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 $C$.
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,
$\begin{aligned} n(U) & =10000, n(A)=40 \% n(B)=20 \% \text { and } n(C)=10 \% \\ n(A \cap B) & =5 \% \\ n(B \cap C) & =3 \% \\ n(A \cap C) & =4 \% \\ n(A \cap B \cap C) & =2 \%\end{aligned}$
(i) Number of families which buy newspaper $A$ only
$$\begin{aligned} & =n(A)-n(A \cap B)-n(A \cap C)+n(A \cap B \cap C) \\ & =(40-5-4+2) \%=33 \% \\ 10000 \times 33 / 100 & =3300 \end{aligned}$$
(ii) Number of families which buy none of $A, B$ and $C$
$$\begin{aligned} = & n(U)-n(A \cup B \cup C) \\ = & n(U)-[n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C) \\ = & -n(A \cap C)+n(A \cap B \cap C)] \\ = & 100-[40+20+10-5-3-4+2] \\ = & 100-60\%=40\%\\ = & 10000 \times \frac{40}{100}=4000 \end{aligned}$$
In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows French =17, English = 13, Sanskrit $=15$ French and English = 09, English and Sanskrit $=4$, French and Sanskrit = 5, English, French and Sanskrit $=3$. Find the number of students who study
(i) only French.
(ii) only English.
(iii) only Sanskrit.
(iv) English and Sanskrit but not French.
(v) French and Sanskrit but not English.
(vi) French and English but not Sanskrit.
(vii) atleast one of the three languages.
(viii) none of the three languages.
Let $F$ be the set of students who study French, $E$ be the set of students who study English and $S$ be the set of students who study Sanskrit.
Then,
$$\begin{aligned} n(U)=50, n(F) & =17, n(E)=13, \text { and } n(S)=15 \\ n(F \cap E) & =9, n(E \cap S)=4, n(F \cap S)=5 \\ n(F \cap E \cap S) & =3 \\ \because \quad n(F) & =17 \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad a+b+e+f & =17 \quad \text{.... (i)}\\ n(E) & =13 \\ \Rightarrow \quad b+c+d+e & =13 \quad \text{.... (ii)}\\ n(S) & =15 \\ \Rightarrow \quad d+e+f+g & =15 \quad \text{.... (iii)}\\ n(F \cap E) & =9 \\ \Rightarrow \quad b+e & =9 \quad \text{.... (iv)}\\ n(E \cap S) & =4 \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad e+d & =4 \quad \text{... (v)}\\ n(F \cap S) & =5 \\ \Rightarrow \quad f+e & =5 \quad \text{.... (vi)}\\ n(F \cap E \cap S) & =3 \\ \Rightarrow \quad e & =3\quad \text{.... (vii)} \end{aligned}$$
From Eqs. (vi) and (vii), $f=2$
From Eqs. (v) and (vii), $\quad d=1$
From Eqs. (iv) and (vii), $b=6$
On substituting the values of e, $f$ and $d$ in Eq. (iii), we get
$$\begin{aligned} 1+3+2+g & =15 \\ g & =9 \end{aligned}$$
On substituting the values of $b, d$ ande in Eq. (ii), we get
$$\begin{aligned} 6+c+1+3 & =13 \\ c & =3 \end{aligned}$$
On substituting the values of $b, e$ and $f$ in Eq. (i), we get
$$\begin{aligned} a+6+3+2 & =17 \\ a & =6 \end{aligned}$$
(i) Number of students who study French only, $a=6$
(ii) Number of students who study English only, $c=3$
(iii) Number of students who study Sanskrit only, $g=9$
(iv) Number of students who study English and Sanskrit but not French, $d=1$
(v) Number of students who study French and Sanskrit but not English, $f=2$
(vi) Number of students who study French and English but not Sanskrit, $b=6$
(vii) Number of students who study atleast one of the three languages
$\begin{aligned} & =a+b+c+d+e+f+g \\ & =6+6+3+1+3+2+9=30\end{aligned}$
(viii) Number of students who study none of three languages $=$ Total students $-$ Students who study atleast one of the three languages $=50-30=20$
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