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

Answer: (B)

3300

Answer 2

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