Question 427227
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Here are two informal solutions using logical reasoning instead of formal algebra.<br>
(1)<br>
At the start, the 25-quart radiator is full of 60% antifreeze, so it contains .60(25) = 15 quarts of antifreeze.<br>
In the end, we want it to be full of 40% antifreeze, so it will contain .40(25) = 10 quarts of antifreeze.<br>
The amount of antifreeze we want to finish with (10 quarts) is 2/3 of the 15 quarts we started with; since we are adding water which contains no antifreeze, we want the radiator to finish with 2/3 of the original antifreeze mixture, which means we want to drain 1/3 of the original mixture and replace it with water.  1/3 of 25 quarts is  8 1/3 quarts.<br>
ANSWER: 8 1/3 quarts<br>
(2)<br>
We are mixing 60% antifreeze with 0% antifreeze to obtain a mixture that is 40% antifreeze.  40% is twice as close to 60% as it is to 0%, so the amount of the original antifreeze mixture must be twice as much as the added water -- i.e., 2/3 of the final mixture must be the original 60% antifreeze and 1/3 must be the added water.  Again, 1/3 of 25 quarts is 8 1/3 quarts.<br>
ANSWER (again, of course): 8 1/3 quarts<br>