You have 12 similar looking coins. 11 of them
weigh the same. One of them is heavier than the
others. You also have a scale. You can put coins
on both sides of the scale and it'll tell you which
side is heavier or will stay in the middle if both
sides weight the same. What is the minimum
number of weighing required to find out the odd
coin.
Answers
Answer:
1
Step-by-step explanation:
it's one because all are same in weight
It gives same all time
Answer:
4
Step-by-step explanation:
Divide the coins into three groups of 4 each Take any two groups and weigh them. If they are equal, then the third group contains the counterfeit coin. If they are not equal then one of it contains the counterfeit coin. Now, in the first case, compare the weights of the first (or the second group) with the third group, if the former weighs less than the later then the weight of the counterfeit coin is more than the weight of the normal coins, else it weighs less. Similarly we can find out for the second case. Thus in the first stage it will require 2 weighings. Now in each successive stage only one comparison of two groups with equal number of coins are required to find out the counterfeit coin. Thus first divide 4 into 2, 1 ,1 parts, then if the false coin is in 2 then we require one more weighing. Hence, a total of 4 weighings are required to identify the counterfeit coin.