In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy news-paper C. 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three mewspapers, find the number of families which buy (i) A only

Answer

To find the number of families that buy only newspaper A, we can follow these steps:

### Step 1: Identify the total number of families and percentages
- Total families = 10,000
- Percentage of families buying:
- Newspaper A = 40% of 10,000 = 4,000 families
- Newspaper B = 20% of 10,000 = 2,000 families
- Newspaper C = 10% of 10,000 = 1,000 families

### Step 2: Identify the percentages of families buying combinations of newspapers
- Families buying both A and B = 5% of 10,000 = 500 families
- Families buying both B and C = 3% of 10,000 = 300 families
- Families buying both A and C = 4% of 10,000 = 400 families
- Families buying all three newspapers A, B, and C = 2% of 10,000 = 200 families

### Step 3: Use the principle of inclusion-exclusion to find families that buy only A
To find the number of families that buy only newspaper A, we can use the formula:

\[ \text{Families buying A only} = N(A) - N(A \cap B) - N(A \cap C) + N(A \cap B \cap C) \]

Where:
- \(N(A)\) = Families buying A = 4,000
- \(N(A \cap B)\) = Families buying A and B = 500
- \(N(A \cap C)\) = Families buying A and C = 400
- \(N(A \cap B \cap C)\) = Families buying A, B, and C = 200

### Step 4: Substitute the values into the formula
\[ \text{Families buying A only} = 4000 - 500 - 400 + 200 \]

### Step 5: Calculate the result
\[ \text{Families buying A only} = 4000 - 500 - 400 + 200 = 4000 - 900 + 200 = 3300 \]

### Step 6: Conclusion
The number of families that buy only newspaper A is **3,300**.

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