You have been given 8 Balls in a box. From that only one odd ball being heavier or Lighter. So, you have been asked to find out the minimum number of weighings required to spot the odd heavier/lighter ball among all the 8 identically looking balls using a common balance.
Please take some time to think and come up with a logical approach before looking at the solution.
This puzzle is not a direct one, but is a little trickier to solve. Let us label our 8 balls as B1,B2,B3, C1,C2,C3, D1 and D2. Let us weigh B1, B2, B3 on one side of the balance and C1, C2, C3 on the other side. If both the sides are equal, then we can say that odd ball is one in D1 or D2.
Taking this case, since we know that all B’s and C’s are normal balls, so lets weigh D1 with either of B’s or C’s (lets say with B1). With this way if D1 is equal to B1, then we can say that D2 is the odd ball or, D1 itself was odd.
B1 B2 B3 === C1 C2 C3
D1 == B1
D2 –> odd ball.
So, 3 weighings in this case.
Now, lets take other scenario: if B’s weighed more than C’s then we are sure that D’s are normal ones.
B1 B2 B3 > or < C1 C2 C3 ==> D1 and D2 are normal.
Let’s assume B’s are heavier than C’s and we still don’t know whether the odd is among B’s or C’s.
We know : B1 B2 C3 > C1 C2 C3
Now we compare B1 C2 C3 to C1 D1 D2 . Now if the B1 C2 C3 is heavier than C1 D1 D2.
Then it means the odd ball is among B1 and C1.
If B1 C2 C3 is lighter than C1 D1 D2 then the odd ball is among C2 and C3.
So we have zeroed down to 2 balls.
Now its very easy as we can compare any of the normal ball with one of the 2 balls.So answer is 3,i.e in 3 weighings we can find out the odd ball.