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Weighing V.


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23 replies to this topic

#1 rookie1ja

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Posted 30 March 2007 - 05:51 PM

Weighing V. - Back to the Water and Weighing Puzzles
On a Christmas tree there were two blue, two red, and two white balls. All seemed same. However, in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls.
Using a balance scale twice, identify the lighter balls.

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
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#2 Nogumption

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Posted 28 June 2007 - 05:59 PM

your solution will not work all the time, because if there is equilibrium it could be either of the balls on each side that could be the heavy one. meaning even if you changed the color balls you could get the same configuration again when replaced with the ones left.
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#3 dsu

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Posted 28 June 2007 - 10:27 PM

your solution will not work all the time, because if there is equilibrium it could be either of the balls on each side that could be the heavy one. meaning even if you changed the color balls you could get the same configuration again when replaced with the ones left.



Here is how it works:

Case I (first weighing gives equilibrium)
2 possibilities: Bw rW or bW Rw
in your next weighing, you get w W or W w
You have now identified all the other balls - those you have weighed and those you have not.

Case IIa (first weighing left side heavier)
BW rw you instantly identify all the heavier and lighter balls

Case IIb (first weighing right side heavier)
bw RW you instantly identify all the heavier and lighter balls
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#4 earl11

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Posted 04 July 2007 - 12:49 PM

(Upper case letters denote heavier balls and lower case letters denote lighter balls)

1st weighing - In the left pan put one red and one white ball and in the right pan, put one blue and one white ball.

Case 1. Both pans are equal. This can mean

Rw and bW
OR
rW and Bw

Next, weigh the two white balls. If the left pan is heavier, then the red ball left out is the heavier one and the blue ball left out is the lighter one. Vice versa if the right pan is heavier.

Case 2. Left pan is heavier. This can mean

(i) rW and bw
OR
(ii) RW and Bw
OR
(iii) RW and bw

In all 3 cases, the heavier white ball has to be in the left pan.
Now, weigh the coloured ball in the heavier pan (i.e. red) with the ball which is left out of the other colour (i.e. blue)
Case a. They are equal - This is possible if we weigh R with B (iii).
Case b. Red is heavier than blue - This is possible if we weigh R with b (ii).
Case c. Red is lighter than blue - This is possible if we weigh r with B (i).

Case 3. Right pan is heavier. In this case, the heavier white ball is in the right pan. We then weigh the blue ball from the heavier pan with the red one which has been left out and come up with similar conclusions as in Case 2.
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#5 rubarb1

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Posted 07 August 2007 - 08:48 PM

This problem could also be solved in a similar manner by weighing all six balls at the same time.

On the left scale you have two blue balls and one white.
On the right scale you have two red balls and the other white.

BbW - Rrw or Bbw - RrW

The white balls act like a control. Meaning that you will always have two heavy balls and one light ball on one side, and two light balls and one heavy ball on the other side. This means that the scales will always tilt toward the heavy white ball. Once this happens, you simply remove one blue ball and one red ball.

For the sake of example, lets say the heavy white ball is on the same side as the red balls (Bbw - RrW). If the scales stay where they are when you remove the blue and red ball, then you know that you have left the light blue ball and the heavy red ball on the scales (bw - RW). If the scales equalize, then you have left the heavy blue ball and the light red ball (Bw - rW)
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#6 Jsquare

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Posted 15 August 2007 - 04:00 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.
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#7 cjk

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Posted 05 September 2007 - 06:13 PM

This comes down to symantics! If you mean a pair of scales that weighs 2 objects at the same time showing each weight then the answer would be easy. If you mean a set of scales which only weighs 1 object at a time then I can't see a possible answer.
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#8 rookie1ja

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Posted 05 September 2007 - 06:46 PM

This comes down to symantics! If you mean a pair of scales that weighs 2 objects at the same time showing each weight then the answer would be easy. If you mean a set of scales which only weighs 1 object at a time then I can't see a possible answer.


I mean a pair of scales that weighs 2 objects at the same - comparing their weight ... either the 2 objects weigh the same or they don't (so no numbers
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#9 adamgram

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Posted 16 September 2007 - 04:01 PM

1- place one white and one blue on the left, one white and one red on the right

If scales are balanced...
2- remove red and blue balls to find the heavier white ball, the ball that shared the scale with the heavier white ball is light, and the other ball of that color is heavy. The ball that shared the scale with the lighter white ball is heavy, and the other ball of that color is light

If scales are not balanced...
2- say the right hand side was heavier, than the heavier white ball was on the right. Remove white balls from scale and move the remaining ball from the left scale to the the right. place the red and white balls not used in the first step on the left.

If scales are balanced, we have one heavy and one light on the right hand side, so whichever color was on the right with the heavier white ball in step one must be the heavy one, and the other ball on the right is the light one.

If scales are not balanced, the 2 non-white balls in step one were the same weight, and they are now both on the right scale being weighed against the red and blue ball of the other weight, so step 2 reveals whether these are the 2 heavy balls or the 2 light balls.
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#10 bartman

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Posted 24 September 2007 - 07:56 PM

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
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