How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 55% antifreeze?

To find this solution, we have two fractions: .55/1 = 1/x. Cross-multiplying will give us a value of .55x = 1. Dividing out the .55 to isolate the x term (the amount of the total solution, water plus antifreeze) gives us x = (1/0.55). This improper fraction is approximately equal to 1.818 gallons in total. This means that approximately 0.818 gallons of water needs to be added to the antifreeze to dilute it to a 55% concentration.