**Question:**

A can do a piece of work in 14 days while B can do it in 21 days. They began together and worked at it for 6 days. Then, A fell ill and B had to complete the remaining work alone. In how many days was the work completed?

**Solution:**

Time taken by $\mathrm{A}$ to complete the work $=14$ days

Work done by $\mathrm{A}$ in one day $=\frac{1}{14}$

Time taken by $\mathrm{B}$ to complete the work $=21$ days

Work done by $\mathrm{B}$ in one day $=\frac{1}{21}$

Work done jointly by $\mathrm{A}$ and $\mathrm{B}$ in one day $=\frac{1}{14}+\frac{1}{21}=\frac{3+2}{42}=\frac{5}{42}$

Work done by $\mathrm{A}$ and $\mathrm{B}$ in 6 days $=\frac{5}{42} \times 6=\frac{5}{7}$

Work left $=1-\frac{5}{7}=\frac{2}{7}$

With B working alone, time required to complete the work $=\frac{2}{7} \div \frac{1}{21}=\frac{2}{7} \times 21=2 \times 3=6$ days

So, the total time taken to complete the work $=6+6=12$ days