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ATQ, Let the total work be 80 units (LCM of 10, 16, and 20). Combined efficiency of 'A' and 'B' =(80/10) = 8 units/day Combined efficiency of 'B' and 'C' = (80/16) =5 units/day Combined efficiency of 'A' and 'C' =(80/20) = 4 units/day So, combined efficiency of 'A', 'B', and 'C' = [(8+5+4)/2] = 8.5 units/day And, efficiency of 'B' = 8.5 - 4 = 4.5 units/day Therefore, required time ={(0.5×80)/4.5} = 9 days
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