Question

Subtract the sum of $3\frac{5}{9}$ and $3\frac{1}{3}$ from the sum of $5\frac{5}{6}$ and $4\frac{1}{9}$.

Answer 1

According to the question, we have to find the value of $(5\frac{5}{6}+4\frac{1}{9})-(3\frac{5}{9}+3\frac{1}{3})$

$=(\frac{35}{6}+\frac{37}{9})-(\frac{32}{9}+\frac{10}{3})$

On rearranging, we get,

$=\frac{35}{6}-\frac{10}{3}+\frac{37}{9}-\frac{32}{9}$

$=(\frac{35}{6}-\frac{10}{3})+(\frac{37}{9}-\frac{32}{9})$

Now, as $6$ is the multiple of $3$, therefore their LCM will be $6$.

$=(\frac{35}{6}-\frac{10}{3}\times \frac{2}{2})+\left(\frac{37-32}{9}\right)$

$=\left(\frac{35-20}{6}\right)+\left(\frac{37-32}{9}\right)$

$=\left(\frac{15}{6}\right)+\left(\frac{5}{9}\right)$

$=\frac{5}{2}+\frac{5}{9}$

As $2$ and $9$ are co prime numbers, therefore, their LCM will be their product ie. $18$

$=\frac{5}{2}\times \frac{9}{9}+\frac{5}{9}\times \frac{2}{2}$

$=\frac{45+10}{18}$

$=\frac{55}{18}$

$=3\frac{1}{18}$

Hence, $(5\frac{5}{6}+4\frac{1}{9})-(3\frac{5}{9}+3\frac{1}{3})=3\frac{1}{18}$