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 B and C and 4% buy A and C. If 2%families buy all the three newspapers, then find the number of families which buy A only

Options

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

Answer

To solve the problem step by step, we will use the principle of inclusion-exclusion to find the number of families that buy only newspaper A.

### Step 1: Determine the total number of families buying each newspaper
- Total families = 10,000
- Families buying newspaper A = 40% of 10,000 = 0.40 × 10,000 = 4,000
- Families buying newspaper B = 20% of 10,000 = 0.20 × 10,000 = 2,000
- Families buying newspaper C = 10% of 10,000 = 0.10 × 10,000 = 1,000

**Hint:** Convert percentages into actual numbers by multiplying the percentage (as a decimal) by the total number of families.

### Step 2: Determine the number of families buying combinations of newspapers
- 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

**Hint:** Again, convert the given percentages for intersections into actual numbers using the same method.

### Step 3: Use the formula for families buying only newspaper A
The formula to find the number of families that buy only newspaper A is:

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

Substituting the values we calculated:

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

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

1. Subtract families buying A and B:
\[ 4000 - 500 = 3500 \]
2. Subtract families buying A and C:
\[ 3500 - 400 = 3100 \]
3. Add families buying all three newspapers:
\[ 3100 + 200 = 3300 \]

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

**Final Answer:** 3,300 families buy only newspaper A.