Pink Poogle Toy Forum

The official community of Pink Poogle Toy
Main Site
NeoDex
It is currently Fri Nov 01, 2024 1:05 am

All times are UTC




Post new topic This topic is locked, you cannot edit posts or make further replies.  [ 10 posts ] 
Author Message
 Post subject: Monkey Puzzle
PostPosted: Fri Oct 01, 2004 1:31 pm 
Honorary Member
Honorary Member
User avatar

Posts: 2116
Joined: Mon May 31, 2004 10:25 am
Location: Melb, Australia
Okay my dad gave me this a few months ago and I just found it, so i'll give brownie points to whoever actually solves this ;)

Quote:
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?


:roflol:


Do you miss Bill? I miss Bill. Go to this thread to petition Bill Back!


Top
 Profile  
 
 Post subject:
PostPosted: Fri Oct 01, 2004 4:26 pm 
PPT God
PPT God
User avatar

Posts: 1349
Joined: Sun Sep 26, 2004 6:55 am
Location: Over There!! :)
o_O :roll: *head explodes* :x


Image


Top
 Profile  
 
 Post subject:
PostPosted: Fri Oct 01, 2004 6:55 pm 
Honorary Member
Honorary Member
User avatar

Posts: 2456
Joined: Mon May 31, 2004 5:54 am
Location: THE HEART OF DARKNESS. *ominous drums* Haha, just kidding, I'm stalking you. :D
Gender: Female
How long? Or how much it weighs?

I can't answer the latter, but there's inadequate information to be able to determine how <i>long</i> the rope is. I think. oO


can't find the sig from this set, so instead, you get a <3 .


Top
 Profile  
 
 Post subject:
PostPosted: Fri Oct 01, 2004 10:27 pm 
PPT God
PPT God
User avatar

Posts: 1277
Joined: Mon May 31, 2004 5:39 pm
Location: Thanks Darklegendary for the Secret Santa set!
Ok I have seriously been trying this for the past 2 hours, maybe more. I wouldn't have made it this far without calculus and a graphing calculator. I hit the last sentence and I think it is mistyped or something.

Quote:
the difference in weight between the weight of the weigh and the weight of the weight and the weight of the monkey


I seriously think you made a copy and paste error here. the word "weight" (or one of it's forms) is used 6 times in a 22 word statement!

the weight of the weigh and the the weight of the weight?


Edit: I decided to run the problem anyway and just finish it.

The answer I got was 1.7049 feet.

Here is the text document that I did my calculations in:

Quote:
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.


Image
I'm finally back for the summer, although hopefully I'll spend a little more time away from this screen.


Top
 Profile  
 
 Post subject:
PostPosted: Fri Oct 01, 2004 10:56 pm 
PPT Trainee
PPT Trainee
User avatar

Posts: 696
Joined: Thu Jun 17, 2004 2:15 pm
Location: In Runescape =O
This may or may not be right, but I strained my brain trying to figure it out, so I don't really care now. :roll:

The length of the rope is 7 feet 4 inches.

I'm going to give this to one of my friend who derives a sick pleasure from giving ME these kinds of problems :P


Image
~ By Optimus ~


Top
 Profile  
 
 Post subject:
PostPosted: Sat Oct 02, 2004 4:33 am 
PPT God
PPT God
User avatar

Posts: 1122
Joined: Mon May 31, 2004 4:25 am
Location: Wishing I was in a play
Gender: Female
Just looking at this, I get the feeling it's a trick question. I can't even begin to imagine how this would be realistically solved, but I can't find a trick in it, either. However, one has to assume two things: One, both the monkey and the weight weigh the same amount, or else the rope would not be in equilibruim, and Two, most of that information is designed to throw you off. Now, it says the rope combined with the weight is half as much again as the difference between the wieght of the monkey and the weight. I don't see how that could be possible, seeing as they both weigh they same. The rope cannot be very long, if it exists at all. ;) My guess, 2 feet long.


Image


Top
 Profile  
 
 Post subject:
PostPosted: Sat Oct 02, 2004 4:44 am 
Honorary Member
Honorary Member
User avatar

Posts: 2116
Joined: Mon May 31, 2004 10:25 am
Location: Melb, Australia
Erm I had to type this up from a very old newspaper article, from 1977 :)


Do you miss Bill? I miss Bill. Go to this thread to petition Bill Back!


Top
 Profile  
 
 Post subject:
PostPosted: Sat Oct 02, 2004 5:01 am 
PPT Warrior
PPT Warrior
User avatar

Posts: 838
Joined: Mon May 31, 2004 6:40 am
Location: Cybertron, Australia
Man, that's a crazy puzzle. I couldn't even get past the second line without my head exploding. o_O


Image


Top
 Profile  
 
 Post subject:
PostPosted: Sat Oct 02, 2004 7:34 pm 
PPT Trainee
PPT Trainee
User avatar

Posts: 696
Joined: Thu Jun 17, 2004 2:15 pm
Location: In Runescape =O
zorg wrote:
Erm I had to type this up from a very old newspaper article, from 1977 :)

Wow! That is old! :o
I did enjoy solving this, even if I did have to put an ice pack on my head for the rest of the day :P
Zorg, will you tell us what the answer is?


Image
~ By Optimus ~


Top
 Profile  
 
 Post subject:
PostPosted: Sun Oct 03, 2004 11:19 am 
Honorary Member
Honorary Member
User avatar

Posts: 2116
Joined: Mon May 31, 2004 10:25 am
Location: Melb, Australia
Fizzy wrote:
zorg wrote:
Erm I had to type this up from a very old newspaper article, from 1977 :)

Wow! That is old! :o
I did enjoy solving this, even if I did have to put an ice pack on my head for the rest of the day :P
Zorg, will you tell us what the answer is?
.

If I remember right, I got

7 feet 4 inches too :)

But that may be wrong, the article didn't say the answer


Do you miss Bill? I miss Bill. Go to this thread to petition Bill Back!


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic This topic is locked, you cannot edit posts or make further replies.  [ 10 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 27 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group