<PRE><B>Question 1100319</B>
&lt;br&gt;Tutor ikleyn is primarily a scientist, so she always starts these mixture problems with the &quot;concentration equation&quot; -- which is fine for someone who already knows how to solve this kind of problem and is familiar with that equation.
But I don&#39;t find that approach very useful for a student show is trying to LEARN HOW to solve this kind of problem.
So let me  offer a different approach.&lt;br&gt;
And the method I use is NOT the traditional algebraic approach taught in most beginning algebra classes.&lt;br&gt;
Step 1: Analyze the problem and understand what is really significant.  In this problem, it is irrelevant that some amount of the original 25% antifreeze solution is removed and replace with pure antifreeze.  In the end, the problem simply involves mixing 25% antifreeze solution with pure antifreeze to get a 40% solution.&lt;br&gt;
Step 2: Make a very rough estimate of what the answer should be.  You will be mixing 25% antifreeze with 100% antifreeze to get a 40% antifreeze solution.  Clearly you will need to use far more of the 25% solution than the pure 100% antifreeze.&lt;br&gt;
Step 3: This is where my method is very different from the traditional method....&lt;br&gt;
In step 2 we found by logical analysis that we need to use more of the 25% solution than the pure antifreeze to get a 40% solution.  We can refine that analysis to solve the problem exactly.  Specifically, the ratio in which the two ingredients must be mixed is exactly determined by where the 40% of the final mixture lies between the 25% and 100% of the two ingredients.&lt;br&gt;
The 40% is 60% away from 100%; it is 15% away from the 25%.
Those differences of 60 and 15 mean that the two ingredients must be mixed in the ratio 60:15, or 4:1.&lt;br&gt;
Since the radiator holds 20L, we must use 16L of one ingredient and 4L of the other.
And since we already logically concluded that we need more or the 25% than of the 100%, the answer is that we need 4L of the pure antifreeze and 16L of the original 25% antifreeze.&lt;br&gt;
Then finally we need to go back and answer the exact question that was asked:
4L of the original antifreeze solution must be drained and replaced with pure antifreeze to get 20L of 40% antifreeze.&lt;br&gt;
With all the words of explanation, this sounds like a long and tedious process.  But without all the words, here is all that is involved in solving this problem:&lt;br&gt;
100-40 = 60; 40-25 = 15
60:15 = 4:1 
So 16L of the original and 4L of the pure antifreeze</PRE>