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Posers and Puzzles

Posers and Puzzles

  1. 23 May '07 19:04
    I don't know if this has already been done, but I was asked this riddle in college, and just cannot work out the answer. Any ideas, clues or solutions would be appreciated

    There was a rope hanging over a pulley with a weight on one end and a monkey on the other. The rope weighs 400g/m; the age of the monkey and the age of it's mother were together equal to 4 years; the weight of the monkey was as many kg as it's mother was years old; the mother was twice as old as the monkey was when the mother was half as old as the monkey will be when the monkey is 3 times as old as the mother was when the mother was 3 times as old as the monkey. The weight of the weight and the weight of the rope were half as much as the difference between the weight of the weight and the weight of the weight and the weight of the monkey.
    What was the length of the rope?
  2. 23 May '07 20:56 / 1 edit
    Originally posted by Darlek2354
    I don't know if this has already been done, but I was asked this riddle in college, and just cannot work out the answer. Any ideas, clues or solutions would be appreciated

    There was a rope hanging over a pulley with a weight on one end and a monkey on the other. The rope weighs 400g/m; the age of the monkey and the age of it's mother were together equa and the weight of the weight and the weight of the monkey.
    What was the length of the rope?
    Well, it all seems to depend on mother's age... let's call the variables:

    A=age
    m=monkey
    W=weight
    Ww=weight that's hanging on the rope
    Wr=weight of the rope
    Lr=length of the rope
    mm=monkey's mother

    Am1 is the age of the monkey when it was 3 times younger than the mm. Then, 3*Am1=Amm1 (mm's age at that time).
    Thus 9*Am1=monkey's age when it will be 3 times as old as the mother at thet time. Then Amm2 (mm's age, when it is half as old as m will be)should be 9*Am1/2=4.5*Am1

    In the first line of the previous paragraph I expressed the Amm1 in terms of Am1. The difference in their ages thus wil be (3-1)Am1=2Am1. From here continuing the last paragraph, the monkey's age at that time will be 2.5*Am1

    So, 5*Am1=mm's age now, and 3*Am1=m's age now. Thus 5*Am1+3*Am1=8 yrs=>Am1=1year (age of the monkey when the mother was three times as old)

    So, the m (and thus the Ww) = 5 kg

    Set the two parts of the pulley equal:

    Wr+5=1.5(5+5-5)

    Wr=2.5 kg

    Now, factor label: 2.5kg=2500g*1m/400g=6.25m

    I guess, that's right, but I'm not sure.

    P. S. Sorry for my disorganization
  3. 24 May '07 02:07 / 2 edits
    the age of the monkey and the age of it's mother were together equal to 4 years

    So, 5*Am1=mm's age now, and 3*Am1=m's age now. Thus 5*Am1+3*Am1=8 yrs

    I'm not sure where you got the 8 yrs from, but here's how I would do it:

    Set up three simultaneous equations with three variables (x = monkey's current age; y = monkey's mother's current age; z will be defined later). The first is easy: x + y = 4. The last two need to be worked backwards:

    when the mother was 3 times as old as the monkey.
    So the mother was 3 times his age then.

    when the monkey is 3 times as old as the mother
    When the mokey is 9 times his age back then.

    when the mother was half as old as the monkey will be
    When the mother was 4.5 times the monkey's age back then.

    as old as the monkey was
    When the monkey was 2.5 times his age back then (he grew by 1.5 times his age; the mother was 3 times it, now she's 4.5 times)

    the mother is twice
    The mother is now five times his age back then, and the monkey is now 2.5 times his age back then.

    Since it was the same amount of years earlier, let's call it z. Now we have:

    x + y = 4
    x = 2.5(x - z)
    y = 5(x - z)

    I'll just skip to the solution: (x, y, z) = (4/3, 8/3, 4/5) [not that x or z matter]. Anyway, this means that the monkey weighs 8/3 kilograms.

    The weight of the weight and the weight of the rope were half as much as the difference between the weight of the weight and the weight of the weight and the weight of the monkey.

    I'll assume that the part in bold shouldn't be there (if it really should then you'll need to clarify what two values the difference is between). At this point, I must ask what you meant by the first clause of that sentence. Do you mean that the sum of the weights of the rope and the weight is half that difference, or that each one is individually? If you mean their sum, then I see no way of deducing how long the rope is. There has to be one more piece of information. Are the weight and the monkey level? Do we know the amount of rope on one side of the pulley relative to the other side?
  4. 24 May '07 02:23
    you're right, my explanation is based on an assumption that net acceleration of the system is 0 (pulley is at equilibrium). If it is not - the problem is not soluble without new piece(s) of data.

    P. S. Bold is a mistake.
  5. 24 May '07 02:32
    you're right, my explanation is based on an assumption that net acceleration of the system is 0 (pulley is at equilibrium).

    Not only that; even if we assume the pulley is at equilibrium (which I think is a safe assumption), we still don't know the amount of rope on one side relative to the amount of rope on other side (i.e. how far one hangs down relative to the other). The solution depends on this.