MCQ Questions for 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.

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

$={29 – (4 + 8 + 6)} = 11$

1460243_1199498_ans_24387ca5a0fc4a569288365f3a4fedcf.PNG

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

0%
$10$
0%
$12$
0%
$15$
0%
$none\ of\ these$

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

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 =

$^{9}C_3 + ^{9}C_4 + .... + ^{9}C_9$

$= (^{9}C_0 + ^{9}C_1 + ..... + ^{9}C_9) - ^{9}C_0 - ^{9}C_1 - ^{9}C_2$

$=2^9 - 46$

$=512 -46$

$=466$

$A=\left\{3, \left\{ 4, 5\right\}, 6\right\}$ , then the statement is true or false

$\left\{ 3\right\} \subseteq A$

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

0%
$45,10$
0%
$10,45$
0%
$25,30$
0%
$15,25$

Explanation

Given

$n\left(X\right) = 50$ ,

$n\left(A\right) = 35$ ,

$n\left(B\right) = 20$

$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = 5$

$(i)\,n\left(A \cup B\right)$

$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = n\left(X\right) - n\left(A\cup\,B\right)$

$n\left(A\cup\, B\right) = n\left(X\right) - n\left({A}^{\prime}\cap\,{B}^{\prime}\right)$

$n\left(A\cup\, B\right) = 50 - 5= 45$

$(ii) n\left(A \cap B\right)$

$n\left(A\cup B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cap B\right)$

$n\left(A\cap\, B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cup\, B\right)$

$n\left(A\cap\, B\right) = 35 + 20 - 45= 10$

$p$ and

$q$ are two proposition, then

$\sim(p\leftrightarrow q)$ is

0%
$\sim p \wedge \sim q$
0%
$\sim p \vee \sim q$
0%
$(p \wedge \sim q) \vee (\sim p \wedge q)$
0%
$none of these$

$A$ and

$B$ are two non empty sets then

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

0%
$A ^ { C } \cup B ^ { C }$
0%
$A ^ { C } \cap B ^ { C }$
0%
$A \cup B ^ { C }$
0%
$A ^ { C } \cap B$

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.....

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?

0%
$p\vee q$
0%
$p\Rightarrow q$
0%
$p\wedge (\sim q)$
0%
$q\Rightarrow p$

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

0%
$5$
0%
$\dfrac{20}{7}$
0%
$4$
0%
$\dfrac{10}{3}$

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

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

0%
$B\cap C^1$
0%
$A\cap C$
0%
$B\cap C^1$
0%
$A\cap C^1$

If two sets

$P$ and

$Q,n\left(P\right)=5,n\left(Q\right)=4$ then

$n\left(P\times Q\right)=$

0%
$20$
0%
$9$
0%
$25$
0%
$5/4$

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

0%
$p\Longrightarrow q$ is equivalent to $p\wedge -q$
0%
$p\vee q$ and $p\wedge q$ have the same truth value
0%
The converse of $tanx=0\Longrightarrow x=0$ is $x\neq 0\Longrightarrow tanx=0$
0%
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 :

0%
There is no even number in the set $Q$
0%
Every $a \in Q$ is an odd number.
0%
$(a)$ and $(b)$ both
0%
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 :

0%
$\{1, 2, 5, 9\}$
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

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

0%
$A\subset B$
0%
$B\subset A$
0%
$A\bigcup B$
0%
$A\cap B=\phi$

Explanation

$\begin{array}{l} A=\left\{ { \left( { x,\, y } \right) \, :\, y={ e^{ 2x } },\, x\in R } \right\} \, and \\ B=\left\{ { \left( { x,\, y } \right) \, :\, y={ x^{ 2 } },\, x\in R } \right\} \\ then, \\ A=\left( { { e^{ \circ } },\, { e^{ 1 } },\, { e^{ 2 } },....,\, { e^{ \infty } } } \right) \\ =\left( { 1,\, { e^{ 1 } },\, { e^{ 2 } },....... } \right) \\ and, \\ B=\left( { 0,\, 1,\, 4,\, ..... } \right) \\ \therefore A\subset B \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

$A=\left\{1,2,3,4\right\}; B=\left\{2,4,6,8\right\}; C=\left\{3,4,5,8\right\}$ then

$A\cap B\cap C=$

0%
$\phi$
0%
${4}$
0%
$\mu$
0%
${2,4}$

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:

$\left\{b, c\right\}\subset \left\{a, \left\{b, c\right\}\right\}$

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 ?

0%
$100$
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

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

0%
540
0%
440
0%
810
0%
620

if universal set

$\sum { =\left\{ a,b,c,d,e,f,g,h, \right\} A=\left\{ b,c,d,e,f \right\} } and\quad C=\left\{ c,d,e,f,g \right\} ,\quad then\quad find\quad B-A$

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

0%
336
0%
8
0%
36
0%
320

Explanation

$\begin{array}{l} Given \\ P\left( s \right) =\left\{ { 1,\, 2,\, 3 } \right\} \\ P\left( s \right) =all\, subset\, of\, 5 \\ =\left\{ { \phi ,\left\{ 1 \right\} ,\left\{ 3 \right\} ,\left\{ { 1,2 } \right\} ,\left\{ { 2,3 } \right\} ,\left\{ { 3,1 } \right\} ,\left\{ { 1,2,3 } \right\} } \right\} \\ No.\, of\, element\, of\, 5=0\left( 5 \right) =3 \\ similarly \\ O\left( { p\left( s \right) } \right) =8 \\ Total\, no.\, of\, one-one\, function \\ =\frac { { n! } }{ { \left( { n-m } \right) ! } } \, ,\, where\, n=o\left( { p\left( s \right) } \right) \, and\, m=o\left( 5 \right) \\ =\frac { { 8! } }{ { \left( { 8-3 } \right) ! } } =\frac { { 8! } }{ { 5! } } \\ =336\, \, \\ No.\, of\, one-one\, function\, is\, 336 \\ Hence,\, \, option\, \, A\, \, is\, correct, \end{array}$
1217888_1309290_ans_148476409ff94e659de01d6734148e9a.PNG

1217888_1309290_ans_148476409ff94e659de01d6734148e9a.PNG

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

0%
$\dfrac{1}{2}$
0%
$1$
0%
$\dfrac{7}{4}$
0%
$\dfrac{-3}{2}$

Explanation

Putting

$x=2$ in

$\left( x-2 \right) f\left( x \right) +2f\left( \dfrac { 1 }{ x } \right) =2$

$\left( 2-2 \right) f\left( 2 \right) +2f\left( \dfrac { 1 }{ 2 } \right) =2$

$2f\left( \dfrac { 1 }{ 2 } \right) =2$

$f\left( \dfrac { 1 }{ 2 } \right) =1$

$A=\left\{1, 2, 3, 4\right\}$ , then the number of subsets of

$A$ that contain the element

$2$ but not

$3$ , is

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

0%
$\displaystyle p\wedge \sim q$
0%
$\displaystyle \sim p\vee q$
0%
$\displaystyle \sim p\wedge q$
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)

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

$a$ .

$A=\left\{ 1,3,5,7,9,11,13,15,17 \right\} ,B=\left\{ 2,4,....,18 \right\}$ and N is the universal set, then

$A'\cup \left( \left( A\cup B \right) \cap B' \right)$

0%
A
0%
N
0%
B
0%
R

Explanation

$A=\left\{1,3,5,7,9,11,13,15,17\right\}$

${A}^{\prime}=N-A=N-\left\{1,3,5,7,9,11,13,15,17\right\}=\left\{2,4,6,8,10,12,14,16,18\right\}$

$A\cup B=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}$

${B}^{\prime}=N-B=N-\left\{2,4,6,8,10,12,14,16,18\right\}=\left\{1,3,5,7,9,11,13,15,17\right\}$

$\left(A\cup B\right)\cap {B}^{\prime}$

$=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}\cap\left\{1,3,5,7,9,11,13,15,17\right\}$

$=\left\{1,3,5,7,9,11,13,15,17\right\}$

${A}^{\prime}\cup \left(\left(A\cup B\right)\cap {B}^{\prime}\right)$

$=\left\{2,4,6,8,10,12,14,16,18\right\}\cup\left\{1,3,5,7,9,11,13,15,17\right\}$

$=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}=N$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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