Let us name the pipes as A, B and C
\(T_A = 12\) hours
\(T_B = 18\) hours
\(T_C = 36\) hours
We know Rate, \(R = \frac{1}{T}\)
Therefore,
\(R_A = \frac{1}{12}\)
\(R_B = \frac{1}{18}\)
\(R_C = \frac{1}{36}\)
contractor opens the first pipe and after 2 hours, the second pipe. At the end of 6 hours, he realized that the third pipe is open tooi.e. A worked for 6 hours, B worked for 4 hours and C worked for 6 hours too
So, their combined work = \(\frac{6}{12} + \frac{4}{18} - \frac{6}{36} = \frac{5}{9}\)
(NOTE: -ve sign indicates that pipe C is doing opposite work)Since, these pipes have filled \(\frac{5}{9}^{th}\) of the tank in 6 hours, the remaining work is \(1 - \frac{5}{9} = \frac{4}{9}\)
i.e. \(\frac{4}{9}^{th}\) of the work is to be completed by A and B together
\(R_{A+B} = \frac{1}{12} + \frac{1}{18} = \frac{5}{36}\)
\(Rate = \frac{Work}{Time} = (\frac{4}{9})(\frac{36}{5}) = \frac{16}{5}\) hours
TOTAL time required to fill the tank = 6 hours + \(\frac{16}{5}\) hours = \(\frac{46}{5}\) hours = \(\frac{46}{5}(60)\) minutes = \(552\) minutes = \(9\) hours \(12\) minutes
_________________
I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/