A cistern has 3 pipes A, B and C. A and B call fill it in 3 and hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at

part of the cistern filled by A in 2 hrs = $$ \frac{1}{3} \times 2 = \frac{2}{3} $$

part of cistern filled by B in 1 hr = $$ \frac{1}{4} $$

when all the pipes are opened, net part filled in 1 hr = $$ \frac{1}{3} + \frac{1}{4} - 1 = \frac{4 + 3 - 12}{12} = \frac{5}{12} $$

$$ \frac{5}{12} $$ part is emptied per hour

time taken to empty $$ \frac{11}{12} $$ part = $$ \frac{11}{12} \times \frac{12}{11} = \frac{11}{5} $$ hrs

                                         = 2 hrs 12 min

required time = 5 + 2 : 12 = 7 : 12 pm