Interview Questions And Answers 
-
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 )
406 Views |
(17 votes, avg: 3.24)
Loading ...
Answers »
Our Sponsors
Our Sponsors
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.
Leave an Answer/Comment