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.