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Weighing V.

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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.

Pls visit New Puzzles section to see always fresh brain teasers.

Weighing V. - solution

Lay one red and one white ball on left pan and one blue and the other white ball on the right pan. If there is equilibrium, then it is clear that there is one heavier and one lighter ball on each side. That’s why comparing white balls is enough to learn everything.

However, if at first weighing one side is heavier, then there must be a heavier white ball on that side. The next reasonable step is to compare the already weighed red ball and yet not weighed blue ball. After that, the character of each ball is clear, isn’t it?

On a Christmas tree there were two blue, two red and two white balls. All seemed the same, however in each colour pair one ball was heavier. All three lighter balls were the same weight, just like all three heavier balls. Using a pair of scales twice, identify the lighter balls.

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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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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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(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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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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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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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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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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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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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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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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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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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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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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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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Posted (edited) · Report post

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

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

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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

This does not consider the case where your second weighing has equilibrium. Then you don't know if they're heavy or light, just that they're the same.

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I got almost the same thing as the spoiler except for the 2nd weighing, I'd weigh the red and blue balls from the first weighing against the two white balls because from the first weighing, we can deduce that those red and blue balls are both light, both heavy, or the one on the heavier side is heavy and the one on the lighter side is light. You can find out which one applies by weighing it against the two weight balls, and then you know everything.

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Put one white, red, and blue ball on each side. One side will be heavier, so you know that that side has two heavy balls. Take two of the 3 balls from the heavy side and weigh them against each other. If they balance, they are the heavy balls, and the other one must be light. If they don't balance, you know which of the three is the light ball, and therefore the other two are heavy. As you then know one of each pair, you automatically know the others.

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Posted (edited) · Report post

easy.

first balls: a b

second balls: c d

third balls: e f

now we weight, b and c.. d and e only..

if b = c, d = e then we can conclude b c f / a d e

if b =/c , d =/e then we can conclude b d f / a c e ( if b and d heavier )

if b = c, d =/e then b c e / a d f.

you dont have to know which one heavier or lighter, you can simply group them up with simple measurement

question?

Edited by kyky
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