Question

In a town of $1000$ 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 $B,3\%$ families buy $B$ and $C$ and $4\%$ families buy $A$ and $C$. If $2\%$ families buy all the three newspapers, then, number of families which buy $A$ only is

• A. $340$
• B. $330$
• C. $420$
• D. $225$

Answer 1

The correct option is B $330$

Total no. of Families $=1000$

A be the no of families buy newspaper $A=\frac{40}{100}\times 1000=400$

B be the no. of families buy newspaper $B=\frac{20}{100}\times 1000=200$

C be the no. of farilies buy newspaper $C=\frac{10}{100}\times 1000=100$

$n\left(A\right)=400,n\left(B\right)=200,n\left(C\right)=100$

$n(A\cap B)=50,n(B\cap C)=30,n(A\cap C)=40$

$n(A\cap B\cap C)=20$

Number of families which buy only $A$ is

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

$=400-50-40+20=330$