Word Problems from History


Here are a few word problems that come from ancient time. Although the wording may be a little tricky sometimes, they are fun to solve!

v  This problem describes the life of Diophantus, a scholar from Alexandria. Not much is known about his personal life, but this “riddle” from Greek Anthology gives a short summary of his life.

The riddle reads, “'Here lays Diophantus,' the wonder behold. Through art algebraic, the stone tells how old: 'God gave him his boyhood one-sixth of his life, One twelfth more as youth while whiskers grew rife; And then yet one-seventh ere marriage begun; In five years there came a bouncing new son. Alas, the dear child of master and sage After attaining half the measure of his father's life chill fate took him. After consoling his fate by the science of numbers for four years, he ended his life.'

Since these words are very overwhelming we could re-write the riddle as follows: “Diophantus passed 1/6 of his life in childhood, 1/12 in youth, and 1/7 more as a bachelor. Five years he married was born a son who died 4 years before his father, at ½ his father’s final age.”

How old was Diophantus when he died?

Here are just two possible solutions:

System of Equations:

Straight algebra: if y is the number of years Diophantus lived, then we can say that

y=(1/6)y+(1/12)y+(1/7)y+(1/2)y+5 +4

We can find a common denominator: y= (14/84)y+(7/84)y+(12/84)y+(42/84)y+5+4

                                                                          y= (75/84)y+9



                                                                            y=84 years






v  This problem is from Mahavira, a Hindu Mathematicias from around 850 A.D.

“A powerful, unvanquished, excellent black snake which is 80 angulas in length enters into a hole at the rate of 7 ½ angulas in 5/14 of a day; in the course of ¼ of a day, its tail grows 11/4 of an angula. O ornament of the arithmeticians, tell me by what time the serpent enters fully into the hole?”

Again, if we look past the unfamiliar wording, the problem can become fairly easy. First, we can fgure out how much the snake grows in a day. Then, we can find how much of the snake enters the hole in one day.

If the snake grows 7 ½ angulas in 5/14 of a day, then it grows (7.5)/(5/14)=21 angulas in 1 day.

If the (11/4)  of the snake enters the hole in ¼ of a day, then (11/4)/(1/4)=11 angulas gives us the amount of the snake that enters the hole in a day.

So, after one day, 21-11 angulas of the snake enters the hole. So each day, 10 angulas of the snake enters the hols. Since the snake is 80 angulas long, we can say

10x=80 where x represents the number of days it will take for the snake to be completely in the hole. Therefore, it will take 8 days. 

A wolf, a goat and a cabbage must be taken to the other side of the river. You have a boat, which is not large enough to take more than one of them.
If you leave the wolf with the goat, the wolf will eat the goat.


If you leave the goat with the cabbage, the cabbage will be eaten.

v  The next problem comes from a work known as Problems for the Quickening of the Mind, which was written during the Dark Ages.

A wolf, a goat and a cabbage must be taken moved across a river in a boat holding only one besides the ferryman. How must he carry them across so that the goat shall not eat the cabbage, nor the wolf eat the goat?

Although this situation is not something that is very likely to occur, it still takes a problem solvers mind to figure this one out.

First, you take the goat to the other side.

Then you get the wolf, take it to the other side but bring the goat back with you.

Then, get the cabbage and take it to the other side.

Lastly, return to get the goat and all three have made it to the other side… safely!




Source for explorations and historical explanations: Source: Eves, H. (1990). An introduction to the history of mathematics, 6th Edition. Pacific Grove: Thompson Learning, Inc.


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