There are 12 balls. One of them is slightly heavier or lighter than the others. You have a classical balance with two pans (which only indicates which pan is heavier/lighter). You are allowed to weigh only three times. Can you find out the ball which is different, and also whether it's heavier or lighter?
Hint: Taking six balls on each pan (for the first weighing) won't work.
2007-03-18 22:24:14 · 5 answers · asked by Ankit 2 in Science & Mathematics ➔ Mathematics
source: http://ankit-mathur.blogspot.com
2007-03-18 22:31:46 · update #1
Note that the ball may either be heavier or lighter. You also have to spot whether the defective ball is heavier or lighter.
2007-03-18 22:39:00 · update #2
Let's label the balls as follows:
o = unknown
H = potentially heavy
L = potentially light
N = definitely normal
V = "the scale"
1) ooooVoooo
If this balances, then we're left with 4 o and 8 N, so
2a) oooVNNN
If this balances, then there is only 1 o left, so
3a) oVN determines heavier/lighter
If left side of 2a is heavier, then
3b) HVH will determine which it is (if it balances, then it's the remaining H)
If left side of 2a is lighter, then
3c) LVL (same as HVH, only now light side will be the bad boy)
Now, if 1st weighing results in imbalance, we're left with:
4N, 4H and 4L, so...
2b) HHHL V NNNH
If it balances, we're left with LLL, so same as 3c
If left side heavier, we're left with HHH, so same as 3b
If right side heavier, we're left with LH, so...
3d) LH V NN, determines which it is
2007-03-18 22:46:25 · answer #1 · answered by blighmaster 3 · 0⤊ 0⤋
Put four balls in each pan. If one pan is heavier than the other, then that pan contains the heavier ball. If they are equal, then the remaining four balls contain the heavier ball. Once you've found the group of four balls that has the heavier ball, put one of those in each pan. If one is heavier, then you have found the heavier ball. If they are the same, then put the other two balls in the pan, and one should be heavier.
2007-03-19 05:33:51 · answer #2 · answered by Anonymous · 1⤊ 0⤋
Take 4balls and 4 balls and weigh them. If both sides are equal, you know the different one isn't there.
Weigh the other two, 1 with 1. If the first scale is balanced, weigh the other two. The different ball will be spotted.
2007-03-19 05:28:23 · answer #3 · answered by peteryoung144 6 · 1⤊ 1⤋
...I would fill each pan to the top with water... then, simultaneously, add 4 balls into each pan... the balls would displace the water (or one might float)... that would be try number one... now I've only got 4 balls left to try... try number 2... taking 2 ball simultaneously, placing them into the water. If the pans have not tip'd yet... the final ball is the culprit.
2007-03-19 05:35:18 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Browse through the questions I've answered recently to see the answer.
It's long. I don't want to type it out again.
2007-03-19 05:43:34 · answer #5 · answered by Curt Monash 7 · 0⤊ 0⤋