Question

From a rope 11 m long, two pieces of lengths $2\frac{3}{5}\mathrm{m}\mathrm{and}3\frac{3}{10}\mathrm{m}$ are cut off. What is the length of the remaining rope?

Answer 1

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,

$$\begin{gathered} 2\frac{3}{5}+3\frac{3}{10}\Rightarrow \left(2+\frac{3}{5}\right)+\left(3+\frac{3}{10}\right) \\ =\frac{13}{5}+\frac{33}{10} \end{gathered}$$
LCM of 5 and 10 is 10, i.e., $\left(5\times 1\times 2\right)$.

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \frac{\left(13\times 2\right)+\left(33\times 1\right)}{10} \\ =\frac{26+33}{10} \\ =\frac{59}{10} \end{gathered}$$

∴​ $2\frac{3}{5}+3\frac{3}{10}=\frac{59}{10}$
Length of the remaining rope $=11-\frac{59}{10}$

$$\begin{gathered} =\frac{110-59}{10} \\ =\frac{51}{10} \\ =5\frac{1}{10}\text{m} \end{gathered}$$

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