Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at markdotmath@gmail.com. I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

Mixing it up with Alligation

Posted by mark schwartz on March 23, 2016

Introduction:

A common question asked in an Algebra class is : “Isn’t there an easier way to do this?” I happen to find an easier way in an 1864 text and with a few small changes, I presented it in my Algebra classes. Some students who previously had used the traditional method found that this was easier. They then asked why the method isn’t still in use and I had no answer. It simply has dropped from use. I have found it in several other texts of that period (1850-1875) but having not seen all current texts, I suspect that it’s not in any because if it were, I believe it would be in all of them.

The Story:

In 1864, in “Eaton’s Common School Arithmetic” (pg. 268) is a presentation of a process called “alligation”. There are two types of alligation, this one simply called alligation, the other called alligation alternate. Both address “the process of mixing quantities of different prices so as to obtain a mixture of a required intermediate price.” This process enables the student to solve problems with 2 or more quantities, just like the problems in today’s texts. Further, this alligation process can be used not only for problems with quantities of different prices but also quantities with different percent solutions, or the value of the “intermediate” solution, or the quantity of both of the solutions mixed together, or the quantities of nuts, or coffee, or candy. Take a look.

Take the following problem: How much of a 2% solution must be mixed with 15 gallons of a 9% solution to get a 7% solution? Traditionally, the following equation would be generated:

.02(x) + .09(15) = .07(x + 15)        – – – – –   Solving, 6 gallons of 2% solution are needed.

Now look at the alligation process. First, the target percent solution is 7. Then, the 2% solution is 5 less than 7% and the 9% solution is 2 more than the 7%. Eaton notes the 2% solution as a “deficit” of 5, expressed as −5 and notes the 9% solution as a “surplus”, expressed as +2. Eaton did this so that the deficit and surplus combined with their quantities would “balance” at zero.

Using Eaton’s balancing idea but modifying his language to state both differences from 7 as positives gives the following picture, which by the way, is a further modification of his presentation while maintaining the essence of his idea. Two percent is 5 “units” from 7% and 9% is 2 “units” from 7, It’s important to put the percents in increasing order from left to right and doing this, the picture is:

Given percents                                                            2%               7%           9%

Find units from the  target                                (7%   ̶ 2%)               (9% ̶ 7%)

Units from the target of 7%                                     5                               2

Just a brief note before continuing … the following explanation of the problem is what I call slow-motion math. Students need not take down every detail of this explanation and certainly aren’t required to repeat every detail when working through a solution to a problem, but I have found this to be an effective first-time way to step students through the understanding of the relationships found in problems. And now to continue …

Clearly, 5 units from the target and two units from the target don’t balance; something 5 units from the target won’t balance something 2 units from the target. But what are these “somethings”? They are the number of gallons of the solution and can be used to bring about a balance. Looking back at the problem, it was stated that there were 15 gallons of the 9% solution, and these 15 gallons are 2 units from the target. This combination of gallons and units is to balance a certain number of gallons 5 units from the target. Designating these units as “x”, we have 5x and 2(15) and these can be seen as an equation:

5x  =  2•15

the target has been replaced with an equal sign, so now we have an equation (TA DA!) which describes the relationships stated in the problem. So, what Eaton has given us is a way to structure an equation in a much simpler way than the traditional approach.

The alligation equation is         5x = 2•15

The traditional equation is     .02(x) + .09(15) = .07(x + 15)

The answer is 6 gallons of the 2% solution.

Now look at this problem: Thirty percent of the dairy cattle in Washington County are Holsteins, whereas 60% of the dairy cattle in neighboring Cade County are Holsteins. In the combined two-county area, 50% of the 3600 dairy cattle are Holsteins. How many dairy of each kind is there? You can table it or simply build the equation without it. The following table shows how to build the equation, using slow-motion math. Follow the “if you do this” and “you get this” statements.

If you do this:                                                                 You get this:

put the percents in increasing order, left to right        30%         50%,       60%

 

Use 50% as the target. 30% is 20 units from the target and 60% is 10 units from the target                  20             10
Let x be the cattle from Washington County and x is 20 units from the target                  20 x
Let (3600 ̶     x) be the cattle from Cade county and (3600 ̶   x) is 10 units from the target             10(3600   ̶   x)
Use the last 2 terms in the “you get this” column                  20 x = 10(3600 ̶     x)

 

The alligation equation is:           20 x = 10 (3600 ̶   x)

The traditional equation is:         .30x + .60 (3600 ̶   x) = .50 (3600)

There are 1200 cattle in Washington County and 2400 cattle in Cade County. Be careful here. If you look at the percents for Washington and Cade Counties, the ratio is 1:2, which is the same ratio for the number of cattle in each … but again, this is a nice coincidence so don’t rely on that relationship in this type of problem.

As I was trolling in a 1959 text for a procedure, I came across the following problem and thought about solving it with alligation. Here, alligation will be used to find one of the percents, whereas in the usual mixture problem, 3 percents are given and you are to find a quantity.

Here’s the problem: there is a uniform mix of clover and sesame seeds, of which 1/25th is replaced with clover seeds, resulting in a mix which contains 5% clover seeds. What was the original percent clover seed?

Several things have to be considered. First, the clover seed that replaces the 1/25th is 100% clover seed. Second, the quantity to which the 100% clover seed is added is 24/25th of the initial uniform mixture. Then, the key to this problem is to think of this as adding two different quantities together to get a third quantity. Doing this frames this as a standard mixture problem. One quantity is the 24/25th initial mix and the other quantity is 1/25th clover seed which is added. Further, the reason that we know that what is to be found – the “x” – is less than the goal of 5% is because, in mixture problems, the target is always between the other two given percents in a problem and since 100% is greater than the 5% goal the x% must be less than 5%. From this analysis comes the following alligation set-up:

If you do this:                                                                           You get this:

put the percents in increasing order, left to right    x%           5%,         100%

 

Use 5% as the target, x% is 5 ̶ x   units from the target and 100% is 95 units from the target              5 ̶   x         95
The quantity 95 units from the target is 1/25th of the original mixture, recalling that 1/25th was replaced.                                95(1/25)
The quantity (5 ̶ x) units from the target is 24/25ths, recalling that this was what existed before the replacement.    24/25)(5 ̶ x)
Use the last 2 terms in the “you get this” column (24/25)(5 ̶ x) = 95(1/25)

 

The alligation equation would be                            (24/25)(5 ̶ x) = 95(1/25)

The traditional equation, if you prefer, would be   1.00(1/25) +x(24/25) = .05(1)

The answer is 1 and 1/24th percent.

Both equations are fussy: the traditional equation has a mix of decimal and fractional values – convert everything to decimals or fractions. Without a calculator, this is even fussier.

The alligation equation can resolve a little quicker because multiplying both sides by 25 clears the fractions.

The choice is yours (I still choose alligation.)

As students get more comfortable with this procedure they may be able to just build the equation, but it need not be done with the table. If not, they can always “picture” it first and then build the equation.

This is an interesting and relatively simple process when compared to the current traditional procedure in today’s texts. This alligation procedure was presented in several classes, simultaneously with the traditional procedure, and students preferred alligation. Wouldn’t you?

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