Question

In a town of $10000$ families, it was found that $40\%$ families buy newspaper $A,20\%$ families buy newspaper $B$ and $10\%$ families buy newspaper $C,5\%$ buy $A$ and $B,3\%$ buy $B$ and $C$ and $4\%$ buy $A$ and $C$. If $2\%$ families buy all of three newspapers, then the number of families which buy A only, is

• A. 4400
• B. 3300
• C. 2000
• D. 500

Answer 1

Answer: (B)

$n(A)=40\%$ of $10000=4000,n(B)=2000$
$n(C)=1000,n(A\cap B)=500,n(B\cap C)=300$
$n(C\cap A)=400,n(A\cap B\cap C)=200$
$\therefore n(A\cap \bar{B}\cap \bar{C})=n\left\{A\cap (B\cup C{)}^{\prime }\right\}$
$=n(A)-n\{A\cap (B\cup C)\}$
$=n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$
$=4000-500-400+200=3300$