Question 163719
R = radiator = 36 Liters of mixture.
A = Antifreeze = .04 * 36 = 1.44 Liters of antifreeze
W = Water = .96 * 36 = 34.56 Liters of Water.
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a mixture of .20 Antifreeze would take 7.2 Liters of Antifreeze (.2*36).
this mixture would have 28.8 Liters of Water (.8*36).
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you need to add (7.2 - 1.44) Liters of Antifreeze which equals 5.76 Liters of Antifreeze.
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you also need to remove (34.56 - 28.8) Liters of Water which equals 5.76 Liters of Water.
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In order to remove 5.76 Liters of Water, you have to remove 5.76/.96 Liters of the original mixture.  This equals 6 Liters of the original mixture.
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Removing 6 Liters of the original mixture also removes .04*6 Liters of Antifreeze.  This equals .24 Liters of Antifreeze.
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Now you have to add 5.76 Liters of Antifreeze plus .24 Liters of Antifreeze that you removed.  This equals 6 Liters of Antifreeze which just happens to be the amount of mixture that you removed.
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Your new mixture contains the following:
6 Liters of Antifreeze you added.
30 Liters of the original mixture.
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30 Liters of the original mixture contains .04 * 30 Liters of Antifreeze which equals 1.2 Liters of Antifreeze.
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6 Liters of Antifreeze you added plus 1.2 Liters of Antifreeze still in there makes 7.2 Liters of Antifreeze.
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You now have .2 * 36 Liters of Antifreeze = 7.2 Liters.
The rest is water which is .8 * 36 Liters of Water = 28.8 Liters.
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if there's an easier way to do this i haven't figured it out yet.
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Answer is you have to remove 6 Liters of the original mixture and you have to add 6 Liters of pure Antifreeze.
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