Question
Individual βMβ can do a piece of work in 40 days,
while βMβ and βNβ together can do it in 16 days. If βNβ worked alone for 10 days and left, how many days will βMβ take to complete the remaining work?Solution
Let the total work = 80 units {L.C.M of 40 and 16}
Then, efficiency of βMβ = 80 Γ· 40 = 2 units/day
Combined efficiency of βMβ and βNβ = 80 Γ· 16 = 5 units/day
So, efficiency of βNβ = 5 β 2 = 3 units/day
So, work done by βNβ alone in 10 days = 3 Γ 10 = 30 units
Remaining work = 80 β 30 = 50 units
Time taken by 'M' alone to complete the remaining work = 50 Γ· 2 = 25 days
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