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 / abhishek gupta
[REQUEST- PLEASE READ IT ATLEAST TWICE]
4 ITERATIONS OF WEIGHING....
DIVIDE THE 9 COINS INTO 3 BATCHES (4, 4 AND 1).
1) FIRST WEIGH THE FIRST BATCH WITH THE SECOND ONE. IF THE
BALANCE DOESN'T SHOW ANY MOVEMENT, THEN THE LEFT OUT COIN IS
THE ODD COIN.
2) IF THERE IS ANY MOVEMENT, THEN TEST THE 2 BATCHES
INDIVIDUALLY, i.e., 2 ON EACH SIDE OF THE BALANCE FOR BOTH
THE BATCHES. OUT OF THE 2 BATCHES, 1 WOULD SHOW MOVEMENT.
THEN IT IS CLEAR THAT THE ODD COIN IS FROM EITHER OF THE 4
COINS (SAY, THE SECOND BATCH).
3) [A] THEN, BALANCE ANY 2 COINS FROM THE SECOND BATCH
(HAVING THE ODD COIN) WITH 2 FROM THE FIRST BATCH (NOT
HAVING ANY ODD COIN).
4) IF THE BALANCE SHOWS ANY MOVEMENT, THEN AGAIN BALANCE 2
COINS (1 OF SECOND BATCH FROM THE BALANCE + 1 FROM THE FIRST
BATCH) WITH 2 MORE COINS FROM THE FIRST BATCH. IF THE
BALANCE DOESN'T SHOW ANY MOVEMENT, THEN THE LEFT OUT COIN ON
THE BALANCE [A] OF SECOND BATCH IS THE ODD ONE. BUT IF THE
BALANCE SHOWS SOME MOVEMENT THEN THE COIN WHICH WE TOOK OF
SECOND BATCH FROM THE BALANCE [A] IS THE ODD ONE OUT.
NOTE--- IN ALL OTHER ANSWERS, YOU ARE NOT ABLE TO FIND "THE
1 ODD COIN" TILL THE STEP. BUT HERE YOU GET IT CORRECT IN
ANY WAY, IN MAX-4 STEPS AND NOT 9 STEPS.
| Is This Answer Correct ? | 3 Yes | 21 No |
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