Divide the 12 ball into three groups: 1 \ 2 \ 3 \ 4, 5 \ 6 \ 7 \ 8, 9 \10/1\12.

Do the following: the first group (1\2\3\4) and the second group are placed at both ends of the balance.

The results are as follows:

1. Equilibrium. This means that the defective products are in the third group.

The following operations are performed: 1\2 and 9\ 10 are placed at both ends of the balance.

A. balance. The defective product is in 1 1- 12.

Put 1 and 1 1 on the balance. Balance, 12 is defective; If it is not balanced, 1 1 is defective.

B. imbalance. The defective products are at 9%+00.

Put 1 and 9 on the balance. Balance, 10 is defective; If it is unbalanced, 9 is defective.

2. 1 \ 2 \ 3 \ 4 & gt； 5\6\7\8 means that the first group is more important than the second group. Explain that the defective products are in these two groups.

The operation is as follows: put 1\2\3\5 and 9 \10 \1\ 4 at both ends of the balance.

A.1\ 2 \ 3 \ 5 = 9 \10 \1\ 4. The defective product is in June \ 7 \ 8, and it is relatively light.

Put 6 and 7 at both ends of the balance. Balance 8 is defective; The lighter imbalance is the defective products.

b. 1\2\3\5 >9\ 10\ 1 1\4。

The analysis shows that: 1. The defective product is in1\ 2 \ 3 \ 4 \ 5; 2. The defective products cannot be 4\5. Because 4\5 is a defective product, it can't meet 1\2\3\4 > 5\6\7\8,1\ 2 \ 3 \ 5 > 9\ 10\ 1 1\4。

Therefore, the defective product is in 1\2\3, and the defective product is heavier.

Put 1 2 at both ends of the balance. Balance 3 is defective; The serious imbalance is the defective products.

c . 1 \ 2 \ 3 \ 5 & lt； 9\ 10\ 1 1\4。

The analysis shows that: 1. The defective product is in1\ 2 \ 3 \ 4 \ 5; 2. The defective product cannot be 1\2\3. Because 1\2\3 is a defective product, 1\2\3\4 > 5\6\7\8,1\ 2 \ 3 \ 5 <; 9\ 10\ 1 1\4。

Therefore, the defective products are in 4\5, 4 is heavier and 5 is lighter.

Put 1 4 at both ends of the balance. Balance 5 is defective; If it is not balanced, 4 indicates a defect.

3. 1 \ 2 \ 3 \ 4 & lt； 5\6\7\8, indicating that the defective products are in these two groups.

The next operation is the same as example 2, and the analysis is similar, so it is easy to get the final operation.

(Reprinted)

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