Q: If \$A = \{x : x\$ is a natural number \$\}, B = \{x : x\$ is an even natural number\$\} C = \{x : x\$ is an odd natural number\$\}\$ and \$D = \{x : x\$ is a prime number \$\},\$ find(i) \$A \cap B\$ (ii) \$A \cap C\$ (iii) \$A \cap D\$ (iv) \$B \cap C\$ (v) \$B \cap D\$ (vi) \$C \cap D\$

\$A = \{x:x \text {is a natural number}\}=\{1,2,3,4,5.....\}\$\$B = \{x:x \text{ is an even natural number}\}=\{2,4,6,8 .....\}\$\$C = \{x:x \text{ is an odd natural number}\}=\{1,3,5,7,9 .......\}\$\$D = \{x:x \text {is a prime number}\}=\{2,3,5,7 ....\}\$(i) \$A\cap B = \{x:x \text{ is a even natural number}\} = B\$(ii) \$A\cap C =\{x:x \text { is an odd natural number}\} = C\$(iii) \$A\cap D= \{x:x \text { is a prime number}\}=D\$(iv) \$B\cap C=\phi\$(v) \$B\cap D=\left \{2\right \}\$(vi) \$C\cap D= \{x:x \text {is odd prime number}\}\$

Q: If the universal set \$U =\{1, 2, 3, 4, 5, 6, 7 \}, A = \{ 1, 2, 5, 7 \} , B = \{3, 4, 5, 6\}\$. Verify \$\left( A\cup B \right) '=A'\cap B'\$

Given, universal set, \$U=\{1, 2, 3, 4, 5, 6, 7\}\$\$A=\{1, 2, 5, 7\}\$\$B=\{3, 4, 5, 6\}\$\$(A\cup B)'=U-(A\cup B)\$\$=\{1, 2, 3, 4, 5, 6, 7\}-\{1, 2, 3, 4, 5, 6, 7\}\$\$=\phi\$\$A'\cap B'=(U-A)n(U-B)\$\$=\{3, 4, 6\}n\{1, 2, 7\}\$\$=\phi\$Hence \$(A\cup B)'=A'\cap B'\$.

Q: If \$A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6, 8\}\$ and C= \$\{3,4,5,6\}\$, then verify : \$A - (B \cup C) = (A - B) \cap (A - C)\$.

Given, \$A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6, 8\}\$ and \$C=\{3,4,5,6\}\$For the LHS:Union of two sets will have the elements of both sets.So, \$ B \cup C = \{2,3,4,5,6,8 \}\$ \$ A - (B \cup C) \$ will have elements of \$A\$ which are not in \$ (B \cup C) \$So, \$ A - (B \cup C) = \{ 1 \}\$ ..... \$(1)\$For the RHS:\$ A - B \$ will have elements of \$A\$ which are not in \$B\$.So, \$ A - B = \{ 1,3,5 \}\$ \$ A - C \$ will have elements of \$A\$ which are not in \$C\$.So, \$ A - C = \{ 1,2 \}\$ Intersection of two sets has the common elements of both the sets. \$\Rightarrow (A - B) \cap (A - C) = \{1\}\$ ..... \$(2)\$From \$(1)\$ and \$(2),\$ we have\$ A - (B \cup C) =(A - B) \cap (A - C) \$Hence, the given expression is true.

Q: If \$A\$ and \$B\$ are two sets, then \$A\cap (A\cup B)\$ equals

\$A\cap(A\cup B)\$\$=(A\cap A) \cup (A\cap B)\$ \$=A \$Hence answer is Option A.

Q: Write power sets of the following \$\left \{ 1, 2, 3 \right \}\$

\$\left \{ 1, 2, 3 \right \} \$Subsets of \$ \left \{ 1, 2, 3 \right \} \$ are \$\phi ,\left \{ 1 \right \}, \left \{ 2 \right \}, \left \{ 3 \right \}, \left \{ 1, 2 \right \} , \left \{ 1, 3 \right \} , \left \{ 2, 3\right \} \$ and \$\left \{ 1, 2, 3 \right \}\$Hence , \$P \left \{ 1 , 2, 3 \right \} =\{ \phi , \{1\} , \{2\}, \{3\} , \{1,2\} , \{1,3\} , \{2,3\} , \{ 1, 2, 3 \} \}\$

Q: Let \$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\$. Then the number of subsets of \$A\$ containing exactly two elements is

Number of elements in \$A= 10\$ Number of subsets of \$A\$ containing exactly two elements \$=\$ Number of ways we can select \$ 2 \$ elements from \$10\$ elements \${}^{ 10 }{ C }_{2}=\dfrac { 10\times 9 }{ 2 } =45\$\$\therefore\$ Number of subsets of \$A\$ containing exactly two elements \$=45\$

Question 1

Let $A,B$ and $C$ be the sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C.$ Show that $B=C$

Accepted Answer

${\textbf{Step-1:Prove using suitable formula of sets.}}$&#10;&#10;                $\text{It is given that, }$                ${\text{A}} \cup {\text{B = A}} \cup {\text{C}}$         ${\text{&#8230;(1)}}$                $\text{and, }$&#10;&#10;                ${\text{A}} \cap {\text{B = A}} \cap {\text{C}}$          ${\text{&#8230;(2)}}$   &#10;&#10;                ${\text{Taking}}$  $\text{'}\cap {\text{ C'}}$ ${\text{on both sides in equation (1)}}$&#10;&#10;                $\left( {{\mathrm{A\cup B}}} \right) \cap {\text{C =&#10;}}\left( {{\mathrm{A\cup C}}} \right) \cap {\text{C}}$&#10;&#10;                ${\text{We know that,}}$&#10;&#10;                $\left( {{\mathrm{A\cup B}}} \right) \cap {\text{C =&#10;}}\left( {{\text{A}} \cap {\text{C}}} \right)\cup \left( {{\text{B}} \cap&#10;{\text{C}}} \right)$ ${\text{and}}$ $\left(&#10;{{\mathrm{A\cup C}}} \right) \cap {\text{C = C}}$&#10;&#10;                ${\text{So,}}$&#10;&#10;                $\left( {{\text{A}} \cap {\text{C}}}&#10;\right)\cup \left( {{\text{B}} \cap {\text{C}}} \right) = C$&#10;&#10;                $\left( {{\text{A}} \cap \text{B}} \right)\cup \left(&#10;{{\text{B}} \cap {\text{C}}} \right) = C$            ${\text{&#8230;(3)}}$                       ${\textbf{[From (2)]}} $              &#10;&#10;                 ${\text{Again,}}$                 ${\text{Taking}}$  $\text{'}\cap{\text{ B'}}$ ${\text{on both side in equation&#10;(1)}}$&#10;&#10;                 $\left(&#10;{{\mathrm{A\cup B}}} \right) \cap B{\text{ = }}\left( {{\mathrm{A\cup C}}} \right) \cap B$&#10;&#10;                 ${\text{B&#10;= }}\left( {{\text{A}} \cap {\text{B}}} \right)\cup\left( {{\text{C}}&#10;\cap {\text{B}}} \right)$&#10;&#10;                 ${\text{B = }}\left( {{\text{A}} \cap {\text{B}}}&#10;\right)\cup\left( {{\text{B}} \cap {\text{C}}} \right)$&#10;&#10;                 ${\text{B&#10;= C}}$                        ${\textbf{[From (3)}]}$&#10;&#10;${\textbf{Hence, proved.}}$&#10;&#10;

Question 2

In a survey of 200 students of a school it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. The number of student who study all the three subject is

Accepted Answer

Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd $U$ represents total studentsSo, $n(U) = 200$,$n(M) = 120, n(P) = 90$$n(C)= 70, n(M \cap P) = 40, n(P\cap C)=30$$n(M \cap C)=50, n(M\cup P \cup C)' = 20$$\therefore n(M \cup P \cup C)' = n(U) - n(M\cup P \cup C)$$\Rightarrow 20=200-n(M\cup P \cup C)$$\Rightarrow n(M \cup P \cup C) = 180$$\Rightarrow n(M \cup P \cup C) = n(M) + n(P) + n(C)$$-n(M\cap P)-n(P\cap C) - n(C\cap M)+n(C\cap M \cap P)$$\therefore 180 = 120 + 90+70-40-30-50+n(C\cap M \cap P)$$\Rightarrow 180=280-120+n(C\cap M \cap P)$$\Rightarrow n(P \cap C \cap M) = 300- 280 = 20$Hence, the number of students studying all three subjects is $20$.

Question 3

If $A = \{x : x$ is a natural number $\}, B = \{x : x$ is an even natural number$\} C = \{x : x$ is an odd natural number$\}$ and $D = \{x : x$ is a prime number $\},$ find(i) $A \cap B$ (ii) $A \cap C$ (iii) $A \cap D$ (iv) $B \cap C$ (v) $B \cap D$ (vi) $C \cap D$

Accepted Answer

$A = \{x:x \text {is a natural number}\}=\{1,2,3,4,5.....\}$$B = \{x:x \text{ is an even natural number}\}=\{2,4,6,8 .....\}$$C = \{x:x \text{ is an odd natural number}\}=\{1,3,5,7,9 .......\}$$D = \{x:x \text {is a prime number}\}=\{2,3,5,7 ....\}$(i) $A\cap B = \{x:x \text{ is a even natural number}\} = B$(ii) $A\cap C =\{x:x \text { is an odd natural number}\} = C$(iii) $A\cap D= \{x:x \text { is a prime number}\}=D$(iv) $B\cap C=\phi$(v) $B\cap D=\left \{2\right \}$(vi) $C\cap D= \{x:x \text {is odd prime number}\}$

Question 4

If the universal set $U =\{1, 2, 3, 4, 5, 6, 7 \}, A = \{ 1, 2, 5, 7 \} , B = \{3, 4, 5, 6\}$. Verify $\left( A\cup B \right) '=A'\cap B'$

Accepted Answer

Given, universal set, $U=\{1, 2, 3, 4, 5, 6, 7\}$$A=\{1, 2, 5, 7\}$$B=\{3, 4, 5, 6\}$$(A\cup B)'=U-(A\cup B)$$=\{1, 2, 3, 4, 5, 6, 7\}-\{1, 2, 3, 4, 5, 6, 7\}$$=\phi$$A'\cap B'=(U-A)n(U-B)$$=\{3, 4, 6\}n\{1, 2, 7\}$$=\phi$Hence $(A\cup B)'=A'\cap B'$.

Question 5

If $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6, 8\}$ and C= $\{3,4,5,6\}$, then verify : $A - (B \cup C) = (A - B) \cap (A - C)$.

Accepted Answer

Given, $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6, 8\}$ and $C=\{3,4,5,6\}$For the LHS:Union of two sets will have the elements of both sets.So, $ B \cup C = \{2,3,4,5,6,8 \}$ $ A - (B \cup C) $ will have elements of $A$ which are not in  $ (B \cup C) $So, $ A - (B \cup C)   = \{ 1 \}$ ..... $(1)$For the RHS:$ A - B $ will have elements of $A$ which are not in $B$.So, $ A - B   = \{ 1,3,5 \}$  $ A - C $ will have elements of $A$ which are not in $C$.So, $ A - C   = \{ 1,2 \}$  Intersection of two sets has the common elements of both the sets. $\Rightarrow (A - B) \cap (A - C) = \{1\}$ ..... $(2)$From $(1)$ and $(2),$ we have$ A - (B \cup C) =(A - B) \cap (A - C) $Hence, the given expression is true.

Question 6

If $A$ and $B$ are two sets, then $A\cap (A\cup B)$ equals

Accepted Answer

$A\cap(A\cup B)$$=(A\cap A) \cup (A\cap B)$ $=A $Hence answer is Option A.

Question 7

In a class of $60$ students, $25$ students play cricket and $20$ students play tennis, and $10$ students play both the games, then the number of students who play neither are

Accepted Answer

Let 'A' be the set of students who play cricket, 'B' be the set of students who play tennis.It is given that, $n(A)=25$, $n(B)=20$ and $n(A\cap B)=10=$ students who play both the games Thus $n(A\cup B) =n(A)+n(B)-n(A\cap B)=20+25-10=35$Therefore the students who play neither of the cricket and tennis $=60-35=25$

Question 8

Write power sets of the following $\left \{ 1, 2, 3 \right \}$

Accepted Answer

$\left \{ 1, 2, 3 \right \} $Subsets of $ \left \{ 1, 2, 3 \right \} $ are $\phi ,\left \{ 1 \right \}, \left \{ 2 \right \}, \left \{ 3 \right \}, \left \{ 1, 2 \right \} , \left \{ 1, 3 \right \} , \left \{ 2, 3\right \} $ and $\left \{ 1, 2, 3 \right \}$Hence , $P \left \{ 1 , 2, 3 \right \} =\{ \phi  , \{1\} , \{2\}, \{3\} , \{1,2\} , \{1,3\} , \{2,3\} , \{ 1, 2, 3 \} \}$

Question 9

Show that the relation R defined by $\left( a,b \right) R \left( c,d \right)$ such that $a+d=b+c$ on AXA, where $A=\left\{ 1,2,....10 \right\}$ is an equivalence relation. Hence write the equivalence class $\left[ 3,4 \right] ;a,b,c,d\in A.$

Accepted Answer

R is an equivalance relation if R is reflexive,symmetric and transitive. a)checking if it is reflexive;Given R in $A\times A\quad and\quad \left( a,b \right) R\left( c,d \right) \quad such\quad that\quad a+d=b+c$For reflexive,consider $\left( a,b \right) R\left( a,b \right) \quad \quad \left( a,b \right) \in A$and applying given condition$\Rightarrow a+b=b+a$;which is true for all A$\therefore\quad R\quad is\quad reflexive.$b)checking if it is symmetric;given$\left( a,b \right) R\left( c,d \right) \quad such\quad that\quad a+d=b+c$consider $\left( c,d \right) R\left( a,b \right) \quad on\quad A\times A$applying given condition$\Rightarrow c+b=d+a\quad which\quad satisfies\quad given\quad condition$Hence R is symmetric.c)checking if it is transitive;$Let\quad \left( a,b \right) R\left( c,d \right) \quad and\quad \left( c,d \right) R\left( e,f \right) $$and\quad \left( a,b \right) ,\left( c,d \right) ,\left( e,f \right) \in A\times A$applying given condition:$\Rightarrow a+d=b+c\quad \rightarrow 1\quad and\quad c+f=d+e\quad \rightarrow 2$equation 1$\Rightarrow a-c=b-d$$now\quad add\quad equation\quad 1\quad and\quad 2;$$\Rightarrow a-c+c+f=b-d+d+e$$\Rightarrow a+f=b+e$$\therefore\quad \left( a,b \right) R\left( e,f \right) \quad also\quad satisfies\quad the\quad condition$Hence R is transitive.Therfore by above inspection we can say that R is an equivalence relation.$\Rightarrow steps\quad for\quad finding\quad equivalence\quad class\quad \left[ 3,4 \right] :$Let $\left( 3,4 \right) R\left( a,b \right) \quad on\quad A\times A\quad where\quad A=\left\{ 1,2,3,.......10 \right\} $$\Rightarrow 3+b=4+a$$let\quad a=1\Rightarrow b=2$therfore one pair $\left( a,b \right) =\left( 1,2 \right) $similarly we can find the pairs $\left( a,b \right) $Therefore equivalence class of $\left[ 3,4 \right] =\left\{ \left( 1,2 \right) ,\left( 2,3 \right) ,\left( 3,4 \right) ,\left( 4,5 \right) ,\left( 5,6 \right) ,\left( 6,7 \right) ,\left( 7,8 \right) ,\left( 8,9 \right) ,\left( 9,10 \right)  \right\} $

Question 10

Let $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Then the number of subsets of $A$ containing exactly two elements is

Accepted Answer

Number of elements in $A= 10$ Number of subsets of $A$ containing exactly two elements $=$ Number of ways we can select $ 2 $ elements from $10$ elements ${}^{ 10 }{ C }_{2}=\dfrac { 10\times 9 }{ 2 } =45$$\therefore$ Number of subsets of $A$ containing exactly two elements $=45$