Mixture Problems

You know this one too:

Given:

Solution A contains 20% acid

Solution B contains 60% acid

Find: The number of mL of each necessary to obtain 450mL a solution that contains 30% acid.

I usually teach this the same way I do work and rate problems. Find what we call a "guiding equation" that is based on the units available in the problem and set up some sort of table to organize the information. Depending on the given information, we may use two variables and solve a system of equations. Maybe not. But at the end of the day, this is going to be some complex variation of a rate problem.

I have a colleague who sees these problems in terms of proportions and our recent conversations have inspired this GeoGebra applet.



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There may be some rounding error, but I think this gives an interesting conceptual spin to a problem that can very easily become an algorithmic nightmare.

We can make some quick adjustments for the good ol' price per pound problems as well.




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Download % Mixture applet
Download Price per Pound applet

Work Problems

Y'all know the drill.

Person A does a job in A hours, Person B does the same job in B hours.  How long does it take them to do the job together?

Teaching the work problem can turn into a debacle fast.  Really fast.  Most teachers boil these down to an algorithm they themselves might not understand.

"Here's the formula, kids.  Learn it.  Know it.  Use it.  Now let's move on to something else."

That's never set well and a couple of years ago, it dawned on me that a work problem is nothing more than a complex rate problem.

Person A can do 1/A of the job per hour and person person B can do 1/B of the job per hour.  How much of the job do they complete per hour when working together?  

This has been a game changer.  Find the common denominator and add 'em up.  Now we have a common rate.  From there it's pretty easy to determine how long it takes to complete the job.  We give no formulas and kids understand how to find an answer and why it works.

We had kids derive two different algorithms for work problems.

x = (AB)/(A+B)

and

1/A + 1/B = 1/X

Sure, kids.  Use it.  G'ahead.

But then we added the dreaded twist.

Person A can do the job in A hours.  If person B helps, they can do the job in X hours.  How long would it take person B to do the job when working alone?

I give you, Clinton's Theorem:

I kid you not,  this kid came up with this all himself.  By "integer portion of X" he means, everything to the left of the decimal.  He's truncated the number to make things "simpler."  

As he was presenting this to us before we left for break, I was baffled.  It works, but why?  So today, I sat down next to him and had him explain the formula to me again, I took notes and then simplified the expression.  I won't deprive you the joy of doing so yourself.

Seriously, have at it.