Monkey Word Problem
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 in inches?
The answer is 5.75.
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1 solution
My strategy is to work backward through the part of the problem dealing with ages of the monkey and mother. Most of the way through the very lengthy sentence, the various monkey ages and mother ages alternate. I will let the various monkey ages be x i . Where the subscript i counts the number of times, working backward, that the monkey is mentioned. Similarly, the various mother ages will be y i , working backward. The present age of the monkey is mentioned first (remember we are working backward), and will be x 1 . Next mentioned is the monkey again. So x 2 = (1/4) x 1 . Next the mother is mentioned so y 1 = 3 x 2 =3(1/4) x 1 = (3/4) x 1 . Notice how each y is related to the previous x. Next the monkey is mentioned so x 3 = y 1 = (3/4) x 1 . Continuing, always substituting to get the latest x or y in terms of x 1 , y 2 = (1/3) x 3 = (1/4) x 1 , x 4 = 2 y 2 = (1/2) x 1 , y 3 = 4 x 4 = 2 x 1 , x 5 = y 3 =2 x 1 , y 4 =(1/2) x 5 = x 1 , x 6 = 3 y 4 = 3 x 1 , y 5 =(1/2) x 6 = (3/2) x 1 . Notice that the very long sentence about the ages begins with the mother and ends with the monkey. (The same result can be obtained by multiplying all the numbers together that occur in the sentence, but the justification for doing so is the process we just went through for each relationship.) What we have found so far is that the present age of the mother is one and one half times the present age of the monkey. The problem tells us that the sum of the present ages is 30. So x 1 + y 5 =(5/2) x 1 = 30. Therefore, the present age of the monkey is 2/5 of 30, or 12, i.e., x 1 = 12. The mother’s present age is y 5 = (3/2) x 1 = 18. The problem tells us that the length of the rope in feet is the same as the monkey’s age in years. So the rope is 12 feet long. The rope weighs 1/3 of a pound per foot; so the rope weighs 4 pounds = 64 ounces. The monkey weighs 18 ounces (same as the mother’s age in years). Let w = weight of banana. From the statement of the problem, half the weight of the monkey + w = (1/4)(weight of rope + attached weight on other side). Therefore, 9 + w = (1/4)(64 + 18), where from the problem, the attached weight is the same as the monkey’s weight. Solving for the weight of the banana gives w = 23/2 ounces. Since the banana weighs 2 ounces per inch, the length of the banana is 23/4 = 5 ¾ inches.