Two pipes can fill a tank in 5 and 6 hours respectively. A third pipe will empty it in 12 hours. If all the pipes are opened, then find the time taken to either fill or empty the tank.

$$ work = efficiency \times time $$

P1,P2 and P3 be the pipes

P1 fill the tank in 5 hrs

P2 fill the tank in 6 hrs

P3 empty the tank in 12 hrs

work done = LCM of P1,P2 and P3 = LCM (5,6,12) = 60

efficiency of P1 = $$ \frac{60}{5} = 12 $$

efficiency of P2 = $$ \frac{60}{6} = 10 $$

efficiency of P3 = $$ \frac{60}{12} = 5 $$

efficiency of P1 + P2 + P3 = 12 + 10 - 5 = 17

$$ 60 = 17 \times time $$

$$ time = \frac{60}{17} hrs $$