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Each problem that I solved became a rule which served afterwards to solve other problems.

Discours de la Methode. 1637.

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# Using Historical Problems in the Middle School

## Problems from a 19th Century Armenian Textbook

The third group of problems was designed to take advantage of the unique nature of the particular middle school where these were used, namely, the Armenian Sister’s Academy, Radnor, Pa.. In addition to the standard curriculum, all students study the Armenian language and culture. Thus, these problems come from a nineteenth century Armenian arithmetic textbook.

The students first had to translate the problem from Armenian and then solve the roblem. The translation from the Armenian was originally conceived of as a family project. Students were encouraged to involve their parents and grandparents in the translation. It turned out that the students were able to do a lot of the translation on their own.

The following summary describes our good fortune in discovering an Armenian arithmetic textbook that was both relevant and meaningful. The 19th century Armenian textbook, Tvapanootyoon (Arithmetic), published in 1887 in what was at the time Constantinople, Turkey, and is today Istanbul, was written to teach primary and secondary students the basics of numbers. Written by mathematician Hagop Bohajian, the book is written to accommodate the needs of beginner students, while it is graduated to challenge intermediate and advanced students as well. In an introduction to his book, Mr. Bohajian described it as comprehensive yet concise, covering a full range of computational principles and formulas. This book belonged to Mikael Markaridian, the great-grandfather of a student at the school, who resided in Turkey in the former Armenian state of Garin (Erzerum) at the time of his youth. Academy student Katrina Selverian discovered the book in her grandfather’s basement.

The book offers theoretical prose and textual instructions to guide the performance of mathematical functions, while providing problem solving sections to encourage practice and learning. Answers are printed in a final section.  Mr. Bohajian warned teachers to be careful not to allow students to hasten forward through the book if they had not mastered the previous sections. He assured them, simultaneously, that any student who mastered the entire book would be able to accomplish any mathematical task encountered in business with facility and ease.

The problems here are presented in the English translation by the students.

1.   A leaves a city at 9:15 and goes to another city going 4 mph. B leaves this second city at 9:30 and goes to the first city going 3 1/2 mph. The 2 cities are 21 miles apart. At what time are A and B going to meet?

Solution.

At 9:15 A is moving 4mph and B is not moving yet. We then found that at 9:30, the two people were 20 miles apart. At 10:30 they were 12 1/2 miles apart. We added 4 and 3 1/2 and got 7 1/2.   It was 20 miles at 9:30, which is 7 1/2.  So 20 mi / 7 1/2 mph = 2 2/3 hours = 2 hr. and 40 min. after 9:30, which equals 12:10. Therefore A and B will meet at 12:10.

2.   There is a garden with fruit trees, 1/2 of them are apple trees, 1/4 of them are peach trees, and 1/6 of them are plum trees. The remaining trees are 200 cherry trees. How many trees are there in the garden?

Solution.

We wrote out all of the fractions 1/2, 1/4, and 1/6 and found a common denominator which was 12. We then added up all the fractions to get 11/12. We saw that if you added 1/12, you would get 1. 1/12 is the number of cherry trees which is 200. Multiply 200 by 12 which equals 2400 which is the number of trees in the garden.

3.   A man who sells eggs, sells half his eggs and one more to the first buyer. To the second buyer he sells half of the remaining eggs and one more. The third buyer buys half of the remaining eggs and one more, and there are no more eggs left. How many eggs were there in the beginning?

Solution.

I did this problem by working backwards.

(((((0 + 1) x 2) + 1) x 2) + 1) x 2  = 14, the number of eggs he started with.

4.    If 30 men need 40 days to complete a job, how many men will it take to complete a job that is 5 times bigger in one-fifth the amount of time?

Solution.

There are 30 men, 40 days, and 1 project, (30,40,1). Since the project is 5 times as big, it becomes (30,40,5). Since it must take 1/5 of 40 days or 8 days, it then becomes (30,8,5). You multiply 5 (from the 1/5) by 5 ( the size of the new project) , which equals 25. You then multiply 25 by 30 (men) which equals the answer. 750 men are needed to finish 5 times the project in 8 days.

5.    A man left 38,000 dollars that was to be divided  among his three sons and his three daughters. Each son is going to get 33 1/3 % more than the oldest daughter; and the younger daughters are going to get 33 1/3 % less than the oldest daughter. What is everyone’s portion?

Solution.

Let S = son, D = youngest daughter, X = oldest daughter. Then S = (1/3)X + X and D = X – (1/3)X.   The equation for the entire amount is 3S + 2D + X = 38,000. Substituting, 3((1/3)X + X) +2(X–(1/3)X) +X =38,000. Simplifying we got (6 1/3)X = 38,000. X= 6000 which is what the oldest daughter gets. We substituted and got S = 8000 and D = 4000.