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Weighing the balls 
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#1
Feb1606, 10:26 AM

P: 398

Żou have 12 balls.
All of them have the same mass. Except one. You don't know if it weighs more or less than the other 11 balls. Your mission is to find the odd one out, and say if it weighs more or less than the 11 others. You have a balancescale. You may use it for three weighings. The technique must work no mmater where the odd ball is. EDIT: Added the balance part 


#2
Feb1606, 12:10 PM

P: 15,319

I think there's one item missing from the givens. Do you know the weight of the balls, or is that unknown?



#3
Feb1606, 12:23 PM

P: 398

There is no need to know the masses of the balls in order to complete the brain teaser.
The correct answer will fill many A4:s of possible weighings. ex. If i weigh those balls against those, which one will weigh the most? How and in which weighingss do I get controlballs etc. 


#4
Feb1606, 12:41 PM

P: 657

Weighing the balls
If it's really a *scale*... then... hm. Not sure how many weighs you'd need, but I'd bet a bit more than 3. I get between 3 and 6 weighs using a scale, 3 if you make lucky picks, 6 if you're unlucky. Better than 6, anyone? DaveE 


#5
Feb1606, 01:11 PM

P: 398

Oh, you meant it that way! Yes it is a balance scale. You get the masses of the balls relative to each other and nothing else. Sorry for the confusion.



#6
Feb1606, 04:26 PM

P: 657

DaveE 


#7
Feb1606, 04:29 PM

P: 398

It is a balance.



#8
Feb1606, 04:39 PM

P: 367




#9
Feb1706, 05:05 AM

P: 398

The question is with a balance. Solve it. If you want a scale post another brain teaser



#10
Feb1706, 11:25 AM

P: 657

The "findthe1of12objectsthatweighsadifferentamount" problem has been done a bunch of times on this forum, so it's entirely possible that you won't see much of a response. The last time someone posted it, you can see the response they got: http://www.physicsforums.com/showthread.php?t=105118 So, don't feel too bad if there isn't a reply. I don't remember seeing a *scale* question before, though, so I know at least I was interested (and apparently Moo, too) in seeing a new problem. But, anyway, I'm done. Unless someone wants to discuss the scale problem further... DaveE 


#11
Feb1706, 11:31 AM

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#12
Feb1706, 11:57 AM

P: 398

Ok, i didn't realize that it had been posted to death.
Ok it is a scale instead of a balance 


#13
Feb1706, 01:22 PM

P: 160




#14
Feb1706, 01:26 PM

P: 398

ok, post the answer in the colour ;P



#15
Feb1706, 02:42 PM

P: 160

Eleven balls have mass M, the target ball has a greater or lesser mass. First, balance four balls against four balls. Case 1: the balance is level so the target is in the remaining goup of 4. Label these four balls 1, 2, 3 and 4. Set aside ball 4 and balance group {1,2,3} against known balls {M,M,M} from the remaining eight. Case 1.1: the balance is level so ball 4 is the target. A third balance determines if it is heavy or light. Case 1.2: the balance tips towards {1,2,3} so one of these balls is the target, which is now known to be heavy. Balance ball 1 against 2 to find out which of these three is the heavy one. Case 1.3: the balance tips towards {M,M,M} so one of {1,2,3} is the target, which is now known to be light. Balance ball 1 against 2 to find out which of these three is the light one. Case 2: the first balance was not level so the target is in this group of 8 balls. Label balls on the heavier side {1,2,3,4} and balls on the lighter side {5,6,7,8}. Now balance balls {1,2,M} against balls {3,4,5}. Case 2.1: the balance is level so the target is in remaining set {6,7,8} and is known to be light (from the first balance). Balance 6 against 7 to know which of these three is the light one. Case 2.2: the balance tips towards {1,2,M} so either a ball in {1,2} is heavy or ball 5 is light. Balance ball 1 againt ball 2 to know if one is heavy and which one, or to know that ball 5 is light if the balance is level. Case 2.3: the balance tips towards {3,4,5} so either ball 3 or 4 is heavy. Balance 3 against 4 to know which one. I hope I didn't screw up anywhere. Someome will let me know for sure. 


#16
Feb1906, 02:23 PM

P: 126

i propose a solution shud be right ...enjoy
i will use W for the word weigh..u remove 2 coins .so u W the 2 piles of 5 coins .if they r the same then the 2 coins contain 1 odd coin so u take them and wiegh them against each other now u no 1 is light or normal or the other is heavy or normal, so now u choose any of them and weigh them against a normal coin. if the piles of 5 r difirent u name 1 pile heavy pile and 1 pile light similar to wat we did with the 2 coins. we remove a light coin put it aside and we form groups HHL HHL HLL we weigh HHL vs HHL if they weigh diffirent then odd coin is either H ,H or L , ( call normal coin N) so u do HL vs NN leaving out L .. if HHL vs HHL shows they weigh same then odd coin in HLL with 1 wiehing left u do HL vs NN leaving out L ...QED tell me if u find me incorect.(btw i didnt write all details in last part u can easily fill in the gaps) 


#17
Jul106, 03:08 AM

P: 8




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