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?
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
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