There are 12 black balls of equal size and shape.One of them is heavier than the other 11 balls. You are given a weighing balance.Now using not more than 3 weighing you have to find out which is the defective ball in the lot.

Well this is a enhancement to the already discussed 8 ball puzzle in puzzle arena. To start with remember that you can weigh maximum 3 times to arrive at the correct decision.

Divide the 12 balls into group of three. So we will have 4 balls in each group. Let’s name the groups as A,B and C.
Now weigh any two groups one weighing balance. Let us take A and B for this scenario.
So there will be two possibilities.
1. Both A and B are equal.
2. One of them is Heavy.

Let’s take the first case. If both are equal this means that the defective ball is in groupĀ  C. so take Group C and divide it into two groups of two balls each and weigh them. One of the group will weigh heavier. So for the third weighing weigh the two balls against each other and you can proceed.

Similarly for the second case.

You will know whether A is heavy or B is heavy. just proceed as done in case 1 and you will get the solution in 3 weighing.