There are nine bags of rice looking alike, eight of which have equal and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance the minimum number of weighting required to identify the heavier bag is
2
Minimum Weighings for Nine Bags of Rice
The problem asks us to find the minimum number of weighings required to identify one slightly heavier bag among nine identical-looking bags of rice, using a weighing balance of unlimited capacity.
Weighing Balance Principle
A standard weighing balance (like a two-pan scale) is ideal for this type of problem. The most efficient strategy is to divide the items into three groups as equally as possible for each weighing. By doing so, you can eliminate two-thirds of the possibilities in a single weighing.
- If the two weighed groups balance, the heavier item must be in the unweighed group.
- If one side is heavier, the heavier item is on that side.
Step-by-Step Solution to Identify the Heavier Bag
First Weighing: Dividing the Nine Bags
We begin with 9 bags. To utilize the weighing balance effectively, we divide the bags into three groups:
- Group A: 3 bags
- Group B: 3 bags
- Group C: 3 bags
Now, we place Group A on one pan of the balance and Group B on the other pan.
| Weighing Result | Inference (Heavier Bag is in...) | Bags Remaining to Check |
|---|---|---|
| Group A balances Group B ($\text{A} = \text{B}$) | Group C | 3 bags |
| Group A is heavier than Group B ($\text{A} > \text{B}$) | Group A | 3 bags |
| Group B is heavier than Group A ($\text{B} > \text{A}$) | Group B | 3 bags |
After the first weighing, regardless of the outcome, we have successfully narrowed down the location of the heavier bag to a group of 3 bags.
Second Weighing: Pinpointing the Heavier Bag
Now, we have a group of 3 bags that contains the heavier one. Let's call these three bags $\text{Bag}_1$, $\text{Bag}_2$, and $\text{Bag}_3$. We perform a second weighing using these three bags.
Place $\text{Bag}_1$ on one pan and $\text{Bag}_2$ on the other pan.
| Weighing Result | Inference (Heavier Bag is...) |
|---|---|
| $\text{Bag}_1$ balances $\text{Bag}_2$ ($\text{Bag}_1 = \text{Bag}_2$) | $\text{Bag}_3$ (the unweighed bag) |
| $\text{Bag}_1$ is heavier than $\text{Bag}_2$ ($\text{Bag}_1 > \text{Bag}_2$) | $\text{Bag}_1$ |
| $\text{Bag}_2$ is heavier than $\text{Bag}_1$ ($\text{Bag}_2 > \text{Bag}_1$) | $\text{Bag}_2$ |
In all possible outcomes of this second weighing, we are able to definitively identify the single heavier bag.
Conclusion on Weighings Required
Based on the two-step process described above, the minimum number of weighings required to identify the heavier bag among nine bags is 2.
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