2 answers
(click or scroll down)
if x liters are drained, then you want
0.20(30-x) + x = 0.40*30
has 30 liters of 20% solution. How many liters of this should be drained and replaced with 100%
antifreeze to get the desired strength?
0.20(30-x) + x = 0.40*30
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.