A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work?
It is given that A can finish the work in 20 days and B can finish the same work in 15 days.
$\therefore$ Work done by A in 1 day $=\frac{1}{20}$
Work done by B in 1 day $=\frac{1}{15}$
$\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 1 day $=\frac{1}{20}+\frac{1}{15}$
$=\frac{3+4}{60}=\frac{7}{60}$
$\therefore$ Work done by $(\mathrm{A}+\mathrm{B})$ in 2 days $=\frac{14}{60}=\frac{7}{30}$
Remaining work $=1-\frac{7}{30}=\frac{23}{30}$
It is given that the remaining work is done by B.
$\because$ Complete work is done by B in 15 days.
$\therefore \frac{23}{30}$ of the work will be done by $\mathrm{B}$ in $\left(15 \times \frac{23}{30}\right)$ days or $\frac{23}{2}$ days or $11 \frac{1}{2}$ days.
Thus, the remaining work is done by B in $11 \frac{1}{2}$ days.