ATQ, Let the total work be 90 units (LCM of 18 and 30) So, efficiency of 'C' = (90/18) = 5 units/day And efficiency of 'D' = (90/30) = 3 units/day New efficiency of 'C' = 60% of 5 = (0.6) X 5 = 3 units/day New efficiency of 'D' = 50% of 3 = (0.5) X 3 = 1.5 units/day So, required time =(90/(3+1.5) = 90/4.5 = 20 days
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