The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 3.6 L, how much coolant should be drained and replaced by water to reduce the antifreeze concentration to the recommended level?

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Irene Simon · Accepted Answer

Since the radiator is filled with 3.6L already with 60%antifeeze and 40% water, you simply multiply   3.6  L  ×  .60  =  2.16  L of antifreeze. You multiply   3.6  L  ×  .40  =  1.44  L of water. With percentages, you simply put themover 100 and get a decimal. Since you want a 50% solution ofantifreeze and water, you multiply   3.6  L  ×  .50  =  1.8  L of each solution. From here, you can subtract 2.16L - 1.8L to find the amount of antifreeze you need to drain,and this is the same amount of water you need to replace the antifreeze with to obtain a 50%-50% solution of antifreeze and water.

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