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Logic Problem for Corona Virus Shutins!

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H

Harold

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You have 12 steel balls---all identical--but one is either lighter or heavier than the other 11. You don't know which-but your job is to find out through Logic.
You have a balance scale--small device with a bowl on one side ()left)--bowl on the other side (right), and a pointer in the middle--Put a small
something or other in one bowl --it goes down and the pointer deflects that way--same for the other side bowl. I have to explain it this way for the
mechanically deficient among us! So here's the problem-- You put something on one of the bowls, and the
pointer deflects--that's 1 weighing--Go to it folks--With 3 weighings, you have to figure out which of the 12 steels balls is either lighter or heavier
than the other 11---This isn't trickery--just pure logic!
 
Step one: Six in each pan. You see which six is lighter and divide in two.
Step two: put three of the six in each pan. One pan will be lighter.
Step 3. Put one ball from the lighter 3 in each pan. If one is lighter it will show. If they are equal, the third is lighter.
 
The premiss of this "Locgic Problem:", is that the ball could be either lighter OR heavier than the other 11. After 3 weighings, you
should be able to pick up a particular ball, and say this ball is "heavier" than the other 11, or you can say this ball is "lighter"
than the other 11--
 
What if the odd ball out is heavier? This solution only works if the odd ball is lighter than the others. If the odd ball is heavier you lost it after step one.
 
waltesefalcon said:
What if the odd ball out is heavier? This solution only works if the odd ball is lighter than the others. If the odd ball is heavier you lost it after step one.
Click to expand...
Actually, it works the same way. Just put in heavy or heavier instead of light or lighter.
 
Right but that'll take more than three steps.
 
So, if a weighing technically only occurs when the pointer deflects you could just start off by putting a ball on each side then adding a ball to each side one by one until the pointer deflects one way or the other. One of the last balls added would have to be of a different weight. Then all you'd have to do is determine which is the odd ball by weighing each against one of the other balls.
 
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