Question 454937
Six hundred liters of 50% alcohol solution was diluted to make a 40% alcohol solution.
 How many liters of the 50% solution needed to be replaced by pure water in
 order to bring the alcohol concentration down to 40%?
:
Let x = amt or solution removed and amt of pure water added to replace it
:
Remember pure water is a 0% solution
:
.50(600-x) = .40(600)
300 - .50x = 240
-.50x = 240 - 300
-.50x = -60
x = {{{(-60)/(-.50)}}}
x = +120 liters of 50% solution removed and 120 liters of pure water added.