When I was in high school taking Intermediate Algebra, now called Algebra 2, we learned how to solve mixture problems. At that time, I discovered a way of solving them that was simpler and quicker than the method I was being taught, and a year later, I was accosted by my algebra teacher in the hall and he asked me to teach him my method, as he wanted to teach it in his classes. This was quite a coup for a 16-year-old. Some time later, I came up with another method. By the time I started teaching high school algebra, it was no longer required, but the curriculum changes going on now may bring about the return of the mixture problem, so we will now present a typical mixture problem and solve it three ways and you can decide which way you prefer.
If two quarts of a 40% saline solution and 3 quarts of a 60% saline solution are mixed together, what percent saline solution will result?
The traditional method: A “40% saline solution” means that 40% of the two quarts of solution is salt, so in the 40% solution there are (.40)(2) = .8 quarts of salt, and in the 60% solution there are (.60)(3) = 1.8 quarts of salt, for a total of .8 + 1.8 = 2.6 quarts of salt in a total of 2 + 3 = 5 quarts of solution. Therefore, the result is 2.6 5 = .52 = 52% saline solution. Not being a chemist, I don’t know if it is even possible to have saline solutions that are as concentrated as the ones in this problem, but the math is valid. (What calculation would you make if it were 25 quarts of 28% solution and 50 quarts of 46% solution?)
Alternative method #1: If the amounts of the two solutions were the same, we could take the mean of the two percentages as the answer, but since there are different amounts, we need to find the weighted mean: , so the answer is 52%. (What calculation would you make if it were 25 quarts of 28% solution and 50 quarts of 46% solution?)
Alternative method #2 (This is the method I figured out when I was in high school.): Intuitively, it seemed to me that if we used, for example, one half as much of Q % solution as P % solution, the Q % solution would have one half as much influence on the result as the P % solution, so the resultant solution would have a percentage that is half as far from P % as from Q %. Example: Mixing 20 ounces of 30% solution with 10 ounces of 45% solution results in a 35% solution, because 35 is half as far from 30 as it is from 45. In our problem, the amount of the 60% solution is the amount of the 40% solution, so the resultant solution should have a percentage that is as far from 40% as it is from 60%. Therefore the ratio of these two distances is 3:2. Using the standard method of solving ratio problems, divide 60 – 40 = 20 into 3 + 2 = 5 parts and add 3 of them to 40, resulting in 40 + 3 = 52. For those of you who like formulas, if m measures of a P % solution and n measures of a Q % solution are mixed, and R represents the resultant percentage, then R = P + . Note: It’s easier for most students to understand this formula if they choose P to be the smaller of the two percentages, but it works whether P < Q or not.
All this seemed intuitively plausible to me, but in math, plausibility is insufficient justification for acceptance of a conclusion; we must logically prove that the conclusion cannot be other than it is. The formula that is used in the traditional method (and implicitly in alternative method #1), R = , is valid by definition, i.e., it is an application of the definition of a solution percentage, and the formula that we are using in this new method can be derived from that formula, using the same variables:
Which of these formulas is easier to use in a particular problem depends on the values of the variables and your own predilections. Of course, if you’re using a calculator to do the arithmetic, it’s all equally easy.