Let's define some variables 1) V = volume of water the radiator can hold = 3.6L 2) CWinter = concentration of antifreeze for Winter = 65% = .65*3.6 = 2.34L 2a) Concentration of water in Winter = 100-65 = 35% = .35*36 = 1.26L 3) CSummer = concentration of antifreeze for Summer = 50% = .50*3.6 = 1.80L 3a) Concentration of water in Winter = 100-50 = 50% = 1.80L Concentration = Liters of Antifreeze/Volume in radiator For summer we want 1.8L of antifreeze and 1.8L of water. To get 1.8L of antifreeze from the winter mixture we need to drain the radiator to the volume of antifreeze from 2.34L to 1.8L or a ratio of 1.8/2.34 = .769, or a 23.1% reduction. Since you can't separate the water and antifreeze you must reduce the same percent of water, thus you must drain 23.1% of the winter mixture or, 3.6*.231 = .832L. Since you wish the radiator to be full during the summer too, you must add back .832L of water.

Let x be the amount you need to drain and replace with pure water.

65(3.6-x) = 50(3.6)

Why does this set-up work? 

I have faith you can complete the rest...have a great day.