It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now
has 30 liters of 20% solution. How many liters of this should be drained and replaced with 100%
antifreeze to get the desired strength?
if x liters are drained, then you want
0.20(30-x) + x = 0.40*30
what number/30 L = 40%. So you want all of this to add up to 1200 for the number.
Let X = volume of 20% stuff to drain.
So you want (30L-XL)*20% + XL*100% = 1200
Solve for X L and I get 7.5 L. So you drain 7.5 L from the radiator. That leaves 22.5 L of 20% and you add 7.5 L of the 100% stuff to get 30 L of 40% in radiator. Not nearly as neat as oobleck's solution but both get the same answer.