The correct option is (a) - 2.
[as per provisional answerkey]Solution
We have 7 cubes in total: 6 identical (heavy) and 1 lighter cube. We need to find the minimum number of weighings required to identify the light cube with 100% certainty using a common balance (two pans).
Step 1: First Weighing
Divide the 7 cubes into three groups: Group A (3 cubes), Group B (3 cubes), and Group C (1 cube).
Place Group A on the left pan and Group B on the right pan.
- Case 1: If the pans balance, the light cube is the one remaining (Group C). We found it in 1 weighing. However, we need "certainty" for the worst-case scenario.
- Case 2: If the pans do not balance, the light cube is in the group that goes up (the lighter side). Let's assume Group A is lighter.
Step 2: Second Weighing
Take the 3 cubes from the lighter group (Group A) and repeat the process.
Place 1 cube on the left pan and 1 cube on the right pan, leaving 1 cube aside.
- Case 1: If the pans balance, the cube left aside is the lightest one.
- Case 2: If the pans do not balance, the cube on the side that goes up is the lightest one.
In all possible scenarios, the light cube is identified in at most 2 weighings.
Why the other options are incorrect
- Option (b) - 3: This is a common mistake if one uses a binary search (splitting into 2 groups) rather than a ternary search (splitting into 3 groups). While 3 weighings will work, it is not the minimum number required.
- Option (c) - 4: This is far beyond the required number of attempts. Even a simple one-by-one comparison would identify the cube in fewer steps.
- Option (d) - 1: One weighing is only sufficient if we are lucky (i.e., if the 7th cube left aside happens to be the light one). It does not provide "certainty" for all 7 cubes.
Key Concept
The maximum number of items (N) that can be tested in 'n' weighings using a balance scale is given by the formula N=3n; since 7 is less than 32 (9), two weighings are sufficient.