## In 1877, Mr. Ray Reasons with Fractions

Posted by mark schwartz on September 8, 2016

__Introduction__

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.

- 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.

- 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.

- 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.

- 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.

- 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.

- 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.

- 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.

## Leave a Reply