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 $2 %$ families buy all the three newspapers, find the number of families which buy B only

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Solution

Number of families who buys only newspaper $B$

$= n (B) - n (B \cap A) - n (B \cap C) + n (A \cap B \cap C)$

$= 20 - 5 - 3 + 2 = 14$

$∴ \frac{14}{100} \times 10, 000 = 1400$

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In a town of

10, 000

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% families buy newspaper

A

20

% buy newspaper

B

and

10

% 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 three newspapers, find number of families which buy None of

A, B, C

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

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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% families buy all the three newspapers, find the number of families which buy B only

Number of families who buys only newspaper B =n(B)−n(B∩A)−n(B∩C)+n(A∩B∩C)=20−5−3+2=14∴14100×10,000=1400

Number of families who buys only newspaper $B$

$= n (B) - n (B \cap A) - n (B \cap C) + n (A \cap B \cap C)$

$= 20 - 5 - 3 + 2 = 14$

$∴ \frac{14}{100} \times 10, 000 = 1400$