Question

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy
newspaper B, 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. Find the number of families who buy
A and C but not B​

Answer 1

Answer:

$number \: of \: families \: who \: buy \: a \: and \: c \: but \: not \: b = 10000 - 10000 \times \frac{20}{100} - 10000 \times \frac{5}{100} - 10000 \times \frac{3}{100} \\ = 10000 - 2000 - 500 - 300 \\ = 7800$