SOLUTION: If Solution A contains 45% hydrogen and Solution B contains 69% hydrogen. How much of each solution should be mixed in order to create 150 liters of a solution that contains 61% hy

Question 1132117: If Solution A contains 45% hydrogen and Solution B contains 69% hydrogen. How much of each solution should be mixed in order to create 150 liters of a solution that contains 61% hydrogen?

Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
One way: there is a 24% difference absolutely between the two solutions. The 61% is 2/3 of the way to 69%, so it should be 2/3 of the amount.
x=45%
150-x=69%
.45x+(150-x)*.69=150*0.61=91.5
.45x+103.5-.69x=91.5
-.24x=-12
x=50 liters of 45%
150-x=100 liters of 69%
(100/150) is 2/3.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Let me expand a bit on the very good solution provided by tutor @boreal.

In his response, he showed two different ways to get the answer.

In his first line, he got almost to the solution in just that one line; he could have gotten all the way to the solution with one line:

61% is 2/3 of the way from 45% to 69%; therefore 2/3 of the mixture should be the 69%; 2/3 of 150 liters is 100 liters. So 100 liters of 69% and 50 liters of 45%.

Then in his next 7-8 lines he shows a traditional algebraic solution for mixture problems.

Look at the amount and complexity of the work in the algebraic solution method compared to the informal first method.

The lesson here is that, if a formal algebraic solution is not required, there is a much faster and easier path to the answer.
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