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Home » In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 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 (a) The number of families which buy newspaper A only. (b) The number of families which buy none of A, B and C
In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 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 (a) The number of families which buy newspaper A only. (b) The number of families which buy none of A, B and C
CBSE | Class 11 | Maths | NCERT Exemplar | Sets
In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 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 (a) The number of families which buy newspaper A only. (b) The number of families which buy none of A, B and C

Solution:

As per the question,

=10,000 = Total number of families

\mathrm{A}=\mathrm{n}(\mathrm{A})=40 \% = Number of families buying newspaper

B=n(B)=20 \% = Number of families buying newspaper

\mathrm{C}=\mathrm{n}(\mathrm{C})=10 \% = Number of families buying newspaper

=n(A \cap B)=5 \% = Number of families buying newspaper A and B

=n(B\capC) = 3 \% = Number of families buying newspaper B and C

=n(A \cap C)=4 \% = Number of families buying newspaper A and C

=n(A \cap B \cap C)=2 \% = Number of families buying all three newspapers

Let U be the total number of families

Let A be the number of families buying newspaper A

Let B be the number of families buying newspaper B

Let C be the number of families buying newspaper C

(a) No. of families that buy A newspaper only

Percentage of families that buy A newspaper only

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

=40-5-4+2

=33 \%

No. of families which buy A newspaper only

\begin{array}{l} =((33 / 100) \times 10000) \\ =3300 \end{array}

As a result, there are 3300 families which buy A newspaper only

(b) The number of families which buy none of the newspaper A, B and C.

Percentage of families which either buy A, B and C.

\begin{array}{l} =n(A \cup B \cup C) \\ =n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(A \cap C)+n(A \cap B \cap C) \\ =40+20+10-5-3-4+2 \\ =60 \% \end{array}

Percentage of families which buy none of A, B and C.

= Total percentage – No. of students who play either

\begin{array}{l} =100 \%-60 \% \\ =40 \% \end{array}

No. of families which buy none of A, B and C.

\begin{array}{l} =((40 / 100) \times 10000) \\ =4000 \end{array}

As a result, there are 4000 families which buy none of A, B and C.

i

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