**Question:**

*Tick (✓) the correct answer:*

A can do a piece of work in 25 days, which B alone can do in 20 days. A started the work and was joined by B after 10 days. The work lasted for

(a) $12 \frac{1}{2}$ days

(b) 15 days

(c) $16 \frac{2}{3}$ days

(d) 14 days

**Solution:**

(c) $16 \frac{2}{3}$ days

A can do the piece of work in 25 days.

Work done by A in 1 day $=\frac{1}{25}$

B can do the same work in 20 days.

Work done by B in 1 day $=\frac{1}{20}$

A alone completes $\frac{10}{25}, \mathrm{i}, \mathrm{e} ., \frac{2}{5}$ of the work in 10 days.

Now, work remaining $=1-\frac{2}{5}=\frac{3}{5}$ Work done by $(\mathrm{A}+\mathrm{B})$ in 1 day $=\frac{1}{25}$$+\frac{1}{20}=\frac{9}{100}$

$\therefore$ Time taken if they work together $=\frac{3}{5} \div \frac{9}{100}=\frac{3}{5} \times \frac{100}{9}=\frac{20}{3}=6 \frac{2}{3}$ days