A large petrol tank has two inlet tubes (a large one and a small one) and one outlet tube. It takes 3 hours to fill the tank with the large inlet tube. On the other hand, it takes 6 hours to fill the tank with the small inlet tube. The outlet tube 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 tubes are in operation? Give your answer to two decimal places (e.g., 0.25, 0.5, or 0.75).
Solution:
In one hour,
Soln 1:
the large inlet tube fills 1 / 3 of the tank;
the small inlet tube fills 1 / 6 of the tank;
the outlet tube empties 1 / 9 of the tank; and therefore
all three tubes 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 tube) = V / 3
Flow Rate (small inlet tube) = V / 6
Flow Rate (outlet tube) = 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.