Answer / sabarish

ans ) 2 iterations
arrange 9 coins into 3 batches each of 3.

first weigh : weigh 2 batches and find the defective. if
the balance shows no movement then the 3rd batch is under
defect.

second weigh : from this batch , take any 2 balls out out
of 3, and weigh it.if there is movement in balance then
defective ball is found. if not , then the 3rd ball tht is
outside is the defective.

Answer / samveg gupta

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.
1. Take 8 coins and weigh 4 against 4.
o If both are not equal, goto step 2
o If both are equal, goto step 3

2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.
o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
 If both are equal, L4 is the odd coin and is lighter.
 If L2 is light, L2 is the odd coin and is lighter.
 If L3 is light, L3 is the odd coin and is lighter.

o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
 If both are equal, there is some error.
 If H1 is heavy, H1 is the odd coin and is heavier.
 If H2 is heavy, H2 is the odd coin and is heavier.

o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
 If both are equal, L1 is the odd coin and is lighter.
 If H3 is heavy, H3 is the odd coin and is heavier.
 If H4 is heavy, H4 is the odd coin and is heavier.

3. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.
o If both are equal, there is some error.
o If X is heavy, X is the odd coin and is heavier.
o If X is light, X is the odd coin and is lighter.

Answer / 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.

Answer / vikram

ans: 3 iterations

divide coins in batches of 3

do 2 iterations of the batches u would know which is
defective batch and also wheather its lighter r heavier.

in another iteration of defective batch u would find
defective coin.

Answer / 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.

Answer / nishant

1 itteration coz you need to test it maximum 9
times .....but the answer you will get in 1 time only....
when you place the wrong coin

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