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

$n\left(A\right)=40\%of10,000=4,000$

$n\left(B\right)=20\%of10,000=2,000$

$n\left(C\right)=10\%of10,000=1,000$

$n(A\cap B)=5\%of10,000=500$

$n(B\cap C)=3\%of10,000=300$

$n(C\cap A)=4\%of10,000=400$

$n(A\cap B\cap C)=2\%of10,000=200$

We want to find the number of families which buy only A $=n\left(A\right)-\left[n\right(A\cap B)+n(A\cap C)-n(A\cap B\cap C\left)\right]$

$=4000-[500+400-200]=4000-700=3300$