we have twelve identical balls and among them one is defective as it weigh more than the rest.suppose you are provided with a physical balance,what will be the minimum no of times you should use the physical balance to figure out the defective ball
Answers
Answer:
I will weigh it four times 4 times
Step-by-step explanation:
The task can be completed in 5 weighings (or 3 if no off ball exists).
Note that the questions say a mechanical balance scale and not a counter balance scale, so we cannot weight one group against another in one weighing.
The problem would be easier if we knew whether the off ball (if there is one) is heavier or lighter. Also, if a counter balance were used, we could narrow it down to 4 balls in a single weighing. But this is not the scenario.
Spitting the balls into two groups of 6 and weighing both will not work. If they weigh the same then there is no off ball, but it not, we don't know it one ball is heavier in one group or one is lighter in the other group.
So, we arbitrarily split the balls into 3 groups of 4, weighting each group separately. For each group we divide the weight by 4 to get the average weight. If all averages are the same, there is no off ball. Otherwise, we know the group of 4 which contains the off ball, the weight of a good ball and the weight of the off ball.
We have now made 3 weighings. If no off ball exists, we are done; otherwise we continue with the group of 4 balls known to contain the off ball.
We now arbitrarily weight 2 of the remaining 4 balls. Since we know the weight of the good and off balls, we know whether the off ball is in the group of 2 we just weighted or in the remaining two.
A fifth and final weighing is required to arbitrarily weight 1 of the 2 balls known to be off.
Perhaps someone can come up with a more clever way to find the off ball in less weighings