A and B can complete a piece of work in 12 days and 18 days respectively. They work together for 4 days, after which A leaves the work. B continues alone for 6 days, and then C joins B. If C alone can complete the whole work in 9 days, find the total time (in days) taken to complete the work.
\(10\frac{2}{3}\)
Take the whole work as 36 units (LCM of 12, 18 and 9). Then A does 3 units/day, B does 2 units/day and C does 4 units/day.
A and B together for 4 days: 4 × (3 + 2) = 20 units.
B alone for 6 days: 6 × 2 = 12 units. Work done so far = 20 + 12 = 32 units, so 36 – 32 = 4 units remain.
B and C together do 2 + 4 = 6 units/day, so they need \(\frac{4}{6} = \frac{2}{3}\) day for the rest.
Total time = 4 + 6 + \(\frac{2}{3} = 10\frac{2}{3}\) days.
Check: \(20 + 12 + 6 \times \frac{2}{3} = 20 + 12 + 4 = 36\) units, which is the whole work.
Hence, \(10\frac{2}{3}\).
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