Question 846874
Let C = capacity and amount of volume of solution in the radiator.
Let H = fraction of antifreeze solution as pure antifreeze the radiator HAS initially;
Let W = fraction of antifreeze solution WANTED.
Let v = volume of antifreeze solution to drain and replace with pure antifreeze.
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The values of the described variables are:
Let C = 25 L
Let H = 1/5
Let W = 3/5
Let v = unknown to solve, Liters
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{{{highlight((CH-Hv+1*v)/C=W)}}}.
If you can understand how to obtain this equation, then that is most of the solution taken care of.  You then just solve for v, and substitute the values for the other variables and compute the value.
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If you do not understand the equation, then some help is needed for it.  The equation simply is the ratio of the amount of pure antifreeze to the quantity of antifreeze solution to be held in the radiator.