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A Weighty ball problem?
Okidoki 8 ball bearings, one of which is slightly different from the rest.
You cannot visually see the difference its only in its weight. You do not know if its heavier or lighter. You have a pair of balance scales, you are only allowed to make 3 balance tests. How do you tell which is the odd bearing? Something else thats being discussed in the office. :) Quick replies wanted so I can be the one to answer it :D :D |
Re: A Weighty ball problem?
2^3 = 8 so, yes, it can be done in three tests.
1) ABC vs DEF 2) ABF vs DEC 3) AEF vs DBC then just follow the letters... Edit Just read the bit about not knowing if it's heavier of lighter :dunce: mmmm... thinking cap on, instead. |
Re: A Weighty ball problem?
Put 2 bearings on each side of the scale. If they are not equal in weight, the odd one is in this lot. Now check two of these, if these are equal in weight, check one of these against one of the other two. If these are equal in weight, the remaining one is the odd one out. If not, it's the 'new one'. If the first 4 are equal in weight, do the last steps with the other 4.
Edit: Or, in letters. Check AB vs CD. If unequal, check A v B, and the B v C. If AB = CD repeat last steps with E, F ,G and H. |
Re: A Weighty ball problem?
eh?
what about g & h? |
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If the first 4 are equal, you do step 2 and 3 with EFGH |
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Big hint here : Start with weighing 2 vs 2 Full answer here http://www.mathnerds.com/mathnerds/b...t/solution.asp |
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Re: A Weighty ball problem?
Too much haste - ignore me I'm talking balls!
I think Danielf has the answer: 1) AB vs CD, if they balance goto 5 2) A vs B, if they balance goto 4 3) A vs C, if they are unbalanced then answer is A, balanced it's B 4) A vs C, if they are unbalanced then answer is C, balanced it's D 5) AB vs EF, if they balance goto 7 6) A vs E, if they are unbalanced then answer is E, balanced it's F 7) A vs G, if they are unbalanced then answer is G, balanced it's H. |
Re: A Weighty ball problem?
1: Weigh 2 against 2 - if this group are equal, they become your reference, the other group the sample, otherwise reverse it.
2: Weigh 2 of the sample against 2 of the reference - if equal, they move to the reference, otherwise the other two move to the reference. 3. Now out of the remaing two left in the sample, weigh ONE against one of the reference - if unequal, you've found it, otherwise it's the other one |
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Re: A Weighty ball problem?
The links given are for a 12 ball similar problem.
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I'll have to see what the lads here make of it. cheers. |
Re: A Weighty ball problem?
Now I've had time to take lunch, and write it up here's a solution
I think It's more or less what Danielf had, but with more detail. STEP 1 Weigh 2 v 2 Possible results A) pairs are unequal (heavier or lighter not known) B) pairs are equal Thus we have 4 balls known to be "good" and 4 balls still "suspect". STEP 2) Weigh 2 "suspect" balls v 2 "known good" balls Possible results C) pairs are unequal (heavier or lighter is now known) D) pairs are equal (discard these) STEP 3C Weigh one of the "suspect" balls tested in STEP 2 against a "known good" ball Possible Results E) Balance equal (the other suspect is guilty) F) Balance unequal (the chosen suspect is guilty) STEP 3D Weigh one of the "suspect" balls not tested in STEP 2 against a "known good" ball G) Balance equal (the other suspect is guilty) H) Balance unequal (the chosen suspect is guilty) NOTE In G) it is impossible to tell if the guilty ball is heaver or lighter without a further weighing |
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