Question
βPβ, βQβ, and βRβ can
complete a task in 10 days, 15 days, and 20 days, respectively. βPβ starts the task alone and is joined by βQβ on the second day. On the third day, βQβ is replaced by βRβ, and again on the fourth day, βPβ works alone. The process repeats in the same manner. Find the total time taken to complete the task. (Round off to two decimal places)Solution
ATQ,
Let the total work be the LCM of [10, 15, 20] = 60 units. Efficiency of βPβ = 60/10 = 6 units/day. Efficiency of βQβ = 60/15 = 4 units/day. Efficiency of βRβ = 60/20 = 3 units/day. Work done on the first day = 6 Γ 1 = 6 units. Work done on the second day = (6 + 4) Γ 1 = 10 units. Work done on the third day = (6 + 3) Γ 1 = 9 units. Total work done in the first three days = 6 + 10 + 9 = 25 units. Remaining work = 60 β 25 = 35 units. Time taken to complete the remaining work = (6/6) + (10/10) + (9/9) + (10/17) β 3 + (10/17) β 3.59 days. Total time = 3 + 3.59 = 6.59 days.
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