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Good one

Nice one

4 times

i Think 11 times

Whats the right answeR?

in 4 attempts we can find out.

Divide them into 3 groups - 4 each lets call them A, B ,C

Weigh - A against B and C.

two of them would be equal and one would differ.Lets assume B is different.

4 coins in B would be a,b,c,d

weigh
a and b
a and c

if the they weigh the same then answer id D

If a and b are diff the a and c are equal then b is answer.
if a and b are equal and a and c are diff then c is the answer.
if a and b are diff and a and c are also diff then a is the answer.

I will not give the complete solution but anser of question is 3.
Now think how its possible in 3 chances

in 4 attempts we can find out.

Divide them into 3 groups - 4 each lets call them A, B ,C

Weigh - A against B and C.

two of them would be equal and one would diff.Lets assume B is different with more weight.

now divide the 4 in B in 2 groups a, b

weigh a and b pick the one which is more lets assume b is more

now weight the 2 coins in b separately and pick the one with more weight .

The same can be done for less weight also.

Jaya
in problem its not mentioned fake coin is heavier or lighter than other. so in ur solution we need one more chance to find out that. so we need 4 chances as per ur solution

sorry I posted it wrong ....... it only takes 3 attempts. As we are weighing 3 groups in the first 2 weighs , 2 would be equal and one would be more or less. Hence we can determine from there if the defecttive coin weighs more or less.

So i would still need only 3 chances

Divide them into 3 groups - 4 each lets call them A, B ,C

Weigh - A against B and C.

two of them would be equal and one would diff.Lets assume B is different with more weight.

now divide the 4 in B in 2 groups a, b

weigh a and b pick the one which is more lets assume b is more

now weight the 2 coins in b separately and pick the one with more weight .

The same can be done for less weight also.

jaya
how many times u have done iight
first u weight A and B
then u weight B and C
then u weight ur group a and b
and then u weight 2 coins
so 4 times

Let divide 12 coins in 6 groups.
It takes minimum 6 attempts to identify FAKE coin. I don't think less than 6 attempts could be possible.

Divide the 12 coins into 2 parts i.e. 6 of each. Lets say Set A and B

Take Set A & weigh 3 at one side & 3 on another side.

If they are equal than we have to take other set i.e. B of 6 else we will take the same set i.e. A.

Now the 6 coins divide them into 2 parts of 3 coins.

Weigh one set of three coin with the 3 coin of from the other set of 6. If this is equal than the coin should be in Balance 3 coin if one side is getting higher or lower than that side need to be taken out.

From 3 coins now weigh 1,1 coin if the result is balance the left one is the coin. If not we can take higher or lower one as we already know after weighing through 3 coin from other set the wait is higher or lower.

Great ans by Jaya.. kep it up..

It takes 3 times only.I assume fake coin weighs less.
First attempt make 12 coins as two parts by 6 each then weigh them.And take the part which has less weight.
Second attempt take the less weight part and make two parts as 3 each then weigh them n take the less weight part.
Third attempt take 2 coins from the three coins that you have got and weigh them if the two coins weigh equal then the third one is the fake coin r else if the two coins wont match their weight equal then the lesser weight coin is the fake coin.

I forgot to mention that Fake coin never be heavier as why would some one make a fake coin which is heavier as that would be loss as requires more ,material.

Now divide in 2 parts of 6 each lets say A & B

Divide A in to 2 parts again of 3 each.

Weigh these 3 each of A, if they are equal take other set B if not use the same.

Now if this is equal the other set will have fake coin & vice verse.

Take 3 coins from set which is having fake which is known as unequal. Weigh against the 3 equal of other set. This we are weighing 2nd time. Now what ever result comes we get 3 coins in which fake coin is other. weigh 3 rd time 1 on one side and another on one side we will get the fake coin.

Divide into groups of 4; A, B, and C

Put A on one side and B on other.

If both sides are balanced fake coin is in C.

Put 2 coins from A on one side and 2 from C on the other. If balanced, weigh one of the 2 coins left in C with one of the correct coins. If not balanced note whether two coins of C are more or less than 2 coins of A. Put one of these coins in C on one side and another on the other side. The side that has the same sign (more or less) has the fake coin.

If A and B are not balanced, note which one is weighs more and which one less. Let us say A is more. Fake coin is either in A or in B and C has all correct coins.
Take 2 coins of A and 2 of B and put them on one side and 4 coins of C on the other side. If both sides are not in balance (note A+B is more or less than C) then take one coin of A and one coin of B and put them on one side and put 2 coins of C on the other side. If they are not in balance and weight is more than that of C then fake coin is the one in A; if less then fake coin is one in B. If 2-2 coins are in balance then the fake coin is in the remaining two. Similarly if weight is more it belongs to A and if less (refer bracket above) then fake coin belongs to B.

If there is balance when we take 2+2 from A and B and 4 from C then fake coin is in the remaining 2 of A or B. Take 1 of A and 1 of B and weigh with 2 C. If not in balance and the weight is more then 2C then fake coin is the one in A if less it is in B. If 1A and 1B are in balance with 2C then the fake coin is in remaining A or B. If A+B (refer bracket) was more then it is A or if it is less then it is B.

light or heavy .. it will take 4 attempts at the max & 3 minimum ...

divide 3 coins each & make 4 sets ABCD

1st weigh - A & B
result - THEY ARE SAME
INFERENCE - C OR D IS BAD

2ND WEIGH - A & C
RESULT - THEY ARE SAME
INFERENCE - D IS BAD


3RD WEIGH - D1 & D2
RESULT 1 - THEY ARE SAME
INFERENCE - D3 IS BAD : THIS IS THE MINIMUM 3 WEIGH TO DETERMINE


RESULT 2 - D1 & D2 WEIGH DIFFERENT
THEN WE GO FOR 4TH WEIGH ..
4TH WEIGH - D1 & D3
RESULT 1 >> SAME, THEN D2 IS BAD
RESULT 2 >> DIFF, THEN D1 IS BAD

GREAT U MR.dsraman ,

thanks abishek ,,, looking for some nice good puzzles ... there seems to be no additions

Here is the solution to the 12 coin problem - Divide the 12 coins into three main groups of 4 coins, and then each of these into sub-groups of 1 and 3 coins. Place two main groups on the pans of the balance leaving the third main group on the table. Observe the condition of the balance.(first weighing). Rotate the sub-groups of 3 coins and observe the condition of the balance. (second weighing)If it changes, it will identify the 3 coin group that contains the odd coin and its relative weight. In this case, split this group into single coins and put two on the pans of the balance and one on the table. This will identify the odd coin. Problem solved.
If the condition of the balance did not change, rotate the single coins. This will identify the odd weight coin and its relative weight. Problem solved.

7 times
5 times when da coin weighs equal n 6th time when da coinz weight unequal den one of da coin is again weighed with a previous coin if da weight is same den da leftout coin is fake else that very coin is fake......