Question
'A' and 'B' together can do a piece of work in 12 days
while 'B' and 'C' together can do the same work in 15 days. 'A' started the work alone and worked on it for 1 day, then 'B' alone worked on it for 8 days and 'C' alone completed the remaining work in 16 days. Find the time taken by 'B' alone to complete the whole work.Solution
Let the total work be 60 units. {LCM of 12 and 15} Combined efficiency of 'A' and 'B' = 60 ÷ 12 = 5 units/day Combined efficiency of 'B' and 'C' = 60 ÷ 15 = 4 units/day Amount of work done by 'A' in 1 day = 1 × a units
Amount of work done by 'B' in 8 days = 8 × b units
Amount of work done by 'C' in 16 days = 16 × c units We have,
a + b = 5
b + c = 4 From these, a = 5 − b and c = 4 − b Total work done = 60 units
So,
1(5 − b) + 8b + 16(4 − b) = 60
5 − b + 8b + 64 − 16b = 60
69 − 9b = 60
9b = 9
b = 1 unit/day Time taken by 'B' alone to complete the work = 60 ÷ 1 = 60 days
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