Question

In a town of 10000 families, it was found that 40% families buy a newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy both A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then the number of families which buy A only.

• A. $3300$
• B. $3500$
• C. $3600$
• D. $3700$

Answer 1

The correct option is A $3300$

$n\left(A\right)=40$

$n\left(B\right)=20$

$n\left(C\right)=10$

$n(A\cap B)=5$

$n(B\cap C)=3$

$n(C\cap A)=4$

$n(A\cap B\cap C)=2$

We want to find $n\left(Aonly\right)=n\left(A\right)–\left[n\right(A\cap B)+n(A\cap C\left)\right]+n(A\cap B\cap C)$

$n\left(Aonly\right)=4000–[500+400]+200=4000–700=3300$