## Monkeys and Bananas

By: Mr. Wilson on
January 8, 2014

*correct*answer, but eventually I understood that all those numbers, letters, and symbols proved his point. I want to share the problem with you. Can you figure it out? Here it is: A rope over the top of a fence has the same length on each side, and weighs one-third of a pound per foot. On one end hangs a monkey holding a banana, and on the other end a weight equal to the weight of the monkey. The banana weighs 2 ounces per inch. The length of the rope in feet is the same as the age of the monkey, and the weight of the monkey in ounces is as much as the age of the monkey's mother. The combined ages of the monkey and its mother are 30 years. One-half the weight of the monkey, plus the weight of the banana is one-fourth the sum of the weights of the rope and the weight. The monkey's mother is one-half as old as the monkey will be when it is three times as old as its mother was when she was one-half as old as the monkey will be when it is as old as its mother will be when she is four times as old as the monkey was when it was twice as old as its mother was when she was one-third as old as the monkey was when it was as old as its mother was when she was three times as old as the monkey was when it was one-fourth as old as it is now. How long is the banana?

## Comments

See what your friends and neighbors have to say about this.

Nobody told me there would be math.

<raises hand>

Mr. Wilson, can we assume that the simian-banana-rope-weight assembly is static? If so, piece of cake.

Yes Eddie, for the purposes of this problem assume a spherical cow.

This was the kind of BS story problem my 7th grade math teacher would throw at us for “fun” on Friday puzzles day. Maybe two kids loved this. The rest of us withdrew into sullen silence, knowing we were too stupid to ever be worthy of decent careers. I can pretty much trace my lack of success and general life incompetence to 7th grade math at Pound Junior High.

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