If $$A = \{2, 3\}$$ and $$B = \{1, 2\}$$, find $$A \times B$$.

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

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

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

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

MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 8

$\displaystyle n\left ( P\times Q \right )=0$ then n(P) can possibly be

0%
0
0%
10
0%
20
0%
Any value

Explanation

$n(P)$ can be of any value as we are not sure of

$n(Q)$
Hence,

$n(P)$ can take any of the given values.

Let R be a relation from N to N defined by

$R=\{(a, b) :a, b\in N$ and

$a=b^2\}$ . Is the following statement true for

$(a, b)\in R\Rightarrow (b, a)\in R ?$ .

0%
True
0%
False

Explanation

No, because

$(4, 2)\in R$ but

$(2, 4)\not{\in}R$ .
1499275_1666348_ans_07c1082c3b7c4a0a985ad773ac7ff5f7.jpg

If a R b, b R c and a R c, and a,b,c,

$\displaystyle \epsilon$ A, then the relation R on the set A is said to be a/an_____ relation

0%
reflexive
0%
symmetric
0%
transitive
0%
equivalence

Explanation

If in a relation, an element

$a$ is related to

$b$ , and

$b$ is related to a third element

$c$ , the relation

$R$ is said to be transitive if

$a$ is also related to

$c$

Let

$R$ be the relation in the set

$\left\{1, 2, 3, 4\right\}$ given by

$R=\left\{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\right\}$ . Choose the correct answer.

0%
$R$ is reflexive and symmetric but not transitive.
0%
$R$ is reflexive and transitive but not symmetric.
0%
$R$ is symmetric and transitive but not reflexive.
0%
$R$ is an equivalence relation.

Explanation

$R=\left\{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\right\}$
It is seen that

$(a, a)\in R$ , for every

$a\in \left\{1, 2, 3, 4\right\}$ .

$\therefore$

$R$ is reflexive.
It is seen that

$(1, 2)\in R$ , but

$(2, 1)\notin R$ .

$\therefore$

$R$ is not symmetric.
Also, it is observed that

$(a, b), (b, c)\in R\Rightarrow (a, c)\in R$ for all

$a, b, c \in \left\{1, 2, 3, 4\right\}$

$\therefore$

$R$ is transitive.
Hence,

$R$ is reflexive and transitive but not symmetric.
The correct answer is B.

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

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

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

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

Explanation

Given,

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

$B=\left \{3,8\right \}$ ,

Therefore,

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

$A\cap B=\left \{1,2,3\right \}\cap \left \{3,8\right \}=\left \{3\right \}$

$\therefore (A\cup B)\times (A\cap B)=\left \{1,2,3,8\right \}\times \left \{3\right \}$

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

$R$ is a relation from a set

$A$ to the set

$B$ and

$S$ is a relation from

$B$ to

$C,$ then the relation

$SoR$

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

Explanation

Since

$R\subseteq A\times B$ and

$S\subseteq B\times C$ , we have
So

$R\subseteq A\times C$ .

$\therefore$ So R is a relation from

$A$ to

$C.$

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

$B = \left \{a, b, c\right \}$ . The relations from

$A$ to

$B$ is

0%
$\left \{(1, 2), (1, 3), (2, 3), (2, 4), (3, 4), (3, 1)\right \}$
0%
$\left \{(a, b), (a, c), (b, a), (b, c), (c, a)\right \}$
0%
$\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$
0%
$\left \{(a, 1), (1, 3), (b, 2), (c, 3), (b, 3), (b, 4)\right \}$

Explanation

$A=\{1,2,3,4\}$

$B=\{a,b,c\}$

Possible Relation

$R:A\to B$

$=\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$

Option C contains Relation

$R:A\to B$

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

$B = \left \{5, 7, 9\right \}$ . The relation from

$B$ to

$A$ is

0%
$\left \{(a, 5), (a, 7), (b, 7), (c, 9)\right \}$
0%
$\left \{(5, 7), (9, 9), (7, 5)\right \}$
0%
$\left \{(5, a), (5, b), (5, c)\right \}$
0%
$\left \{(5, b), (7, c), (7, a), (9, b)\right \}$

Explanation

$A=\{a,b,c\}$

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

$R:B \longrightarrow A$

$R=\{(5,a),(5,b),(5,c),(7,a),(7,b),(7,c),(9,a),(9,b),(9,c)\}$

Option C and D contain relations in R.

What is the first component of an ordered pair

$(1, -1)$ ?

0%
$1$
0%
$-1$
0%
$2$
0%
$0$

Explanation

In an ordered pair

$(x,y)$ , the first component is

$x$ and the second component is

$y$ .

Therefore, in an ordered pair

$(1,-1)$ , the first component is

$1$ .

$R$ is a relation defined in

$R\times T$ by

$(a,b) R (c,d)$ iff

$a-c$ is an integer and

$b=d$ . The relation

$R$ is

0%
an identity relation
0%
an universal relation
0%
an equivalence relation
0%
None of these

Explanation

We have

$R=\left \{((a,b), (c,d)):a-c\in Z\ and \ b=d; a,b,c,d\in R\right \}$ .

Let

$(a,b)\in R\times R$

$\therefore (a,b)R(a,b)$ , because

$a-a=0\in Z$ and

$b=b$

$\therefore$ R is reflexive.

Let

$(a,b) R (c,d).\Rightarrow a-c\in Z$ and

$b=d$

$\Rightarrow c-a\in Z$ and

$d=b$

$\Rightarrow (c,d) R (a,b)\Rightarrow R$ is symmetric.

Let

$(a,b)R(c,d)$ and

$(c,d)R(e,f)$ .

$\Rightarrow a-c\in Z, b=d, c-e\in Z, d=f$

$\Rightarrow (a-c)+(c-e)\in z, b=f$

$\Rightarrow a-e\in z, b=f\Rightarrow (a,b)R(e,f)$

$\Rightarrow R$ is transitive.

$\therefore R$ is an equivalence relation.

$\therefore$ the correct answer is (c).

Let a relation

$R$ be defined by

$R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$ . The relation

$R^{-1}\circ R$ is given by

0%
$\left \{(1,1), (4,4), (7,4), (4,7), (7,7)\right \}$
0%
$\left \{(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right \}$
0%
$\left \{(1,5), (1,6), (3,6)\right \}$
0%
None of these

Explanation

We have

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

$\therefore R^{-1}=\left \{(5,4), (4,1), (6,4), (6,7), (7,3)\right \}$

$(4,4)\in R^{-1}\circ R$ because

$(4,5)\in R$ and

$(5,4)\in R^{-1}$

$(1,1)\in R^{-1}\circ R$ because

$(1,4)\in R$ and

$(4,1)\in R^{-1}$

$(4,4)\in R^{-1}\circ R$ because

$(4,6)\in R$ and

$(6,4)\in R^{-1}$

$(4,7)\in R^{-1}\circ R$ because

$(4,6)\in R$ and

$(6,7)\in R^{-1}$

$(7,4)\in R^{-1}\circ R$ because

$(7,6)\in R$ and

$(6,4)\in R^{-1}$

$(7,7)\in R^{-1}\circ R$ because

$(7,6)\in R$ and

$(6,7)\in R^{-1}$

$(3,3)\in R^{-1}\circ R$ because

$(3,7)\in R$ and

$(7,3)\in R^{-1}$

$\therefore R^{-1}\circ R=\left \{(4,4), (1,1), (4,7), (7,4), (7,7), (3,3)\right \}$ .

$\therefore$ The correct answer is

$B$ .

Given

$(a - 2, b + 3) = (6, 8)$ , are equal ordered pair. Find the value of

$a$ and

$b$ .

0%
$a = 8$ and $b = 5$
0%
$a = 8$ and $b = 3$
0%
$a = 5$ and $b = 5$
0%
$a = 8$ and $b = 6$

Explanation

By equality of ordered pairs, we have

$(a - 2, b + 3) = (6, 8)$
On equating we get

$a - 2 = 6$

$a = 8$

$b + 3 = 8$

$b = 5$
So, the value of

$a = 8$ and

$b = 5.$

What is the second component of an ordered pair

$(3, -0.2)$ ?

0%
$3$
0%
$0.2$
0%
$1$
0%
$-0.2$

Explanation

In an ordered pair

$(x,y)$ , the first component is

$x$ and the second component is

$y$ .

Therefore, in an ordered pair

$(3,-0.2)$ , the second component is

$-0.2$ .

What is the Cartesian product of

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

$B = \left \{a, b\right \}$ ?

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

Explanation

$A$ and

$B$ are two non empty sets, then the Cartesian product

$A \times B$ is set of all ordered pairs

$(a,b)$ such that

$a\in A$ and

$b\in B$ .

Given

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

$B = \{a,b\}$

Hence

$A\times B = \{(1,a),(1,b),(2,a),(2,b)\}$

The minimum number of elements that must be added to the relation

$R =\{(1,2)(2,3)\}$ on the set of natural numbers so that it is an equivalence is

0%
$4$
0%
$7$
0%
$6$
0%
$5$

Explanation

For

$R$ to be equivalence relation, it should be reflexive, symmetric and transitive.

Now, for

$R$ reflexive:

$\Rightarrow (x, x) \in R$ for all

$x \in \{1, 2, 3\}$

$\Rightarrow (1,1), (2,2), (3,3) \in R$

For

$R$ is symmetric:

$\Rightarrow (1,2), (2,3) \in R$

$\Rightarrow (2,1), (3,2) \in R$

For

$R$ is transitive:

$\Rightarrow (1,2), (2,3) \in R$

$\Rightarrow (1,3) \in R$

Also

$(3,1) \in R$ as

$R$ is symmetric.

So, the total number of elements is

$9$ .

Hence, minimum

$7$ elements must be added.

The relation

$R$ defined on the set

$A=\left \{1,2,3,4,5\right \}$ by

$R=\left \{(x,y):|x^2-y^2| < 16\right \}$ is given by

0%
$\left \{(1,1), (2,1), (3,1), (4,1), (2,3)\right \}$
0%
$\left \{(2,2), (3,2), (4,2), (2,4)\right \}$
0%
$\left \{(3,3), (4,3), (5,4), (3,4)\right \}$
0%
None of these

Explanation

We have

$R=\left \{(x,y):|x^2-y^2| < 16\right \}$ .

Take

$x=1,$ we have

$|x^2-y^2| < 16\Rightarrow |1-y^2| < 16$

$\Rightarrow |y^2-1| < 16\Rightarrow y=1,2,3,4$ .

Take

$x=2,$ we have

$|x^2-y^2| < 16 \Rightarrow |4-y^2| < 16$

$\Rightarrow |y^2-4| < 16\Rightarrow y=1,2,3,4$ .

Take

$x=3,$ we have

$|x^2-y^2| < 16\Rightarrow |9-y^2| < 16$

$\Rightarrow |y^2-9| < 16\Rightarrow y=1,2,3,4$ .

Take

$x=4,$ we have

$|x^2-y^2| < 16\Rightarrow |16-y^2| < 16$

$\Rightarrow |y^2-16| < 16\Rightarrow y=1,2,3,4,5$ .

Take

$x=5,$

$|x^2-y^2| < 16 \Rightarrow |25-y^2| < 16$

$\Rightarrow |y^2-25|<16 \Rightarrow y = 4,5$

$\therefore R=\left \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), \\(3,3), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (5,4), (5,5)\right \}$

$\therefore$ The correct answer is

$D$ .

$(x, y)$ and

$(p, q)$ are two ordered pairs. Find the values of

$p$ and

$y$ , if

$(4y + 5, 3p - 1) = (25, p + 1)$

0%
$p = 0, y = 5$
0%
$p = 1, y = 5$
0%
$p = 0, y = 1$
0%
$p = 1, y = 1$

Explanation

Given

$(x,y)=(p,q)$

$(4y + 5, 3p - 1) = (25, p + 1)$
On equating we get

$4y + 5 = 25$

$4y = 25 - 5$

$4y = 20$

$y = 5$

$3p - 1 = p + 1$

$3p - p = 1 + 1$

$2p = 2$

$p = 1$
So, the value of

$p = 1, y = 5$

$(x, y)$ and

$(p, q)$ are two ordered pairs. Find the values of

$x$ and

$p$ , if

$(3x - 1, 9) = (11, p + 2)$

0%
$x = 4, p = 9$
0%
$x = 6, p = 7$
0%
$x = 4, p = 5$
0%
$x = 4, p = 7$

Explanation

Given

$(x,y)=(p,q)$

$(3x - 1, 9) = (11, p + 2)$
By equating

$3x - 1 = 11$

$3x = 12$

$x = 4$

$9 = p + 2$

$p = 7$
So, the value of

$x = 4, p = 7.$

$A \times B = \{(3, a), (3, -1), (3, 0), (5, a), (5, -1), (5, 0)\}$ , find

$A$ .

0%
$\{a, 5\}$
0%
$\{a, -1\}$
0%
$\{0, 5\}$
0%
$\{3, 5\}$

What is the relation for the following diagram?

0%
$R =\{(2, 5), (2, 6), (3, 7)\}$
0%
$R = \{(1, 5), (2, 6), (3, 7)\}$
0%
$R =\{(1, 5), (2, 6), (1, 7)\}$
0%
$R = \{(1, 5), (2, 6), (2, 7)\}$

Explanation

$R:A\to B$

$A\times B = \{(1,5),(2,6),(3,7)\}$

Which of the following do(es) not belong to

$A \times B$ for the sets

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

$B =\{0, 2\}$ ?

0%
$R = \{(1, 0), (2, 2)\}$
0%
$R = \{(1, 1), (2, 1)\}$
0%
$R = \{(1, 0), (1, 2)\}$
0%
$R = \{(1, 2), (2, 2)\}$

Explanation

A relation for the sets

$A$ and

$B$ from

$A$ to

$B$ will be a subset of the

$A\times B.$
Now the ordered pair belonging to

$A\times B$ are

$\{1,2\}\times\{0,2\}$

$=\{(1,0),(1,2),(2,0),(2,2)\}$
Hence,

$\{(1,1)\}$ and

$\{(2,1)\}$ do not belong to

$A\times B$ .

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

$B = \{1, 2\}$ , find

$A \times B$ .

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

$A \times B =$

${(2, 4), (2, a), (2, 5), (1, 4), (1, a), (1, 5)}$ , find

$B$ .

0%
$\{4, 2, 5\}$
0%
$\{4, a, 5\}$
0%
$\{4, 1, 5\}$
0%
$\{2, a, 5\}$

What is the relation for the set

$A =\{-1, 0, 3\}$ and

$B =\{1, 2, 3\}$ ?

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

Explanation

$A\times B=\{(-1,1),(-1,2),(-1,3),(0,1),(0,2),(0,3),(3,1),(3,2),(3,3)\}$

R is subset of

$A\times B$

So option B is ans.

Ordered pairs

$(x, y)$ and

$(3, 6)$ are equal if

$x = 3$ and

$y = ?$

0%
$3$
0%
$6$
0%
$-6$
0%
$-3$

Explanation

Given

$(x,y)= (3,6)$

$x=3$

$y=6$

The value of

$x=3$ and

$y=6$ .

Determine whether each of the following relations are reflexive, symmetric and transitive.

Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}

0%
R is reflexive
0%
R is symmetric
0%
R is transitive
0%
None of these

Let

$S=\{(a, b, c) \in N \times N\times N:a+b+c=21, a \le b \le c\}$ and

$T=\{(a, b, c)\in N\times N\times N: a, b, c\, are\, in\,AP\}$ , where

$N$ is the set of all natural numbers. Then, the number of elements in the set

$S\cap T$ is

0%
$6$
0%
$7$
0%
$13$
0%
$14$

Explanation

We have,

$a+b+c=21$ and

$\dfrac{a+c}{2}=b$

$\Rightarrow a+c+\dfrac{a+c}{2}=21$

$\Rightarrow a+c=14\Rightarrow \dfrac{a+c}{2}=7$

$\Rightarrow b=7$
So, a can take values from

$1$ to

$6$ and

$c$ can have values from

$8$ to

$13$

$a=b=c=7$

$[\because a\le b \le c]$
So, there are

$7$ such triplets.

$x^2=xy$ is a relation which is:

0%
Symmetric
0%
Reflexive
0%
Transitive
0%
All of these

Explanation

$xRx \Rightarrow x^2=x.x$ Hence reflexive

$xRy \Rightarrow x^2=xy. \Rightarrow x=y$

$yRx \Rightarrow y^2=xy \Rightarrow y=x$ Hence symmetric

$xRy$ and

$yRz$

$\Rightarrow x^2=xy\Rightarrow x=y.$ Also

$y^2=yz\Rightarrow y=z$

Since

$x = z, x.x=z.x \Rightarrow x^2=xz \Rightarrow xRz$ . Hence transitive.

Let

$A = \left \{1, 2, 3\right \}$ . Then number of relations containing

$(1, 2)$ and

$(1, 3)$ which are reflexive and symmetric but not transitive is

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

Explanation

Total possible pairs

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

Reflexive means

$(a,a)$ should be in relation.So

$(1,1),(2,2),(3,3)$ should be in a relation.

Symmetric means if

$(a,b)$ is in relation,then

$(b,a)$ should be in relation.

So,since

$(1,2)$ is in relation,

$(2,1)$ should be in relation and since

$(1,3)$ should be in relation.

Transitive means if

$(a,b)$ is in relation,and

$(b,c)$ is in relation,then

$(a,c)$ is in relation.

If we add

$(2,3)$ ,we need to add

$(3,2)$ for symmetric,but it could become transitive then

Relation

$R1=\{ (1,1),(1,2),(1,2),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$

So,there is only

$1$ possible relation.

The relation

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

$\{1, 2, 3\}$ is

0%
Symmetric only
0%
Reflexive only
0%
An equivalence relation
0%
transitive only

Explanation

Given Relation

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

Reflexive: If a relation has

$\{(a,b)\}$ as its element, then it should also have

$\{(a,a), (b,b)\}$ as its elements too.

Symmetric: If a relation has

$(a,b)$ as its element, then it should also have

$\{(b,a)\}$ as its element too.

Transitive: If a relation has

$\{(a,b), (b,c)\}$ as its elements, then it should also have

$\{(a,c)\}$ as its element too.

Now, the given relation satisfies all these three properties.

Therefore, its an equivalence relation.

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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