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?

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Solution

Time taken by A to complete the work $= 14$ days
i.e., Work done by A in one day $= \frac{1}{14}$
Time taken by B to complete the work $= 21$ days
i.e., Work done by B in one day $= \frac{1}{21}$
Work done jointly by A and B in one day $= \frac{1}{14} + \frac{1}{21}$
$= \frac{3 + 2}{42} = \frac{5}{42}$
Work done by A and B in $6$ days $= \frac{5}{42} \times 6 = \frac{5}{7}$
Work left $= 1 - \frac{5}{7} = \frac{7 - 5}{7} = \frac{2}{7}$
With B working alone, time required to complete the work
$= \frac{2}{7} \div \frac{1}{21}$ [Since, Work done by B in one day is $\frac{1}{21}$ ]
$= \frac{2}{7} \times 21$
$= 2 \times 3$
$= 6$ days
Hence, B complete the remaining work in $6$ days, previous work done By A and B in $6$ days.
So, the total time taken to complete the work $= 6 + 6 = 12$ days

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