Interview Questions And Answers 
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You have 12 identical looking coins. Only one coin out of the 12 coins is definitely bad, it can be heavier or lighter, we don’t know at this stage.
You have a ( two ) pan balance and are allowed only three tries. You can use any mix of coins in the tries. You can mark the coins to identify them and to not let them get mixed up.At the end of the third weighing on the pan balance you have to conclusively identify the bad coin and state if it was heavier or lighter than the rest of the coins. ( It may take you days to solve it, there is only one solution )
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1. Divide the coins in two parts 6 each and weight them.
2. take that part which is lighter as it must have the bad coin. again divide them in 3coins each and weight.
3. take that part which is lighter. so you have 3 coins in which one is bad coin. now take one coin seperate and weight remaining two coins. If any of the coin weights lighter, that is the bad coin. If both coins are of equal weight the coin you have seperated is the bad coin.
We can divided into three part two coins ,five coins ,and five coins. We will put five coins on the one pan balance and we will put five coins on other the pan balance. We will ckech and talk heavier or lighter than pan balance. Then we will divided three part again one coin , two coins , and coins. We will put two coins on the one pan balance and we will put two coins on other the pan balance. We will ckech and talk heavier or lighter than pan balance. Then put two coins different pan balance .We will ckech and talk heavier or lighter than pan balance.
easy if your not unlucky
3 groups of 4 coins: A B C
measure A against B
- if even then its in C
- then measure 2 C’s with 2 C’s
- then replace one C with a normal
- if you check all your results theres only 1 coin that matches the result each time
- if uneven lets say A is heavier, then replace 2 coins from A with B coins, fill in 2 C coins into B.
- if even then one of the two missing A coins are heavy and just test it against a normal
- If uneven with the two A coins heavier, one of the A coins is the heavier one or on the other side one of the B coins are lighter
- take 1 mabye heavier B and put it with 1 mabye lighter A and a normal: balance this with 2 normals and a mabye lighter A
- if balanced the remaining B is heavier
- If unbalanced having the BAand normal lgihter then the A in that group is lighter
- If unbalanced with the A and two normals lighter then tough luck ahha cause this is a really low probablity. either the opposite B is heavy or that A is light.
- If uneven with with the side of B coins and normals then the B coins on the AB side are lighter.
and you just test them with normal.
Divide the coins into 3 groups of 4 and:
1. Weigh 4 against 4 - outcome is either they are equal and hence the coin is in the 3rd group not weighed or one of the weighed group is heavier.For any of the outcomes proceed to
2.Weigh 2 against 2 and find the heavier side
3. Weigh 1 against one and you have your coin
Out of 12 coins,
Do weighing of 5 ,5 coins . –> 1st weighing
If (5,5) is balanced then
do balancing between last 2 coins.
Else
Select 5 coins which are less heavier in prevoius measurement.
Do weighing of any (2,2) coins out of 5. –>2nd weighing
If (2,2) is balanced then
coin which is left is the most lighter
else
do again weighing of lighter 2 coins —> 3rd weighing
1. 4 coins in left pan, 4 in right, assume unbalance,
1-4 is heavy, 5-8 light, 9-12 good. (If balanced, one in 9-12 is bad, easy to solve)
2. 1,2,5,6 in left pan, 3,7,9,10 in right.
Assume unbalance. if left pan is light, narrow down to 5,6,3, if left is heavy, narrow down to 1,2,7. If balance narrow down to 4,8
3. for case 5,6,3, take 3,5 in left, 9,10 right
for case 1,2,7, take 1,7 left, 9,10 right
The bad coin can be haeavier or can be lighter. All the answers have assumed that the bad coin is lighter, which is incorrect.
I wonder if there is any possible solution for the problem.
Divide 12 coins into 3 groups of 4 coins each.
1111 2222 3333 (mark these numbers to show which group they belong to)
step 1) weigh 1111 (pan X) and 2222 (pan Y)
Case A:
If they are the same then:
- take 3 coins from third group
- take 3 coins from first (or second) group
step 2) weigh 333 and 111
Case A-1:
if they are same
step 3) weigh remaining coin in group 3 with any other coin.
So now you have the wrong coin and whether it is lighter or heavier
Case A-2:
if they are different, you know whether the “black sheep” is heavier or lighter”
step 3) weigh one coin from group 3 on each scale
if they are the same, the culprit is the other coin.
if they are different, you can identify the culprit because you know whether he is lighter or heavier.
Case B (after step1):
If they are different, note whether the scales with 1s were lighter or heavier (observation A).
step 2)
weigh 1123(pan X) 1133(pan Y)
if same:
- problem is with one of the three remaining 2s
- step 3) weigh 2 and 2 (trivial with observation A)
if different:
- if pan X shows same result as observation A, then problem is with one of the 1s on pan X. therefore, step 3 is weigh 1 and 1.
- if pan X shows different result from observation A, problem is with the 2 on pan X (or) on the 1s on pan Y. therefore, step 3 is weigh 1 and 1 from pan Y. if same, then problem is with the 2. otherwise use observation A to get the problematic 1.
my apologies for the long answer
Divide the 12 coins into 4 groups {(A1, A2, A3), (B1, B2, B3), (C1, C2, C3), and (D1, D2, D3),
1. First Weighing:
Put group A and B in pan 1 and 2 respectively and weigh.
If {(group A greater than or less than group B) OR (group A equals group B)} then:
First Case: - It means that groups C and D are all good coins.
Second Case: It means that groups A and B are all good coins.
Whichever way the bad coin lies in two groups, so let’s take the first case,
(i.e. group A and group B are not equal) for the second weighing.
2. Second Weighing:
Reweigh group A and B with some additions and subtractions from good coins as follows:
Put (A1, A2, C1) in pan 1 and put (B1, C2, C3) in pan 2, and do the second weighing. Note that A3 was removed from pan1 and B2 & B3 were removed from pan 2.
If {(A1, A2, C1) greater than or less than (B1, C2, C3)} OR {(A1, A2, C1) equals (B1, C2, C3) } then:
First Case: - It means that A3, B2, and B3 are all good coins. It also means that the inequality in the weights is as a result of any of coin A1, A2 or B1.
Second Case: - It means that A1, A2 and B1 are all good coins. It also means that the inequality in the weights is as a result of any of coin A3, B2 or B3.
Whichever way we will be left with 3 coins to do the third weighing, so let’s use A3, B2, and B3 for the third weighing.
3. For Third Weighing:
Put A1 in pan 1 and A2 in pan 2 and weigh:
If (A1 greater than or less than A2) then
This means that if Pan 1 has been greater since second weighing, then A1 is the bad coin and it is heavier. If Pan 1 were less since second weighing, then A1 is definitely the bad coin but this time, lighter.
Else if (A1 equals A2) then:
It means that B1 is the bad coin and it is either lighter or heavier depending on the outcome of the second weighing. i.e. if Pan 2 was greater then B1 was heavier otherwise it is lighter.
Hey,
1. Take 6 coins out of equision. Divide the other 6 on 3 + 3. If they are equal, than the bad coin is among the other 6 coins, if not equal, than it is among our measured six coins.
2. Since we don’t know if the bad one is lighter or heavier. Let’s note which 3 are lighter and which are heavier (the ones on the left or right). Let’s assume that coins on the right (1,2,3) are heavier than coins on the left (4,5,6)
3. We know that the 6 we didn’t measure are all good coins, so if we substitute either (1,2,3) on the right or (4,5,6) on the left with 3 from the good bunch of 6 coins we didn’t measure (on either left side or right side), there are two possible outcomes. If after substituting heavier coins (1,2,3) on the right with 3 good coins - they all equalize, then we know (1. (1,2,3) are good coins, 2. bad coin is amongh (4,5,6) and 2. bad coin is lighter than the rest) and visa versa, if we substitute lighter coins (4,5,6) on the left with 3 good coin and they equalize, then we know (1. (4,5,6) are good coins, 2. bad coin is among (1,2,3) and 2. bad coin is heavier than the rest).
4. Now we are left with 3 coins that have a bad one among them. Let’s assume they are (4,5,6), then we know the bad coin must be lighter than the rest (see above). We take one coin out (4) and only weigh 5 and 6 against each other, if they equalize, then a bad one is 4. If 5 is lighter than 6, than the bad one is 5.
Hope this all makes sense.
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