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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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Number of days A required do a piece of work : 14 days 

Number of days B required do a piece of work : 21 days

Work done by A in one day: \(\frac{1}{14}\)

Work done by B in one day: \(\frac{1}{21}\)

Work done by A and B together in one day: \(\frac{1}{14}+\frac{1}{21}=\frac{5}{42}\)

They can do the work together in \(\frac{42}{5}\) days .

A and B worked together for 6 days, so work completed by A and B in 6 days : 6 x \(\frac{5}{42}=\frac{5}{7}\)

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

Number of days taken by B to complete the left over work : \(\frac{2}{7}\times21\) = 6

6 ( here 21 is days required by B to complete a piece of work).

∴ Time taken to finish the work: 6 + 6 = 12 days.

∴ Total time taken to finish the work: 12 days

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