If $$A=\{1, 2, 3\}$$ and $$B=\{3, 8\}$$, then $$(A\cup B)\times (A\cap B)$$ is

$$\{(3, 1), (3, 2), (3, 3), (3, 8)\}$$

$$\{(1, 2), (2, 2), (3, 3), (8, 8)\}$$

$$\{(8, 3), (8, 2), (8, 1), (8, 8)\}$$

If $$A = \{5, 7\}, B= \{7, 9\}$$ and $$C = \{7, 9, 11\},$$ find $$(A \times B) \cup (A \times C)$$

$$\{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

$$\{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$

If $$A=\{2, 4\}$$ and $$B=\{3, 4, 5\}$$ then $$(A\cap B)\times (A\cup B)$$ is

$$\{(2, 2), (3, 4), (4, 2), (5, 4)\}$$

$$\{(2, 4), (3, 4), (4, 4), (4, 5)\}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 5

$ABC\sim PQR$ , then AB:PQ =

0%
$AC:PB$
0%
$AC:PR$
0%
$AB:PR$
0%
$AC:RQ$

Let

$A=\left \{ 1, 2, 3 \right \}$ . Then number of equivalence relations containing (1, 2) is:

0%
1
0%
2
0%
3
0%
4

Explanation

Observe that 1 is related to 2. So, we have two possible cases.

Case 1: When 1 is not related to 3, then the relation.

$R_{1}=\left \{ (1, 1),(1, 2),(2, 1),(2, 2),(3, 3) \right \}$ is the only equivalence relation containing

$(1, 2)$ .

Case 2: When 1 is related to 3, then

$A \times A$ =

$\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)\}$ is the only equivalence relation containing

$(1,2)$

$\therefore$ There are two required equivalence relations.

Let A and B be finite sets containing m and n elements respectively. The number of relations that can be defined from A and B is:

0%
mn
0%
$2^{mn}$
0%
$2^{m+n}$
0%
$n^{m}$

Explanation

Here,

$O(A)=m$ and

$O(B)=n$ .

Hence

$O(A×B)=mn$

Since every subset of

$A×B$ is a relation from

$A$ to

$B$ , therefore, number of relations from A to B is equal to the number of the subsets of

$A×B$ , i.e.,

$2^{m{n}}$

Which of the following are not equivalence relations on I?

0%
a R b if $a+b$ is an even integer
0%
a R b if $a-b$ is an even integer
0%
a R b if $a< b$
0%
a R b if $a= b$

Explanation

For Option

$[A]$

aRb if

$a+b$ is an even integer

$Reflexive:$

$a+a=2a\Rightarrow$ is

$even$

$number$

$Symmetric$

$a+b=b+a\Rightarrow$ is

$even$

$number$

$Transitive$

$a+b=2k;b+c=2p$ then

$a+c=2k+2p-2b=2(k+p-b)\Rightarrow$ is an

$even$

$number$

$\therefore$ this is a equivalence relation

For option

$[B]$

aRb if

$a-b$ is an even number

$Reflexive:$

$a-a=0\Rightarrow$ is

$even$

$number$

$Symmetric$

$a-b=2k;b-a=-2k\Rightarrow$ is

$even$

$number$

$Transitive$

$a-b=2k;b-c=2p$

$a-b+b-c=a-c=2k+2p=2(k+p)$ is

$even$

$number$

$\therefore$ this is a equivalence relation

For option

$[C]$

$Reflexive:$

$a\not <a\Rightarrow$

$not$

$Reflexive:$

$Symmetric:$

$a<b$ but

$b\not< a$

$\Rightarrow$

$not$

$Symmetric$

$Transitive$

$a<b;b<c\Rightarrow a<c$

$\Rightarrow$

$Transitive$

$\therefore$ not a Equivalence relation

For option

$[D]$

aRb if a=b

$\Rightarrow$

$Reflexive,Symmetric,Transitive$

$\therefore$ this is a equivalence relation

$X=\left \{ 1, 2, 3, 4, 5 \right \}$ and

$Y=\left \{ 1, 3, 5, 7, 9 \right \}$ , determine which of the following sets represent a relation and also a mapping.

0%
$R_{1}=\left \{ (x, y):y=x+2, x\in X, y\in Y \right \}$
0%
$R_{2}=\left \{ (1, 1),(2, 1),(3, 3),(4, 3),(5, 5) \right \}$
0%
$R_{3}=\left \{ (1, 1),(1, 3),(3, 5),(3, 7),(5, 7) \right \}$
0%
$R_{4}=\left \{ (1, 3),(2, 5),(4, 7),(5, 9),(3, 1) \right \}$

Explanation

$R_1={(1,3),(2,4),(3,5),(4,6),(5,7)}$

Since

$4$ and

$6$ do not belong to

$Y$

$(2,4),(4,6)∉R_1$

$R_1={(1,3),(3,5),(5,7)}⊂A×B$

Hence

$R_1$ is a relation but not a mapping as the elements

$2$ and

$4$ do not have any image.

$R_2$ : It is certainly a mapping and since every mapping is a relation, it is a relation as well.

$R_3$ : It is a relation being a subset of

$A×B$ but the elements

$1$ and

$3$ do not have a unique image and hence it is not mapping.

$R_4$ : It is both a mapping and a relation. Each element in A has a unique image. It is also one-one and onto mapping and hence a bijection

Let R be a reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then:

0%
$m\geq n$
0%
$m\leq n$
0%
$m=n$
0%
$m=-n$

Explanation

The set consists of n elements and for relation to be reflexive it must have at least n ordered pairs. It has m ordered pairs therefore

$m≥n$ .

Let A be a finite set containing n distinct elements. The number of relations that can be defined on A is:

0%
$2^{n}$
0%
$n^{2}$
0%
$2^{n^{2}}$
0%
2n

Explanation

Since

$A$ contains

$n$ distinct elements, therefore,

$A×A$ contains

$n×n=n^2$ distinct elements. since every subset of

$A×A$ is a relation on

$A$ ., therefore, number of relations on

$A$ is equal to the order of the power set of

$A×A$ , is equal to the order of the power set of

$A×A$ , i.e.,

$2^{n^{2}}$

Let

$A=\left \{ 1, 2, 3 \right \}$ . Which of the following is not an equivalence relation on A?

0%
$\left \{ (1, 1),(2, 2),(3, 3) \right \}$
0%
$\left \{ (1, 1),(2, 2),(3, 3),(1, 2),(2, 1) \right \}$
0%
$\left \{ (1, 1),(2, 2),(3, 3),(2, 3),(3, 2) \right \}$
0%
$\left \{ (1, 1),(2, 1) \right \}$

Explanation

every relation in options follow the equivalence property.

but in fourth option this option only contains two elements

$[[{1,1}],[{2,1}]]$

this option doesn't satisfy reflexive i.e., doesn't contain all

$(a,a)$ type

$(2,2),(3,3)$

Not symmetric i.e., doesn't contain this type of elements

$(1,2)$ missing

and as well transitive.

and all others options satisfy these relations.

"Every reflexive relation is an anti-symmetric relation", Is it true?

0%
True
0%
False

The relation "congruence modulo

$m$ " is:

0%
reflexive only
0%
symmetric only
0%
transitive only
0%
an equivalence relation

Explanation

$R$ be the relation, then

$x R y$

$\Leftrightarrow x-y$ is divisible by

$m$ .

$x R x$ because

$x-x=0$ is divisible by

$m$ .

So,

$R$ is reflexive.

$xRy\Rightarrow x-y$ is divisible by

$m$

$y-x=-(x-y)$ is divisible by

$m$

$\Rightarrow yRx$

So,

$R$ is symmetric

Let

$x R y$ and

$y R z$

$\Rightarrow x-y=k_{1}m, y-z=k_{2}m$

$\therefore x-z=x-y+y-z=k_{1}m+k_{2}m=(k_{1}+k_{2})m$ .

So,

$R$ is transitive.
As R is reflexive, symmetric and transitive, it is an equivalence relation.

Let R be the relation in the set N given by

$=\left \{ (a, b):a=b-2, b >6 \right \}$ . Choose the correct answer.

0%
$(2, 4)\in R$
0%
$(3, 8)\in R$
0%
$(6, 8)\in R$
0%
$(8, 7)\in R$

Explanation

$(a,b)∈R$ only if

$a=b−2$ and

$b>6$

$(6,8)∈R$ as

$6=8−2$ and

$8 >6$

Hence

$(c)$ is the correct alternative

Find the number of relations from

$\left \{ m, o, t, h, e, r \right \}$ to

$\left \{ c, h, i, l, d \right \}$

0%
$30$
0%
$30^{2}$
0%
$2^{30}$
0%
$60$

Explanation

$n \{m.o,t,b,e,r\} = 6 = m$ (let)

$n \{c,h,i,l,d\} = 5 = n$ (let)

$\therefore$ numbers of Relations

$= { 2 }^{ mn }$

$= { 2 }^{ 6\times 5 }$

$= { 2 }^{ 30 }$

Which of the following statements is true?

0%
The x-axis is a vertical line
0%
The y-axis is a horizontal line
0%
The scale on both axes must be the same in a Cartesian plane
0%
The point of intersection between the x-axis and the y-axis is called the origin

Explanation

Since, Origin is the point of intersection of

$x$ and

$y,$ we can say that option

$D$ is correct.

$R$ is a relation from a set

$A$ to a set

$B$ and

$S$ is a relation from

$B$ to a set

$C$ , then the relation

$SOR$

0%
is from $A$ to $C$
0%
is from $C$ to $A$
0%
does not exist
0%
none of these

Explanation

Domain of

$SOR$ is the domain of

$R$ and the range is the range of

$S$ . Hence the mapping is from set

$A$ to set

$C$ .

Let

$A=\left \{ 1,2,3 \right \}$ . The total number of distinct relations that can be defined over

$A$ is:

0%
$2^{9}$
0%
$6$
0%
$8$
0%
None of the above

Explanation

Total number of distinct binary relations over the set

$A$ will be

$2^{n^2}$

$=2^{3^{2}}$

$=2^{9}$

$=512$

Let

$X=\{1, 2, 3, 4, 5 \}$ and

$Y=\{1, 3, 5, 7, 9\}$ . Which of the following is/are relations from

$X$ to

$Y$ ?

0%
$R_1=\{(x, y):y=2+x, x\in X,y\in Y\}$
0%
$R_2=\{(1, 1), (2, 1), (3, 3), (4, 3), (5, 5)\}$
0%
$R_3=\{(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)\}$
0%
$R_4=\{(1, 3), (2, 5), (2, 4), (7, 9)\}$

Explanation

$R_1=\{(x, y):y=2+x, x\in X,y\in Y\}$

$x=1 \rightarrow y =1+2=3$

$x=2 \rightarrow y =2+2=4$

$x=3 \rightarrow y =3+2=5$

$x=4 \rightarrow y =4+2=6$

$x=5 \rightarrow y =5+2=7$

$R=\{(1,3),(2,4),(3,5),(4,6),(5,7)\}$

Here in

$(4,6)$ ,

$6$ does not belong to either

$X or Y$

$R_1$ is not a relation between

$X$ and

$Y$

$R_{4}$ , since it contains an element

$(7,9)$ which relates set Y to set Y, while rest elements relate set X to Y.

$\therefore R_4$ is not a relation between

$X$ and

$Y$

Let

$R=\{(1, 3), (2, 2), (3, 2)\}$ and

$S=\{(2, 1), (3, 2), (2, 3)\}$ be two relations on set

$A=\{1, 2, 3\}$ .Then

$R$ o

$S$ is equal

0%
$\{(2, 3), (3, 2), (2, 2)\}$
0%
$\{(1, 3), (2, 2), (3, 2), (2 ,1), (2, 3)\}$
0%
$\{(3, 2), (1, 3)\}$
0%
$\{(2, 3), (3, 2)\}$

Explanation

$f\epsilon X\times Y$ and

$g\epsilon Y\times Z$

Then

$gof\epsilon X\times Z$ i.e

$(x,z)\epsilon gof$

Applying the above concept, we gives us

$RoS=\{(2,1)|(1,3),(3,2)|(2,2),(2,3)|(3,2)\}$

$=\{(2,3),(3,2),(2,2)\}$

Give a relation R={(1,2), (2,3)} on the set of natural numbers, add a minimum number of ordered pairs.

0%
enlarged relation is symmetric
0%
enlarged relation is transitive
0%
enlarged relation is reflexive.
0%
enlarged relation is equivalence relation

Explanation

$R=\{(1,2), (2,3)\}$

A)To make the relation symmetric, we have to add (2,1) and (3,2) in R.
Enlarged symmetric relation

$R'=\{(1,2), (2,3),(2,1), (3,2)\}$

B) To make enlarged relation transitive , we have to add (1,3), (1,1),(2,2) in R'.
Also, we need to add (3,1) to make it symmetric
So, the new enlarged relation is

$R''=\{(1,2), (2,3),(2,1), (3,2),(1,1),(2,2),(3,1)\}$

C) For enlarged relation to be symmetric, we need to add (1,1) in R''

$R''=\{(1,2), (2,3),(2,1), (3,2),(1,1),(2,2),(3,1),(1,1)\}$

D) For enlarged relation to be equivalence,

$A=\{1, 2, 3\}$ and

$B=\{3, 8\}$ , then

$(A\cup B)\times (A\cap B)$ is

0%
$\{(3, 1), (3, 2), (3, 3), (3, 8)\}$
0%
$\{(1, 3), (2, 3), (3, 3), (8, 3)\}$
0%
$\{(1, 2), (2, 2), (3, 3), (8, 8)\}$
0%
$\{(8, 3), (8, 2), (8, 1), (8, 8)\}$

Explanation

$A\cup B=\{1,2,3,8\}$

$A\cap B=\{3\}$

$\therefore (A\cup B)\times (A\cap B)$

$=\{(1,3),(2,3),(3,3),(8,3)\}$

$\displaystyle :n(A)= m,$ then number of relations in

$A$ are

0%
$\displaystyle 2^{m}$
0%
$\displaystyle 2^{m}-2$
0%
$\displaystyle 2^{m^{2}}$
0%
None of these

Explanation

Given

$\displaystyle :n(A)= m$ Relation in a set

$A$ is subset of

$\displaystyle A\times A$ For Number of subset of

$A$ We have

$\displaystyle n(A\times A) = n(A)\times n(A)= m^{2}$

$\displaystyle \therefore$ Number of subset of

$\displaystyle A\times A = 2^{m^{2}}$

every subset of

$\displaystyle A\times A$ is a relation

$\displaystyle \therefore$ Number of relations in A is

$\displaystyle 2^{m^{2}}= 2^{m\times m}$

$\displaystyle X= \left \{ 1,2,3,4,5 \right \}$ and

$\displaystyle Y= \left \{ 1,3,5,7,9 \right \}$ then which of the following sets are relation from

$X$ to

$Y$

0%
$\displaystyle R_{1}= \left \{ (x,a):a= x+2,x\in X,a\in Y \right \}$
0%
$\displaystyle R_{2}= \left \{ (1,1),(2,1),(3,3),(4,3),(5,5) \right \}$
0%
$\displaystyle R_{3}= \left \{ (1,1),(1,3),(3,5),(3,7),(5,7) \right \}$
0%
$\displaystyle R_{4}= \left \{ (1,3),(2,5),(4,7),(5,9),(3,1) \right \}$

Explanation

${ R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 2,4 \right) ,\left( 3,5 \right) ,\left( 4,6 \right) ,\left( 5,7 \right) \right\}$
since

$4$ and

$6$ do not belong to

$y$ .

$\therefore \left( 2,4 \right) ,\left( 4,6 \right) \notin { R }_{ 1 }$

$\therefore { R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 3,5 \right) ,\left( 5,7 \right) \right\} \subset X\times Y$
Hence,

${ R }_{ 1 }$ is a relation but not a mapping as the elements

$2$ and

$4$ do not have any image.

${ R }_{ 2 }:$ It is certainly a mapping and since every mapping is a relation, it is a relation as well.

${ R }_{ 3 }:$ It is a relation being a subset of

$X\times Y$ .

${ R }_{ 4 }:$ It is both mapping and a relation. Each element in

$X$ has a unique image. It is also one-one and on-to mapping and hence a bijection.

In order that a relation

$R$ defined on a non-empty set

$A$ is an equivalence relation.
It is sufficient, if

$R$

0%
is reflexive
0%
is symmetric
0%
is transitive
0%
possesses all the above three properties.

Let

$A= \{ 1,2,3,.......50\}$ and

$B=\{2,4,6.......100\}$ .The number of elements

$\left ( x, y \right )\in A\times B$ such that

$x+y=50$

0%
$24$
0%
$25$
0%
$50$
0%
$75$

Explanation

The elements will be

$(2,48),(48,2)$

$(4,46), (46,4)$
:
:

$(2n,50-2n), (50-2n,2n)$
Now we have

$2,4,6,8...$ upto

$48$
This forms an A.P
The number terms is

$24$ .

Let

$\displaystyle A= \left \{ a,b,c \right \}$ and

$\displaystyle B= \left \{ 4,5 \right \}$ Consider a relation

$R$ defined from set

$A$ to set

$B$ then

$R$ is subset of

0%
$A$
0%
$B$
0%
$\displaystyle A\times B$
0%
$\displaystyle B\times A$

Explanation

If A and B are two non empty sets, the relation from set

$A$ to set

$B$ is denoted as

$\displaystyle A\times B$

$A = \{5, 7\}, B= \{7, 9\}$ and

$C = \{7, 9, 11\},$ find

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

0%
$\{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$
0%
$\{(5, 7), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$
0%
$\{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$
0%
none of these

Explanation

Product of two sets is found by forming ordered pairs by multiplying every element of the first set with the second set.
So,

$A \times B =$ {

$(5,7), (5,9), (7,7),(7,9)$ }
And

$A \times C =$ {

$(5,7), (5,9), (5,11), (7,7),(7,9), (7,11)$ }
Union of two sets has all the elements of both the sets.
So,

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

$(5,7), (5,9), (5,11), (7,7),(7,9), (7,11)$ }

Let

$A=\left \{ 1,2,3 \right \}$ and

$B=\left \{ a,b \right \}$ .Which of the following subsets of

$A\times B$ is a mapping from

$A$ to

$B$

0%
$\left \{ \left ( 1,a \right ),\left ( 3,b \right ),\left ( 2,a \right ),\left ( 2,b \right ) \right \}$
0%
$\left \{ \left ( 1,b \right ),\left ( 2,a \right ),\left ( 3,a \right ) \right \}$
0%
$\left \{ \left ( 1,a \right ),\left ( 2,b \right ) \right \}$
0%
none of these

Explanation

$A=\left\{ 1,2,3 \right\} \\ B=\left\{ a,b \right\} \\ A\times B=\left\{ \left( 1,a \right) ,\left( 2,a \right) ,\left( 3,a \right) ,\left( 1,b \right) ,\left( 2,b \right) ,\left( 3,b \right) \right\} .$

$\left\{ \left( 1,a \right) ,\left( 3,b \right) ,\left( 2,a, \right) \left( 2,b \right) \right\} \subset A\times B$

$\left\{ \left( 1,b \right) ,\left( 2,a \right) ,\left( 3,a \right) \right\} \subset A\times B$

$\left\{ \left( 1,a \right) ,\left( 2,b \right) \right\} \subset A\times B$

$A=\{2, 4\}$ and

$B=\{3, 4, 5\}$ then

$(A\cap B)\times (A\cup B)$ is

0%
$\{(2, 2), (3, 4), (4, 2), (5, 4)\}$
0%
$\{(2, 3), (4, 3), (4, 5)\}$
0%
$\{(2, 4), (3, 4), (4, 4), (4, 5)\}$
0%
$\{(4, 2), (4, 3), (4, 4), (4, 5)\}$

Explanation

$A\cap B=\{4\}$

$A\cup B=\{2,3,4,5\}$
Hence

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

$=\{(4,2),(4,3),(4,4),(4,5)\}$

$R$ is an anti symmetric relation in

$\displaystyle A$ such that

$(a,b),(b,a)\:\in\: R$ then

0%
$\displaystyle a\leq b$
0%
$\displaystyle a= b$
0%
$\displaystyle a\geq b$
0%
None of these

Explanation

For

$(a,b), (b,a)$ belongs to the same relation which is anti symmetric that is possible only if the relation is reflexive i.e

$a=b$

$R$ is a relation from

$\displaystyle \left \{ 11,12,13 \right \}$ to

$\displaystyle \left \{ 8,10,12 \right \}$ defined by

$y=x-3$ then the

$R^{-1}$ .

0%
$\displaystyle \left \{ (11,8),(13,10) \right \}$
0%
$\displaystyle \left \{ (8,11),(10,13) \right \}$
0%
$\displaystyle \left \{ (8,11),(9,12),(10,13) \right \}$
0%
$None\ of\ these$

Explanation

$x=\{11,12,13\}, y=\{8,10,12\}$
Now,

$y=x-3$

So, the relation

$R$ becomes

$\{(11,8),(13,10)\}$
So,

$R^{-1}$ becomes

$\{(8,11),(10,13)\}$

Let

$A$ and

$B$ be two finite sets having

$m$ and

$n$ elements respectively. Then the total number of mapping from

$A$ to

$B$ is

0%
$mn$
0%
$\displaystyle 2^{mn}$
0%
$\displaystyle m^{n}$
0%
$\displaystyle n^{m}$

Explanation

Consider an element

$\displaystyle a\in A,$ it can be assigned to any of the

$n$ elements of

$B$ i.e. it has

$n$ images. Similarly each of the

$m$ elements of

$A$ can have

$n$ images in

$B.$ Hence the number of mapping is

$\displaystyle n\times n\times n\times ...m \mbox{times}=n^{m}.$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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