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 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 the number of families which only buy newspaper A is

To solve this problem, we need to break it down into several steps:

1. First, we need to find out the total number of families that buy each newspaper.

   • 40% of 10,000 families buy newspaper A, which is 4,000 families.
   • 20% of 10,000 families buy newspaper B, which is 2,000 families.
   • 10% of 10,000 families buy newspaper C, which is 1,000 families.

2. Then, we need to find out the number of families that buy more than one newspaper.

   • 5% of 10,000 families buy both A and B, which is 500 families.
   • 3% of 10,000 families buy both B and C, which is 300 families.
   • 4% of 10,000 families buy both A and C, which is 400 families.
   • 2% of 10,000 families buy all three newspapers, which is 200 families.

3. To find out the number of families that only buy newspaper A, we need to subtract the number of families that buy A and another newspaper from the total number of families that buy A.

   • The number of families that buy A and B or A and C is 500 + 400 = 900 families.
   • But since we've counted the families that buy all three newspapers twice, we need to add them back once. So the total number of families that buy A and another newspaper is 900 - 200 = 700 families.

4. Finally, subtract the number of families that buy A and another newspaper from the total number of families that buy A to get the number of families that only buy A.

   • The number of families that only buy A is 4,000 - 700 = 3,300 families.