Let $C$ be the set of students who play cricket and $T$ be the set of students who play tennis.

Then, $$n(U)=60, n(C)=25, n(T)=20, \text { and } n(C \cap T)=10$$

$$\therefore \quad n(C \cup T)=n(C)+n(T)-n(C \cap T)$$

$=25+20-10=35$

$$\begin{aligned} \therefore \text { Number of students who play neither } & =n(U)-n(C \cup T) \\ & =60-35=25 \end{aligned}$$

Let $M$ be the set of students who study Mathematics, $P$ be the set of students who study Physics and $C$ be the set of students who study Chemistry.

Then,

$$\begin{aligned} n(U) & =200, n(M)=120, n(P)=90 \\ n(C) & =70, n(M \cap P)=40, n(P \cap C)=30 \\ n(C \cap M) & =50, n\left(M^{\prime} \cap P^{\prime} \cap C^{\prime}\right)=20 \end{aligned}$$

$$\begin{aligned} & & n(U)-n(M \cup P \cup C) & =20, \\ & & n(M \cup P \cup C) & =200-20=180 \\ \because & & n(M \cup P \cup C) & =n(M)+n(P)+n(C)-n(C \cap M)+n(M \cap P \cap C) \\ \Rightarrow & & 180 & =120+90+70-40-30-50+n(M \cap P \cap C) \\ \Rightarrow & & 180 & =160+n(M \cap P \cap C) \\ \Rightarrow & & n(M \cap P \cap C) & =180-160=20 \end{aligned}$$

So, the number of students who study all the three subjects is 20.

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 $C$

$$\begin{aligned} = & n(U)-n(A \cup B \cup C) \\ = & n(U)-[n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C) \\ = & -n(A \cap C)+n(A \cap B \cap C)] \\ = & 100-[40+20+10-5-3-4+2] \\ = & 100-60\%=40\%\\ = & 10000 \times \frac{40}{100}=4000 \end{aligned}$$

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 $=50-30=20$