Thinking Process

Solve the equation and get the value of $x$.

Solution

(i) We have,

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

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

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

$ \therefore \quad C=\lbrace1, p\rbrace $

Thinking Process

Solve the given equation and get the value of respective variable.

Solution

(i) We have,

$ \begin{matrix} \text { We have, } & D =\lbrace t \mid t^{3}=t, t \in R\rbrace \\ \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 =\lbrace-1,0,1\rbrace \\ \end{matrix} $

$ \begin{aligned} & \begin{matrix} \therefore & t^{3}=t \\ \Rightarrow & t^{3}-t=0 \Rightarrow t(t^{2}-1)=0 \end{matrix} \\ & \therefore \quad D=\lbrace-1,0,1\rbrace \end{aligned} $

(ii)

$ \begin{aligned} & \text { We have, } E =\lbrace w \lvert, \frac{w-2}{w+3}=3., w \in R\rbrace \\ & \therefore \quad \frac{w-2}{w+3}=3 \\ & \Rightarrow \quad w-2=3 w+9 \Rightarrow w-3 w=9+2 \\ & \Rightarrow \quad-2 w=11 \quad \Rightarrow \quad w=\frac{-11}{2} \\ & \therefore \quad E=\lbrace \frac{-11}{2} \rbrace \end{aligned} $

(iii) We have,

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

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

$ \Rightarrow x^{4}-3 x^{2}-2 x^{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 =\lbrace -\sqrt{3},-\sqrt{2}, \sqrt{2}, \sqrt{3}\rbrace $

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

Thinking Process

First, write all the factors of $2^{p-1}$, where $p=1,2,3, \ldots, p$ and then get $y$.

Solution

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

$ \therefore \quad Y=\lbrace1,2,2^{2}, 2^{3}, \ldots, 2^{p-1}, 2^{p}-1\rbrace $

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{aligned} \therefore \quad \text { Sum of factors } & =1+2+4+8+16+32+64+128 \\ & =255 \neq 2 \times 128 \end{aligned} $

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

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

$\therefore$ For $x=3$,

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

$\Rightarrow$ $ -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.

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

$ =992=2(496) $

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

Hence, statement (iv) is false.

Solution

Given, $\quad L=\lbrace1,2,3,4\rbrace, M=\lbrace3,4,5,6\rbrace$ and $N=\lbrace1,3,5\rbrace$

$\therefore \quad M \cup N=\lbrace1,3,4,5,6\rbrace$

$ L-(M \cup N)=\lbrace2\rbrace $

Now,

$ L-M=\lbrace1,2\rbrace, L-N=\lbrace2,4\rbrace $

$\therefore \quad(L-M) \cap(L-N)=\lbrace2\rbrace$

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

Solution

(i) Let $ x \in A $

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

Hence, $\subset A \cup B$

(ii) If

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

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 $\quad A \cup B = B$

Let $\quad y \in A$

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

$\Rightarrow A \subset B$

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=\lbrace1,2,3,4, \ldots, 100\rbrace $

(i) Required subset $=\lbrace2,4,6,8, \ldots, 100\rbrace$

(ii) Required subset $=\lbrace1,4,9,16,25,36,49,64,81,100\rbrace$

Solution

Given,

$ X=\lbrace1,2,3\rbrace $

(i) $\lbrace4 n \mid n \in X\rbrace=\lbrace4,8,12\rbrace$

(ii) $\lbrace n+6 \mid n \in X\rbrace=\lbrace7,8,9\rbrace$

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

(iv) $\lbrace n-1 \mid n \in X\rbrace=\lbrace0,1,2\rbrace$

Solution

Given,

$ Y=\lbrace 1,2,3, \ldots, 10\rbrace $

(i) $\lbrace a: a \in Y.$ and $.a^{2} \notin Y\rbrace=\lbrace4,5,6,7,8,9,10\rbrace$

(ii) $\lbrace a: a+1=6, a \in Y\rbrace=\lbrace5\rbrace$

(iii) is less than 6 and $a \in Y=\lbrace1,2,3,4,5\rbrace$,

Solution

Solution

Thinking Process

To prove this we have to show that $(A-B) \cap(A-C) \subseteq A-(B \cup C)$ and $A-(B \cup C) \subseteq(A-B) \cap(A-C)$

Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

From Eqs. (i) and (ii),

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

Thinking Process

To solve the above problem, use distributive law on sets

i.e., $\quad A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$.

Solution

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

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, $A-(B-C)\neq (A-B)-C$

Hence, given statement is false.

Solution

Let

$x \in A \cap C$

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

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

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

Hence, given statement is true.

Solution

Let

$x \in A \cup C$

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

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

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

Hence, given statement is true.

Solution

Let $x \in A \cup B$

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

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

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

Hence, given statement is true.

Thinking Process

To solve the above problem, use distributive law i.e., $A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$.

Solution

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

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

$ \begin{matrix} =(A \cup B) \cap(A \cup A^{\prime})=(A \cup B) \cap U & {[\because A \cup A^{\prime}=\cup]} \\ =A \cup B=R H S & {[\because A \cap U=A]} \end{matrix} $

Solution

$ \begin{aligned} 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)=\varphi \cup(A \cap B) \\ & =A \cap B=RHS \end{aligned} $

Solution

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

Solution

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

Thinking Process

First of all solve the given equation and get the value of $x$.

Solution

Since

$ \begin{aligned} T=\lbrace x \lvert, \frac{x+5}{x-7}-5. & =\frac{4 x-40}{13-x} \rbrace\\ \because \frac{x+5}{x-7}-5 & =\frac{4 x-40}{13-x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{x+5-5(x-7)}{x-7}=\frac{4 x-40}{13-x} \\ & \Rightarrow \quad \frac{x+5-5 x+35}{x-7}=\frac{4 x-40}{13-x} \\ & \Rightarrow \quad \frac{-4 x+40}{x-7}=\frac{4 x-40}{13-x} \\ & \Rightarrow \quad-(4 x-40)(13-x)=(4 x-40)(x-7) \\ & \Rightarrow \quad(4 x-40)(x-7)+(4 x-40)(13-x)=0 \\ & \Rightarrow \quad(4 x-40)(x-7+13-x)=0 \\ & \Rightarrow \quad 4(x-10) 6=0 \\ & \Rightarrow \quad 24(x-10)=0 \\ & \Rightarrow \quad x=10 \\ & \therefore \quad T=\lbrace10\rbrace \end{aligned} $

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{aligned} 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{aligned} $

$ \because \quad 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 $

$\Rightarrow \quad g=1$

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

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{aligned} 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{aligned} $

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

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

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

$ \text { i.e., } \quad 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., } \quad \begin{aligned} 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{aligned} $

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

$ \text { i.e., } \quad \begin{aligned} n(E \cap M)+n(M & \cap S)+n(E \cap S)-3 n(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{aligned} $

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

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

$ =60-35=25 $

Thinking Process

To solve this problem, use the formula for all the three subjects $n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(C \cap A)+n(A \cap B \cap C)$.

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{matrix} 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{matrix} $

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{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 \end{aligned} $

$ \because \quad n(F)=17 $

$ \begin{matrix} \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{matrix} $

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

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{aligned} 1+3+2+g & =15 \\ \Rightarrow g & =9 \end{aligned} $

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

(ii), we get

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

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

(i), we get

$ a+6+3+2=17 $

$ \Rightarrow \quad 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{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 $

Thinking Process

First find the total number of elements for the both sets, then compare them.

Solution

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

But each element is used 10 times, so

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

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

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

• Option (a) 15: This option is incorrect because if ( n = 15 ), then the total number of elements in ( B_1, B_2, \ldots, B_{15} ) would be ( 3 \times 15 = 45 ). Since each element is repeated 9 times, the total number of unique elements would be ( \frac{45}{9} = 5 ), which contradicts the given condition that ( S ) has 15 unique elements.

• Option (b) 3: This option is incorrect because if ( n = 3 ), then the total number of elements in ( B_1, B_2, \ldots, B_3 ) would be ( 3 \times 3 = 9 ). Since each element is repeated 9 times, the total number of unique elements would be ( \frac{9}{9} = 1 ), which contradicts the given condition that ( S ) has 15 unique elements.

• Option (d) 35: This option is incorrect because if ( n = 35 ), then the total number of elements in ( B_1, B_2, \ldots, B_{35} ) would be ( 3 \times 35 = 105 ). Since each element is repeated 9 times, the total number of unique elements would be ( \frac{105}{9} \approx 11.67 ), which is not an integer and contradicts the given condition that ( S ) has 15 unique elements.

Thinking Process

We know that, if a set A contains n elements, then the number of subsets of $A$ is equal to $2^{n}$.

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 $

• Option (b) 7,4: This option is incorrect because if ( m = 7 ) and ( n = 4 ), then the number of subsets of the first set would be ( 2^7 = 128 ) and the number of subsets of the second set would be ( 2^4 = 16 ). The difference between the number of subsets would be ( 128 - 16 = 112 ), which matches the given condition. Therefore, this option is actually correct, not incorrect.

• Option (c) 4,4: This option is incorrect because if ( m = 4 ) and ( n = 4 ), then the number of subsets of both sets would be ( 2^4 = 16 ). The difference between the number of subsets would be ( 16 - 16 = 0 ), which does not match the given condition of 112 more subsets.

• Option (d) 7,7: This option is incorrect because if ( m = 7 ) and ( n = 7 ), then the number of subsets of both sets would be ( 2^7 = 128 ). The difference between the number of subsets would be ( 128 - 128 = 0 ), which does not match the given condition of 112 more subsets.

Solution

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

$ \begin{aligned} \therefore \quad & =(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{aligned} $

• Option (a): $A^{\prime} \cup B \cup C$ is incorrect because the original expression $(A \cap B^{\prime})^{\prime} \cup (B \cap C)$ does not include the union with $C$ outside of the intersection with $B$. The correct simplification only involves $A^{\prime} \cup B$.

• Option (c): $A^{\prime} \cup C^{\prime}$ is incorrect because the original expression does not involve the complement of $C$. The correct simplification involves the union of $A^{\prime}$ and $B$, not $C^{\prime}$.

• Option (d): $A^{\prime} \cap B$ is incorrect because the original expression involves the union of $A^{\prime}$ and $B$, not their intersection. The correct simplification is $A^{\prime} \cup B$.

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

• (a) $F_2 \cap F_3$: This option is incorrect because the intersection of rectangles ($F_2$) and rhombuses ($F_3$) is the set of squares ($F_4$), not the set of all parallelograms ($F_1$). Not all parallelograms are squares.

• (b) $F_3 \cap F_4$: This option is incorrect because the intersection of rhombuses ($F_3$) and squares ($F_4$) is the set of squares ($F_4$), not the set of all parallelograms ($F_1$). Not all parallelograms are squares.

• (c) $F_2 \cup F_5$: This option is incorrect because the union of rectangles ($F_2$) and trapeziums ($F_5$) does not cover all parallelograms ($F_1$). Specifically, it misses out on rhombuses that are not rectangles and parallelograms that are neither rectangles nor trapeziums.

Solution

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

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

• (a) $S \cap T \cap C=\phi$: This option is incorrect because the intersection of the triangle and the circle is not necessarily empty. Since both the triangle and the circle are contained within the square, there can be points that lie within all three sets $S$, $T$, and $C$.

• (b) $S \cup T \cup C=C$: This option is incorrect because the union of the square, triangle, and circle is not necessarily equal to the circle. The square $S$ contains points that are outside the circle, and the triangle $T$ also contains points that are outside the circle. Therefore, $S \cup T \cup C$ is larger than just $C$.

• (d) $S \cup T=S \cap C$: This option is incorrect because the union of the square and the triangle is not necessarily equal to the intersection of the square and the circle. The union $S \cup T$ includes all points inside the square and the triangle, while the intersection $S \cap C$ includes only the points that are inside both the square and the circle. These two sets are not equivalent.

Solution

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

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

• Option (a) is incorrect because it includes the boundary points of the rectangle, i.e., it includes points where ( x = 0 ), ( x = a ), ( y = 0 ), and ( y = b ). The problem specifies points inside the rectangle, which excludes the boundary.

• Option (b) is incorrect because it includes the boundary points where ( y = 0 ) and ( y = b ). The problem specifies points inside the rectangle, which excludes the boundary.

• Option (c) is incorrect because it includes the boundary points where ( x = 0 ) and ( x = a ). The problem specifies points inside the rectangle, which excludes the boundary.

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 $=n(H^{\prime} \cap F^{\prime})$

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

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

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

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

$=840-550=290$

• Option (a) 210: This option is incorrect because the calculation for the number of persons who read neither Hindi nor English is based on the principle of inclusion-exclusion. The correct calculation shows that 290 persons read neither, not 210. The error in this option likely arises from a miscalculation or misunderstanding of the principle.

• Option (c) 180: This option is incorrect for the same reason as option (a). The principle of inclusion-exclusion correctly calculates the number of persons who read neither Hindi nor English as 290. The number 180 does not fit the correct calculation and is therefore incorrect.

• Option (d) 260: This option is also incorrect because, according to the principle of inclusion-exclusion, the number of persons who read neither Hindi nor English is 290. The number 260 is a result of an incorrect calculation or misunderstanding of the principle.

Thinking Process

If every element of $A$ is an elements of $B$, then $A \subseteq B$.

Solution

(a)

$ \begin{aligned} & X=\lbrace8^{n}-7 n-1 \mid n \in N\rbrace=\lbrace0,49,490, \ldots\rbrace \\ & Y=\lbrace49 n-49 \mid n \in N\rbrace=\lbrace0,49,98,147, \ldots\rbrace \end{aligned} $

Clearly, every elements of $X$ is in $Y$ but every element of $Y$ is not in $X$.

$ \therefore \quad X \subset Y $

• (b) $Y \subset X$: This option is incorrect because not every element of $Y$ is in $X$. For example, 98 is in $Y$ but not in $X$.

• (c) $X=Y$: This option is incorrect because the sets $X$ and $Y$ are not equal. While every element of $X$ is in $Y$, not every element of $Y$ is in $X$. For instance, 98 is in $Y$ but not in $X$.

• (d) $X \cap Y=\varphi$: This option is incorrect because the intersection of $X$ and $Y$ is not empty. For example, 0 and 49 are common elements in both $X$ and $Y$.

Solution

(c) Let $A$ be the set of percentage of those people who watch a news channel and $B$ be the set of percentage of those people who watch another channel.

$ n(A)=63, n(B) =76, \text { and } n(A \cap B)=x $

$ \because n(A \cup B) \leq 100 $

$ \Rightarrow n(A)+n(B)-n(A \cap B) \leq 100 $

$ \Rightarrow 63+76-x \leq 100 \Rightarrow 139-x \leq 100 $

$ \Rightarrow 139-100 \leq x \Rightarrow 39 \leq x $

$ \Rightarrow n(A) =63 $

$ \therefore x(A \cap B) \leq n(A) \Rightarrow x \leq 63 39 \leq x \leq 63 $

• Option (a) $x=35$: This is incorrect because the minimum value of $x$ is 39, as derived from the inequality $139 - x \leq 100 \Rightarrow x \geq 39$.

• Option (b) $x=63$: This is incorrect because while 63 is the maximum possible value for $x$, it is not the only possible value. The correct range for $x$ is $39 \leq x \leq 63$.

• Option (d) $x=39$: This is incorrect because while 39 is the minimum possible value for $x$, it is not the only possible value. The correct range for $x$ is $39 \leq x \leq 63$.

Solution

(c) Let $x \in R$

$ \begin{matrix} \text { We know that, } & -x \neq \frac{1}{x} \\ \therefore & A \cap B=\phi \end{matrix} $

• (a) $A \cap B \neq A$: The intersection of sets $A$ and $B$ cannot be equal to $A$ because not all elements of $A$ are in $B$. Specifically, $A$ consists of points where $y = \frac{1}{x}$, while $B$ consists of points where $y = -x$. These two conditions are not generally satisfied simultaneously, so $A \cap B$ cannot be $A$.

• (b) $A \cap B \neq B$: Similarly, the intersection of sets $A$ and $B$ cannot be equal to $B$ because not all elements of $B$ are in $A$. The points in $B$ satisfy $y = -x$, which is not the same as $y = \frac{1}{x}$ for any $x \in R$. Therefore, $A \cap B$ cannot be $B$.

• (d) $A \cup B \neq A$: The union of sets $A$ and $B$ cannot be equal to $A$ because $B$ contains points that are not in $A$. Specifically, $B$ includes points where $y = -x$, which are not included in $A$ where $y = \frac{1}{x}$. Therefore, $A \cup B$ cannot be $A$.

Solution

(a) $\because$ $ A \cap(A \cup B)=A $

• Option (b) $B$ is incorrect because the intersection of $A$ with the union of $A$ and $B$ will always include only the elements that are in $A$, not $B$.

• Option (c) $\phi$ is incorrect because the intersection of $A$ with the union of $A$ and $B$ will never be an empty set unless $A$ itself is an empty set, which is not specified in the problem.

• Option (d) $A \cap B$ is incorrect because the intersection of $A$ with the union of $A$ and $B$ will include all elements of $A$, not just the elements that are common to both $A$ and $B$.

Thinking Process

To solve this problem, use the distributive law i.e., $A \cap(B \cup C)=(A \cap B) \cup(A \cap C$.).

Solution

(b)

$ \begin{matrix} A^{\prime} \cup[(A \cup B) \cap B] \quad[\because A \cap(B \cup C)=(A \cap B) \cup(A \cap C)] \\ =A^{\prime} \cup[(A \cap B^{\prime}) \cup(B \cap B^{\prime})] \\ =A^{\prime} \cup[(A \cap B^{\prime}) \cup \phi]=A^{\prime} \cup(A \cap B^{\prime}) \\ =(A^{\prime} \cup A) \cap(A^{\prime} \cup B^{\prime}) \\ =A^{\prime} \cup B^{\prime}=N \cap(A^{\prime} \cup B^{\prime}) \\ =A^{\prime} \cup B^{\prime}=(A \cap B)^{\prime} & \\ =\phi=N & {[\because A \cap B=\phi]} \end{matrix} $

• Option (a) $\phi$: This option is incorrect because the expression $(A^{\prime} \cup(A \cup B) \cap B^{\prime})$ simplifies to the universal set $N$, not the empty set $\phi$. The empty set would imply that there are no elements in the resulting set, which is not the case here.

• Option (c) A: This option is incorrect because the expression $(A^{\prime} \cup(A \cup B) \cap B^{\prime})$ does not simplify to the set $A$. The set $A$ is a specific subset of natural numbers, whereas the expression simplifies to the universal set $N$.

• Option (d) B: This option is incorrect because the expression $(A^{\prime} \cup(A \cup B) \cap B^{\prime})$ does not simplify to the set $B$. The set $B$ is another specific subset of natural numbers, whereas the expression simplifies to the universal set $N$.

Solution

(d) $\because \quad S=\lbrace x \mid x$ is a positive multiple of 3 less than 100$\rbrace$

$ \therefore \quad n(S)=33 $

and $P=\lbrace x \mid x$ is a prime number less than 20$\rbrace$

$ \begin{aligned} n(P) & =8 \\ n(S)+n(P) & =33+8=41 \end{aligned} $

• Option (a) 34: This option is incorrect because it assumes that the total number of elements in sets ( S ) and ( P ) is 34. However, the correct total is 41, as ( n(S) = 33 ) and ( n(P) = 8 ), making ( n(S) + n(P) = 41 ).

• Option (b) 31: This option is incorrect because it assumes that the total number of elements in sets ( S ) and ( P ) is 31. However, the correct total is 41, as ( n(S) = 33 ) and ( n(P) = 8 ), making ( n(S) + n(P) = 41 ).

• Option (c) 33: This option is incorrect because it assumes that the total number of elements in sets ( S ) and ( P ) is 33. However, the correct total is 41, as ( n(S) = 33 ) and ( n(P) = 8 ), making ( n(S) + n(P) = 41 ).

Solution

(c)

$ \begin{aligned} X \cap(X \cup Y)^{\prime} & =X \cap(X^{\prime} \cap Y^{\prime}) \\ & =(X \cap X^{\prime}) \cap(X \cap Y^{\prime}) \\ & =\phi \cap(X \cap Y^{\prime})=\phi \end{aligned} $

$ [\because(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}] $

$[\because \phi \cap A=\phi]$

• Option (a) $X$: This option is incorrect because the expression $X \cap(X \cup Y)^{\prime}$ simplifies to $\phi$, not $X$. The complement of $X \cup Y$ is $X^{\prime} \cap Y^{\prime}$, and the intersection of $X$ with $X^{\prime} \cap Y^{\prime}$ results in the empty set $\phi$.

• Option (b) $Y$: This option is incorrect because the expression $X \cap(X \cup Y)^{\prime}$ does not simplify to $Y$. The intersection of $X$ with the complement of $X \cup Y$ results in the empty set $\phi$, not $Y$.

• Option (d) $X \cap Y$: This option is incorrect because the expression $X \cap(X \cup Y)^{\prime}$ simplifies to $\phi$, not $X \cap Y$. The complement of $X \cup Y$ is $X^{\prime} \cap Y^{\prime}$, and the intersection of $X$ with $X^{\prime} \cap Y^{\prime}$ results in the empty set $\phi$.

Solution

The set $\lbrace x \in R: 1 \leq x<2\rbrace$ can be written as $(1,2)$.

Solution

$\therefore$ $ \begin{aligned} A & =\phi \Rightarrow n(A)=0 \\ n\lbrace P(A)\rbrace & =2^{n(A)}=2^{0}=1 \end{aligned} $

So, number of element in $P(A)$ is 1 .

Solution

If $A$ and $B$ are two finite sets such that $A \subset B$, then $n(A \cup B)=n(B)$.

Solution

If $A$ and $B$ are any two sets, then $A-B=A \cap B^{`}$

Thinking Process

We know that, the power set is a collection of all the subset of a set. To solve this problem, write the all subset of the given set.

Solution

$\therefore A=\lbrace1,2\rbrace$

So, the subsets of $A$ are $\phi\lbrace1\rbrace,\lbrace2\rbrace$ and $\lbrace1,2\rbrace$.

$\therefore \quad P(A)=\lbrace\phi,\lbrace1\rbrace,\lbrace2\rbrace,\lbrace1,2\rbrace\rbrace$

Solution

Universal set for $A, B$ and $C$ is given by $U=\lbrace0,1,2,3,4,5,6,8\rbrace$

Solution

If

$U=\lbrace1,2,3,4,5, \ldots, 10\rbrace$,

$A=\lbrace1,2,3,5\rbrace, B=\lbrace2,4,6,7\rbrace$ and $C=\lbrace2,3,4,8\rbrace$

$\therefore \quad B \cup C=\lbrace2,3,4,6,7,8\rbrace$

(i) $(B \cup C)^{\prime}=U-(B \cup C)=\lbrace1,5,9,10\rbrace$

(ii) $C-A=\lbrace4,8\rbrace$

$\therefore \quad(C-A)^{\prime}=U-(C-A)=\lbrace1,2,3,5,6,7,9,10\rbrace$

Solution

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

Solution

(i) $[(A^{\prime} \cup B^{\prime})-A]^{\prime}=[(A^{\prime} \cup B^{\prime}) \cap A^{\prime}]^{\prime}$

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

$=[(A \cap B)^{\prime} \cap A^{\prime}]^{\prime}$ $[\because(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}]$

$=[(A \cap B)^{\prime}]^{\prime} \cup(A^{\prime})^{\prime}=(A \cap B) \cup A$

$=A$

(ii) $[B^{\prime} \cup(B^{\prime}-A)]^{\prime}=[B^{\prime} \cup(B^{\prime} \cap A^{\prime})]^{\prime}$

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

$=B^{\prime} \cup(B \cup A^{\prime}]^{\prime}.$

$[\because A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}]$

$=(B^{\prime})^{\prime} \cap[(B \cup A)^{\prime}]^{\prime}$ $[\because(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}]$

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

$[\because(A^{\prime})^{\prime}=A]$

$=B$

(iii) $(A-B)-(B-C)=(A \cap B^{\prime})-(B \cap C^{\prime})$

$=(A \cap B^{\prime}) \cap(B \cap C^{\prime})^{\prime}$

$=(A \cap B^{\prime}) \cap[B^{\prime} \cup(C^{\prime})^{\prime}]$

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

$=[A \cap(B^{\prime} \cup C)] \cap[B^{\prime} \cap(B^{\prime} \cup C)]$

$=[A \cap(B^{\prime} \cup C)] \cap B^{\prime}$

$=(A \cap B^{\prime}) \cap[(B^{\prime} \cup C) \cap B^{\prime}]$

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

Alternate Method

It is clear from the diagram, $(A-B)-(B-C)=A-B$.

(iv) $(A-B) \cap(C-B)$ $\Rightarrow$ $\Rightarrow$ $(A \cap B^{\prime}) \cap(C \cap B^{\prime})$ $[\because A-B=A \cap B^{\prime}]$ $\Rightarrow \quad(A \cap C)-B$ $[\because A \cap B^{\prime}=A-B]$

(v) $A \times B \cap C=(A \times B) \cap(A \times C)$

(vi) $A \times(B \cup C)=(A \times B) \cup(A \times C)$

Hence, the correct matches are

(i) $\rightarrow$ (b),

(ii) $\rightarrow$ (c),

(iii) $\rightarrow(a)$,

(iv) $\rightarrow$ (f),

(v) $\rightarrow(d)$,

(vi) $\rightarrow(e)$

Solution

True

Since, every set is the subset of itself.

Therefore, for any set $A, A \subset A$.

Solution

False

$ \begin{aligned} & M=\lbrace1,2,3,4,5,6,7,8,9\rbrace \\ & B=\lbrace1,2,3,4,5,6,7,8,9\rbrace \end{aligned} $

Since, every elements of $B$ is also in $M$

$ \therefore \quad B \subset M $

Solution

False

Solution

True

Since, every integer is also a rational number, then $Z \subset Q$

where, $Z$ is the set of integer and $Q$ is the set of rational number.

$\therefore$ $Q \cup Z=Q$

Solution

True

$ \begin{aligned} & R=\lbrace x \in Z \mid x \text { is divisible by } 2\rbrace=\lbrace\ldots-6,-4,-2,0,2,4,6, \ldots\rbrace \\ & T=\lbrace x \in Z \mid x \text { is divisible by } 6\rbrace=\lbrace\ldots,-12,-6,0,6,12, \ldots\rbrace \end{aligned} $

Thus, this every elements of $T$ is also in $R$

$\therefore \quad T \subset R$

Solution

False

$ \begin{aligned} A=\lbrace0,1,2\rbrace, \text { and } B & =\lbrace x \in R \mid 0 \leq x \leq 2\rbrace \\ n(A) & =3 \end{aligned} $

So, $A$ is finite. Since, there are infinite real numbers from 0 to 2 . So, $B$ is infinite.

$\therefore \quad A \neq B$