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 RainyBear
 
posted on October 23, 2001 01:03:02 PM new
I found this on the PayPal site... and don't know the anwser (if there is one). Anybody?

You are given twelve balls. All are the same size and weight, except one which is heavier or lighter than the others. Using a balance scale, can you determine the odd ball in three weighings?

 
 sadie999
 
posted on October 23, 2001 01:08:07 PM new
These puzzles always drive me nuts.

Besides, the minute I saw "twelve balls," I thought I'd clicked the erotica thread and got distracted.


 
 internationalgolf
 
posted on October 23, 2001 01:41:48 PM new
Take the 12 balls and divide them into 3 groups of four and number them from 1-12

Step 1. - Compare 1,2,3,4 to 5,6,7,8. If they are equal, then the varying ball is in the third group of 9,10,11,12.

Step 2. - Compare 1,2,3 with 9,10,11 if equal, the answer is 12 and compare it to 1 to find out if it is heavier or lighter.

Step 3. - If step 2 is not equal then compare 9 and 10, if equal answer is 11. If step 3 is not equal choose the one that matches the fulcrum tilt in step 2.

Step 4. - If step 1 is not equal compare balls 1,2,3,5 to 4,10,11,12. You know from step one weather the tilt was heavier or lighter.

Step 5. - If step 4 is equal then compare 6 to 7, if it is equal the answer is 8 and it is the light one. If it is not equal take the lighter one of 6 & 7.

Step 6. - If step 4 was not equal and tilted down, compare 1 to 2, if equal the answer is 3 and it is heavy. If not equal choose the heavier of 1 & 2

Step 7. - If step 4 tilted down on compare 4 to 9, if equal then 5 is the light ball, if not then 4 is the heavy ball.


This is one of several solutions. My advice – go list some auctions and don’t bother with this stuff.



 
 breinhold
 
posted on October 23, 2001 01:45:07 PM new
aaaaa...never mind.

 
 RainyBear
 
posted on October 23, 2001 01:56:23 PM new
Thanks, internationalgolf. I was soooo curious!

 
 sadie999
 
posted on October 23, 2001 02:17:33 PM new
People really don't get "word problems," do they? LMAO


 
 RainyBear
 
posted on October 23, 2001 02:30:03 PM new
I barely got it even after internationalgolf's explanation... lol.

 
 thedewey
 
posted on October 23, 2001 03:01:12 PM new
I've always liked this one:

There are 4 men standing on one side of a bridge. It's dark, and between them, they only have one flashlight. The bridge is in bad condition, and it's only strong enough to hold two men at a time. In other words, two men go across, and one of them must bring the flashlight back to the remaining men.

Each of the men walks at a different speed.

Man #1 can cross the bridge in 1 minute
Man #2 can cross the bridge in 2 minutes
Man #3 can cross the bridge in 5 minutes
Man #4 can cross the bridge in 10 minutes

If, for example, Man #2 (2 minutes) and Man #4 (10 minutes) go across the bridge, they can only walk as fast as the slowest man. This pair would take 10 minutes to cross, then if Man #2 came back alone with the flashlight, that's 2 more minutes.

There is a 17 minute time limit to get all the men across.

There is no trick -- no one carries anyone else, or anything like that. And yes, it can be done.

(spelling)
[ edited by thedewey on Oct 23, 2001 03:03 PM ]
 
 RainyBear
 
posted on October 23, 2001 04:21:22 PM new
Can two men meet in the middle of the bridge and hand off the flashlight?

 
 thedewey
 
posted on October 23, 2001 04:33:33 PM new
RainyBear -- Nope. Two men must cross completely, then one man comes all the way back.


 
 RainyBear
 
posted on October 23, 2001 04:40:28 PM new
Well, color me dumb! I can't figure it out.

 
 thedewey
 
posted on October 23, 2001 05:00:17 PM new
It's all a matter of getting the right combinations.

 
 RainyBear
 
posted on October 23, 2001 05:18:56 PM new
I cheated... went and looked it up elsewhere on the web. Shame on me.

 
 
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