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