A CLEVER RIDDLE

[M2]

A census taker calls at a house, he asks the woman living there the ages of her three daughters. The woman says "If you multiply their ages, the total is 72. If you add together their ages, the total is the same as the number on my front door, which you can see."

The census taker says "That is not enough information for me to calculate their ages."  The woman says "Well, my eldest daughter has a cat with a wooden leg." The census taker replies "Ah, now I know their ages."

What are the ages of the three girls?

[sfortescue]

The statement of the riddle is flawed since the two oldest could be the same number of years old without being the same age.  So technically, the riddle has no solution.

The solution of an appropriately rephrased version of the problem can be found fairly easily.

[Robb]

The seminars got much easier to both interpret and even sit through once I realized that the high-sounding language was just a sham.  Otherwise, they were usually just basic points somewhat poorly jumbled together.

I know exactly how that is.  I've got a good friend who is deaf.   She and I used to work together, and during staff meetings, when the . . execute directly. . executive director would get long winded, it was amazing how quickly the interpreter would just cut to the chase and give my friend the essence of what was being said and then, for the last fifteen day. . .minutes, they (the interpreter and my friend) would chat about almost anything other than what was going on.  The director didn't get it, since her sign language wasn't all that good.  Sounds like we couldv'e used someone like that during those long, boring seminars with GG.  I know I could have.

[sfortescue]

While we're talking about more serious problems, here's one that doubles as a joke simply because it's funny.  It's in one of the several small booklets that I have that were published by Litton called "Problematical Recreations".  Solving the problem is actually fairly straightforward.

A rope over the top of a fence has the same length on each side.  Weighs 1/3 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 rope is as long 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 monkey and mother are 30 years.  1/2 the weight of the monkey, plus the weight of the banana, is 1/4 as much as the weight of the weight and the weight of the rope.  The monkey's mother is 1/2 as old as the monkey will be when it is 3 times as old as its mother was when she was 1/2 as old as the monkey will be when it is as old as its mother will be when she is 4 times as old as the monkey was when it was twice as old as its mother was when she was 1/3 as old as the monkey was when it was as old as its mother was when she was 3 times as old as the monkey was when it was 1/4 as old as it is now.  How long is the banana?

-- UCLA Engineering Student Newsletter

[al Hartman]

A rope over the top of a fence has the same length on each side.  Weighs 1/3 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 rope is as long 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 monkey and mother are 30 years.  1/2 the weight of the monkey, plus the weight of the banana, is 1/4 as much as the weight of the weight and the weight of the rope.  The monkey's mother is 1/2 as old as the monkey will be when it is 3 times as old as its mother was when she was 1/2 as old as the monkey will be when it is as old as its mother will be when she is 4 times as old as the monkey was when it was twice as old as its mother was when she was 1/3 as old as the monkey was when it was as old as its mother was when she was 3 times as old as the monkey was when it was 1/4 as old as it is now.  How long is the banana?

Stephen,
     i know you can solve these problems in your head while doing a crossword puzzle & eating a meatball sandwich, but just reading that gave me an Excedrin headache.
     Could we just stick to something within the scope of the average person, such as calculating the dimensions of the Giant Hand, before & after a manicure?
     Please excuse me, i have to go lie down now...

al

[vernecarty]

While we're talking about more serious problems, here's one that doubles as a joke simply because it's funny.  It's in one of the several small booklets that I have that were published by Litton called "Problematical Recreations".  Solving the problem is actually fairly straightforward.

A rope over the top of a fence has the same length on each side.  Weighs 1/3 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 rope is as long 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 monkey and mother are 30 years.  1/2 the weight of the monkey, plus the weight of the banana, is 1/4 as much as the weight of the weight and the weight of the rope.  The monkey's mother is 1/2 as old as the monkey will be when it is 3 times as old as its mother was when she was 1/2 as old as the monkey will be when it is as old as its mother will be when she is 4 times as old as the monkey was when it was twice as old as its mother was when she was 1/3 as old as the monkey was when it was as old as its mother was when she was 3 times as old as the monkey was when it was 1/4 as old as it is now.  How long is the banana?

-- UCLA Engineering Student Newsletter

 I assume the weight W, on the other side is equal to the monkey's, M plus the banana B's weight combined? The statment of the problem says the weight is equal to only the monkey's weight...I also assume we ignore friction? I would set L as the length of the rope, W as the weight M as the weight of the monkey and B as the banana's weight...I think we have enough info to solve for B...
Verne

[editor]

While we're talking about more serious problems, here's one that doubles as a joke simply because it's funny.  It's in one of the several small booklets that I have that were published by Litton called "Problematical Recreations".  Solving the problem is actually fairly straightforward.

A rope over the top of a fence has the same length on each side.  Weighs 1/3 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 rope is as long 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 monkey and mother are 30 years.  1/2 the weight of the monkey, plus the weight of the banana, is 1/4 as much as the weight of the weight and the weight of the rope.  The monkey's mother is 1/2 as old as the monkey will be when it is 3 times as old as its mother was when she was 1/2 as old as the monkey will be when it is as old as its mother will be when she is 4 times as old as the monkey was when it was twice as old as its mother was when she was 1/3 as old as the monkey was when it was as old as its mother was when she was 3 times as old as the monkey was when it was 1/4 as old as it is now.  How long is the banana?

-- UCLA Engineering Student Newsletter

 I assume the weight W, on the other side is equal to the monkey's, M plus the banana B's weight combined? The statment of the problem says the weight is equal to only the monkey's weight...I also assume we ignore friction? I would set L as the length of the rope, W as the weight M as the weight of the monkey and B as the banana's weight...I think we have enough info to solve for B...
Verne

I deal with this sort of thing in real life, quite often.  

On a sailboat, we use sealed torlon bearings, in blocks (pulleys for you landlubbers) to guide lines (ropes) over and around things.  "Monkeys" and their mothers, frequently hang on these, and often with bananas, even bottles of beer from time to time. (The father and mother, not the little monkeys)

Friction is the real problem.  If you don't find out the friction of the fence/rope interface, the rest of the problem is moot.  A good engineer wouldn't agree to solve the problem, because the friction is the largest force in the equation, (a knot is just friction) and is unaccounted for.

Sorry, this is a bad problem with no solution except to chase the monkeys away from the fence, being careful not to get bitten by these nasty creatures.

Brent

Brent

[d3z]

Read it carefully.  Friction doesn't come into play, since it tells you that the weight on the other side weighs the same as the monkey and the banana.  It isn't really a physics problem, just a convoluted algebra story problem.

[editor]

Read it carefully.  Friction doesn't come into play, since it tells you that the weight on the other side weighs the same as the monkey and the banana.  It isn't really a physics problem, just a convoluted algebra story problem.

I rebuke you in the name of Clifford Ashley, the guru of lines, splices, bends, hitches and knots.  

Friction is literally a life and death issue.  Ask any mountain climber, who has his own weight on one end of the rope, and the weight of a knot on the other.  Whether of not the weight is equal has no bearing whatsoever of the fact that friction is the main factor in the equation!  

Now, if the problem said to ignore friction,  a foolish thing to do, then you would be 100% correct.  However, since I don't feel like trying to express this problem algebraicly,  I am going to take the easy and correct way out by playing my friction argument.

Like I said, I do this in real life all the time.

Brent

[vernecarty]

Read it carefully.  Friction doesn't come into play, since it tells you that the weight on the other side weighs the same as the monkey and the banana.  It isn't really a physics problem, just a convoluted algebra story problem.

Actually David, that point is not clear; the problem states that the weight on the other side is the same  as the weight of the monkey! To solve the problem we must assume your above statement is true.  If we consider the weight of the monkey and the banana to be equal to the weight on the other side, the solution is of course trivial...

m + b = w

m/2 + b = (R_w + W)/4

m/2 + b =  (R_w + m + b)/4  etc. etc.

R_w = weight of rope


Verne

[enchilada]

Since the banana is related to the length of the rope, there is no way that the riddle can be answered.  We need to know all the elastic, inelastic as well as the time-dependent stress/strain relationships of the rope and how long the monkey was sitting on the rope with his/her banana.  Then there's temperature effects, so we need to know the variation of temperature of the rope during the time considered, as well as what time domain to use for this problem.   blah blah blah....

So, I'll say the banana is between 4 and 8 inches.

[vernecarty]

Since the banana is related to the length of the rope, there is no way that the riddle can be answered.  We need to know all the elastic, inelastic as well as the time-dependent stress/strain relationships of the rope and how long the monkey was sitting on the rope with his/her banana.  Then there's temperature effects, so we need to know the variation of temperature of the rope during the time considered, as well as what time domain to use for this problem.   blah blah blah....

So, I'll say the banana is between 4 and 8 inches.

I say we just eat the banana...
Verne

[sfortescue]

A census taker calls at a house, he asks the woman living there the ages of her three daughters. The woman says "If you multiply their ages, the total is 72. If you add together their ages, the total is the same as the number on my front door, which you can see."

The census taker says "That is not enough information for me to calculate their ages."  The woman says "Well, my eldest daughter has a cat with a wooden leg." The census taker replies "Ah, now I know their ages."

What are the ages of the three girls?

Since there was ambiguity before the eldest daughter was mentioned, that means that there was an alternative answer before that in which there were at least two oldest daughters.  The product of the three ages would in that case be a square number multiplied by the youngest age.

The square numbers that divide into 72 are 1, 4, 9 and 36.  The only choice is 36 since the others would make the third age to be the oldest.  So the alternative answer would have been 6, 6 and 2.  These add up to 14.  Finding the real answer, also with total 14, is left for the reader.

Unfortunately, the statement that there is an eldest daughter doesn't exclude the possibility that the two oldest daughters might be of ages 6 years and 11 months, and 6 years and 1 month.  They would both be said to be 6 years old, yet one is clearly older than the other.

If fractions are allowed, there may be more possibilities.  Something to investigate.

[sfortescue]

A computer search found more than 40 variations on the census taker problem.

Instead of 72, suppose the kids' ages multiply to make 720?

or 900?  or 1008?  or 2800?  or 2880?  or 4032?  or 8640?

Each of these variations on the problem has a unique solution.

(Well, one of them has a second solution that is unrealistic.)

[sfortescue]

Since there are 8 bits in a dollar:

1 Baud = 450 Dollars per Hour.

(Sorry, this is a joke rather than a riddle.)