MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 1

What is general representation of ordered pair for two variables

$a$ and

$b$ ?

0%
$a, b$
0%
$(a, b)$
0%
$(a), b$
0%
$a, (b)$

Explanation

An ordered pair is written in the form

$(x-$ coordinate,

$y-$ coordinate

$)$ .

Ordered pairs

$(x, y)$ and

$(-1, -1)$ are equal if

$y = -1$ and

$x =$ _____

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

Explanation

Given ,

$(x-y)=(-1,1)$

$x=-1, y=-1$

The value of

$x=-1$ .

Is the following statement is true for

$(a, a)\in R$

$\forall a\in N?$

0%
True
0%
False

Explanation

$(2,2) \in N$

But Not

$\in R$

$2^2 \neq 2$

So given statement is False.

If R={

$(x,y)/3x+2y=15$ and x,y

$\displaystyle \epsilon$ N}, the range of the relation R is________

0%
$\{1,2\}$
0%
$\{1,2,..,5\}$
0%
$\{1,2,.,7\}$
0%
$\{3,6\}$

Explanation

Given,

$3x + 2y = 15$

$=> y = \dfrac {15-3x}{2}$
As both

$x , y$ are natural numbers, we should find

$x$ such that

$15-3x$ is divisible by

$2$

We can see that when

$x =1$ or

$3$ , we get

$y = 6$ or

$3$

Hence, the range is

${3,6}$

$x$ co-ordinate of a point is

$2$ and

$y$ co-ordinate is

$0$ , then ordered pair for its coordinate on

$XY$ plane is

0%
$(0, 0)$
0%
$(2, 2)$
0%
$(0, 2)$
0%
$(2, 0)$

Explanation

Ordered pair for co-ordinate of a point in

$XY$ plane is written as

$(x, y).$

So, option D is correct.

Let

$R$ be the relation in the set

$N$ given by

$R=\left\{(a, b): a=b-2, b>6\right\}$ . 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

$R=\left\{(a, b):a=b-2, b>6\right\}$
Now, since

$b>6, (2, 4)\notin R$
Also, as

$3\neq 8-2, (3, 8)\notin R$
And, as

$8\neq 7-2$

$\therefore (8, 7)\notin R$
Now, consider

$(6, 8)$ .
We have

$8 > 6$ and also,

$6=8-2$ .

$\therefore (6, 8)\in R$
Hence the correct answer is C.

If ordered pair

$(a, b)$ is given as

$(-2, 0)$ , then

$a =$

0%
$-2$
0%
$0$
0%
$2$
0%
None of the above

Explanation

$a$ will be first entry in ordered pair

$(a, b).$

So, option A is correct.

$x$ and

$y$ coordinate of a point is

$(3, 10)$ , then the

$x$ co-ordinate is

0%
$0$
0%
$3$
0%
$10$
0%
None of the above

Explanation

Here,

$x$ co-ordinate will be the first entry in ordered pair.

So, option B is correct.

What are the ways of representing a relation?

0%
Set builder form
0%
Roster form
0%
Arrow diagram
0%
All of these

$x$ and

$y$ co-ordinate of a point is

$(3, 10)$ , then

$y$ co-ordinate is

0%
$0$
0%
$3$
0%
$10$
0%
None of the above

Explanation

$y$ coordinate will be second entry in ordered pair.

So, option C is correct.

The first component of all ordered pairs is called

0%
Range
0%
Domain
0%
Function
0%
None of these

$a$ and

$b$ are two variables and

$(a, b)=(b, a)$ , then

0%
$a = 0$
0%
$b = 0$
0%
$a = b$
0%
$\displaystyle a\pm b$

Explanation

$(a, b) = (b, a)$ only if

$a = b.$ .

So, option C is correct.

The second component of all ordered pairs of a relation is

0%
Range
0%
Domain
0%
mapping
0%
none of these

Which of the ordered pair satisfies the relation

$x<y$ ?

0%
$(1,2)$
0%
$(2,1)$
0%
$(1,1)$
0%
None of these

Explanation

$(A) (1,2)$

$x<y$

$1<2$

True ,

$x<y$

$(B) (2,1)$

$2>1$

$(C) (1,1)$

$x=y$

$y$ is second entry and

$x$ is first entry then its ordered pair will be ............

0%
$(x, y)$
0%
$(y, y)$
0%
$(y, x)$
0%
None of the above

Explanation

$x$ is first entry and

$y$ is second, then its ordered pair will be

$(x, y)$ .

So, option A is correct.

x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x =If y = 16 and z = 7 what is x?

0%
180
0%
160
0%
280
0%
200

Explanation

Given x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x = 10. If y = 16 and z = 7

Then

$x=k\frac{y}{z^{2}}$

$\Rightarrow 10=k\times \frac{4}{14\times14}$

$\Rightarrow k=490$

Thus

$x=490\times \frac{16}{7\times 7}=160$

Let N denote the set of all natural numbers and R a relation on

$N\times N$ . Which of the following is an equivalence relation?

0%
$(a, b) R (c, d)$ if $ad(b+c)=bc(a+d)$
0%
$(a, b) R (c, d)$ if $a+d=b+c$
0%
$(a, b) R (c, d)$ if $ad=bc$
0%
$(a, b) R (c, d)$ if $a-d=b-c$

Explanation

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

$ad(b+c)=bc(a+d)$ defined on

$N\times N$ is reflexive, symmetric and transitive. So, it is equivalence relation

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

$a+d=b+c$ and

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

$ad=bc$ is also equivalence relation.

$A$ and

$B$ are two sets having

$3$ and

$4$ elements respectively and having

$2$ elements in common. The number of relations which can be defined from

$A$ to

$B$ is:

0%
$2^{5}$
0%
$2^{10}-1$
0%
$2^{12}-1$
0%
$2^{12}$

Explanation

Given:

$n(A) = 3 = m$

$n(B) = 4 = n$

Number of relation

$= { 2 }^{ mn }$

$= { 2 }^{ 3\times 4 }$

$= { 2 }^{ 12 }$ .

$N$ is the set of positive integers. The relation

$R$ is defined on N x N as follows:

$(a,b) R (c,d)\Longleftrightarrow ad=bc$ Prove that

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

Explanation

Reflexive: let

$(a,b)$ be an ordered pair of positive integer. to show R is reflexive we must show

$\left( \left( a,b \right) ,\left( a,b \right) \right) \in R$ . Multiplication of integer is commutative, so

$ab=ba$ . Thus

$\left( \left( a,b \right) ,\left( a,b \right) \right) \in R$

Symmetric: Let

$(a,b)$ and

$(c,d)$ be an ordered pair of positive integer such that

$\left( a,b \right) R\left( c,d \right)$ . Then

$ad=bc$ . This equation is equivalent to

$cb=da$ , so

$\left( c,d \right) R\left( a,b \right)$ . This shows R is symmetric

Transitive: Let

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

$(e,f)$ be ordered pairs of positive integer such that

$\left( a,b \right) R\left( c,d \right)$ and

$\left( c,d \right) R\left( e,f \right)$ . Then

$ad=bc$ and

$cf=de$ .
Thus

$adf=bcf$ and

$bcf=bde$ , which implies

$adf=bde$ .
Since

$d\neq 0$ , we can cancel it from both sides of this equation to get

$af=be$ .
This shows

$\left( a,b \right) R\left( e,f \right)$ and R is transitive

Let

$R$ be a relation from a set

$A$ to a set

$B$ , then:

0%
$R=A\cup B$
0%
$R=A\cap B$
0%
$R\subseteq A\times B$
0%
$R\subseteq B\times A$

Explanation

$R :$

$A\rightarrow B$

then

$R$ is a subset

$A\times B$

$\therefore$

$R\quad \subseteq \quad A\times B$

Let

$x$ be a real number

$\left [ x \right ]$ denotes the greatest integer function, and

$\left \{ x \right \}$ denotes the fractional part and

$(x)$ denotes the least integer function,then solve the following.

$\left [ 2x \right ]-2x=\left [ x+1 \right ]$

0%
$\left \{ -1,\displaystyle-\frac{1}{2} \right \}$
0%
$\left \{ -1,\displaystyle \frac{1}{2} \right \}$
0%
$\left \{ 1,\displaystyle-\frac{1}{2} \right \}$
0%
$\left \{ 1,\displaystyle \frac{1}{2} \right \}$

Explanation

$\left [ 2x \right ]-2x=\left [ x+1 \right ]$

$\Rightarrow \left \{2x\right\}=\left [ x \right ]+1$ (i)

But range of fractional part is

$[0,1)$

$\Rightarrow 0\leq [x]+1 < 1$

$\Rightarrow -1 \leq [x] < 0$

$\Rightarrow -1 \leq x < 0$

$\Rightarrow [x] =-1$

Thus (i) becomes

$\{2x\} =0$ it means that fraction part is zero for that,

$\therefore x = -\cfrac{1}{2}, -1$ in

$[-1,0)$

$A = \{x, y\}$ and

$B = \{3, 4, 5, 7, 9\}$ and

$C = \{4, 5, 6, 7\}$ , find

$\displaystyle A\times (B\cap C)$

0%
$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$
0%
$\displaystyle \left \{ (x,3),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$
0%
$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,9) \right \}$
0%
none of the above

Explanation

$\displaystyle B\cap C=\left \{ 3,4,5,7,9 \right \}\cap \left \{ 4,5,6,7 \right \}=\left \{ 4,5,7 \right \}$

$\displaystyle \therefore A\times (B\cap C)=\left \{ x,y \right \}\times \left \{ 4,5,7 \right \}$
=

$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$

$\displaystyle x^{2} = xy$ is a relation (defined on set R) which is

0%
Symmetric
0%
Reflexive
0%
Transitive
0%
None of these

Explanation

Given , R:

$x^{2}=xy$

Symmetric: If

$(x,x) \in$ R

$x^{2}=x \times x$

$x^{2}=x^{2},(x,x) \in$ R

So the relation is symmetric

Which one of the following relations on R (set of real numbers) is an equivalence relation

0%
$\displaystyle aR_{1}b\Leftrightarrow \left | a \right |= \left | b \right |$
0%
$\displaystyle aR_{2}b\Leftrightarrow a\geq b$
0%
$\displaystyle aR_{3}b\Leftrightarrow a$ divides $b$
0%
$\displaystyle aR_{4}b\Leftrightarrow a< b$

Explanation

$a R_1 b \iff |a| = |b|$

$a=b$

$R_1$ is a reflexive symmetric and transitive.( as we know

$a=b$ )

$\therefore (a,a)$ is present.

$\therefore (a,b) (b,a)$ will also be present.

So it is an equivalence relation .

Suppose y is equal to the sum of two quantities of which one varies directly as x and the other inversely as x If y = 6 when x = 4 and

$\displaystyle y=\frac{10}{3}$ when x = 3 then what is the relation between x and y?

0%
$\displaystyle y=x+\frac{4}{x}$
0%
$\displaystyle y+2x=\frac{4}{x}$
0%
$\displaystyle y=2x+\frac{8}{x}$
0%
$\displaystyle y=2x-\frac{8}{z}$

Explanation

(A)If

$y=X+\frac{4}{x}\Rightarrow 4+\frac{4}{4}=3 \neq 6=6$

(B)

$y+2x=\frac{4}{x}\Rightarrow 6+2(4)=\frac{4}{4}\Rightarrow 6+8\neq 1$

(C)

$y=2x+\frac{8}{x}\Rightarrow 6=2(4)+\frac{8}{4}\Rightarrow 6\neq 8+2$

(D)

$y=2x-\frac{8}{x}\Rightarrow 6=2(4)-\frac{8}{4}=8-2=6$ If x=4 and y=6

$y=2x-\frac{8}{x}\Rightarrow \frac{10}{3}=2\times 3-\frac{8}{3}\Rightarrow \frac{10}{3}=\frac{10}{3}$ If x=3 and y=

$y=\frac{10}{3}$

Then option (D)

$Y=2x-\frac{8}{x}$

If X is brother of the son of Y's son. How is X related to Y?

0%
Son
0%
Brother
0%
Cousin
0%
Grandson
0%
Uncle

$\displaystyle A\prime$ is symmetric to A and

$\displaystyle B\prime$ is symmetric to B with respect to a line of symmetry, then

0%
$\displaystyle AB>A'B'$
0%
$\displaystyle AB=A'B'$
0%
$\displaystyle AB\neq A'B'$
0%
None of these

$A=\left \{x:x^2-3x+2=0\right \}$ and

$B=\left \{x:x^2+4x-5=0\right \}$ then the value of A-B is

0%
$\left \{1, 2\right \}$
0%
$\left \{2\right \}$
0%
$\left \{1\right \}$
0%
$\left \{5, 2\right \}$

Explanation

$A=\left \{x:x^2-3x+2=0\right \}\Rightarrow A=\left \{1, 2\right \}$

$B=\left \{x:x^2+4x-5=0\right \}\Rightarrow A=\left \{1, -5\right \}$

$\therefore A-B=\left \{2\right \}$

The number of reflexive relations of a set with four elements is equal to

0%
2 $^{16}$
0%
2 $^{12}$
0%
2 $^{8}$
0%
2 $^{4}$

Explanation

Total number of reflexive relations in a set with n elements = 2

$^n$ Therefore, total number of reflexive relations set with 4 elements = 2

$^4$

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

Total possible pair=

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

Reflexive means

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

So,

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

Transitive means if

$(a,b)$ is in relation and

$(b,c)$ is in relation.

So if

$(1,2)$ is in relation and

$(2,1)$ is in relation,then

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

Relative

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

So,smallest relation is

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

If we add

$(2,3)$ ,then we have to add

$(3,2)$ also,as it is symmetric but as

$(1,2)\&(2,3)$ are there,we need to add

$(1,2)$ also,as it is transitive.

As we are adding

$(1,3)$ we should add

$(3,1)$ also,as it is symmetric.

Relation

$R2=\{(1,2),(2,1),(1,1),(2,2),(3,3),(2,3),(1,3),(3,1)\}$

Hence only two possible relation are there which are equivalence.

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.