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Let total work = 20 units (LCM of 20 and 10) Efficiency of ‘A’ = (20/20) = 1 units/day Efficiency of ‘B’ = (20/10) = 2 units/day Combined efficiency of ‘A’, ‘B’ and ‘C’ = (20/5) = 4 units/day Efficiency of ‘C’ = 4 – (1 + 2) = 1 units/day Required time taken = (20/1) = 20 days