P, Q, and R can complete a work in 12 days. 15 days and 30 days, respectively, working alone. How soon can the work be completed if P is assisted by Q and R on alternate days?
Total work = 60 (LCM of 12, 15 and 30) P's efficiency = 60/12 = 5 Q's efficiency = 60/15 = 4 R's efficiency = 60/30 = 2 Pattern of work = PQ, PR, PQ, PR... Work done by PQ and PR in 2 days = (5 + 4) + (5 + 2) = 16 2 days => 16 units 2 × 3 = 6 days => 16 × 3 = 48 units remaining in 12 units done by PQ in = work/efficiency = 9/ (5 + 4) = 1 days Total time taken = 6 + 1 = 7 days Now remaining 3unit work done by PR = 3/5+2=3/7day 7+3/7 =52/7days.
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