Question

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% 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, then number of families which buy A only is

• A. 3100
• B. 3300
• C. 2900
• D. 1400

Answer 1

The correct option is B 3300

n(A) = 40% of 10,000 = 4,000
n(B) = 20% of 10,000 = 2,000
n(C) = 10% of 10,000 = 1,000
$n(A\cap B)$ = 5% of 10,000 = 500
$n(B\cap C)$ = 3% of 10,000 = 300
$n(C\cap A)$ = 4% of 10,000 = 400
$n(A\cap B\cap C)$ = 2% of 10,000 = 200
We want to find $n(A\cap B^c\cap C^c)=n[A\cap (B\cup C{)}^c]$
$=n\left(A\right)-n[A\cap (B\cup C\left)\right]=n\left(A\right)-n\left[\right(A\cap B)\cup (A\cap C\left)\right]=n\left(A\right)-\left[n\right(A\cap B)+n(A\cap C)-n(A\cap B\cap C\left)\right]$
= 4000 - [500 + 400 - 200] = 4000 - 700 = 3300