**I am full of holes
but still hold water.
What am i?**

## Defective ball

You have 12 balls all look identical (in shape, color etc.). All of them have same weight except one defective ball. You don’t know that the defective one is heavier or lighter than other balls. You can use a two sided balance system (not the electronic one). It is given that the minimum no. of measures required to separate the defective ball is three. Find the way you separate the defective ball.

See Solution : Defective ballDivide the balls into 3 groups of 4 balls.

weigh 2 groups, one on each side.

There will be two cases

a) The weight on both side is equal i. e. these two groups don’t have the defective ball.

b) One side has less weight than the other side.

**Case (a) :**

You know 8 balls are of equal weight and one of the remaining 4 balls have a defective one. Name these four as B1, B2, B3, B4.

Second weigh :

Take B1, B2 and weight them.

(i) If they are unequal then either B1 is defective or B2. Compare B1 with one of eight balls. If B1 is equal to that then B2 is defective otherwise B1.

Total measurements in this case : 1 (first weigh) + 1 (second weigh) + 1 (B1 with other ball) = 3

(ii) If B1 and B2 are equal then either B3 is defective or B4. Compare B3 with one of eight balls. If B3 is equal to that then B4 is defective otherwise B3.

Total measurements in this case : 1 (first weigh) + 1 (second weigh) + 1 (B3 with other ball) = 3

**Case (b) :**

Mark the balls in the side with less weight as L and with more weight as M. We get 4L and 4M.

Second weigh :

Take 2L and 1M in One side say A and take 2L and 1M in Other side say B of balance system. 2M are reserved for now.(i) If side A is down and next side goes up then it has two possibilities.

1. One of 2L in B has less weight than other 7 balls

2. The 1M in A has more weight(ii) If side B is down then it also has two possibilities.

1. One of 2L in A has less weight than other 7 balls

2. One of 1M in B has more weight than other 7 balls(iii) Both sides are balanced

Third weigh :

In cases (i) or (ii) we will get 3 balls (2L and 1M) after the second weigh.

For case (i) and (ii)of (b) :

Weigh two L balls with each other, if they are equal then the 1M is heavier and if they are not then the ball with less weight is defective.

For case (iii) of (b) :

In this case one of the two reserved balls is defective. We have 2M balls. Weigh them, the one which is heavier is defectives. because we know that issue is with the ball with more weight.