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
(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
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