Question
‘R’ and ‘S’ can do a piece of work in 20 days
and 30 days, respectively. ‘R’ started working alone and was replaced by ‘S’ after 10 days. In how many days can ‘S’ complete the remaining amount of work alone?Solution
Let the total work be 60 units (LCM of 20 and 30) Efficiency of ‘R’ = 60/20 = 3 units/day Efficiency of ‘S’ = 60/30 = 2 units/day Let the number of days taken by ‘S’ to finish the remaining amount of work be ‘x’ ATQ, (3 × 10) + 2x = 60 Or, 2x = 60 – 30 Or, 2x = 30 Or, x = 30/2 = 15 Therefore, time taken by ‘S’ to complete the remaining work alone = 15 days
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