Question

In a town of $10,000$ families it was found that $40\%$ families buy newspaper A, $20\%$families 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 $2\%$ buy all the three newspapers, the number of families which buy none of A, B, C is

• A. $3000$
• B. $4000$
• C. $4500$
• D. $3500$

Answer 1

The correct option is A $3000$

Total number of families $=10000$

Families buy newspaper $A=\frac{40}{100}\times 10000=4000$

Families buy newspaper $B=\frac{20}{100}\times 10000=2000$

Families buy newspaper $C=\frac{10}{100}\times 10000=1000$

$\therefore$ The number of families which buy none A,B,C is $=10000-(4000+2000+1000)=10000-7000=3000$