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#1 User is offline   rookie1ja Icon

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Posted 30 March 2007 - 05:48 PM

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?



Spoiler for Solution:
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.



Spoiler for old wording:
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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#2 User is offline   Jumpinp Icon

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Posted 22 July 2007 - 07:51 PM

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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#3 User is offline   kylestaggs Icon

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Posted 17 September 2007 - 02:07 AM

It said that they were the same weight.
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#4 User is offline   Martini Icon

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Posted 20 September 2007 - 12:28 AM

Quote

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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#5 User is offline   slmo Icon

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Posted 10 October 2007 - 08:56 PM

I like this puzzle, similar to the 27 tennis balls...keep 'em coming
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#6 User is offline   hipowertech Icon

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Posted 01 November 2007 - 09:38 PM

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.
Many people try to be funny,
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Lighten up!
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#7 User is offline   St0n3d Icon

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Posted 03 January 2008 - 06:13 PM

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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#8 User is offline   LogicAL Icon

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Posted 18 January 2008 - 12:44 AM

View PostSt0n3d, on Jan 3 2008, 12:13 PM, said:

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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#9 User is offline   hampydance Icon

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Posted 01 March 2008 - 02:27 AM

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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#10 User is offline   bradthepilot Icon

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Posted 10 March 2008 - 07:48 AM

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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