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cheechee

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#1
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

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#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!
 

Adam Goi

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#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.
 

meng

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#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)
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

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

Pokka

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#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.
heh, nice try. i understand. but what if the ball tat is different is heavier and not lighter than the other 11?
 

meng

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

Jester

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#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 :D
 

meng

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#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 :D
 

meng

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#12
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! :D
:eek:
 

chaotic

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#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 ;)
 

cheechee

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