The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 3.6L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?
We start with 3.5 liters of liquid
We drain out x liters of the liquid.
That leaves 3.5-x liters of the liquid.
Those 3.5-x liters of the liquid contains 60% antifreeze.
Therefore those 3.5-x liters of the liquid left in the radiator contain
.60(3.5-x) liters of antifreeze.
Now we fill up the radiator with water, and when we do we will have
3.5 liters of liquid, and 50% of those 3.5 liters, or .50(3.5) liters,
of liquid now will be the .60(3.5-x) liters of antifreeze left in the
radiator.
So we set those equal:
.50(3.5) = .60(3.5-x)
Solve that and get 7/12 of a liter or about .58 liters.
Edwin