<PRE><B>Question 1207753</B>
&lt;br&gt;
Your setup is NOT correct.&lt;br&gt;
Your setup says you are mixing 15 liters of 40% antifreeze with an unknown amount of pure (100%) antifreeze to get (15+x) liters of 60% antifreeze.&lt;br&gt;
The capacity of the cooling system is 15 liters, so that can&#39;t be right.&lt;br&gt;
The x liters of pure antifreeze can only be put into the cooling system after x liters of the 40% antifreeze is drained out.  So you are mixing (15-x) liters of the 40% antifreeze with x liters of pure antifreeze to get 15 liters of 60% antifreeze:&lt;br&gt;
(15-x)(.40)+x(1.00)=15(.60)&lt;br&gt;
Solve as shown in the response from the other tutor.&lt;br&gt;
If formal algebra is not required, here is a solution using a quick and easy informal method that can be used to solve any 2-part mixture problem like this.&lt;br&gt;
You are mixing 40% antifreeze with 100% antifreeze to get 60% antifreeze.
Look at the three percentages (on a number line, if it helps) and observe that 60% is 1/3 of the way from 40% to 100%.
That means 1/3 of the mixture needs to be the higher percentage antifreeze.&lt;br&gt;
1/3 of 15 liters is 5 liters, so the mixture needs to use 5 liters of pure antifreeze.  That of course means 5 liters of the 40% antifreeze must be drained to make room for the pure antifreeze.&lt;br&gt;
ANSWER: 5 liters&lt;br&gt;
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