Seven cubes are identical in shape. Out of these, the weight of each of the six cubes is equal and the weight of the remaining cube is less than the weight of any other cube. A balance is used to identify the lightest cube. What is the minimum number of attempts required to distinguish the odd cube with certainty?

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 $N=3^n$; since 7 is less than $3^2$ (9), two weighings are sufficient.