# How long is the banana?

Found at The Universe of Discourse : How long is the banana? where there is a program that determines the solution.

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 of the rope 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 monkeyâ€™s mother.

The combined ages of the monkey and its mother is 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 monkeyâ€™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 its is now.

How long is the banana?

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The comedic value alone more than justifies this story problemâ€™s existence.

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Ha! So @bee and I solved this for â€śfunâ€ť last night. (PS: I forgot to mention we also had a mathingbeemergency, i.e., Beeminder said we had to do 1 page of math for our mathingbee goal.) It actually wasnâ€™t too bad! We got an answer of 5+3/4 inches which I see matches the solution at the link.

The trick was to take that ridiculous paragraph and parse it out starting from the end:

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The monkeyâ€™s mother is one-half as old as \phantom{ii}\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot 1\cdot 4\cdot 2\cdot \tfrac{1}{3}\cdot 1\cdot 3\cdot \tfrac{a}{4}
the monkey will be when it is three times as old as \phantom{xx\tfrac{1}{2} \cdot}3\cdot \tfrac{1}{2}\cdot 1\cdot 4\cdot 2\cdot \tfrac{1}{3}\cdot 1\cdot 3\cdot \tfrac{a}{4}
its mother was when she was one-half as old as \phantom{xx\tfrac{1}{2} \cdot 3\cdot}\tfrac{1}{2}\cdot 1\cdot 4\cdot 2\cdot \tfrac{1}{3}\cdot 1\cdot 3\cdot \tfrac{a}{4}
the monkey will be when it is as old as \phantom{xx\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot}1\cdot 4\cdot 2\cdot \tfrac{1}{3}\cdot 1\cdot 3\cdot \tfrac{a}{4}
its mother will be when she is four times as old as \phantom{xx\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot 1\cdot}4\cdot 2\cdot \tfrac{1}{3}\cdot 1\cdot 3\cdot \tfrac{a}{4}
the monkey was when it was twice as old as \phantom{xx\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot 1\cdot 4\cdot}2\cdot \tfrac{1}{3}\cdot 1\cdot 3\cdot \tfrac{a}{4}
its mother was when she was one-third as old as \phantom{xx\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot 1\cdot 4\cdot 2\cdot}\tfrac{1}{3}\cdot 1\cdot 3\cdot \tfrac{a}{4}
the monkey was when it was as old as \phantom{xx\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot 1\cdot 4\cdot 2\cdot \tfrac{1}{3}\cdot}1\cdot 3\cdot \tfrac{a}{4}
its mother was when she was three times as old as \phantom{xx\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot 1\cdot 4\cdot 2\cdot \tfrac{1}{3}\cdot 1\cdot}3\cdot \tfrac{a}{4}
the monkey was when it was one-fourth as old as its is now \phantom{xx\tfrac{1}{2} \cdot 3\cdot \tfrac{1}{2}\cdot 1\cdot 4\cdot 2\cdot \tfrac{1}{3}\cdot 1\cdot 3\cdot}\tfrac{a}{4}

So the whole paragraph is just saying that the monkeyâ€™s mother is \tfrac{3\cdot 4\cdot 2\cdot 3}{2\cdot 2\cdot 3\cdot 4} = \tfrac{3}{2} as old as the monkey. Then if you know that there are 16 ounces in a pound and 12 inches in a foot, youâ€™re home free.

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I donâ€™t understand the problem.

Is the idea of the problem that the two weights balance? If so, shouldnâ€™t this say â€śa weight equal to the weight of the monkey + bananaâ€ť?

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@bee and I took that as a total red herring. Whether any weights balance never enters the picture!

PS: You might be right though, that they meant to say that the weight on the other side weighed as much as the monkey plus the banana. I think that would give a slightly different answer. We took the description at face value though, assuming the monkey-rope-banana system was just not in equilibrium. Or maybe it is, with the friction of the rope on the fence? But since no kinematics are actually involved, it doesnâ€™t matter.

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Ohh, I see!

From that hint I started from the end of the comedy sentence, similar to how you did. Then I wrote down equations from the rest of the text. I, er, ended up with -3/2 as the weight of the monkey. Then I remembered Iâ€™m not in school any more so now Iâ€™m going to do something totally different instead of fix my mistake

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