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The problem in the image involves a set of families buying different combinations of three newspapers \( A \), \( B \), and \( C \). Let's break it down:

### Problem:
In a town of 10,000 families, we are given the following information:
- 40% of families buy newspaper \( A \),
- 20% buy newspaper \( B \),
- 10% buy newspaper \( C \),
- 5% buy both \( A \) and \( B \),
- 4% buy both \( A \) and \( C \),
- 2% buy all three newspapers \( A \), \( B \), and \( C \).

The goal is to find how many families buy only newspaper \( A \).

### Approach:
We will use the principle of inclusion-exclusion and Venn diagrams to solve this.

Let:
- \( |A| \) represent the number of families buying newspaper \( A \),
- \( |B| \) represent the number of families buying newspaper \( B \),
- \( |C| \) represent the number of families buying newspaper \( C \),
- \( |A \cap B| \), \( |A \cap C| \), and \( |B \cap C| \) represent the number of families buying combinations of two newspapers,
- \( |A \cap B \cap C| \) represent the number of families buying all three newspapers.

1. **Convert percentages to actual numbers**: Since there are 10,000 families, we convert each percentage into the number of families:
   - \( |A| = 40\% \times 10,000 = 4000 \),
   - \( |B| = 20\% \times 10,000 = 2000 \),
   - \( |C| = 10\% \times 10,000 = 1000 \),
   - \( |A \cap B| = 5\% \times 10,000 = 500 \),
   - \( |A \cap C| = 4\% \times 10,000 = 400 \),
   - \( |A \cap B \cap C| = 2\% \times 10,000 = 200 \).

2. **Find number of families buying only newspaper \( A \)**: 
   To find the families that buy only \( A \), we need to subtract the families that buy \( A \) and other combinations (i.e., \( A \cap B \), \( A \cap C \), and \( A \cap B \cap C \)) from the total number of families that buy \( A \).

   The formula for families buying only \( A \) is:
   \[
   |A \text{ only}| = |A| - (|A \cap B| + |A \cap C| - |A \cap B \cap C|)
   \]

3. **Substitute the known values**:
   \[
   |A \text{ only}| = 4000 - (500 + 400 - 200)
   \]
   Simplifying:
   \[
   |A \text{ only}| = 4000 - (700) = 3300
   \]

Thus, the number of families that buy only newspaper \( A \) is **3,300 families**.

Would you like further explanation or have any questions?

### Additional Questions:
1. How would the number of families buying only newspaper \( B \) be calculated?
2. Can you calculate how many families buy at least one newspaper?
3. What would be the approach to calculate the number of families buying only \( B \) and \( C \), but not \( A \)?
4. How would the solution change if the percentage of families buying all three newspapers was increased?
5. What role does the principle of inclusion-exclusion play in solving this type of problem?

### Tip:
When working with set problems involving percentages or counts, always convert percentages to actual numbers before applying any formulas.

In a town of 10000 families, it was found that 40% of families buy newspaper A, 20% buy newspaper B, and 10% of families buy newspaper C, 5% of families buy A and B, 4% buy A and C, and 2% of the families buy all the three newspapers, then find the number of families which buy A only.

In this problem, you will learn how to find the number of families buying only newspaper A in a town of 10,000 families, using the inclusion-exclusion principle. The problem involves percentages and set theory. Suitable for students in Grades 9-12.

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