Question 803689
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?
 .6(w+1) = 1
.6w + .6 = 1
.6w = 1 - .6
.6w = .4
w = .4/.6
w = {{{2/3}}} gal of water required to have 60% solution