In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy Band Cand 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buyA only is

Options

A. 3100
B. 3300
C. 2900
D. 1400

Answer

To find the number of families that buy only newspaper A, we can use the principle of inclusion-exclusion. Let's break down the problem step by step.

### Step 1: Identify the total number of families and percentages
- Total families = 10,000
- Families buying newspaper A = 40% of 10,000 = 0.4 * 10,000 = 4,000
- Families buying newspaper B = 20% of 10,000 = 0.2 * 10,000 = 2,000
- Families buying newspaper C = 10% of 10,000 = 0.1 * 10,000 = 1,000

### Step 2: Identify the overlaps
- Families buying both A and B = 5% of 10,000 = 0.05 * 10,000 = 500
- Families buying both B and C = 3% of 10,000 = 0.03 * 10,000 = 300
- Families buying both A and C = 4% of 10,000 = 0.04 * 10,000 = 400
- Families buying all three newspapers A, B, and C = 2% of 10,000 = 0.02 * 10,000 = 200

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

\[ \text{Families buying only A} = (\text{Families buying A}) - (\text{Families buying A and B}) - (\text{Families buying A and C}) + (\text{Families buying all three}) \]

Substituting the values we calculated:

\[ \text{Families buying only A} = 4000 - 500 - 400 + 200 \]

### Step 4: Calculate the number of families buying only A
Now, we perform the calculation:

\[ \text{Families buying only A} = 4000 - 500 - 400 + 200 = 4000 - 900 + 200 = 4000 - 700 = 3300 \]

### Step 5: Final calculation
Thus, the number of families that buy only newspaper A is 3300.

### Conclusion
The number of families which buy newspaper A only is **3300**.

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