From a rope 11 m long, two pieces of lengths

Solution:

Length of the rope when two pieces of lengths $2 \frac{3}{5} \mathrm{~m}$ and $3 \frac{3}{10} \mathrm{~m}$ are cut off $=$ Total length of the rope - Length of the two cut off pieces

$\therefore 11-\left(2 \frac{3}{5}+3 \frac{3}{10}\right)$

Now,

$2 \frac{3}{5}+3 \frac{3}{10} \Rightarrow\left(2+\frac{3}{5}\right)+\left(3+\frac{3}{10}\right)$

LCM of 5 and 10 is 10 , i.e., $(5 \times 1 \times 2)$.

We have:

$\frac{(13 \times 2)+(33 \times 1)}{10}$

$=\frac{26+33}{10}$

$=\frac{59}{10}$

$\therefore 2 \frac{3}{5}+3 \frac{3}{10}=\frac{59}{10}$

Length of the remaining rope $=11-\frac{59}{10}$

$=\frac{110-59}{10}$

$=\frac{51}{10}$

$=5 \frac{1}{10} \mathrm{~m}$

Therefore, the length of the remaining rope is $5 \frac{1}{10} \mathrm{~m}$.