Using Historical Problems in the Middle School - Historical Problems from Various Sources

The following problems come from Robert McGee of Cabrini College.   In each case, he presents the problem and then gives a solution found by the students, in their own words.

The first group of problems come from various times and places:

1.   A bamboo shoot 10 feet tall has a break near the top. The top touches the ground 3 feet from the stem. What is the length of the stem left standing erect ?  (Yang Hui, China, 1261)

Solution:

Facts:     A² + B² = C²

              A² +  9 = C²

              C + A = 10

Try and Check

            Single digits didn’t work ( ex. 4² + 3² is not equal to 6²)

            Decimals with tenths didn’t work (4.6² + 3² is not equal to 5.4²)

            Decimals with hundredths did work .

Answer: 4.55² + 3² = 5.45²

             4.55 is the length.

2.   Ottaviano Sempioni bought three jewels, the second of which cost 4 coins more than the first, the third cost 5 coins more than the first and second together, and all three cost 81 coins. Required the cost of each. (Onofrio, Arithmetica, Italy, 1670)

Solution:

First I drew 3 objects representing each jewel. Then I wrote the equation for each jewel, the first being A, the second A + 4 and the third 2A + 9. Then I put the equations together making 4A + 13 = 81.  Solving,  A = 17, so I went back to each equation and I found my answer. 1st = 17, 2nd = 21, and 3rd = 43.

3.   A hundred bushels of grain are distributed among 100 persons in such a way that each man receives 3 bushels, each woman 2 bushels, and each child half a bushel. How many men, women and children are there?  (Alcuin of York, England, 8th century)

Solution. Method was guess and test.   Here the first three numbers are the number of men, women, and children, respectively, while the last number is the total number of bushels.

12, 14, 74     101                                12, 12, 76    98

11, 13, 76       97                                13, 11, 76    99

10, 14, 76       96                                14, 10, 76  100

Therefore:     Men        = 14

                    Women   = 10

                    Children  = 76

A second solution of 11 men, 15 women and 74 children was also found.

4.   A powerful, unvanquished, excellent black snake, which is 80 angulas in length, enters into a hole at the rate of 7 1/2  angulas in 5/14 of a day, and in the course of 1/4 of a day, its tail grows 11/4 of an angula. O ornament of arithmeticians, tell me by what time the serpent fully enters into the hole.  (Mahavira, India,  ca. 850)

Solution:

5/14 day loses 7 1/2 angulas, then (5/14)(14/5) = 1 day, loses (7 1/2)(14/5) = 21 angulas.

1/4 day gains 11/4 angulas, (1/4)(4) = 1 day, gains (11/4)(4) = 11 angulas.

   –21 + 11 = –10 angulas per day.

X days = –80 angulas, 1 day = –10 angulas,

8 days = –80 angulas

Therefore in 8 days the snake is in the hole.

5.   Divide 60 into four Such parts, that the first being increased by 4, the second decreased by 4, the third multiplied by 4, the fourth part divided by 4, that the Sum, the difference, the product and the Quotient shall be one and the Same number.  (Benjamin Banneker, U.S., 18th century)

Solution:

Let the numbers be X, Y, A, and B. Then

X + 4 = Y – 4 or X + 8 = Y. Also

X + 4 = 4A or (1/4)X + 1 = A. Also

X + 4 = B/4 or 4X + 16 = B. Then

X + X + 8 + (1/4)X + 1 + 4X + 16 = 60

             (6 1/4)X + 25 = 60

             (6 1/4)X = 35  and X = 5 3/5

Solution:

X = 5 3/5

Y = 13 3/5

A = 2 2/5

B = 38 2/5