Solution

(i) We have,

$\begin{array}{rl}A& =\{x:x\in R,2x+11=15\}\\ \therefore 2x+11& =15\\ 2x& =15-11\Rightarrow 2x=4\\ x& =2\\ A& =\{2\}\\ \text{(ii) We have},B& =\{x\mid x^2=x,x\in R\}\\ x^2& =x\\ \Rightarrow x^2-x& =0\Rightarrow x(x-1)=0\\ \Rightarrow x& =0,1\\ \Rightarrow B& =\{0,1\}\end{array}$

(iii) We have, $C=\{x\mid x$ is a positive factor of prime number $p\}$.

Since, positive factors of a prime number are 1 and the number itself.

$\therefore C=\{1,p\}$

Solution

(i) We have,

$\begin{array}{cc}\text{ We have, }& D=\{t\mid t^3=t,t\in R\}\\ \therefore & t^3=t\\ \Rightarrow & t^3-t=0\Rightarrow t(t^2-1)=0\\ \Rightarrow & t(t-1)(t+1)=0\Rightarrow t=0,1,-1\\ \therefore & D=\{-1,0,1\}\end{array}$

$\begin{array}{rl}& \begin{array}{cc}\therefore & t^3=t\\ \Rightarrow & t^3-t=0\Rightarrow t(t^2-1)=0\end{array}\\ & \therefore D=\{-1,0,1\}\end{array}$

(ii)

$\begin{array}{rl}& \text{ We have, }E=\{w|,\frac{w-2}{w+3}=3.,w\in R\}\\ & \therefore \frac{w-2}{w+3}=3\\ & \Rightarrow w-2=3w+9\Rightarrow w-3w=9+2\\ & \Rightarrow -2w=11\Rightarrow w=\frac{-11}{2}\\ & \therefore E=\{\frac{-11}{2}\}\end{array}$

(iii) We have,

$F=\{x\mid x^4-5x^2+6=0,x\in R\}$

$\therefore x^4-5x^2+6=0$

$\Rightarrow x^4-3x^2-2x^2+6=0$

$\Rightarrow x^2(x^2-3)-2(x^2-3)=0$

$\Rightarrow (x^2-3)(x^2-2)=0$

$\Rightarrow x=\pm \sqrt{3},\pm \sqrt{2}$

$\therefore F=\{-\sqrt{3},-\sqrt{2},\sqrt{2},\sqrt{3}\}$

Note in roaster form, the order in which elements are listed is immaterial. Thus, we can also write $F=\{-\sqrt{3},\sqrt{2},-\sqrt{2},\sqrt{3}\}$.

Solution

$Y=\{x\mid x.$ is a positive factor of the number $2^{p-1}(2^p-1)$, where $2^p-1$ is a prime number $\}$. So, the factor of $2^{p-1}$ are $1,2,2^2,2^3,\dots ,2^{p-1}$.

$\therefore Y=\{1,2,2^2,2^3,\dots ,2^{p-1},2^p-1\}$

Solution

(i) Since, the factors of 35 are $1,5,7$ and 35 . So, statement (i) is true.

(ii) Since, the factors of 128 are 1, 2, 4, 8, 16, 32, 64 and 128.

$\begin{array}{rl}\therefore \text{ Sum of factors }& =1+2+4+8+16+32+64+128\\ & =255\ne 2\times 128\end{array}$

So, statement (ii) is false. (iii) We have,

$x^4-5x^3+2x^2-112x+6=0$

$\therefore$ For $x=3$,

$\Rightarrow 81-135+18-336+6=0$

$\Rightarrow$ $-346=0$

which is not true.

Hence, statement (iii) is true.

(iv) $\because$ $496=2^4\times 31$

So, the factors of 496 are $1,2,4,8,16,31,62,124,248$ and 496.

$\therefore$ Sum of factors $=1+2+4+8+16+31+62+124+248+496$

$=992=2(496)$

So, $496\in \{y\mid$ the sum of all the positive factor of $y$ is $2y\}$.

Hence, statement (iv) is false.

Solution

Given, $L=\{1,2,3,4\},M=\{3,4,5,6\}$ and $N=\{1,3,5\}$

$\therefore M\cup N=\{1,3,4,5,6\}$

$L-(M\cup N)=\{2\}$

Now,

$L-M=\{1,2\},L-N=\{2,4\}$

$\therefore (L-M)\cap (L-N)=\{2\}$

Hence, $L-(M\cup N)=(L-M)\cap (L-N)$.

Solution

(i) Let $x\in A$

$\Rightarrow x\in A$ or $x\in B\Rightarrow x\in A\cup B$

Hence, $\subset A\cup B$

(ii) If

$\begin{array}{c}A\subset B\\ x\in A\cup B\end{array}$

Let

$\Rightarrow$ $x\in A\text{ or }x\in B\Rightarrow x\in B$

$\Rightarrow$ $A\cup B\subset B$

But $B\subset A\cup B$

From Eqs. (i) and (ii),

$\Rightarrow A\cup B=B$

If $A\cup B=B$

Let $y\in A$

$\Rightarrow y\in A\cup B\Rightarrow y\in B$

$\Rightarrow A\subset B$

Hence, $A\subset B\Rightarrow A\cup B=B$

(iii) Let $\Rightarrow x\in A\cap B$

$\Rightarrow x\in A\text{and}x\in B\Rightarrow x\in A$

Hence, $A\cap B\subset A$

Solution

We have,

$N=\{1,2,3,4,\dots ,100\}$

(i) Required subset $=\{2,4,6,8,\dots ,100\}$

(ii) Required subset $=\{1,4,9,16,25,36,49,64,81,100\}$

Solution

Given,

$X=\{1,2,3\}$

(i) $\{4n\mid n\in X\}=\{4,8,12\}$

(ii) $\{n+6\mid n\in X\}=\{7,8,9\}$

(iii) $\frac{n}{2}|,n\in X=\frac{1}{2}.,1,\frac{3}{2}$

(iv) $\{n-1\mid n\in X\}=\{0,1,2\}$

Solution

Given,

$Y=\{1,2,3,\dots ,10\}$

(i) $\{a:a\in Y.$ and $.a^2\notin Y\}=\{4,5,6,7,8,9,10\}$

(ii) $\{a:a+1=6,a\in Y\}=\{5\}$

(iii) is less than 6 and $a\in Y=\{1,2,3,4,5\}$,

Solution

Solution

Solution

Let $x\in (A-B)\cap (A-C)$

$\Rightarrow x\in (A-B)$ and $x\in (A-C)$

$\Rightarrow (x\in A$ and $x\notin B)$ and $(x\in A$ and $x\notin C)$

$\Rightarrow x\in A$ and $(x\notin B$ and $x\notin C)$

$\Rightarrow x\in A$ and $x\notin (B\cup C)$

$\Rightarrow x\in A-(B\cup C)$

$\Rightarrow (A-B)\cap (A-C)\subset A-(B\cup C)$

Now, let $y\in A-(B\cup C)$

$\Rightarrow$ $y\in A$ and $y\notin (B\cup C)$

$\Rightarrow y\in A$ and $(y\notin B$ and $y\notin C)$

$\Rightarrow (y\in A$ and $y\notin B)$ and $(y\in A$ and $y\notin C)$

$\Rightarrow y\in (A-B)$ and $y\in (A-C)$

$\Rightarrow y\in (A-B)\cap (A-C)$

$\Rightarrow A-(B\cup C)\subset (A-B)\cap (A-C)$

From Eqs. (i) and (ii),

$A-(B\cup C)=(A-B)\cap (A-C)$

Solution

$\begin{array}{rl}LHS& =(A-B)\cup (A\cap B)\\ & =[(A-B)\cup A]\cap [(A-B)\cup B]\\ & =A\cap (A\cup B)=A=RHS\end{array}$

Hence, given statement is true.

Solution

See the Venn diagrams given below, where shaded portions are representing $A-(B-C)$ and $(A-B)-C$ respectively.

Clearly, $A-(B-C)\ne (A-B)-C$

Hence, given statement is false.

Solution

Let

$x\in A\cap C$

$\Rightarrow x\in A$ and $x\in C$

$\Rightarrow x\in B$ and $x\in C[\because A\subset B]$

$\Rightarrow x\in (B\cap C)\Rightarrow (A\cap C)\subset (B\cap C)$

Hence, given statement is true.

Solution

Let

$x\in A\cup C$

$\Rightarrow$ $x\in A$ and $x\in C$

$\Rightarrow x\in B$ and $x\in C[\because A\subset B]$

$\Rightarrow x\in B\cup C\Rightarrow A\cup C\subset B\cup C$

Hence, given statement is true.

Solution

Let $x\in A\cup B$

$\Rightarrow$ $x\in A$ and $x\in B$

$\Rightarrow x\in C$ and $x\in C$ $[\because A\subset C$ and $B\subset C]$

$\Rightarrow$ $x\in C\Rightarrow A\cup B\subset C$

Hence, given statement is true.

Solution

$LHS=A\cup (B-A)=A\cup (B\cap A^{\prime })$

$[\because A-B=A\cap B^{\prime }]$

$\begin{array}{cc}=(A\cup B)\cap (A\cup A^{\prime })=(A\cup B)\cap U& [\because A\cup A^{\prime }=\cup ]\\ =A\cup B=RHS& [\because A\cap U=A]\end{array}$

Solution

$\begin{array}{rl}LHS& =A-(A-B)=A-(A\cap B^{\prime })\\ & =A\cap (A\cap B^{\prime }{)}^{\prime }=A\cap [A^{\prime }\cup (B^{\prime }{)}^{\prime }]\\ & =A\cap (A^{\prime }\cup B)\\ & =(A\cap A^{\prime })\cup (A\cap B)=\phi \cup (A\cap B)\\ & =A\cap B=RHS\end{array}$

Solution

$\begin{array}{rl}LHS& =A-(A\cap B)=A\cap (A\cap B{)}^{\prime }[\because A-B=A\cap B^{\prime }]\\ & =A\cap (A^{\prime }\cup B^{\prime })[\because (A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }]\\ & =(A\cap A^{\prime })\cup (A\cap B^{\prime })=\varphi \cup (A\cap B^{\prime })[\because (A^{\prime }{)}^{\prime }=A]\\ & =A\cap B^{\prime }\\ & =A-B=RHS\end{array}$

Solution

$\begin{array}{rl}LHS& =(A\cup B)-B=(A\cup B)\cap B^{\prime }[\because A-B=A\cap B^{\prime }]\\ & =(A\cap B^{\prime })\cup (B\cap B^{\prime })=(A\cap B^{\prime })\cup \varphi [\because B\cap B^{\prime }=\varphi ]\\ & =A\cap B^{\prime }[\because A\cup \varphi =A]\\ & =A-B=RHS\end{array}$

Solution

Since

$\begin{array}{rl}T=\{x|,\frac{x+5}{x-7}-5.& =\frac{4x-40}{13-x}\}\\ \because \frac{x+5}{x-7}-5& =\frac{4x-40}{13-x}\end{array}$

$\begin{array}{rl}& \Rightarrow \frac{x+5-5(x-7)}{x-7}=\frac{4x-40}{13-x}\\ & \Rightarrow \frac{x+5-5x+35}{x-7}=\frac{4x-40}{13-x}\\ & \Rightarrow \frac{-4x+40}{x-7}=\frac{4x-40}{13-x}\\ & \Rightarrow -(4x-40)(13-x)=(4x-40)(x-7)\\ & \Rightarrow (4x-40)(x-7)+(4x-40)(13-x)=0\\ & \Rightarrow (4x-40)(x-7+13-x)=0\\ & \Rightarrow 4(x-10)6=0\\ & \Rightarrow 24(x-10)=0\\ & \Rightarrow x=10\\ & \therefore T=\{10\}\end{array}$

Hence, $T$ is not an empty set.

Solution

Let

$\Rightarrow x\in A\cap (B\cup C)$

$\Rightarrow x\in A$ and $x\in (B\cup C)$

$\Rightarrow x\in A$ and $(x\in B$ or $x\in C)$

$\Rightarrow (x\in A$ and $x\in B)$ or $(x\in A$ and $x\in C)$

$\Rightarrow x\in A\cap B$ or $x\in A\cap C$

$x\in (A\cap B)\cup (A\cap C)$

$\Rightarrow A\cap (B\cup C)\subset (A\cap B)\cup (A\cap C)$

Again, let

$\Rightarrow y\in (A\cap B)\cup (A\cap C)$

$\Rightarrow y\in (A\cap B)$ or $y\in (A\cap C)$

$\Rightarrow (y\in A$ and $y\in B)$ or $(y\in A$ and $y\in C)$

$\Rightarrow y\in A$ and $(y\in B$ or $y\in C)$

$\Rightarrow y\in A$ and $y\in B\cup C$

$\Rightarrow y\in A\cap (B\cup C)$

$\Rightarrow (A\cap B)\cup (A\cap C)\subset A\cap (B\cup C)$

From Eqs. (i) and (ii)

$A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

Solution

Let $M$ be the set of students who passed in Mathematics, $E$ be the set of students who passed in English and $S$ be the set of students who passed in Science.

Then, $n(u)=100$

$\begin{array}{rl}n(E)=15,n(M)& =12,n(S)=8,n(E\cap M)=6,n(M\cap S)=7\\ n(E\cap S)& =4,\text{ and }n(E\cap M\cap S)=4\end{array}$

$\because n(E)=15$

$\Rightarrow a+b+e+f=15$

and $n(M)=12$

$\Rightarrow b+c+e+d=12$

On substituting the values of $d,e$ and $f$ in Eq. (iii), we get

$3+4+0+g=8$

$\Rightarrow g=1$

On substituting the value of $b,e$ and $d$ in Eq. (ii), we get

$2+c+4+3=12$

$\Rightarrow c=13$

On substituting $b,e$, and $f$ in Eq. (i), we get

$a+2+4+0=15$

$\Rightarrow$ $a=9$

(i) Number of students who passed in English and Mathematics but not in Science

$=b=2$

(ii) Number of students who passed in Mathematics and Science but not in English

$=d=3$

(iii) Number of students who passed in Mathematics only $=c=3$

(iv) Number of students who passed in more than one subject

$\begin{array}{rl}& =b+e+d+f\\ & =2+4+3+0=9\end{array}$

Alternate Method

Let $E$ denotes the set of student who passed in English. $M$ denotes the set of students who passed in Mathematics. $S$ denotes the set of students who passed in Science.

Now,

$\begin{array}{rl}n(U)& =100,n(E)=15,n(m)=12,n(S)=8\\ n(E\cap M)& =6,n(M\cap S)=7\\ n(E\cap S)& =4,n(E\cap M\cap S)=4\end{array}$

(i) Number of students passed in English and Mathematics but not in Science

$\text{ i.e., }\begin{array}{rl}n(E\cap M\cap S^{\prime })& =n(E\cap M)-n(E\cap M\cap S)[\because A\cap B^{\prime }=A-(A\cap B)]\\ & =6-4=2\end{array}$

(ii) Number of students passed in Mathematics and Science but not in English.

$\text{ i.e., }n(M\cap S\cap E^{\prime })=n(M\cap S)-n(M\cap S\cap E)$

$=7-4=3$

(iii) Number of students passed in mathematics only

$\text{ i.e., }\begin{array}{rl}n(M\cap S^{\prime }\cap E^{\prime })& =n(M)-n(M\cap S)-n(M\cap E)+n(M\cap S\cap E)\\ & =12-7-6+4=3\end{array}$

(iv) Number of students passed in more than one subject only

$\text{ i.e., }\begin{array}{rl}n(E\cap M)+n(M& \cap S)+n(E\cap S)-3n(E\cap M\cap S)+n(E\cap M\cap S)\\ & =6+7+4-4\times 3+4\\ & =17-12+4=5+4=9\end{array}$

Solution

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

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

$\therefore$ $\begin{array}{rl}n(C\cup T)& =n(C)+n(T)-n(C\cap T)\\ & =25+20-10=35\end{array}$

$\therefore$ Number of students who play neither $=n(U)-n(C\cup T)$

$=60-35=25$

Solution

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, $n(U)=200,n(M)=120,n(P)=90$

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

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

Solution

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,

$n(U)=10000,n(A)=40$% n(B)=20% and n(C)=10%

$n(A\cap B)=5$ %

$n(B\cap C)=3$%

$n(A\cap C)=4$ %

$n(A\cap B\cap C)=2$%

(i) Number 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 %

$10000\times 33/100=3300$

(ii) Number of families which buy none of $A,B$ and $C$

$=n(U)-n(A\cup B\cup C)$

$=n(U)-[n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)$

=100-[40+20+10-5-3-4+2] =100-60 %=40 %

$=10000\times \frac{40}{100}=4000$

Solution

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,

$\begin{array}{rl}n(U)=50,n(F)& =17,n(E)=13,\text{ and }n(S)=15,\\ n(F\cap E)& =9,n(E\cap S)=4,n(F\cap S)=5,\\ n(F\cap E\cap S)& =3\end{array}$

$\because n(F)=17$

$\begin{array}{cc}\Rightarrow & a+b+e+f=17\\ \Rightarrow & n(E)=13\\ \Rightarrow & b+c+d+e=13\\ \Rightarrow & n(S)=15\\ \Rightarrow & d+e+f+g=15\\ \Rightarrow & n(F\cap E)=9\\ & b+e=9\\ & n(E\cap S)=4\end{array}$

$\begin{array}{rl}\Rightarrow & e+d=4\\ \Rightarrow & n(F\cap S)=5\\ \Rightarrow & n(F\cap E\cap S)=3\\ \Rightarrow & e=3\end{array}$

From Eqs. (vi) and (vii), $f=2$

From Eqs. (v)and (vii), $d=1$

From Eqs. (iv) and (vii), $b=6$

On substituting the values of $e,f$ and $d$ in Eq. (iii), we get

$\begin{array}{rl}1+3+2+g& =15\\ \Rightarrow g& =9\end{array}$

On substituting the values of $b,d$ ande in E

(ii), we get

$\begin{array}{rl}6+c+1+3& =13\\ c& =3\end{array}$

On substituting the values of $b,e$ and $f$ in E

(i), we get

$a+6+3+2=17$

$\Rightarrow a=6$

(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

$\begin{array}{rl}& =a+b+c+d+e+f+g\\ & =6+6+3+1+3+2+9=30\end{array}$

(viii) Number of students who study none of three languages $=$ Total students - Students who study atleast one of the three languages

$=50-30=20$

Solution

(c) If elements are not repeated, then number of elements in $A_1\cup A_2\cup A_3,\dots \cup A_{30}$ is $30\times 5$.

But each element is used 10 times, so

$S=\frac{30\times 5}{10}=15$

If elements in $B_1,B_2,\dots ,B_n$ are not repeated, then total number of elements is $3n$ but each element is repeated 9 times, so

$\begin{array}{rl}& S=\frac{3n}{9}\Rightarrow 15=\frac{3n}{9}\\ \therefore & n=45\end{array}$

Solution

(a) Since, number of subsets of a set containing melements is 112 more than the subsets of the set containing $n$ elements.

$\because 2^m-2^n=112$

$\Rightarrow 2^n\cdot (2^{m-n}-1)=2^4\cdot 7$

$\Rightarrow 2^n=2^4\text{ and }{2}^{m-n}-1=7$

$\Rightarrow n=4\text{ and }{2}^{m-n}=8$

$\Rightarrow 2^{m-n}=2^3\Rightarrow m-n=3$

$\Rightarrow m-4=3\Rightarrow m=4+3$

$\therefore m=7$

Solution

(b) We know that, $(A\cap B{)}^{\prime }=(A^{\prime }\cup B^{\prime })$ and $(A^{\prime }{)}^{\prime }=A$

$\begin{array}{rl}\therefore & =(A\cap B^{\prime }{)}^{\prime }\cup (B\cap C)\\ & =[A^{\prime }\cup (B^{\prime }{)}^{\prime }]\cup (B\cap C)\\ & =(A^{\prime }\cup B)\cup (B\cap C)=A^{\prime }\cup B\end{array}$

Solution

(d) Every rectangle, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram.

So, $F_1$ is either of $F_1,F_2,F_3$ and $F_4$.

$\therefore F_1=F_2\cup F_3\cup F_4\cup F_1$

Solution

(c) The given sets can be represented in Venn diagram as shown below

It is clear from the diagram that, $S\cup T\cup C=S$.

Solution

(d) Since, $R$ be the set of points inside the rectangle.

$\therefore R=\{(x,y):0<x<a$ and $0<y<b\}$

Solution

(b) Let $H$ be the set of persons who read Hindi and $E$ be the set of persons who read English.

Then, $n(U)=840,n(H)=450,n(E)=300,n(H\cap E)=200$

Number of persons who read neither $=n(H^{\prime }\cap F^{\prime })$

$=n(H\cup E{)}^{\prime }$

$=n(U)-n(H\cup E)$

$=840-[n(H)+n(E)-n(H\cap E)]$

$=840-(450+300-200)$

$=840-550=290$

Solution

(a)

$\begin{array}{rl}& X=\{8^n-7n-1\mid n\in N\}=\{0,49,490,\dots \}\\ & Y=\{49n-49\mid n\in N\}=\{0,49,98,147,\dots \}\end{array}$