Question
Working individually, 'A' can finish 60% of a task in 18
days, and 'B' takes 15 days longer than 'A' to complete the entire task. However, 'B' and 'C' working together can complete the task in 20 days. Now, What will be the amount of time 'C' would need to complete the same task when working alone.Solution
ATQ, Time taken by 'A' alone to complete the entire work = 18 ÷ 0.6 = 30 days Time taken by 'B' alone to complete the entire work = 30 + 15 = 45 days Let the total work = L.C.M of 30, 45 and 20 = 180 units Then, efficiency of 'B' = (180/45) = 4 units/day Combined efficiency of 'B' and 'C' = (180/20) = 9 units/day So, efficiency of 'C' alone = 9 - 4 = 5 units/day So, time taken by 'C' alone to complete the entire work = (180/5) = 36 days
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