Three pipes P, Q and R which are used to fill a tank. Each of the three pipes can be used either as an inlet pipe or as an outlet pipe. Each of the pipes P, Q and R can individually fill/empty the tank in 18 hours, 27 hours and 30 hours respectively. On a given day one pipe becomes outlet pipe and the remaining pipes become inlet and this continues alternatively. If on first day pipe P becomes outlet pipe, next day pipe Q and pipe R the following day, then in how many days will the tank be filled?

Solution

Pipe P, Q and R can fill/empty the tank in 18, 27 and 30 hrs respectively. Let the total capacity of the tank = 270 units (LCM of 18, 27 and 30) Pipe P can fill/empty (270/18) = 15 units/hrs Pipe Q can fill/empty (270/27) = 10 units/hrs Pipe R can fill/empty (270/30) = 9 units/hrs Total work done on day 1 = - 15 + 10 + 9 = + 4 units Total work done on day 2 = 15 – 10 + 9 = + 14 units Total work done on day 3 = 15 + 10 – 9 = + 16 units In 3 days, (4 + 14 + 16) = 34 units work is done. In 3 x 7 = 21 days, (34 x 7) = 238 units work is done. On 22^nd day, Pipe P works as outlet. Therefore, In 22 days, 238 + 4 = 242 units work is done. Remaining work = 270 – 242 = 28 units On 23^rd day, Q works as outlet pipe = 14 units work is done On 24th day, R works as outlet pipe = 14 units work is done Therefore, total time taken = 23 (7/8) days