A, B and C can finish a piece of work in 12 days, 15 days and 20 days respectively. If 1st day A+B, 2nd day B+C, 3rd day C+A work and they work in the same manner, then in how many days the work will be finished ?

\(7\frac{3}{7}\)

1st day work ( A + B ) = 9R

2nd day work ( B + C ) = 7R

3rd day work ( C + A ) = 8R

total 3 days work = 24R

6 days work = 48R

7th day ( A + B ) will work i.e. 9R

remaining work = 60R - 57R = 3R

on next day ( B + C ) will work with 7R efficiency

time = \(\frac{3R}{7R}\) = \(\frac{3}{7}\)

total days taken = 7 + \(\frac{3}{7}\)  = 7\(\frac{3}{7}\) days