Author Topic: Prehistoric maths joke  (Read 770 times)

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Offline mary62

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Prehistoric maths joke
« on: September 17, 2017, 13:56:00 PM »

In a certain tribe, in which polygamy was practiced, a married man’s standing in the tribe depended upon the combined weight of his wives-the greater the combined weight, the more important was the man. Every year, on weighing day and according to custom, the married men would stand their wives on neatly spread animal skins, and the chief of the tribe would come around with a crude seesaw and balance the wives of one man against those of another in order to determine the relative importance of the men. Now Gog had only one wife, who was very heavy, while Gug had two much slenderer wives, and all year the two men argued as to who was the more important, When weighing day arrived, Gog placed his wife on a large hippopotamus skin, and Gug placed his wives on two small gazelle skins
When the weighing was performed, it was found that Gog’s wife exactly balanced against the two wives of Gug. Thus it turned out that the two men were equally important, since, by the chief’s ruling, “the squaw on the hippopotamus is equal to the sum of the squaws on the other two hides.”



Offline Menthol

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Re: Prehistoric maths joke
« Reply #1 on: September 17, 2017, 23:41:20 PM »
Groan .......  ;D




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