Question 1158615
<br>
Algebraically....<br>
x mL of 0% solution, plus 60 mL of 60% solution, equals (x+60) mL of 40% solution:<br>
{{{0(x)+.60(60) = .40(x+60)}}}<br>
Solve using relatively simple algebra.<br>
And here is an informal way to solve this kind of mixture problem quickly and easily.<br>
You are starting with 60% solution and adding 0% solution, stopping when you get to 40%.
40 is one-third of the way from 60 to 0.
Therefore, 1/3 of the mixture is what you are adding.
That means the ratio of 60% solution to 0% solution is 2:1.
Since the amount of 60% solution is 60 mL, the amount of 0% solution required is half of that, which is 30mL.<br>
Another variation of the informal approach is to reason that the target 40% is "twice as close" to 60% as it is to 0%; and therefore the mixture needs to have twice as much of the 60% solution as it has of the 0% solution.  So the 60 mL of the 60% solution is twice as much as the amount you need of the 0% solution, leading again to the answer that you need 30 mL of the 0% solution.<br>