An automobile radiator contrains 10 qt of a mixture of water

An automobile radiator contrains 10 qt of a mixture of water and antifreeze that is 20% antifreeze. How much of this mixture must be drained out and replaced with pure antifreeze in order to obtain a 50% mixture in the radiator? Please explain.

Solution

The automobile radiator contrains 10 qt of a mixture of water and antifreeze that is 20% antifreeze so thatthis mixture has 2qt of antifreeze and 8qt. of water. Suppose x qts of this mixture is drained out to be replaced with pure antifreeze in order to obtain a 50% mixture in the radiator. Then 0.8x qts of water and 0.2x qts of antifreeze have been drained out.The amount of mixture in the radiator is constant at10qts, and since this mixture now contains 50 % antifreeze, The amount of antifreeze now is 5 qts and the amount of water is also 5 qts. Since the original quantity of water in the radiator was 8 qts, and fresh water has not been added, we have 8 - 0.8x = 5 or, 0.8x = 8 -5 = 3 so that x = 3/0.8 = 3.75qts. Thus 3.75 qts of mixture has been replaced with pure antifreeze.

Note:

3.75 qts of mixure that has been drained out contained 3.75 *80 % = 3 qts of water and 0.75 qts of antifreeze. The original quantity of antifreeze was 10*20 % = 2 qts so that the present quantity of anti freeze is 2 -o.75 + 3.75 = 5 qts. Thus, the amount of water and antifreeze are equal.