Interesting question

[cheechee]

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?

 

[Fella]

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!

 

[Adam Goi]

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.

 

[meng]

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)

Originally posted by AdamGoi
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.

 

[Fella]

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

 

[Pokka]

Originally posted by AdamGoi
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?

 

[meng]

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

 

[Adam Goi]

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

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

 

[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

 

[meng]

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

 

[Adam Goi]

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

 

[meng]

good choice!
maybe they can give a better answer??

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

 

[StreetShooter]

*sigh*

This is how I solve all these kinds of problems:

http://www.google.com.sg/search?q=12+balls+balance+11+1&ie=UTF-8&oe=UTF-8&hl=en

 

[chaotic]

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

 

[cheechee]

Originally posted by StreetShooter
*sigh*

This is how I solve all these kinds of problems:

http://www.google.com.sg/search?q=12+balls+balance+11+1&ie=UTF-8&oe=UTF-8&hl=en

This is good... :devil:

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