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Old 13 Dec 2007, 18:49   #1
JanT
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Question Puzzle: Twelve Billiard Balls

In Psychology today, we were given Piaget's Twelve Billiard Balls puzzle to sort out. Why? Ask me another. Surely this would be a Maths problem.


The Twelve Billiard Balls Puzzle

There are twelve billiard balls, all the same size, shape and colour. All weigh exactly the same, except that one ball is slightly different in weight, but not noticeably so in the hand. Moreover, the odd ball might be lighter or heavier than the others.


Your challenge is to discover the odd ball and whether it is lighter or heavier. You must use a beam balance only, and you are restricted to three weighing operations.


Solution here.
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Old 13 Dec 2007, 20:35   #2
R.
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Old 14 Dec 2007, 18:44   #3
Cpl Mickey
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I'll go along with that thought R
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Old 14 Dec 2007, 18:48   #4
Lord Kagan
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surely an easier way would be to counter balance each ball on the beam balance and whatever one isnt balanced contains the heavier/lighter ball by process of elimination?
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Old 14 Dec 2007, 18:54   #5
Cpl Mickey
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Originally Posted by Lord Kagan View Post
surely an easier way would be to counter balance each ball on the beam balance and whatever one isnt balanced contains the heavier/lighter ball by process of elimination?
That sounds too easy. lol !
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Old 14 Dec 2007, 19:32   #6
samurai7
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Quote:
Originally Posted by Lord Kagan View Post
surely an easier way would be to counter balance each ball on the beam balance and whatever one isnt balanced contains the heavier/lighter ball by process of elimination?
yeah, but you're only allowed three weighing processes, and there are twelve balls, innit?
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Old 14 Dec 2007, 19:45   #7
duke knooby
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1st attempt.. 6 on each side
2nd attempt.. 3 on each side
3rd attempt.. 1 on each side.. if balanced the un used ball is the odd one.

only re weigh the heavy side

Last edited by duke knooby; 14 Dec 2007 at 19:50.
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Old 14 Dec 2007, 19:55   #8
AndyK
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Quote:
Originally Posted by duke knooby View Post
1st attempt.. 6 on each side
2nd attempt.. 3 on each side
3rd attempt.. 1 on each side.. if balanced the un used ball is the odd one.

only re weigh the heavy side

What happens if the ball that is a different weight is lighter than the other 11?
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Old 14 Dec 2007, 20:04   #9
duke knooby
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good point...
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Old 14 Dec 2007, 21:13   #10
mszee
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Yeah, ok...I knew that one before...started reading from the link and remembered why I forgot it...

Well, I think lesson well learned here is that it's much worse without any balls...I think so anyway...
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Old 15 Dec 2007, 13:29   #11
Lord Kagan
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ill work my way out later... cant be fecked now
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