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CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Set Theory
Quiz 10
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
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0%
10
0%
11
0%
12
0%
13
Explanation
Let $$A$$, $$B$$, and $$C$$ be the set of people who like $$product\ A$$, $$product\ B,\ and\ product\ C$$ respectively.
Accordingly,
$$ n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12$$,
$$n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8$$
The Venn diagram for the given problem can be drawn as above.
It can be seen that number of people who like product C only is
$$={29 – (4 + 8 + 6)} = 11$$
In a battle $$70\% $$ of the combatants lost one eye, $$80\% $$ an ear, $$75\% $$ an arm, $$85\% $$ a leg and $$x\% $$ lost all the four limbs the minimum value of $$x$$ is
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0%
$$10$$
0%
$$12$$
0%
$$15$$
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$$none\ of\ these$$
The number of subsets $$ R$$ of $$P=(1,2,3,....,9)$$ which satisfies the property "There exit integers a<b<c with a$$\in $$R, b$$\in $$R,c$$\in $$R" is
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0%
$$512$$
0%
$$466$$
0%
$$467$$
0%
None of these
Explanation
Every subset of P containing at least 3 elements from P will always satisfy the required property since each number is distinct in the set P.
So, Total no. of ways = $$^{9}C_3 + ^{9}C_4 + .... + ^{9}C_9 $$
$$= (^{9}C_0 + ^{9}C_1 + ..... + ^{9}C_9) - ^{9}C_0 - ^{9}C_1 - ^{9}C_2$$
$$=2^9 - 46$$
$$=512 -46$$
$$=466$$
If $$A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$$, then the statement is true or false
$$\left\{ 3\right\} \subseteq A$$
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0%
True
0%
False
Explanation
$$A=\{3,\{4,5\},6\}$$
We have to find out if $$\{3\}$$ is subset of $$A$$ or not.
Since, $$3$$ is present in elements of $$A$$, then $$\{3\}\subseteq \{A\}$$
$$\therefore$$ Given statement is true.
In a universal set x,$$n\left( x \right) = 50$$ ,$$n\left( A \right) = 35$$ , $$n\left( B \right) = 20$$ , $$n\left( {A' \cap B'} \right) = 5$$ ,then $$n\left( {A \cup B} \right),n\left( {A \cap B} \right)$$ are repsectively
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0%
$$45,10$$
0%
$$10,45$$
0%
$$25,30$$
0%
$$15,25$$
Explanation
Given $$n\left(X\right) = 50$$,
$$n\left(A\right) = 35$$,
$$n\left(B\right) = 20$$
$$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = 5$$
$$(i)\,n\left(A \cup B\right)$$
$$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = n\left(X\right) - n\left(A\cup\,B\right)$$
$$n\left(A\cup\, B\right) = n\left(X\right) - n\left({A}^{\prime}\cap\,{B}^{\prime}\right)$$
$$n\left(A\cup\, B\right) = 50 - 5= 45$$
$$(ii) n\left(A \cap B\right)$$
$$n\left(A\cup B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cap B\right)$$
$$n\left(A\cap\, B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cup\, B\right)$$
$$n\left(A\cap\, B\right) = 35 + 20 - 45= 10$$
If $$p$$ and $$q$$ are two proposition, then $$\sim(p\leftrightarrow q)$$ is
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$$\sim p \wedge \sim q$$
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$$\sim p \vee \sim q$$
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$$(p \wedge \sim q) \vee (\sim p \wedge q)$$
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$$none of these$$
If $$A$$ and $$B$$ are two non empty sets then $$( A \cup B ) ^ { C } = ?$$
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$$A ^ { C } \cup B ^ { C }$$
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$$A ^ { C } \cap B ^ { C }$$
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$$A \cup B ^ { C }$$
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$$A ^ { C } \cap B$$
Explanation
If A and B are two no empty sets then $${ \left( A\cup B \right) }^{ C }={ A }^{ C }\cap { B }^{ C }$$
Let $$S={1,2,3,.....10}$$ and $$P={1,2,3,4,5}$$ The number of subsets $$'Q'$$ of $$S$$ such that $$p \cup Q=S$$, are.....
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0%
$$128$$
0%
$$256$$
0%
$$32$$
0%
$$64$$
If 'p' is true and 'q' is false, then which of the following statement is not true?
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$$p\vee q$$
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$$p\Rightarrow q$$
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$$p\wedge (\sim q)$$
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$$q\Rightarrow p$$
If the number of $$5$$ elements subsets of the set $$A\left\{\ a_{1},a_{2}.....a_{20}\right\}$$ of $$20$$ distinct elements is $$k$$ times the number of $$5$$ elements subsets containing $$a_{4}$$, then $$k$$ is
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$$5$$
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$$\dfrac{20}{7}$$
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$$4$$
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$$\dfrac{10}{3}$$
Explanation
No of $$5$$ elements subset $$=^{20}C_{5}=\dfrac {20!}{15! \, 15!}=\dfrac {20 \times 19\times 18\times 17\times 16\times 15!}{15! \, 5\times 4\times 3\times 2\times 1}=15504$$
No of $$5$$ elements subset containing $$a_{4}$$ $$=^{19}C_{4}=\dfrac {19!}{4! \, 15!}$$
$$\Rightarrow \dfrac {20!}{15! \, 15!}=k\dfrac {19!}{4!\, 15!}$$
$$\Rightarrow \dfrac {20\times 19!}{5\times 4!}=k\dfrac {19!}{4!}$$
$$\Rightarrow k=4$$
Let $$A$$ and $$B$$ be two sets. $$A \cap B = \{1, 2\}, A \cup B = \{1, 2, 3, 4, 5\}$$ and if $$n_1$$ is the maximum number of function from $$A$$ to $$B$$ and $$n_2$$ is the minimum number of function form $$A$$ to $$B$$ then $$n_1 - n_2$$ equals
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0%
$$32$$
0%
$$56$$
0%
$$64$$
0%
$$81$$
The value of set $$(A\cup B\cup C)\cap(A\cap B^1\cap C^1)^1\cap C^1$$ is equal to
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$$B\cap C^1$$
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$$A\cap C$$
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$$B\cap C^1$$
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$$A\cap C^1$$
If two sets $$P$$ and $$Q,n\left(P\right)=5,n\left(Q\right)=4$$ then $$n\left(P\times Q\right)=$$
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$$20$$
0%
$$9$$
0%
$$25$$
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$$5/4$$
Explanation
$$\begin{aligned} n(P \times Q) &=n(P) \times n(Q) \\ &=5 \times 4 \\ &=20 \end{aligned}$$
$$\therefore$$ option $$A$$ is correct.
The statement that is true among the following is
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$$p\Longrightarrow q$$ is equivalent to $$p\wedge -q$$
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$$p\vee q$$ and $$p\wedge q$$ have the same truth value
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The converse of $$tanx=0\Longrightarrow x=0$$ is $$x\neq 0\Longrightarrow tanx=0$$
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The contrapositive of $$3x+2=8\Longrightarrow x=2$$ is $$x\neq 2\Longrightarrow 3x+2\neq 8$$
Let $$Q$$ be a non empty subset of $$N$$ and $$q$$ is a
statement as given below :
$$q:$$ There exists an even number $$a \in Q$$
Negation of the statement $$q$$ will be :
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There is no even number in the set $$Q$$
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Every $$a \in Q$$ is an odd number.
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$$(a)$$ and $$(b)$$ both
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None of these
Explanation
Let $$Q$$ be a non empty subset of $$N$$ and $$q.$$ There exists even number $$a\in Q$$ negation of statement $$q.$$ There does not exist a even number in set Q.
Hence, the answer is there is no even number in the set $$Q.$$
If A= {1, 2, 5} and B= {3, 4, 5, 9}, then $$A \bigcup B$$ is equal to :
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$$\{1, 2, 5, 9\}$$
0%
$$\{1, 2, 3, 4, 9\}$$
0%
$$\{1, 2, 3, 4, 5, 9\}$$
0%
None of these
Explanation
$$A=\left\{ 1,2,5 \right\} \quad B=\left\{ 3,4,5,9 \right\} $$
$$A\cup B=\left\{ 1,2,3,4,5,9 \right\} $$
Option C.
In a battle $$70$$% of the combatants lost one eye, $$80$$% an ear, $$75$$% an arm, $$85$$% a leg, $$x$$% lost all the four limbs the minimum value of $$x$$ is
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0%
$$10$$
0%
$$12$$
0%
$$15$$
0%
$$5$$
Explanation
$$70\%$$ of the combatants lost one eye, $$80\%$$ an ear, $$75\%$$ an arm and $$85\%$$ a leg.
Now,
$$\Rightarrow$$ The combatants who lost one eye and one ear $$=(70+80-100)\%$$
$$=50\%$$
$$\Rightarrow$$ The combatants who lost one eye, one ear and one erm $$=(50+75-100)\%$$
$$=25\%$$
$$\Rightarrow$$ The combatants who lost one eye, one ear one arm and one leg $$=(25-85-100)\%$$
$$=10\%$$
$$\therefore$$ Combatant who lost all the four limbs $$=10\%$$
$$\therefore$$ $$x=10\%$$
Two sets A and B are defined as follows
$$A=\left\{ \left( x,y \right) :y={ e }^{ 2x },x\in R \right\} $$ and
$$B=\left\{ \left( x,y \right) :y={ x }^{ 2 },x\in R \right\} $$, then
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0%
$$A\subset B$$
0%
$$B\subset A$$
0%
$$A\bigcup B$$
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$$A\cap B=\phi $$
Explanation
$$\begin{array}{l} A=\left\{ { \left( { x,\, y } \right) \, :\, y={ e^{ 2x } },\, x\in R } \right\} \, and \\ B=\left\{ { \left( { x,\, y } \right) \, :\, y={ x^{ 2 } },\, x\in R } \right\} \\ then, \\ A=\left( { { e^{ \circ } },\, { e^{ 1 } },\, { e^{ 2 } },....,\, { e^{ \infty } } } \right) \\ =\left( { 1,\, { e^{ 1 } },\, { e^{ 2 } },....... } \right) \\ and, \\ B=\left( { 0,\, 1,\, 4,\, ..... } \right) \\ \therefore A\subset B \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$$
If $$A=\left\{1,2,3,4\right\}; B=\left\{2,4,6,8\right\}; C=\left\{3,4,5,8\right\}$$ then $$A\cap B\cap C=$$
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0%
$$\phi$$
0%
$${4}$$
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$$\mu$$
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$${2,4}$$
Explanation
$$\begin{matrix} A=\left\{ { 1,2,3,4 } \right\} \\ B=\left\{ { 2,4,6,8 } \right\} \\ C=\left\{ { 3,4,5,8 } \right\} \\ A\cap B\cap C=\left\{ { 4 } \right\} . \\ \end{matrix}$$
Examine whether the following statements are true or false:
$$\left\{b, c\right\}\subset \left\{a, \left\{b, c\right\}\right\}$$
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0%
True
0%
False
In a group of 800 people, 550 can speak Hindi and 450 can speak English. How many can speak both Hindi and English ?
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$$100$$
0%
$$150$$
0%
$$200$$
0%
None of these
Explanation
Let H denote the set of people speaking Hindi and E denote the set of people speaking English. We are given :
$$n(H)=550, n(E)=450, n(H\cup E)=800$$
Now,
$$n(H\cup E)=n(H)+n(E)-n(H\cap E)$$
$$n(H \cap E)=550+450-800=200$$
Hence, 200 persons can speak both Hindi and English.
Let A,B are two sets such that n(A)=4 and n(B)=Then the least possible number of elements in the power set of $$(A\cup B)$$ is
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0%
16
0%
64
0%
256
0%
1024
Explanation
Given,
$$n(A)=4,n(B)=6$$
Then the least number of possible elements in
$$n(A\cup B)=2^{n(A)}.2^{n(B)}=2^4.2^6=2^{10}=1024$$
Left A = {1, 2, 3, 4, 5, 6} and B= {1, 2, 3, 4} be two sets, then the number of functions that can be defined from A to B such that the element '2' in B has exactly 3 pre - images i A, is equal ot
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0%
540
0%
440
0%
810
0%
620
if universal set $$\sum { =\left\{ a,b,c,d,e,f,g,h, \right\} A=\left\{ b,c,d,e,f \right\} } and\quad C=\left\{ c,d,e,f,g \right\} ,\quad then\quad find\quad B-A$$
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0%
$$\left\{ b,c,e,f \right\} $$
0%
$$\left\{ a,b,f,h \right\} $$
0%
$$\left\{ a,g,h \right\} $$
0%
$$\left\{ a,c,e,g \right\}$$
Explanation
Given,
$$ B=$${$$a,b,c,d,e,f,g,h$$}
$$A=$${$$b,c,d,e,f$$}$$ $$,$$C=$${$$c,d,e,f,g$$}
$$\therefore B-A=$${$$a,b,c,d,e,f,g,h$$}-{$$b,c,d,e,f$$}
$$\implies B-A=$${$$a,g,h$$}
If P(S) denotes the set of all subsets of a given set S, then the number of one to one function from the set s={1,2,3} to the set of P(S) is
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0%
336
0%
8
0%
36
0%
320
Explanation
$$\begin{array}{l} Given \\ P\left( s \right) =\left\{ { 1,\, 2,\, 3 } \right\} \\ P\left( s \right) =all\, subset\, of\, 5 \\ =\left\{ { \phi ,\left\{ 1 \right\} ,\left\{ 3 \right\} ,\left\{ { 1,2 } \right\} ,\left\{ { 2,3 } \right\} ,\left\{ { 3,1 } \right\} ,\left\{ { 1,2,3 } \right\} } \right\} \\ No.\, of\, element\, of\, 5=0\left( 5 \right) =3 \\ similarly \\ O\left( { p\left( s \right) } \right) =8 \\ Total\, no.\, of\, one-one\, function \\ =\frac { { n! } }{ { \left( { n-m } \right) ! } } \, ,\, where\, n=o\left( { p\left( s \right) } \right) \, and\, m=o\left( 5 \right) \\ =\frac { { 8! } }{ { \left( { 8-3 } \right) ! } } =\frac { { 8! } }{ { 5! } } \\ =336\, \, \\ No.\, of\, one-one\, function\, is\, 336 \\ Hence,\, \, option\, \, A\, \, is\, correct, \end{array}$$
The function $$f(x)$$ satisfies the condition $$(x-2)f(x)+2f\left(\dfrac{1}{x}\right)=2$$ for all $$x\neq 0$$. Then the value of $$f(2)$$ is
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0%
$$\dfrac{1}{2}$$
0%
$$1$$
0%
$$\dfrac{7}{4}$$
0%
$$\dfrac{-3}{2}$$
Explanation
Putting $$x=2$$ in $$\left( x-2 \right) f\left( x \right) +2f\left( \dfrac { 1 }{ x } \right) =2$$
$$\left( 2-2 \right) f\left( 2 \right) +2f\left( \dfrac { 1 }{ 2 } \right) =2$$
$$2f\left( \dfrac { 1 }{ 2 } \right) =2$$
$$f\left( \dfrac { 1 }{ 2 } \right) =1$$
If $$A=\left\{1, 2, 3, 4\right\}$$, then the number of subsets of $$A$$ that contain the element $$2$$ but not $$3$$, is
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0%
$$16$$
0%
$$4$$
0%
$$8$$
0%
$$24$$
Explanation
The subsets are be $$\left\{1, 2, 4\right\},\left\{1, 2\right\}, \left\{2, 4\right\}, \left\{2\right\}$$
Number of subsets of $$A$$ that contain the element $$2$$ but not $$3$$ is $$4$$
Let $$p$$ and $$q$$ be two statements. amongst the following, the statement is equivalent to $$p\rightarrow q$$ is
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$$\displaystyle p\wedge \sim q$$
0%
$$\displaystyle \sim p\vee q$$
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$$\displaystyle \sim p\wedge q$$
0%
$$\displaystyle p\vee \sim q$$
Explanation
$$ \begin{array}{l} \text { Given, } \quad P \longrightarrow q \\ \text { ne trinow that, } \rightarrow \text { - if then } \\ \Rightarrow \text { if } P \text { then } q \end{array} $$
$$ \begin{array}{lll} p & q & p \rightarrow q \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} $$
$$ \begin{array}{c} \Rightarrow \text { Evaluating option } A(p \wedge \sim q) \\ P \text { iq } p \wedge \sim q \\ 111 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \text { Incorrect assumption } \end{array} $$
$$ \begin{array}{c} \Rightarrow \text { Eualuating } B(\sim p \wedge q) \\ \sim p \quad q \quad \sim p \wedge q \\ 0 \\ 0 \\ 1 \\ 1 \quad 0 \\ \text { Incorrect assumption } \end{array} $$
$$ \begin{array}{rlrl} \Rightarrow \text { Evaluating } & c(\sim p \vee q) \\ & \sim p & q & \sim p v q \\ & 0 & 1 & & \\ 0 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ & \text { correct assumption } \end{array} $$
Let $$\displaystyle S=\left\{a\in N,a\le100\right\}$$ if the equation $$\displaystyle[{\tan}^{2}x]-\tan x-a=0$$ has real roots, then the number of elements $$S$$ is (when $$[.]$$ is greatest integer function)
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0%
$$10$$
0%
$$1$$
0%
$$9$$
0%
$$0$$
Explanation
Given equation is,
$$[tan^2x]-tanx-a=0$$
$$\therefore D=b^2-4ac=(-1)^2-4(1)(-a)$$
$$=1+4a$$
So D must be a odd perfect square
$$\implies \sqrt{1+4a}=2\lambda+1$$
$$\implies 1+4a=4\lambda^2+1+4\lambda$$
$$\implies a=\lambda(\lambda+1)$$
For different values og=f $$\lambda$$ we get
$$a=0,2,6,12,20,30,42,56,72,90$$
Hence there are $$10 $$ values of $$a$$.
If $$A=\left\{ 1,3,5,7,9,11,13,15,17 \right\} ,B=\left\{ 2,4,....,18 \right\} $$ and N is the universal set, then $$A'\cup \left( \left( A\cup B \right) \cap B' \right) $$
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0%
A
0%
N
0%
B
0%
R
Explanation
$$A=\left\{1,3,5,7,9,11,13,15,17\right\}$$
$${A}^{\prime}=N-A=N-\left\{1,3,5,7,9,11,13,15,17\right\}=\left\{2,4,6,8,10,12,14,16,18\right\}$$
$$A\cup B=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}$$
$${B}^{\prime}=N-B=N-\left\{2,4,6,8,10,12,14,16,18\right\}=\left\{1,3,5,7,9,11,13,15,17\right\}$$
$$\left(A\cup B\right)\cap {B}^{\prime}$$
$$=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}\cap\left\{1,3,5,7,9,11,13,15,17\right\}$$
$$=\left\{1,3,5,7,9,11,13,15,17\right\}$$
$${A}^{\prime}\cup \left(\left(A\cup B\right)\cap {B}^{\prime}\right)$$
$$=\left\{2,4,6,8,10,12,14,16,18\right\}\cup\left\{1,3,5,7,9,11,13,15,17\right\}$$
$$=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}=N$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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