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Sagot :
Let V be the volume of the tank. The tap fills the tank at a rate of V/(4 hours), or V/4 per hour. The hole empties the tank at a rate of V per hour. If both the tap and hole work together, the net rate of water flow is V/4 - V or -3V/4 per hour.
Starting with a full tank, the volume v of water at time t is given by the function
v(t) = V + (-3V/4) t
Solve v(t) = 0 to find the time it takes to empty:
V - 3V/4 t = 0
V = 3V/4 t
t = V / (3V/4) = 4/3
So it takes 4/3 hours = 80 minutes to empty a full tank.
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