A large water tank has two inlet pipes (a large one and a small one) and one outlet pipe. It takes 3 hours to fill the tank with the large inlet pipe. On the other hand, it takes 6 hours to fill the tank with the small inlet pipe. The outlet pipe allows the full tank to be emptied in 9 hours.

What fraction of the tank (initially empty) will be filled in 0.64 hours if all three pipes are in operation? Give your answer to two decimal places (e.g., 0.25, 0.5, or 0.75).

In one hour,

Soln 1:

the large inlet pipe fills 1 / 3 of the tank;

the small inlet pipe fills 1 / 6 of the tank;

the outlet pipe empties 1 / 9 of the tank; and therefore

all three pipes together fill [ (1 / 3) + (1 / 6) − (1 / 9) ] of the tank.

Fraction of the tank that will be filled in 0.64 hours =

0.64 [ (1 / 3) + (1 / 6) − (1 / 9) ] = 0.25.

Soln 2:

Let V be the total volume of the tank. From equation (1),

Flow Rate (large inlet pipe) = V / 3

Flow Rate (small inlet pipe) = V / 6

Flow Rate (outlet pipe) = V / 9.

Substituting in equation (2),

Rate of Accumulation in tank = (V / 3) + (V / 6) − (V / 9).

Using the above result in equation (1),

Time required to fill the complete tank = V / [ (V / 3) + (V / 6) − (V / 9) ].

Note that V cancels out on simplifying the above expression.

Fraction of the tank that will be filled in 0.64 hours =

0.64 [ (1 / 3) + (1 / 6) − (1 / 9) ] = 0.25.