CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….
divide the balls into three groups. Let group A contains 3
balls group B contains 3 balls and group C contains 2 balls.
FIRST TIME: BALANCE the groups A and B.
If they are equal then the defective ball is in group C.
case 2: if they are not equal then take the group of 3
balls which has less weight.
IF the defective ball is in group C, we can easily find the
ball which is less.
if the defective ball is in group of 3 which has less
weight, take 2 balls out of this group and find the weight.
if they are equal, then the left out is defective. if not
we can find easily the ball which weigh less.
so we have the following steps:
A (3) B(3) c(2)
first time / |
unequal equal then c has defective.use
| second time to find out
second time take lessweight group
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