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 $n(A\cap B^c\cap C^c)=n[A\cap (B\cup C{)}^c]$.
$=n\left(A\right)-n[A\cap (B\cup C\left)\right]=n\left(A\right)-n\left[\right(A\cap B)\cup (A\cap C\left)\right]$
$=n\left(A\right)-\left[n\right(A\cap B)+n(A\cap C)-n(A\cap B\cap C\left)\right]$
$=4000-[500+400-200]=4000-700=3300$.