Question
'A' and 'B' together can do a piece of work in 15 days
while 'B' and 'C' together can do the same work in 12 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 14 days. Find the time taken by 'B' alone to complete the whole work.Solution
Let the total work be 60 units. {LCM of 15 and 12} Combined efficiency of 'A' and 'B' = 60 ÷ 15 = 4 units/day Combined efficiency of 'B' and 'C' = 60 ÷ 12 = 5 units/day Work by A in 1 day = a units
Work by B in 8 days = 8b units
Work by C in 14 days = 14c units a + b = 4
b + c = 5 So, a = 4 − b and c = 5 − b (4 − b) + 8b + 14(5 − b) = 60
4 − b + 8b + 70 − 14b = 60
74 − 7b = 60
7b = 14
b = 2 units/day Time taken by 'B' alone = 60 ÷ 2 = 30 days
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