## Sheldon’s Compound Proportions

Posted by mark schwartz on April 13, 2016

__Introduction:__

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 19^{th} century compared to the 20 and 21^{st} 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*630 540*15 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 1^{st} 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 1^{st} effect is 160⦁40⦁3 = 19,200 cubic feet. The 2^{nd} 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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