posted by coried on .
Radiator in a certain make needs to contain 50 liters of 40% antifreeze. The radiator now contains 50 liters of 20% antifreeze. How many liters of this solution must be drained and replaced with 100% antifreeze to get the desired strength?
Let X be the number of liters that must be drained and replaced with 100% antifreeze.
[0.2(50-X)+X]/50 = 0.4
Solve that for X.
You want to double the concentration?
If you remove X liters, you remove .2X antifreeze, and .8X water.
You have left in the radiator after draining X, 10-.2X liters of pure antifreeze. You need a total of 20 liters to get 40percent antifreeze.
So you add 10+.2X of antifreeze.
You drain out 10-2X of solution.