Question 919931
A radiator contains 25 quarts of a water and antifreeze solution, of which 20% (by volume) is antifreeze. How much of this solution should be drained and replaced with pure antifreeze for the new solution to be 60% (by volume) antifreeze?
***
let x=amt of 20% antifreeze to be drained and replaced
25-x=amt remaining in radiator.
20%(25-x)+100%x=60%*25
5-.20x+x=15
0.80x=10
x=10/.80=12.5
amt of 20% antifreeze to be drained and replaced=12.5 quarts