Question
βMβ is 60% more efficient than βNβ. βNβ
started the work alone and worked for 20 days and after that βNβ is replaced by βMβ. If the total work is completed in 30 days, find the time taken by βMβ to finish the same work alone?Solution
Let the efficiency of βNβ = β5xβ units/day Efficiency of βMβ = β8xβ units/day Time taken by βMβ to complete the remaining work = 30 β 20 = 10 days Total work = 20 Γ 5x + 10 Γ 8x = 180x units Desired time = 180x/8x = 22.5 days
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