A can do a job in 10 days and B in 12 days. They work on alternate days starting with A. In how many days will the job be finished?
\(10\frac{5}{6}\) days
In alternate-day work we track how much of the job is done in each two-day cycle, then handle the leftover on the next turn.
A finishes the job in 10 days, so A does \(\dfrac{1}{10}\) per day; B finishes in 12 days, so B does \(\dfrac{1}{12}\) per day.
One cycle is A on day 1 and B on day 2, together \(\dfrac{1}{10} + \dfrac{1}{12} = \dfrac{6+5}{60} = \dfrac{11}{60}\) of the work.
After 5 full cycles (10 days) the work done is \(5\times \dfrac{11}{60} = \dfrac{55}{60}\).
The work still left is \(1 - \dfrac{55}{60} = \dfrac{5}{60} = \dfrac{1}{12}\).
Day 11 is A's turn again, and A's rate is \(\dfrac{1}{10}\) per day, so the time to clear the remainder is \(\dfrac{1/12}{1/10} = \dfrac{10}{12} = \dfrac{5}{6}\) day.
Adding this to the 10 days already spent gives \(10 + \dfrac{5}{6} = 10\tfrac{5}{6}\) days.
The job is finished in 10 5/6 days.
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