Question

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, find the number of families which buy B only

Answer 1

Number of families who buys only newspaper $B$

$=n\left(B\right)-n(B\cap A)-n(B\cap C)+n(A\cap B\cap C)$

$=20-5-3+2=14$

$\therefore \frac{14}{100}\times 10,000=1400$