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Weighing VI.

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Weighing VI. - Back to the Water and Weighing Puzzles

There are 9 similar balls. Eight of them weigh the same and the ninth is a bit heavier.

How would you identify the heavier ball if you could use a two-pan balance scale only twice?

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Weighing VI. - solution

Divide the 9 balls into 3 groups of 3. Weigh two groups. Thus you find out which group is the heavier ball in. Choose 2 balls from this group and compare their weights. And that's it.

Having 9 balls, equally big, equally heavy. Only one of them is a bit heavier. How would you identify it if you could use a pair of scales only twice?

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Select 6 balls and Weigh 3 on one side and 3 on the other. Select the pile that is heaviest. of they are the same weight then the Heavy ball is in the pile you didn't weigh.

Now you have 3 balls one of witch is the heavy ball Select 2 balls adn compare there weight. if nether of them are the heavy ball the one not weighed was.

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It said that they were the same weight.

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It said that they were the same weight.

That's true, but the next sentence says, "Only one of them is a bit heavier.".

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I like this puzzle, similar to the 27 tennis balls...keep 'em coming

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I will admit this one seemed simpler than most I have heard here. However, I think I enjoyed it more. As said previously keep them coming, and throw in one or two like this for normal idiots like me. Thanks.

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You could also take 8 of the 9 balls, divide them into 2 groups of 4, wiegh each group of 4. If they weigh the same, then the heaviest is the 9th ball, if they do not weigh the same, the heavier ball is in the heaviest group of 4.

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You could also take 8 of the 9 balls, divide them into 2 groups of 4, wiegh each group of 4. If they weigh the same, then the heaviest is the 9th ball, if they do not weigh the same, the heavier ball is in the heaviest group of 4.

I started out with 4 myself, but what do you do for the second weighing? Once you find the heavier 4, if you weigh 2 and 2, you find the heavier 2, but you need the scale one more time.

The 3 groups of 3 is really the only way to go. :)

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or, you could start by building a small inclined plane (ramp). Allow each of the balls to roll down the ramp without applying any additional force. The added mass of the slightly heavier ball will give it greater inertia, and it should roll a longer distance than the rest of the identical balls. Use scale to confirm. B))

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To heck with the scales. If I had nine balls I am sure my doctor could tell me which one was bigger!

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"You have 9 balls, all of equal size. One ball is slightly heavier than the other 8. Using a set of scales only twice, can you determine which is the heavy ball?" -- better wording to eliminate confusion about the fact that you said all balls were equally heavy

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Roll the balls down an incline at the same time and the one wich reaches the bottom first is the slightly heavier one

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i thought all objects fall at the same speed? except for the ones that are too light

and i also thought the other variable was wind resistance

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Select 6 balls and Weigh 3 on one side and 3 on the other. Select the pile that is heaviest. of they are the same weight then the Heavy ball is in the pile you didn't weigh.

Now you have 3 balls one of witch is the heavy ball Select 2 balls adn compare there weight. if nether of them are the heavy ball the one not weighed was.

Yupp!! thas the right answer.. groups of 3 and proceed.. bout its interesting to note that the question mentioned one ball to be bit heavier. Had it been lighter or heavier.. then finding it out in 2 chances is not possible, unless u get lucky in the first try.

\m/

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Yupp!! thas the right answer.. groups of 3 and proceed.. bout its interesting to note that the question mentioned one ball to be bit heavier. Had it been lighter or heavier.. then finding it out in 2 chances is not possible, unless u get lucky in the first try.

\m/

Ex-squeeze me? Baking powder? What are you trying to say there?

If you don't get the baking powder reference click here.

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divide the balls into groups of 4

put the 2 groups on the scale, if they are of equal weight, then it has to be the last ball. If they are different weight, split the heavier group into 2 groups of 2, then with the heavier group; instead of using the scale; grab the heavier group and its only a matter of weighing them yourselves (use your hands and see which one is heavier) to find the ball; or guess... you have a 50 50 chance.

It said nothing against using yourself as a weight device.

The divide by 3 method is the best because it gives you the answer, but 2 groups of 4 was my first guess so i felt like making it my final and only guess.

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Weigh 3 on each side of the scale...leaving 3 out.

Whichever trio is heavier (the left out set if the two sides of the scale were equal), take 2 of those balls (one on each side of the scale). The heaviest will be either one of those balls, or the left out ball if the two on the scale are equal... :lol:

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