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 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.

Archive for the ‘Proportions’ Category

In 1877, Mr. Ray Reasons with Fractions

Posted by mark schwartz on September 8, 2016


In Mr. Ray’s 1877 Ray’s New Intellectual Arithmetic, an elementary school text, he presents some of the problems with their solution. A sample of these are worth looking at because in every case he shows a solution method which is based on fractions and knowing how to handle a sequence of fractions. But it’s not only the sequence of fraction operations but also the logic of these fraction operations that elementary school children had to follow. This required them to think about the relationships in the problem. I’d like to further note that this method of solution for all 7 problems presented here is seen in many of the texts of that era. It really required students to understand fractions! I’m not proposing that we use this “fractional” method in lieu of solving them by either proportions – the first 4 problems – or simple equations, the last 3 problems.

The Story

All these problems are from his text. Read the solutions slowly to really enjoy the subtlety of the method.

  1. A yard of cloth costs $6, what would 2/3 of a yard cost?  (Pg. 48, # 3)

Solution: 1/3 of a yard would cost 1/3 of $6, which is $2; then, 2/3 of a yard would cost 2 times $2, which are $4.

  1. If 3 oranges are worth 15 cents, what are 2 oranges worth?  (Pg. 49, #19)

Solution: 1 orange is worth 1/3 of 15, or 5 cents; then 2 oranges are worth 2 times 5 cents, which are 10 cents.

  1. At $2/3 a yard, how much cloth can be purchased for $3/4?  (Pg. 75, # 5)

Solution: For $1/3, 1/2 a yard can be purchased, and for $1, 3/2 of a yard; then, for $1/4, 1/4 of 3/2, or 5/8 of a yard can be purchased, and for $3/4, 9/8 = 1 and 1/8.

  1. If 2/3 of a yard o cloth costs $5, what will 3/4 of a yard cost?  (Pg. 101, # 2)

Solution: The cost of 1/3 of a yard will be 1/2 of $5 = $5/2; and a yard will cost 3 times $5/2 = $15/2; then, 1/4 of a yard will cost 1/4 of $15/2 = $15/8; and 3/4 of a yard will cost 3 times $15/8 = $5 and 5/8.

Note that these 4 problems lend themselves well to being solved using proportions. What follows now are 3 more problems, which if presented in today’s texts would likely be solved with simple equations, but again Mr. Ray’s solutions are a sequence of fraction operations.

  1. If you have 8 cents and 3/4 of your money equals 2/3 of mine, how many cents have I? (Pg. 52, #17)

Solution: ¾ of 8 cents = 6 cents; then 2/3 of my money = 6 cents, 1/3 of my money is 1/2 of 6 cents = 3 cents, and all my money is 3 times 3 cents = 9 cents.

  1. Divide 15 into two parts, so that the less part may be 2/3 of the greater.  (Pg. 106, #1)

Solution: 3/3 + 2/3 = 5/3; 5/3 of the greater part = 15; then, 1/3 of the greater part is 1/5 of 15 = 3, and the greater part is 3 times 3 = 9; the less part is 15 ̶ 9 = 6.

  1. A and B mow a field in 4 days; B can mow it alone in 12 days: in what time can A mow it?  (Pg. 110, #14)

Solution: A can mow 1/4 ̶ 1/12 = 1/6 of the field in 1 day; then he can mow the whole field in 6 days.

I hope you appreciate what elementary school students had to do at that time. Since it was elementary school, they weren’t taught proportions and simple equations but they were “exercised” with fractions in a way that I believe could benefit today’s students understanding of fractions.

Posted in algebra, basic math operations, fractions, Historical Math, math instruction, mathematics, proportion, Proportions, remedial/developmental math | Tagged: , , , , , , | Leave a Comment »

The Importance of a Clearly Stated Algorithm

Posted by mark schwartz on August 22, 2016


I posted a piece earlier in this blog titled Sheldon’s Compound Proportions. It describes what Sheldon labels the “cause and effect” method for solving compound proportions, which as far as I can tell, aren’t in todays’ texts. His work was in 1886. You might want to take a look at his idea because this posting talks about other compound proportion procedures at that time and I did it to emphasize the importance of a clearly stated procedure for doing an operation.

The Story

I strolled through my collection of old texts and in quite a few of them found the same prescription for solving compound proportions not using cause and effect. I picked 5 which cover about a 20 year span from 1864 to 1883. They all have the same procedure and what I suspect is that it was the established and accepted solution method at that time. As in todays’ texts, it was just a simple matter of “borrowing” a basic algorithm from someone else’s work. There are other texts of that era which reference Sheldon’s cause and effect method and a few of them introduce it along with the procedure I’ll cite below.

The point is that his method is a much clearer statement of how to handle the information in a compound proportion problem. Further, what I’m suggesting is that we should carefully examine some of our current traditional algorithms to see if the reason students have trouble with them is because of the way they are worded and presented. For example, finding the lowest common denominator (LCD) in order to add/subtract fractions doesn’t require the extended way it’s been typically taught. In fact, I have seen some texts introducing a method which doesn’t require finding an LCD at all. Certain mixture problems can more readily be solved with an 1864 method Mixing it up with Alligation, posted earlier in this blog.

By the way, the 5 texts in which I found this procedure are all arithmetic texts, which indicates to me that this somewhat sophisticated idea of compound proportion was taught in elementary school. I’ll give you example problems from an old text to indicate that, in my view, it was a very handy procedure for the real world experience at that time. Today we call these “application” problems.

Here’s the rule as stated in Greenleaf’s 1881 The Complete Arithmetic, page 235 (the other 4 books are cited below and present the same rule).

Rule for Compound Proportions

“Make that number which is like the answer the third term. Form a ratio of each pair of the remaining numbers of the same kind according to the rule for simple proportion, as if the answer depended on them alone. Divide the product of the means by the product of the given extreme, and the quotient is the fourth term, or answer.”

Embedded in this is reference to “…the rule for simple proportion …” which Greenleaf provides on page 233 and it is:

Rule for Simple Proportions

“Make that number which is of the same kind as the answer the third term. If from the nature of the question the answer is to be larger than the third term, make the larger of the remaining numbers the second and the smaller the first term; but if the answer is to be smaller than the third term, make the second term smaller than the first. Divide the product of the means by the given extreme, and the quotient is the fourth term, or answer.”

Students had to be able to apply this latter rule for simple proportion before being presented compound proportion. There is no conflict between the two rules; in fact, there is some overlap. For simple proportions, the rule directs the student to understand “the nature of the question …” and use that to determine which values go in which of the 4 places in the proportion. The students had to be able to assess and estimate if the answer was going to be larger or smaller and place the correct terms in the first and second places. Wow! There is a lot of estimating and juggling of values and basically it seems that all of this effort is aimed at what we would say today as determining whether it’s a direct or inverse proportion. With problems with simple values, this is a somewhat manageable issue.

For example, a problem from the text is “If a man travel 319 miles in 11 days, how far will he travel in 47 days?” Using the rule for simple proportion, the setup would be:

11/47 = 319/x    (the rule doesn’t use “x”, but I did for demonstration purposes)

The solution is (47×319) ÷ 11 = 1363

However, in today’s approach to simple proportion, the setup (in most cases) simply follows from the order of the information in the problem, giving:

319/11 = x/47

This gives the same answer but notice that the rule states “Divide the product of the means by the given extreme …” and that doesn’t apply here. So, the 1881 rule is quite constraining when it comes to writing the proportion, when indeed there are several ways to set up the proportion for the problem.

Again, there is nothing wrong about the simple or compound proportion rules as provided by Greenleaf. The issue is that the rules are somewhat convoluted and constraining. If a student doesn’t learn this algorithm and follow it precisely, the likelihood is that the correct answer won’t be found. There are a lot of words referring to the terms and judgements that a student must make about which terms go where in the proportion. Further, look at what happens with a compound proportion problem, again from Greenleaf (#67, page 236):

“If 12 men in 15 days can build a wall 30 feet long, 6 feet high, and 3 feet thick, working 12 hours a day, in what time will 30 men build a wall 300 feet long, 8 feet high, and 6 feet thick, working 8 hours a day?”

Now, where does a student begin sorting through all this information if they use the rule above for simple proportion? What’s the “nature of the question”? For example, the rule states “…make the larger of the remaining numbers …” and how is a student to know which number is to be selected? I can visualize the instructor explaining in excruciating terms how all this works. Again, it’s not impossible to apply the rules as stated in 1881 but I urge you to look at Sheldon’s Compound Proportions in this blog and see how much more direct the rule is by framing information as cause and effect.

Briefly, Sheldon’s 1886 statement of the procedure:

“The solution of every example in proportion proceeds on the assumption that effects are in the same ratio as the causes that produce them. Every proportion is the comparison of two causes and two effects. In the method known as Cause and Effect, the causes form one ratio, and the effects the other. The first cause and the first effect are antecedents; the second cause and second effect consequents.”

Notice the simplicity of identifying cause and effect and then the causes forming one ratio and the effects the other. The words” antecedents” and “consequents” could be updated to 1st and 3rd term and 2nd and 4th term, respectively.

Taking the above compound problem the 1st causes are 12 men, 15 days 12 hours a day and the 1st effect is to build the wall 30 feet long, 6 feet high, and 3 feet thick. The 2nd causes are 30 men working 8 hours a day and the 2nd effect is to build a wall 300 feet long, 8 feet high, and 6 feet thick. You are to find “…in what time…” which is a 2nd cause. There is a shortcut that can be used but let me show you – in what I call slow-motion-math – one way to make sure the terms get placed correctly. I typically use the labels and then replace it with the values (for a lot of different types of problems, not just compound proportions). The proportion following Sheldon’s procedure is:

Causes                     Effects

1st       men, days, hours         length, height, thickness

2nd       men, x, hours               length, height, thickness

I used “x” for days in the second cause. If the numbers are substituted, we have:

12•15•12 = 30•6•3
30•x•8     300•8•6

Cross-multiply and divide, solving for x and the answer is 240.

Again, a detailed description of the “cause and effect” is in Sheldon’s Compound Proportions in this blog.

The essence of this posting is to demonstrate the importance of a well thought-out procedure expressed in easily understood language. If you are an instructor, you likely have done this kind of “simplifying” of the algorithm because as stated in the text, it seemed too fussy for students to follow. Not every algorithm can be simplified but I believe it’s an instructor’s responsibility to make math more accessible to students by removing the fog of awkwardly phrased rules and algorithms. Give it a try.

Posted in algebra, basic math operations, Historical Math, math instruction, mathematics, proportion, Proportions, remedial/developmental math | Tagged: , , , , | Leave a Comment »

Rephrase That Impossible Application Problem

Posted by mark schwartz on July 19, 2016


As I was presenting a topic one day, a student said that what I was saying didn’t make any sense and could I please say it differently. My first reaction was to ask the class if it was true for them too; some agreed. It wasn’t a rude statement and I took the comment seriously and did rephrase what I said and asked if that made more sense and apparently I got it right. Then I got to thinking not only about that moment but other moments where what I was saying may not have made sense, but nobody bothered to stop me. As far as I can tell, it wasn’t the math content but the language I used to describe the content that bothered them. Thus the story that follows.

The Story

Question: Let’s say that the first city 4th of July fireworks I attended was in 2005. Since then, I attended the city fireworks every year including this year, 2016. How many fireworks have I attended?

Before considering the answer, consider if that question is the same as: how many years have I attended the city fireworks display on the fourth of July?

If your answer to the first question is 11, you’re wrong and as vague as the second question is, the answer is 11.

In both cases, which seems to be the same case, I suspect you got your answer by simple subtraction, 2016 – 2005 = 11. The thing to consider though is what exactly is being subtracted? Let’s bring this into a more manageable range, like 5 – 1. If you were to do this operation on a number line, you could put your finger on the five and move to the one, counting as you go and thus you would get 4. That four represents the number of movements from point 5 to point 1 on the number line. When you move from 5 to 4, you say “1”, in essence, “scaling” the distance between 5 and 4 as 1 unit, regardless of the actual distance. Given that it’s a number line, the distance between the points on the number line will all be the same. So, when we say 5 – 1, we are asking how many distances are there between 5 and 1. By the way, this distance analogy is similar to the idea of having 5 kittens and giving 1 away – how many kittens have you? In this case, it’s not distance, it’s kittens but conceptually it’s the same. We need not bother with scaling the number of kittens, because it’s a quantity not a distance, although some consider distance a quantity. As far as “how many years have I attended the fireworks?”, what’s being counted here is the number of years – an amount of time scaled rather than a distance. So, 2016 – 2015 is one, etc. as far as counting.

What’s the point? Remember the first question? To repeat: “Let’s say that the first city 4th of July fireworks I attended was in 2005. Since then, I attended the city fireworks every year including this year, 2016. How many fireworks have I attended?”

What is being counted here? Again, consider the number line. We’re not counting the distance between points on the number line rather were counting the number of points. The first question then has to be a subtraction plus one, which really is asking for the inclusive count.

You might say “so what?” to the difference between the first and second questions but looking at them as I did points out that there is a difference. The real issue here is the nature of asking questions in a math class. If we, as instructors, ask ambiguous questions, or questions which require students to reflect on the context of the information as well as the information in the question (and students don’t see the need to reflect on these issues) then we are, in a sense, misleading them and adding to their confusion about math. The context in this case is the words we instructors use.

I’ve seen this in questions in texts. We glibly accept the questions and answers at the end of the chapter and if some of those questions are questionable, we simply don’t assign them. But it’s not just the questions in texts. It’s how we state information, it’s how we use the language to structure questions and present concepts. The difference between the first and second question demonstrates this.

We should be attempting to be better at some precision in our questions and presentations because, like it or not, instructors are math role models for students. If we expect precision and accuracy from students, we should also expect that they can phrase good questions and it’s the instructor that establishes the idea of a well-phrased statement. It also seems it’s a critical component of being able to arrive at a correct answer to a problem. The caution to read “word” problems until you understand it is reasonable, but what if you never “understand” the problem? My thought is that students have to have license to and practice in rephrasing problems, without changing any of the relationships in the problem.

For example, when teaching percent using the percent proportion model (you can see how this is presented in the Percent Proportion posting in this blog), I point out to students that most percent problems can be rephrased. An example: A farmer sold 180 sheep, which represented 16% of all the sheep he had. How many sheep had he after the sale?

There are a lot of extra words in this problem, but only two numbers. I asked the class to rephrase this problem focusing on the relationship between the numbers. The students wrote all their attempts on the board, so that we could discuss the thinking that brought about their answer. What they produced, based on the percent proportion model that I presented to them, were two rephrased problem: First, “180 is 16% of what number of sheep?” to get the total number of sheep and then subtract the 180 he sold. The second was to see that if 180 sheep were subtracted from the total (written as x – 180), this would represent the number of sheep left and the percent left would be 84 (100 – 16). The rephrasing then would be “The number of sheep remaining is 84% of what number of sheep?” The result of this rephrased problem still is the total but again, simply subtract 180 sheep. If you want to play with it, the answer is 945. Even if the percent proportion model isn’t used, this rephrased problem is much more manageable.

But here’s how to set up both results using the percent proportion model.

180  16
 x   100      … solving, x = 1125 … total after sale = 945
x - 180  84
  x      100   … solving, x = 1125 … total after sale = 945

We practiced this rephrasing idea some more and I reminded them that they don’t have to rephrase every problem, but if the problem seems “impossible”, rephrase it.

Posted in algebra, Historical Math, math instruction, mathematics, proportion, Proportions, remedial/developmental math | Tagged: , , , | Leave a Comment »

What? That Much Percent Increase?

Posted by mark schwartz on July 8, 2016


I like coincidences. Particularly when they provide learning opportunities for my students. We had just spent time learning about percent and percent increase and decrease. The problems in the text were good but not really challenging. The coincidence was that I was reading John McPhee’s The Curve of Binding Energy (1974) and I’ll start the story with what he said on page 18.

The Story

“Thousands of miles of tubes, pipes, and other conduits were needed to create a network of flow wherein the gas could now go through a membrane, now return to try again, now go on to a new membrane, gradually advancing, in a process of separation and elimination, until what had begun as seven-tenths of one percent U-235 was more than ninety percent U-235 – fully enriched, weapons-grade uranium.”

I’d never heard this detailed an explanation of how weapons-grade uranium was made. But what really got my attention was that his statement could be a percent increase problem. I worked it out before I gave it to the class, rounding off the initial “more than ninety percent” to a manageable 90%.

Further, I decided that it would be an in-class extra credit exercise and allowed that the students had to first work within their assigned group, but once they had an answer they could discuss it with other groups.

I did this because the percent increase is 12,757% and this size percent increase would cause the groups to question what they did, even if they got that number. There were occasional answers to the problems in the text that resulted in percent increases of more than 100% but nothing quite like this. Once I gave them the problem and answered any preliminary questions and they got to work, I roamed the room listening to the strategies they came up with to do the problem.

The first issue was how to numerically express seven-tenths of one percent. One group asked if they could talk to another group to get help expressing it. So, I stopped the class and said that if they are willing to accept the following condition, they can work as a class to get the answer. The condition was that everyone in the class would get the same grade. They accepted. There was an eruption of conversation and as I roamed around, I was asked if what they got was right. I just referred them to other members of the class.

Once there was consensus on how to represent the initial percent, they simply continued with what they had learned about setting up percent increase problems. By the way, I taught a somewhat non-traditional method that doesn’t use a formula, rather it uses a somewhat modified percent proportion approach. You can look at it in this blog at Percent Proportion.

Several groups quietly called me over to show me their result, asking if they were right. Some were and some weren’t but I wouldn’t say yes or no, reminding them of the condition under which they were working. So, more talk, discussion and exchange of how to set up the problem.

It was interesting to watch the evolution of the shared work – people got up and moved around the room; some asked to and did use the white board; I heard a lot of “show me” and “why did you do that?” and “that doesn’t seem right”. But, ultimately there was class consensus on the right answer.

They did, however, insist that I walk through how I thought about it even though they got it. So, I put on what I call my “slow-motion-math” hat and gave them the following:

Ninety percent is 90/100, so the amount of increase is 90 – .7, or 89.3%. Seven-tenth of one percent is (7/10)(1/100) or 7/1000 (I did this because I saw a lot of questioning on how to express it). This in percent is .7/100 (or if you were sure, you could have just written .7 over 100). So the question can be put in a percent increase frame. First, the amount of increase is 89.3 and since it started at .7, the amount of increase relative to the beginning point can be expressed as 89.3/.7, or 893/7 (they questioned if doing this would give the same answer and we discussed this). Using the proportion statement 893/7 = x/100 gives a percent increase of 12,757%, rounded. So, doing the original problem led to some other related talk about fractions, decimals and rounding. Neat.

After all was said and done, I got questioned about this exercise because there was a sense that it was a trick question. I have noticed that when students feel uncertain about a math problem, the frequently asked question is just that. I then heard stories from the class about their prior math experiences where trick questions unfortunately were used to presumably teach them something about math, but the only learning was frustration because a lot of the tricks were beyond the bounds of what had been taught and in essence they quit. Given what I heard, I may have quit too. Somehow they concluded that math is just knowing the right tricks.

But once they were accepting that it was an interesting problem, I noted to them that as they read books, magazines, watch TV or come across “mathy” stuff, they might play with it as we did with this problem. And of course they noted to me to record the “A” for all of them.

Posted in basic math operations, Historical Math, math instruction, mathematics, percent, proportion, Proportions, remedial/developmental math | Tagged: , , , , | Leave a Comment »

Visually Explaining Shared Work Problems

Posted by mark schwartz on June 3, 2016


In class we had worked through how to solve shared time problems. Some of the students seemed puzzled by the procedure (partly because it has fractions in it) and they expressed it as “I can’t see that”. It struck me that seeing it actually might make a big difference for some students. After all, many of them claim to be visual learners and if I could provide a non-Algebraic approach, it might give students a better way to “see” the relationships. It’s not completely free of Algebra but it did make a difference for many students. One student actually said “Hey, this is Geometry not Algebra” and talking to him later it turns out that in high school he really did well in Geometry but not Algebra and he has always tried to set up Algebra problems geometrically. As it turns out, in Smyth’ 1859 Elementary Algebra (page 39), he cited the importance of seeing the work in one day (or one month) units. Again, the word “seeing”, so I tried to help students see it.

The Story.

Let’s step through this using the following problem:

Danny takes 4 hours to do a job alone. His brother Mike can do the job alone in 6 hours. If they work together, assuming no gain or loss of efficiency, how long will it take them to do the job?

(Before going on, I’d like to note that this will be discussed in slow-motion math which results in a lot of words to describe a somewhat simple procedure. The point is that this is a discussion with the class and most of these words were spoken throughout the demonstration.)

The following represents the amount of time it takes to do the whole job and notice the notation.

|←     time for job done     →|

|←       The whole job          →|

|←     the whole job done  →|

Let’s start with Danny, who can do the whole job by himself in 4 hours. The question is how much of the job can he do in 1 hour? Collectively, the class (well, almost everyone) responded   . Visually, this is

|← 1 hr.→|

|←       The whole job          →|

|← 1/4  →|

Now, Mike can do the whole job by himself in 6 hours. The question is how much of the job can he do in 1 hour? Again, collectively the class responded 1/6. Visually, this is

|←   1 hr.   →|

|←       The whole job          →|

|←   5/12   →|

The 5/12 comes from 1/4 + 1/6 but it’s still a total of 1 hour of time, since they’re working at the same time; we don’t add the time just the amount of job done.

Now what? Consider this: if 5/12 of the work can be done in 1 hour, how long will it take to do the rest of the job? Visually, this is

|←   1 hr.    →|←             x hrs.  →|

|←           The whole job            →|

|←   5/12    →|←          7/12        →|

Amt. of job to do

Notice the additional notation of “Amt. of job to do”. And further, notice (with a little prompting!) that we have a proportion (!), which taken from the figure is

1/(5/12) = x/(7/12)

Yes, I know fractions again but students know how to solve this proportion. Solving for x, the result is 1 2/5 hrs. So, the total time for the job is 1 + 1 2/5 hrs, a total of  2 2/5 hrs. This is the initial hour plus the remaining time to complete the job.

And now, a very interesting visual display of a problem not likely to ever show up in anyone’s text. The problem is

If Adam can do a job in 30 minutes and Bob can do the same job in 20 minutes, how long would it take them to do the job together?

Typically, these kinds of problems are used with reference to hours not minutes, although minutes or hours will work. I’ll use the hours because this introduces fractions and it makes the problem more interesting.  Here’s the questions to consider:

  1. In 1 hour, how much of the job can Adam do?
  2. In 1 hour, how much of the job can Bob do?

Mathematically, Since Adam can do the whole job in 1/2 hour, he can do the job twice in an hour and since Bob can do the job in 1/3 hour, he can do the job three times in an hour. So, a total of 5 jobs can be done in 1 hour and here’s how it looks

|←                       1 hr.                 →|

|←1 job→|←←|←x→|→→|→→|

|←                 5 jobs                     →|

Look at this carefully before you go on. It looks a little weird but it’s correct. It the previous example, there was some part of the whole job left to do, and it was found by setting up a proportion and adding the result to the 1 hour already spent on the job. In this case, we have an excess of 4 “jobs”, which will have to be removed, which means removing the amount of time from the 1 hour it took to do those 4 jobs (breathe…). So, now we have to set up a proportion, get the answer and “add” it to the 1 hour. I stated “add” because the excess has to be removed and it will be identified as a negative amount of time – the amount to be taken away.

Using the same proportion procedure, the proportion looks like this:

1/5 = x/ ̶  4

Solving, x =  ̶  4/5. The amount of time for Adam and Bob to do the job together is 1  ̶  4/5, or  , 12 minutes.

I told you that slow-motion math takes a lot of words, but the point is that the visualization of the problem helps students actually “see” how all the Algebra works, and why. I was asked if Algebra problems can always be set up geometrically, and I said that we’ll see when we get to other kinds of problems.

Posted in algebra, Historical Math, math instruction, mathematics, Proportions | Leave a Comment »

Sheldon’s Compound Proportions

Posted by mark schwartz on April 13, 2016


Here’s another one of those delightful conceptions from one of my old texts, this one dated 1886. It addresses simple and compound proportions. I want to note that compound proportion is presented in many of the other texts in my collection, yet this topic seems non-existent in today’s math curriculum. Also, the authors at that time had a variety of approaches to compound proportion, but I find this one to be the clearest conception, primarily because he uses the terms “cause” and “effect”, which are common terms in today’s conversations and it seems to make it easier to visualize the relationships. Some of the other methods introduced special terms and relationships that I found cumbersome or somewhat vague.

I have on occasion presented it in Algebra classes that collectively were interested in pursuing the idea. They expressed interest because in some of the “application” problems in the text, students had to do 2 proportions serially in order to get an answer, and they found this troublesome. Examining the problems made me aware that Sheldon’s compound proportion method would work.

The Story:

In Sheldon’s Complete Arithmetic (1886) he presents concepts about simple and compound proportions. His method for simple proportions is more complex than today’s method so it won’t be discussed here. If you’re interested, contact me.

Again, as far as I can tell, most current texts that introduce proportions do not include compound proportions. This is an interesting historical shift in the depth of topics covered in mathematics curriculum in the 19th century compared to the 20 and 21st centuries. Other texts of Sheldon’s era routinely address compound proportions using a variety of approaches.

Sheldon has two proposals for working with compound proportions. The first is labelled “Compound Proportion” and the second is labelled “Cause and Effect”. I believe the second one he presents is the most interesting and useful. I worked problems from both sets and interestingly, some of the problems presented for “Compound Proportion” can be solved readily with Cause and Effect but problems in the set for Cause and Effect present difficulties for his “Compound Proportion” method.

So, only his Cause and Effect will be presented here and I believe a good way to introduce compound proportions is to show you the problem he used to demonstrate it, which is:

“If 18 men in 10 days can cut 540 cords of wood, in how many days can 15 men cut 630 cords?”

And now Sheldon’s statement of the procedure:

“The solution of every example in proportion proceeds on the assumption that effects are in the same ratio as the causes that produce them. Every proportion is the comparison of two causes and two effects. In the method known as Cause and Effect, the causes form one ratio, and the effects the other. The first cause and the first effect are antecedents; the second cause and second effect consequents.”

I found that a visual representation of the setup of the proportion helped students understand his method. This includes not using the terms antecedents and consequents, rather beginning the discussion of the method with words, not numbers. Consistent with his statement, the proportion is:

1st cause  =   1st effect
2nd cause      2nd effect

We discussed this in class, emphasizing the relationship of causes to effects.

Although Sheldon’s statement does not include how to solve the proportion, it will be solved just as with simple proportions – cross-multiply and divide. I believe the examples below will clarify how to solve a compound proportion.

Restating the problem:

“If 18 men in 10 days can cut 540 cords of wood, in how many days can 15 men cut 630 cords?”

The first cause is “18 men in 10 days” and the first effect is “540 cords” and the second cause is “in how many days can 15 men” with the effect being “cut 630 cords”.

The “x” will represent “…in how may days”… in the second cause. In “slow-motion math”, here’s how the proportion is set up first with labels, not numbers, and then solved using cross-multiply and divide when the numbers are substituted. The students need not always use the labels but it was found, at first, to be quite helpful in visualizing Sheldon’s method.

Causes     Effects

men*days = 1st cords
men*x      2nd cords

18*10 =    540
15*x       630

540*15x = 18*10*630

x = 18*10*630

x = 14

A brief aside here. Remember that in 1886, there were no calculators, so that most of Sheldon’s examples and problems allowed for a lot of cancellation in the initial ratios, or if the values are carried through rather than doing the indicated multiplications. This isn’t necessary today given students can use calculators. Indeed, from the setup of the problem the student could simply enter (18⦁10⦁630) / (540⦁15). Again, note that the term with the “x” is the divisor.

Look at this and realize that there are two entries in both the first and in the second cause. As in this problem, the conditions of a problem may require multiple entries in cause, or effect, or both (an example is below). When making these entries it’s just a matter of slowly and carefully doing it.

Reflect on the setup and solution for a while before moving onto the next example. I intentionally picked one that on the surface appears very messy but if Sheldon’s method is followed, it becomes clear (at least I think it does!)

The problem: How many men working 10 hours a day will be required to build a wall 160 feet long, 40 feet high and 3 feet thick in 15 days, if 75 men in 12 days of 9 hours each can build a wall 120 long, 30 feet high and 2 feet thick?

I did say it was messy, didn’t I? Some of this can be simplified before putting it in the compound proportion, although all these numbers could be placed according to Sheldon’s method. But, let’s do some preliminary calculations. The 1st cause includes what is being looked for (how many men?), so there is the “x”. The 1st cause also has 10 hours a day for 15 days, or 150 hours. The volume of the wall in the 1st effect is 160⦁40⦁3 = 19,200 cubic feet. The 2nd cause has 75 men and also 12 days of 9 hours each, or 108 hours. The volume of the wall in the 2nd effect is 120⦁30⦁2 = 7200 cubic feet.

So, here’s how the information is placed in the compound proportion …

150*x = 19200
108*75  7200

Cross-multiplying and dividing … (75⦁108⦁19200) / (150⦁72000) … gives 144.

If all the cause and effect values were entered separately – no preliminary calculations – here’s what it would look like (again recall that in 1886, there weren’t calculators but there was simplifying each fraction in the proportion before doing the calculation).

10*15*x  =  160*40*3
12*9*75     120*30*2

Again, the “x” is not included in the calculation but recall that the term with the “x” is the divisor. Below are a few more problems, with answers, for playing with Sheldon’s idea. Just take your time identifying the causes, effects and what you’re looking for. Have fun!

1. If 10 horses in 14 weeks eat 5.65 tons of hay, how long will 11.3 tons last 7 horse?     Answer:   40 weeks

2.  If the freight on 150 cattle averaging 900 pounds is $250 for 100 miles, what should be the freight on 275 cattle averaging 1200 pounds, for 150 miles?   Answer: $916.67

3. If a party of 15 persons pay $690 for 8 weeks’ board, how much should be paid by a party of 12 persons for 16 weeks’ board?      Answer:  $1104

4. If 12 men working 8 hours a day can saw 340 cords of wood in 10 days, how long will it take for 16 men working 9 hours a day to saw 204 cords?              Answer:  4 days

5. When $280 is paid for 175 barrels of apples containing 2 bushels each, how much should be paid for 125 barrels of apples containing 2.5 bushels each?        Answer:   $250

6. … and from Brook’s 1877 The Normal Higher Arithmetic (pg. 432) comes the following problem, which requires you to carefully determine what are the causes and effects.   If 24 pipes, each delivering 6 gal. a minute, fill a cistern 8 ft. long, 6 ft. wide, and 5 ft. deep, in 12 and 36/77 min., how many pipes, each flowing 8 gal. a minute, will a cistern 10 ft. long, 7 ft. wide, and 9 ft. deep, in 21 and 9/11 minutes?         Answer:  27 pipes






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