Question

You have 6 coins of distinct weights and a beam balance with two pans. What is the minimum number of weighings that could identify with certainty the second lightest coin?

Answer 1

Let us assume the six coins be A,B,C,D,E and F. Form the groups with three coins in each and put the groups on the pans (1st weighing). Lets say coins A, C and E are in one group , while coins B, D and F are on the other group.

If Group 1 < Group 2, set aside first , the 2nd group then weigh A, C and E against each other (3 weighing s ) . If the result is A < C < E, choose the second lightest of that group , which is 'C' , then weigh it against each of the coins in Group 2 (3 weighing s).

If C is the lightest against B, D and F, then it is best assured that C is the second lightest of the six coins. If one of the coin in Group 2 , Lets say B, is lightest than C, then B is the second lightest of the six coins.

Therefore , there are 7 minimum weighings in determining the second lightest coin.