What do you think of this approach to solving mixture algebra problems?
There is a category of algebra problem that involves combining two rates to get a third rate. It might involve mixing chemicals with different concentrations or selling two products with different prices or traveling at two different speeds.
There is a single equation involving weighted average that can be used to solve these problems. Let me show the steps to get to this equation. It may look a little scary at first, but what we end up with is very simple.
Suppose we want to solve a problem involving traveling at two different speeds v1 and v2 for corresponding times t1 and t2 for total distance of d.
We have v1 t1 + v2 t2 = d.
Divide both sides by t1 + t2.
t1/(t1+t2) v1 + t2/(t1+t2) v2=d/(t1+t2)
Setting t1/(t1+t2) = w1 and t2/(t1+t2) to w2, we get the universally applicable equation of:
w1 v1 + w2 v2 = v3.
w1 and w2 are two weighting factors representing the portion of the total time traveling at the two speeds. w1 and w2 are fractions between 0 and 1 which add to 1. v3 is just the average overall speed.
We can generalize the equation to:
w1 r1 + w2 r2 = r3 where the r terms stand for rates. Using the fact that w1 + w2 = 1, we can write this using one variable.
w1 r1 + (1-w1)r2 = r3
Here is a video from Khan Academy that shows the standard way of solving a rate problem.
Here is my approach.
We have solutions of 25% concentration and 10% concentration being combined to give a 15% solution.
We get
.25 w + .1(1 – w) = .15
.25w – .1 w + .1 = .15
.15w = .05
w = ⅓
The 25% solution is ⅓ of the total. Since there are 50 ounces of it, the total is 50 / (⅓) = 150 ounces and there are 150 – 50 = 100 ounces of the 10% solution.
Another way of finishing the problem is to say that since the 25% solution is ⅓ of the total, the 10% solution is ⅔ of the total. There is therefore (⅔) / (⅓) * 50 ounces = 100 ounces.
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8 Answers
Perhaps I should have introduced weighted averages with GPAs, which most people are familiar with. We have w1 s1 + w2 s2 + ... + wn sn = s, where the wi weights of tests are determined in advance and the si values are scores and s is the final grade. This is different from a typical mixing problem, where the weights are not known in advance.
A practical GPA problem that comes up frequently is to determine what final grade sn is required to attain a course grade of s. We want to solve for x in w1 s1 + w2 s2 + ... + wn x = s. This is a fairly straightforward calculation.
Here is an interesting site that talks about the solution of this type of problem Check the discussion at about 3 minutes. Applied to the Khan academy problem, it works like this:
The .15 concentration is a balance point between .25 and .10. If we think of a seesaw, the fulcrum (center of gravity) is at .15. When we take each concentration’s difference from .15 and multiply by liquid volume, the two numbers are equal.
50(.25—.15) = x(.15 – .10)
5 = .05x
x = 100
If we were told the total volume was 150 instead of giving the volume of the 25% solution, it is just slightly more difficult. Let x be the volume of the .25 solution.
x(.25 – .15) = (150 – x)(.15 – .1)
.1 x = (150 – x) .05
.15x = 75
x = 50
I am the one lurking, BTW. I can be of absolutely no assistance to you. Just wanted to see what everyone had to say…
My answer is I want to go back to school and enjoy learning Algebra instead of fearing it. Lol
Seems like you are plugging in known values and then solving for the one unknown value, no?
Yes, but there are simpler and more difficult ways of setting up the equations. Thinking of the total volume as a weighted average is both intuitively appealing and requires minimal algebraic manipulation.
My answer is I want to go back to school and enjoy learning Algebra instead of fearing it.
I am doing that. I have long wished I learned calculus. Recently I started re-learning my high school algebra on the free Khan Academy web site, preparing for calculus.
A few times a week sit down with a paper pad and pen and plow through an hour or two of practice and quizzes at a sitting.
@LostInParadise Yeah, cool! I was really good in most math topics, and my core approach was trying to relate to the problem and then relate and/or build the math from that relation. (It also meant that badly-worded math story problems really bothered me.)
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