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There are 9 coins. Out of which one is odd one i.e weight
is less or more. How many iterations of weighing are
required to find odd coin?
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Answer Posted / shashank parulekar
There are 9 coins. Out of which one is odd one i.e. weight
is less or more. How many iterations of weighing are
required to find odd coin?
Divide the coins into 3 groups of 3 each
1. Weigh the first two (Iteration 1)
a. If they are of equal weight, the third group is defective.
i. Weigh two of the third group. (Iteration 2)
1. If they are of equal weight, the third is defective. Weigh it with a normal coin from the first group (Iteration 3). It weighs less or more. That is the final answer.
2. If they are of unequal weight, weigh the heavier one with a normal one in the first group (Iteration 3). If it is heavier, it is the defective one, weighing more than the rest. If it weighs the same as the normal one, the second coin is the defective one, weighing less than the others.
b. If they are of unequal weight, one of them is defective and the third group is normal. Weigh the first of the groups with the third group (Iteration 2). If they weigh equal, the second group is defective. If they do not weigh equal, the first group is defective. The defective group could weigh less or more than normal, depending on whether the defective coin weighs less or more.
i. Weigh coin one and two from the defective group (Iteration 3). If they weigh equal, the third coin is defective.
ii. If they do not weigh equal,
1. the coin that weighs more is defective if the group was defective-heavy.
2. the coin that weighs less is defective if the group was defective-lighter.
| Is This Answer Correct ? | 6 Yes | 4 No |
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