Fri Oct 01, 2004 1:31 pm
A rope passes over a pulley and hangs in equilibrium with a weight on one end and a monkey on the other. The rope, which weighs 4oz per foot , is the same length each side. The ages of the monkey and its mother add up to four, and the monkey weighs as many pounds as its mother is years old. The monkey's mother is twice as old as the monkey was when the monkey's mother was half as old as the monkey will be when the monkey is three times as old as the monkey's mother was when the monkey's mother was three times as old as the monkey.
The combined weights of the weight of the rope and the weight at the end is half as much again as the difference in weight between the weight of the weigh and the weight of the weight and the weight of the monkey.
How long is the ropE?
Fri Oct 01, 2004 4:26 pm
Fri Oct 01, 2004 6:55 pm
Fri Oct 01, 2004 10:27 pm
the difference in weight between the weight of the weigh and the weight of the weight and the weight of the monkey
A rope passes over a pulley and hangs in equilibrium with a weight on one end and a monkey on the other. The rope, which weighs 4oz per foot , is the same length each side. The ages of the monkey and its mother add up to four, and the monkey weighs as many pounds as its mother is years old. The monkey's mother is twice as old as the monkey was when the monkey's mother was half as old as the monkey will be when the monkey is three times as old as the monkey's mother was when the monkey's mother was three times as old as the monkey.
The combined weights of the weight of the rope and the weight at the end is half as much again as the difference in weight between the weight of the weigh and the weight of the weight and the weight of the monkey.
How long is the ropE?
First I defined what I could:
rope=30oz/foot
monk age + mom age = 4
monk weight = mom age
Then I hit a huge snag. I had to label and delete parts of this very confusing sentance just to make sense of it:
The mother is twice as old (a) as the monkey was (b) when the mother was (c) half as old as the monkey will be (d) when the monkey is (e) three times as old as the mother was (f) when the mother was (g) three times as old as the monkey (h).
Then I made truth statements about all the symbols. I couldn't directly link b and c but I was able to use them as a test.
2a=b
c=1/2d
d=e
3e=f
f=g
3g=h
c=1/2e
c=3/2f
c=3/2g
c=9/2h
a=current
b=past
c=past
d=future
e=future
f=past
g=past
h=current
I knew I just had to find the values such that the age difference was the same between b and c and a and h. So I plugged in reasonable values.
a=3
b=1 1/2
c=4 1/2
d=9
e=9
f=3
g=3
h=1
a-h=2 c-b=3
a=3 1/2
b=1 3/4
c=2 1/4
d=4 1/2
e=4 1/2
f=1 1/2
g=1 1/2
h=1/2
a-h=3 c-b=1/2
a=3 1/4
b=1 5/8
c=27/8=3 3/4
d=27/4
e=27/4
f=9/4
g=9=4
h=3/4
a-h=2 1/2 c-b=2 1/8
a=3 1/8
b=1 9/16
c=63/16=3 15/16
d=63/8
e=63/8
f=21/8
g=21/8
h=7/8
a-h=2 1/4 c-b=2 3/8
a=3 1/16
b=1 17/32
c=135/32=4 7/32
d=135/16
e=135/16
f=45/16
g=45/16
h=15/16
a-h=2 1/8 c-b=2 11/16
I finally got sick of plugging in random numbers and started using my graphing calculator and calculus knowledge. For those of you who know cal, I plugged in the values, got a function for the thing and ran a line through it to find where the values are equal to each other.
My calculator came up with 2.393375
So that should be the difference in their ages. So doing that I came up with the equation 2x+2.393375=4 where x=the monkey's current age. Which is .8033125 years. Which means that the mother's age is 3.1966875 years.
As defined above the mother's age equals the monkey's weight on pounds.
Now the second really confusing statement:
The combined weights of the weight of the rope and the weight at the end is half of the difference in weight between the weight of the weigh and the weight of the weight and the weight of the monkey.
weight + weight rope = 1/2 (weight - monk weight)
I took the weight of the weight and the weight of the monkey to be the same because the system is in equilibrium as stated in the first sentance.
And now the whole problem blows up because if I ignore half the statement like I did because I think it was a typing error and the two weights are equal, then subtracting them makes it equal to zero whic will give me a negative weight for the rope.
Errr.... Just to solve it I ran the problem with the negative weight anyway.
3.1966875lbs=51.147oz
With the given 30oz/foot that would put the rope at being 1.7049 feet.
Fri Oct 01, 2004 10:56 pm
Sat Oct 02, 2004 4:33 am
Sat Oct 02, 2004 4:44 am
Sat Oct 02, 2004 5:01 am
Sat Oct 02, 2004 7:34 pm
zorg wrote:Erm I had to type this up from a very old newspaper article, from 1977
Sun Oct 03, 2004 11:19 am
.Fizzy wrote:zorg wrote:Erm I had to type this up from a very old newspaper article, from 1977
Wow! That is old!
I did enjoy solving this, even if I did have to put an ice pack on my head for the rest of the day
Zorg, will you tell us what the answer is?