Question 1118139
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The problem is presented as a drain and replace problem just to confuse you.  What you really have is just a mixture problem where you want to mix 20% antifreeze with 100% antifreeze to get 70 liters of 40% antifreeze.<br>
There are many ways of solving problems like this where you are mixing two ingredients.  The response from the other tutor shows a typical algebraic method.<br>
In my opinion, there is a much easier and faster way to solve this kind of problem, using the ratio in which the two ingredients need to be mixed.<br>
For this problem, we are mixing 20% and 100% antifreeze ingredients to get a 40% antifreeze mixture.  The ratio in which the ingredients must be mixed is directly related to where the 40% lies between the 20% and the 100%.<br>
To be short with the explanation, 40% is 1/4 of the way from 20% to 100%.  (20% to 40% is 20%; 20% to 100% is 80%; 20% is 1/4 of 80%.)  That means 1/4 of the mixture must be the 100% ingredient.<br>
So 1/4 of the 70 liters, or 17.5 liters, is the 100% antifreeze that you added; the remaining 3/4 of the 70 liters, or 52.5 liters, is the 20% antifreeze that was already in the radiator.<br>
To put that result in the terms in which the question was asked, 17.5 liters of the 20% antifreeze must be drained and replaced with 100% antifreeze to get 70 liters of 40% antifreeze.<br>
So in the end here is all that is required to solve this problem:<br>
40-20 = 20; 100-20 = 80; 20/80 = 1/4; 1/4 of 70 = 17.5<br>
If your mental arithmetic is good, it will take you about 10 seconds to work the problem.