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Let the total work = L.C.M of 12, 20 and 5 = 60 units Then, combined efficiency of 'A' and 'B' = 60 ÷ 12 = 5 units/day Combined efficiency of 'B' and 'C' = 60 ÷ 20 = 3 units/day Since, 33(1/3) % =1/3 So, time taken by 'A' and 'C' together to complete the entire work = 5 ÷ (1/3) = 15 days So, combined efficiency of 'A' and 'C' = 60 ÷ 15 = 4 units/day Twice the efficiency of 'C' = (3 + 4) - 5 = 2 units/day So, efficiency of 'C' = 2 ÷ 2 = 1 unit/day So, time taken by 'C' alone to complete the entire work = 60 ÷ 1 = 60 days
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