1. ## Interesting question

Not sure if u heard or even solved this before.

If you are given 12 balls, of which 11 of them are of the same weight whilst 1 is of different weight from the 11.

You have a balancing beam and 3 trials using the balancing beam, remember, only 3 trials. You can try any combination.

How to pick out the 1 ball of different weight?

2. If given the "different weight" ball is lighter or heavier than the rest, it'll be real easy... but in this situation, I'm stumped!

3. Let me give it a go.

1st attempt:

Divide them into 6s and use the balance. The 6 that weigh lesser keep, the heavier ones, discard.

2nd attempt:

Divide them into 3s and use the balance. The 'lighter ones', i.e. the 3, keep and discard the 'heavier ones'.

3rd attempt:
Pick any two of the 3 remaining balls and use the balance. If they weigh the same, then the 3rd ball is the one. If not, the lighter one will show up.

4. Same as mine..but i was pondering how to carry out Step 1 and 2. It is not stated whether the one having a different weight is lighter or heavier than the rest. So there's a possibility that in the 1st attempt, the special one will be in the heavier 6 (that is if you keep the lighter 6)
Let me give it a go.

1st attempt:

Divide them into 6s and use the balance. The 6 that weigh lesser keep, the heavier ones, discard.

2nd attempt:

Divide them into 3s and use the balance. The 'lighter ones', i.e. the 3, keep and discard the 'heavier ones'.

3rd attempt:
Pick any two of the 3 remaining balls and use the balance. If they weigh the same, then the 3rd ball is the one. If not, the lighter one will show up.

5. Problem is, how do you know the "different weight" ball is lighter or heavier than the rest? (step 1)

Let me give it a go.

1st attempt:

Divide them into 6s and use the balance. The 6 that weigh lesser keep, the heavier ones, discard.

2nd attempt:

Divide them into 3s and use the balance. The 'lighter ones', i.e. the 3, keep and discard the 'heavier ones'.

3rd attempt:
Pick any two of the 3 remaining balls and use the balance. If they weigh the same, then the 3rd ball is the one. If not, the lighter one will show up.
heh, nice try. i understand. but what if the ball tat is different is heavier and not lighter than the other 11?

7. my answer is, no, I'll not use the balancing beam.
I'll use electronic one.

8. Yah hor ... I've talked too soon; never really paid attention to Fella's post!

Let me think again ... at the meantime, next better player?

9. me try me try!!!

first try take any 8 and balance 4 each. see which side is more heavy, if both is balance, then "the one ball" is in the other 4.

second split the 4 remaining ball and split into 2 each, then select the more heavy side.

for the last try, you are left with 2 only, so the answer is out

10. same logic...in your 1st and 2nd try, the "special one" is lighter or heavier?
Originally posted by Jester
me try me try!!!

first try take any 8 and balance 4 each. see which side is more heavy, if both is balance, then "the one ball" is in the other 4.

second split the 4 remaining ball and split into 2 each, then select the more heavy side.

for the last try, you are left with 2 only, so the answer is out

11. Hm ... the more I thought of it, the more intriguing it becomes ... I'll try it on my EM1 pupils tomorrow!

12. good choice!
maybe they can give a better answer??
Hm ... the more I thought of it, the more intriguing it becomes ... I'll try it on my EM1 pupils tomorrow!

13. *sigh*

This is how I solve all these kinds of problems:

14. it's amazing how google solves all our problems.. now if only it can solve the world hunger problem, SARs and other things too

15. Originally posted by StreetShooter
*sigh*

This is how I solve all these kinds of problems:

This is good...

Thanks for providing the answer.... Finally can feel relieved knowing this after trying to solve for two days without any result.

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