Question

In a town of 10000 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% families buy all the three newspapers, find the number of families which buy
1)A only
2)B only
3)none of A, B&C
4)exactly two newspapers
5)exactly one newspaper
6)A&C but not B
7)at least one of A, B&C​

Answer 1

Answer:

1) A only:

Total people = 10000

Buying = 40%

$\frac{40}{100} * 10000$

$40 * 100$

$4000$

2) B only:-

Total people = 10000

Buying = 20%

$\frac{20}{100} * 10000$

$20 * 100$

$2000$

3) None of the papers:-

Total people = 10000

Percent of people buying = 84%

Percent not buying = 16%

$\frac{16}{100} * 10000$

$16 * 100$

$1600$

4) Exactly two newspapers:-

Total people = 10000

People buying two papers = 5% + 3% + 4%

= 12%

$\frac{12}{100} * 10000$

$12 * 100$

$1200$

5) Exactly one newspaper:-

Total people = 10000

Percentage buying one paper = 40% + 20% + 10%

= 70%

$\frac{70}{100} * 10000$

$70 * 100$

$7000$

6) A & C but not B:-

Total people = 10000

People buying = 4%

$\frac{4}{100} * 10000$

$4 * 100$

$400$

7) At least one paper:-

The answer will be same as case - 5.

Hope it helps

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