# A can do a piece of work in 40 days and B in 45 days.

Question:

A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work?

Solution:

It is given that A can finish the work in 40 days and B can finish the same work in 45 days.

$\therefore$ Work done by A in 1 day $=\frac{1}{40}$

Work done by B in 1 day $=\frac{1}{45}$

$\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 1 day $=\frac{1}{40}+\frac{1}{45}$

$=\frac{9+8}{360}=\frac{17}{360}$

$\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 10 day $\mathrm{s}=10 \times \frac{17}{360}=\frac{17}{36}$

Remaining work $=1-\frac{17}{36}=\frac{19}{36}$

It is given that the remaining work is done by B.

Complete work is done by B in 45 days.

$\therefore \frac{19}{36}$ of the work will be done by B in $\left(45 \times \frac{19}{36}\right)$ days or $23 \frac{3}{4}$ days.

Thus, the remaining work is done by B in $23 \frac{3}{4}$ days.