Question
A contractor assigned a job to three persons A, B and C.
βAβ which is 25% less efficient than βBβ can complete 20% of a work in 8 days. βCβ takes 10 days more than βBβ to complete the same work. βAβ and βBβ started working together and after 4 days C joined them. Due to some personal emergency βAβ left after 5 more days and rest of the work is completed by 'B' and 'C', together. Find the total time taken to complete the whole work.Solution
Time taken by βAβ to complete the work alone = 8/0.2 = 40 days Time taken by βBβ to complete the work alone = 0.75 Γ 40 = 30 days Time taken by βCβ to complete the work alone = 30 + 10 = 40 days Let, the total work = 120 units Efficiency of βAβ = 120/40 = 3 units/day Efficiency of βBβ = 120/30 = 4 units/day Efficiency of βCβ = 120/40 = 3 units/day Work completed in 4 days = 4 Γ (3 + 4) = 28 units Work completed in next 5 days = 5 Γ (3 + 4 + 3) = 50 units Time taken by βBβ and βCβ to complete remaining work together = (120 β 78)/7 = 6 days Therefore, total time taken = 4 + 5 + 6 = 15 days
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