Question
βXβ and βYβ can do a piece of work in 10 days
and 25 days, respectively. βXβ started working alone and was replaced by βYβ after 5 days. In how many days can βYβ complete the remaining amount of work alone?Solution
Let the total work be 200 units (LCM of 10 and 25) Efficiency of βXβ = 200/10 = 20 units/day Efficiency of βYβ = 200/25 = 8 units/day Let the number of days taken by βYβ to finish the remaining amount of work be βxβ ATQ, (20 Γ 5) + 8x = 200 Or, 8x = 200 β 100 Or, 8x = 100 Or, x = 100/8 = 12.5 Therefore, time taken by βYβ to complete the remaining work alone = 12.5 days
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