solution to puzzle-1

Lets number the balls 1,2,3,…12.

weigh 1 2 3 4 against 5 6 7 8

if equal should be simple…
i.e weigh 9 10 11 aginst 1 2 3
if equal 12 is bad
else by this we will know bad ball is light or heavy
9 against 10 //you know the defect now…

if the fist case was unequal
mark the heavy side balls H1 H2 H3 H4 //if one of these is defective that got to be heavy
lighter side L1 L2 L3 L4 //if one of these is defective that got to be light
rest four are S1 S2 S3 S4 (standard)

2nd weighing should be:
H1 H2 L1 against L2 H3 S3
//we have three distinct groups now
1. if left side goes down then Either H1/H2 is heavy or L2 is light i.e. HHL
2. if left side goes up either L1 is light or H3 is heavy
3. if eaual, the remaining, either H4 is heavy or L3/L4 is light i.e LLH

3rd weighing: say the case is LLH
put one L on either side. The one that goes up is defective (light being defective)
if equal H is defective

If you can find any other technique apart from this, please comment here 🙂


Puzzles – 1

Problem Statement:

You have 12 identical-looking balls. One of these balls has a different weight from all the others. You also have a two-pan balance for comparing weights. Using the balance in the smallest number of times possible, determine which ball has the unique weight, and also determine whether it is heavier or lighter than the others.

I’ll be posting the solution in the next post. 🙂