Comments on: 12 Steel Balls Logic Test http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/ Rida's Zone in Cyberspace Sat, 04 Feb 2012 07:57:58 +0000 http://wordpress.org/?v=2.7.1 hourly 1 By: Rida Al Barazi http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-168844 Rida Al Barazi Mon, 14 Nov 2011 14:36:26 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-168844 @sabeeha, what if the first two groups weren't equal in weight? @sabeeha, what if the first two groups weren’t equal in weight?

]]> By: sabeeha http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-168789 sabeeha Mon, 14 Nov 2011 06:44:51 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-168789 i think i know the answer. here we go! there are 12 balls.divide them into 3 groups.each group will have 4 balls.first wieght 2 groups .if they are the same weight that means the last group has the ball.divide the last group into 4 groups.each group will have 1 ball.weigh 2 balls and if they are the same weight then one of thoeseis heavier or lighter.take one of the ball which you have already weight.weigh the ball then you can find the answer. i hope this is the answer. i think i know the answer.
here we go!

there are 12 balls.divide them into 3 groups.each group will have 4 balls.first wieght 2 groups .if they are the same weight that means the last group has the ball.divide the last group into 4 groups.each group will have 1 ball.weigh 2 balls and if they are the same weight then one of thoeseis heavier or lighter.take one of the ball which you have already weight.weigh the ball then you can find the answer.

i hope this is the answer.

]]> By: Paducahrider http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-130845 Paducahrider Sun, 12 Dec 2010 04:30:38 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-130845 Howdy! More than ten years ago, a classmate told me this teaser one afternoon, as we left a summer class. He was a "puzzle freak" and said he'd been working on it "over a month", and wanted to see how long I'd take. I didn't work on it until that evening. After fiddling around I came to the conclusion that I needed to establish a group of balls that were known to be identical, and use them as "comparison" balls, when needed. Since I had homework, I had to set it aside then get some sleep. The next morning, on forty-plus mile drive, to class, I began the process of solving the problem. While I was involved in doing so, as I drove, I suddenly realized that I had driven 15 miles beyond my turn-off point, and had to turn around and retrace my path, Funny thing was; I couldn't remember anything that had taken place for the previous 20 miles. NOTHING!!! I was late for class, But,,, I solved that D****D teaser. The best advice I will give is that: 1. NEVER ASSUME!! Solve it correctly and everything is irrefutable 2.This is a LOGIC problem more than a math problem. 3. The most common error in solving the problem is ASSUMING that the odd ball is HEAVIER than the others. 4. If this were the case, the problem would be as easy as some seem to feel it is (then proceed to get the wrong answer). I was an HVAC instructor at a Technical College, and would give this problem to my incoming students for extra credit, if they chose to pursue it, without affecting their grade if they failed. No-one ever solved it, during the last decade I taught. Corky. Howdy!
More than ten years ago, a classmate told me this teaser one afternoon, as we left a summer class.
He was a “puzzle freak” and said he’d been working on it “over a month”, and wanted to see how long I’d take.
I didn’t work on it until that evening.
After fiddling around I came to the conclusion that I needed to establish a group of balls that were known to be identical, and use them as “comparison” balls, when needed. Since I had homework, I had to set it aside then get some sleep.
The next morning, on forty-plus mile drive, to class, I began the process of solving the problem.
While I was involved in doing so, as I drove, I suddenly realized that I had driven 15 miles beyond my turn-off point, and had to turn around and retrace my path, Funny thing was; I couldn’t remember anything that had taken place for the previous 20 miles. NOTHING!!!
I was late for class, But,,, I solved that D****D teaser.
The best advice I will give is that:
1. NEVER ASSUME!! Solve it correctly and everything is irrefutable
2.This is a LOGIC problem more than a math problem.
3. The most common error in solving the problem is ASSUMING that the odd ball is HEAVIER than the others.
4. If this were the case, the problem would be as easy as some seem to feel it is (then proceed to get the wrong answer).

I was an HVAC instructor at a Technical College, and would give this problem to my incoming students for extra credit, if they chose to pursue it, without affecting their grade if they failed.
No-one ever solved it, during the last decade I taught.
Corky.

]]> By: Fabian http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-127413 Fabian Tue, 24 Aug 2010 08:06:26 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-127413 Having tried so hard, plaes let me know... Having tried so hard, plaes let me know…

]]> By: James Smythe http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-105275 James Smythe Thu, 27 Nov 2008 15:39:51 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-105275 I really would like the solution to the 12 steel balls puzzle problem as it is beyond me. If you don't want to post it please email it. Thank you very much. I really would like the solution to the 12 steel balls puzzle problem as it is beyond me. If you don’t want to post it please email it. Thank you very much.

]]> By: OC http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-63075 OC Sat, 23 Feb 2008 13:46:06 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-63075 here is how you solve it: first, you divide the balls into 3 sets with 4 balls each. You can number them from 1 to 12 to make things easier to follow. During the first use of the scale, you put the set consisting of balls 1,2,3, and 4 on one side and the set of balls 5,6,7, and 8 on the other side. There are 2 possibilities: the first possibility is that the 2 sets are equal in weight, so you know that the odd ball is in the third set of balls 9,10,11, and 12. For the second use of the scale, you put balls 6,7, and 8 on one side, and balls 9,10, and 11 on the other side. If they are equal, then the odd ball is ball # 12, but if they are not equal, then you also have 2 possibilities, either balls 6,7,and 8 are heavier than balls 9,10, and 11, so you know that the odd ball is either ball 9, 10, or 11, and you also know that the odd ball is lighter than the other balls. So for the third use of the scale, you put ball 9 on one side, and ball 10 on the other side, and if they are equal in weight, then ball 11 is the odd ball, but if they are not equal in weight and one is heavier than the other, then the lighter one is the odd ball whether it is ball 9 or ball 10 (whichever one is the lighter of the two). The other possibility is when you use the scale the second time and weigh balls 6,7,8 and balls 9,10,11, then if the set of balls 9, 10, 11 are heavier, then the odd ball is heavier than the rest, and so for the third time you put ball 9 on one side and ball 10 on the other, and if they are equal in weight then ball 11 is the odd ball, but if they are not equal, then whichever ball of the 2 that is heavier, then that is the odd ball whether it is ball 9 or 10. Now if you go back to the first time we used the scale to see the other possibility which is that the 2 sets of balls are not equal. So set one of balls 1,2,3, and 4 is not equal in weight to the second set of balls 5,6,7,and 8. One set of balls is heavier than the other, and it doesn't really matter at this point which is heavier, so lets assume that set 1 consisting of balls 1,2,3, and 4 is heavier than the other set of balls 5,6,7, and 8. We also know that since sets 1 and 2 are not equal in weight, then one of them contains the odd ball, and therefore balls 9,10,11, and 12 are all of equal weight. Now for the second use of the scale, we put balls 3,5, and 10 on one side, and we put balls 2,4,and 6 on the other side of the scale. We leave balls 1,7 and 8 as a set on the side for now. If the set of balls 3,5, and 10 is equal in weight to the set of balls 2,4, and 6, then that means the odd ball is one of the balls 1,7, and 8. So for the third use of the scale, we put ball 7 on one side and ball 8 on the other, and if they are equal in weight, then we know that ball 1 is the odd ball and that it is heavier than the rest, but if balls 7 and 8 are not equal in weight, then the lighter one of the two is the odd ball which is lighter than the rest of the balls. Now if we go back to the second time we used the scale, if the 2 sets were not equal in weight, then that means the odd ball is one of these balls: 2,3,4,5, and 6. Ball 10 was not included as a possible odd ball since it was already eliminated in the first measurement where we determined that balls 9,10,11, and 12 are normal and equal in weight. Now, for the second measurement, if the set of balls 3,5, and 10 is heavier than the other set of balls 2,4, and 6, then by looking at the first 2 measurements together, we can eliminate balls 2,4, and 5 as possible odd balls. The we are left with balls 3 and 6, one of them being the odd ball. All we have to do now is weigh either one of them, lets say ball 3, weigh it with ball 10(which we know is a normal ball). If balls 3 and 10 are equal in weight, then we know that ball 6 is the odd ball and it is lighter in weight than the rest of the balls. But if balls 3 and 10 were not equal in weight, then we know that ball 3 is the odd ball and it would be heavier than the normal balls. Now to consider the last possibility, lets go back to the second time we used the scale and put balls 3,5, and 10 on one side and balls 2,4, and 6 on the other side. If the set of balls 2,4, and 6 was heavier than the other set, then by looking at both measurements 1 and 2, we can eliminate balls 3 and 6 as possibilities for being the odd ball. so we are left with balls 2,4, and 5 as possibilities for being the odd ball. So for the third use of the scale, we put ball 2 on one side and ball 4 on the other, and if they are equal in weight, then ball 5 is the odd ball and it is lighter than the rest of the balls. But if balls 2 and 4 are not equal in weight and one is heavier than the other, then whichever is the heavier one of the two balls 2 and 4, then that would be the odd ball, and it is heavier than the rest of the normal balls. And that is how we can find which one is the odd ball by using the scale only 3 times. here is how you solve it:
first, you divide the balls into 3 sets with 4 balls each. You can number them from 1 to 12 to make things easier to follow. During the first use of the scale, you put the set consisting of balls 1,2,3, and 4 on one side and the set of balls 5,6,7, and 8 on the other side. There are 2 possibilities: the first possibility is that the 2 sets are equal in weight, so you know that the odd ball is in the third set of balls 9,10,11, and 12. For the second use of the scale, you put balls 6,7, and 8 on one side, and balls 9,10, and 11 on the other side. If they are equal, then the odd ball is ball # 12, but if they are not equal, then you also have 2 possibilities, either balls 6,7,and 8 are heavier than balls 9,10, and 11, so you know that the odd ball is either ball 9, 10, or 11, and you also know that the odd ball is lighter than the other balls. So for the third use of the scale, you put ball 9 on one side, and ball 10 on the other side, and if they are equal in weight, then ball 11 is the odd ball, but if they are not equal in weight and one is heavier than the other, then the lighter one is the odd ball whether it is ball 9 or ball 10 (whichever one is the lighter of the two). The other possibility is when you use the scale the second time and weigh balls 6,7,8 and balls 9,10,11, then if the set of balls 9, 10, 11 are heavier, then the odd ball is heavier than the rest, and so for the third time you put ball 9 on one side and ball 10 on the other, and if they are equal in weight then ball 11 is the odd ball, but if they are not equal, then whichever ball of the 2 that is heavier, then that is the odd ball whether it is ball 9 or 10. Now if you go back to the first time we used the scale to see the other possibility which is that the 2 sets of balls are not equal. So set one of balls 1,2,3, and 4 is not equal in weight to the second set of balls 5,6,7,and 8. One set of balls is heavier than the other, and it doesn’t really matter at this point which is heavier, so lets assume that set 1 consisting of balls 1,2,3, and 4 is heavier than the other set of balls 5,6,7, and 8. We also know that since sets 1 and 2 are not equal in weight, then one of them contains the odd ball, and therefore balls 9,10,11, and 12 are all of equal weight. Now for the second use of the scale, we put balls 3,5, and 10 on one side, and we put balls 2,4,and 6 on the other side of the scale. We leave balls 1,7 and 8 as a set on the side for now. If the set of balls 3,5, and 10 is equal in weight to the set of balls 2,4, and 6, then that means the odd ball is one of the balls 1,7, and 8. So for the third use of the scale, we put ball 7 on one side and ball 8 on the other, and if they are equal in weight, then we know that ball 1 is the odd ball and that it is heavier than the rest, but if balls 7 and 8 are not equal in weight, then the lighter one of the two is the odd ball which is lighter than the rest of the balls. Now if we go back to the second time we used the scale, if the 2 sets were not equal in weight, then that means the odd ball is one of these balls: 2,3,4,5, and 6. Ball 10 was not included as a possible odd ball since it was already eliminated in the first measurement where we determined that balls 9,10,11, and 12 are normal and equal in weight. Now, for the second measurement, if the set of balls 3,5, and 10 is heavier than the other set of balls 2,4, and 6, then by looking at the first 2 measurements together, we can eliminate balls 2,4, and 5 as possible odd balls. The we are left with balls 3 and 6, one of them being the odd ball. All we have to do now is weigh either one of them, lets say ball 3, weigh it with ball 10(which we know is a normal ball). If balls 3 and 10 are equal in weight, then we know that ball 6 is the odd ball and it is lighter in weight than the rest of the balls. But if balls 3 and 10 were not equal in weight, then we know that ball 3 is the odd ball and it would be heavier than the normal balls. Now to consider the last possibility, lets go back to the second time we used the scale and put balls 3,5, and 10 on one side and balls 2,4, and 6 on the other side. If the set of balls 2,4, and 6 was heavier than the other set, then by looking at both measurements 1 and 2, we can eliminate balls 3 and 6 as possibilities for being the odd ball. so we are left with balls 2,4, and 5 as possibilities for being the odd ball. So for the third use of the scale, we put ball 2 on one side and ball 4 on the other, and if they are equal in weight, then ball 5 is the odd ball and it is lighter than the rest of the balls. But if balls 2 and 4 are not equal in weight and one is heavier than the other, then whichever is the heavier one of the two balls 2 and 4, then that would be the odd ball, and it is heavier than the rest of the normal balls. And that is how we can find which one is the odd ball by using the scale only 3 times.

]]> By: spinner http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-57470 spinner Sun, 20 Jan 2008 05:39:49 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-57470 i have the answere to this problem. it is not as ez as it sounds. one clue i will give you is: when using the ballance you must obtain all the information it provides. i have the answere to this problem. it is not as ez as it sounds. one clue i will give you is: when using the ballance you must obtain all the information it provides.

]]> By: Spinner http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-57462 Spinner Sun, 20 Jan 2008 04:27:57 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-57462 Please clarify- how is Grayfox correct? I thought the puzzle states that you do not know if the ball is heavier or lighter.. His equation states you would know which group was heavier. How would you accompish this? You still wouldnt know if the ball was heavier or lighter. I dont think he is correct. Please clarify- how is Grayfox correct? I thought the puzzle states that you do not know if the ball is heavier or lighter.. His equation states you would know which group was heavier. How would you accompish this? You still wouldnt know if the ball was heavier or lighter. I dont think he is correct.

]]> By: Jen http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-33302 Jen Fri, 03 Aug 2007 23:52:59 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-33302 Let's call the 4 balls that weren't used in the first weighing group C. If A & B are unequal, we know that all balls in C are equal, so they are "duds". Simply use one of these in the next weighing to be the 3rd ball. Let’s call the 4 balls that weren’t used in the first weighing group C. If A & B are unequal, we know that all balls in C are equal, so they are “duds”. Simply use one of these in the next weighing to be the 3rd ball.

]]> By: MustAsk http://rida.me/blog/2007/03/18/12-steel-balls-logic-test/comment-page-1/#comment-33056 MustAsk Thu, 02 Aug 2007 17:53:37 +0000 http://www.ridaalbarazi.com/blog/2007/03/18/12-steel-balls-logic-test/#comment-33056 You say, if A and B are unequal, to remove 2 from A and 1 from B and swap the remaining balls from A with one of the remaiing balls from B and to re-weigh. That wont work because I would be weighing 2 balls against 3. You say, if A and B are unequal, to remove 2 from A and 1 from B and swap the remaining balls from A with one of the remaiing balls from B and to re-weigh. That wont work because I would be weighing 2 balls against 3.

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