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# Weighing V.

23 replies to this topic

### #11 Martini

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Posted 24 September 2007 - 08:34 PM

1) Weigh one white ball and one white ball - one on each scale.
- we find out the weight of the light ball
- we find out the weight of the heavy ball

No, the type of scales in the riddle are balance scales where one side is compared to the other. These scales don't tell you how much something weighs when the objects weighed on each side are of unknown weight, therefore your solution won't work.
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### #12 BoilingOil

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Posted 26 September 2007 - 05:55 AM

I originally wrote a complex system with two weighings of 4 balls each. The result was the same as in the spoiler, but required more deductive skills...
Not fair to let you all go through the trouble of reading that dribble.

BoilingOil
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### #13 BoilingOil

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Posted 26 September 2007 - 06:10 AM

Why go to so much work?

It would be much easier to measure a single blue ball against a single red ball, whichever one is lighter move into your light pile, and the heavier one, into your heavy pile, leaving the other one of that color to go into the light pile. Then weigh the white ones against each other to find the lighter one.

There is one problem here... what if in your 1st weighing there is equilibrium? Are they heavy balls or light? You can't say, nor can I

BoilingOil
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### #14 Craig 444

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Posted 11 November 2007 - 10:41 PM

couldnt you tell by how much the branch of the christmas tree is hanging down? but if the heavier ball was heavy enough to do that wouldnt you be able to tell in your hands.
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### #15 broknangel89

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Posted 14 November 2007 - 11:55 PM

why not just pick them up or hang them on the tree and see which ones pull down the limb more?
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### #16 minimeisacutie

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Posted 24 January 2008 - 11:00 PM

1) is the weight difference noticable?
2) is there a significant amount of weight in the heavier ones and a minute( not the word min-ute but the word mi-nute) amount of weight in the lighter ones?

if either of those were true you could just weight them in you hands

Edited by minimeisacutie, 24 January 2008 - 11:01 PM.

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### #17 foamrat

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Posted 25 January 2008 - 07:56 PM

The real puzzle is trying to figure out how to keep the balls from rolling off the scales.
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### #18 ladydulac

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Posted 07 March 2008 - 05:46 AM

There is one problem here... what if in your 1st weighing there is equilibrium? Are they heavy balls or light? You can't say, nor can I <!-- s:) --><!-- s:) -->

BoilingOil

but you can say that they are the equal and the other blue and red are equal. Then if you put one blue and one white on one side and the other blue and white on the other side you should be able to see which three are bunched together.

If you get equilibrium then the whites go to the opposite group if you don't get equilibrium then the whites are in the correct group.
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### #19 exodus

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Posted 07 March 2008 - 09:38 AM

Maybe this is too simple...maybe I'm missing something...you don't need to measure 4 balls at once on 2 scales...

I'd keep it more simple since all light balls weigh the same and all heavy balls weigh the same.

1) Weigh one white ball and one white ball - one on each scale.
- we find out the weight of the light ball
- we find out the weight of the heavy ball
2) Weigh one red ball and one blue ball one on each scale.
- we find out the red ball is heavy or light.
- the other red ball is the opposite
- we find out the blue ball is heavy or light.
- the other blue ball is the opposite

There would be a problem if the red and blue ball you choose are of the same weight (both light or both heavy), then you can't choose between the balls of same color
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### #20 Naruto Uzumaki

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Posted 01 April 2008 - 04:02 PM

why don't you look at the branches that are leaning more than the others and then you would know right?
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