A and B can complete a piece of work in 12 and 18 days
A and B can complete a piece of work in 12 and 18 days respectively. A begins to do the work and they work alternatively one at a time for one day each. The whole work will be completed in
Answer is: \(14 \frac{1}{3} days\)
Explanation:
A can complete total work in 12 days
One day work of A = \(\frac{1}{2}\)part
B can complete work in 18 days
One day work of B = \(\frac{1}{18}\)part
A and B together can complete work in one day,
=\(\frac{1}{12} + \frac{1}{18}\)
=\(\frac{18+12}{216}\)=\(\frac{30}{216}\)=\(\frac{15}{108}\)
=\(\frac{5}{36}\)part
Evidently, the work done by A and B during 7 pairs (14 days)
=\(7 \times \frac{5}{36} = \frac{35}{36}\)
Remaining work = \(\big(1-\frac{35}{36}\big) = \frac{1}{36}\)
Now, on 15th day it is A's turn.
Now \(\frac{1}{12}\)work is done by A in 1 day.
\(\frac{1}{36}\) work will be done by A in \(\big[12 \times \frac{1}{36}\big] = \frac{1}{3}\) day
So, total time taken = \(14 \frac{1}{3}\) days
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