ATQ, Let the total work = 180 units {LCM of 15, 20 and 18} Efficiency of ‘John’ = (180/15) = 12 units/day Efficiency of ‘Emily’ = (180/20) = 9 units/day Efficiency of ‘Michael’ = (180/18) = 10 units/day So, work completed in 3 days = (12 + 9 + 10) = 31 units Work completed in 15 days = 31 × 5 = 155 units Time taken by ‘C’ to complete the remaining work = (180 – 155)/10 = 2.5 days Therefore, total time taken = 15 + 2.5 = 17.5 days
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