P, Q and R can complete a work in 10 days. 20 days and

Solution

P, Q and R can do a piece of work in 10, 20 and 30 days respectively. Total work = 60 (LCM of 10, 20 and 30) P's efficiency = 60/10 = 6 Q's efficiency = 60/20 = 3 R's efficiency = 60/30 = 2 Pattern of work = PQ, PR, PQ, PR... Work done by PQ and PR in 2 days = (6 + 3) + (6 + 2) = 17 2 days => 17 units 2 × 3 = 6 days => 17 × 3 = 51 units remaining 9 units done by PQ in = work/efficiency = 9/(6 + 3) = 1 day Total time taken = 6 + 1 = 7 days ∴ P, Q and R complete the work together when P is assisted by Q and R on every alternate day in 7 days.