer is $3$

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Updated on: 10-Oct-2022

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