Search any question & find its solution
Question:
Answered & Verified by Expert
In a town of 10000 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 newspaper. Find
(i) the number of families which buy newspaper $A$ only.
(ii) the number of families which buy none of $A, B$ and $C$.
(i) the number of families which buy newspaper $A$ only.
(ii) the number of families which buy none of $A, B$ and $C$.
Solution:
2338 Upvotes
Verified Answer
Let $A$ be the set of families which buy newspaper $A, B$ be the set of families which buy newspaper $B$ and $C$ be the set of families which buy newspaper $C$.
Then,
$$
\begin{aligned}
&n(U)=10000, n(A)=40 \% n(B)=20 \% \text { and } n(C)=10 \% \\
&n(A \cap B)=5 \% \\
&n(B \cap C)=3 \% \\
&n(A \cap C)=4 \% \\
&n(A \cap B \cap C)=2 \% \\
&\text { (i) Percentage of families which buy newspaper A only }= \\
&n(A)-n(A \cap B)-n(A \cap C)+n(A \cap B \cap C) \\
&=(40-5-4+2) \%=33 \% \\
&\therefore \text { Number of families which buy newspaper A only - } \\
&10000 \times 33 / 100=3300
\end{aligned}
$$
(ii) Percentage of families which buy none of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ $=n(U)-n(A \cup B \cup C)$
$$
\begin{aligned}
&=n(U)-[n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C) \\
&-n(A \cap C)+n(A \cap B \cap C)] \\
&=100-[40+20+10-5-3-4+2] \\
&=(100-60) \%=40 \% \\
&\therefore \text { Number of families which buy none of } \mathrm{A}, \mathrm{B} \text { and } \mathrm{C} \\
&=\frac{40}{100} \times 10000=4000
\end{aligned}
$$
Then,
$$
\begin{aligned}
&n(U)=10000, n(A)=40 \% n(B)=20 \% \text { and } n(C)=10 \% \\
&n(A \cap B)=5 \% \\
&n(B \cap C)=3 \% \\
&n(A \cap C)=4 \% \\
&n(A \cap B \cap C)=2 \% \\
&\text { (i) Percentage of families which buy newspaper A only }= \\
&n(A)-n(A \cap B)-n(A \cap C)+n(A \cap B \cap C) \\
&=(40-5-4+2) \%=33 \% \\
&\therefore \text { Number of families which buy newspaper A only - } \\
&10000 \times 33 / 100=3300
\end{aligned}
$$
(ii) Percentage of families which buy none of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ $=n(U)-n(A \cup B \cup C)$
$$
\begin{aligned}
&=n(U)-[n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C) \\
&-n(A \cap C)+n(A \cap B \cap C)] \\
&=100-[40+20+10-5-3-4+2] \\
&=(100-60) \%=40 \% \\
&\therefore \text { Number of families which buy none of } \mathrm{A}, \mathrm{B} \text { and } \mathrm{C} \\
&=\frac{40}{100} \times 10000=4000
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.