**Solution:**
In one hour,

the large inlet pipe fills 1 / 2 of the tank;

the small inlet pipe fills 1 / 4 of the tank;

the outlet pipe empties 1 / 6 of the tank; and therefore

all three pipes together fill [ (1 / 2) + (1 / 4) − (1 / 6) ] of the tank.

Fraction of the tank that will be filled in 1.29 hours =

1.29 [ (1 / 2) + (1 / 4) − (1 / 6) ] = 0.75.

**Alternative Solution through Fundamental Equations:**
It it important to note that

Flow Rate = Volume / Time | ... equation (1) |

Rate of Accumulation = Input Rate − Output Rate | ... equation (2) |

Let

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

Flow Rate (large inlet pipe) =

*V* / 2

Flow Rate (small inlet pipe) =

*V* / 4

Flow Rate (outlet pipe) =

*V* / 6.

Substituting in equation (2),

Rate of Accumulation in tank = (

*V* / 2) + (

*V* / 4) − (

*V* / 6).

Using the above result in equation (1),

Time required to fill the complete tank =

*V* / [ (

*V* / 2) + (

*V* / 4) − (

*V* / 6) ].

Note that

*V* cancels out on simplifying the above expression.

Fraction of the tank that will be filled in 1.29 hours =

1.29 [ (1 / 2) + (1 / 4) − (1 / 6) ] = 0.75.

**Food for thought:**
Can you generalize the problem to the case of an arbitrary number of input and output pipes? It's not very difficult!

How realistic is the assumption of constant flow rates? Will the flow rate through the outlet pipe necessarily depend on the level of water in the tank? Does it matter whether the tank is emptied using gravity or with a pump?