MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 9

Determine all ordered pairs that satisfy

$(x - y)^{2} + x^{2} = 25$ , where

$x$ and

$y$ are integers and

$x \geq 0$ . Find the number of different values of

$y$ that occur

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

${ \left( x-y \right) }^{ 2 }+{ x }^{ 2 }=25$
Now,

${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$

$9+16=25$

$\therefore$ There are 2 possibilities:

$I.{ \left( x-y \right) }^{ 2 }=9$ and

${ x }^{ 2 }=16$

$\therefore x=\pm 4$ and

$x-y=\pm 3$

$\left( i \right) .x-y=3\Rightarrow \left( 4,1 \right)$ and

$\left( -4,-7 \right)$

$\left( ii \right) .x-y=-3\Rightarrow \left( 4,7 \right)$ and

$\left( -4,-1 \right)$

$II.{ \left( x-y \right) }^{ 2 }=16$ and

${ x }^{ 2 }=9$

$\therefore x=\pm 3$ and

$x-y=\pm 4$

$\left( i \right) .x-y=4\Rightarrow \left( 3,-1 \right)$ and

$\left( -3,-7 \right)$

$\left( ii \right) .x-y=-4\Rightarrow \left( 3,7 \right)$ and

$\left( -3,-1 \right)$

$\therefore$ Different values of y are

$1,-1,7,-7$

$\therefore 4$ different values of y occur.

Let

$R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$ be a relation on the set

$A = \{1, 2, 3, 4\}$ . The relation

$R$ is

0%
A function
0%
Transitive
0%
Not symmetric
0%
Reflexive

Explanation

Let

$R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$ be a relation on the set

$A = \{1, 2, 3, 4\}$ , then
(a) Since

$(2, 4)$

$\in$

$R$ and

$(2, 3)$

$\in$

$R$ , so

$R$ is not a function.
(b) Since

$(1, 3)$

$\in$

$R$ and

$(3, 1)$

$\notin$

$R$ but

$(1, 1)$

$\notin$

$R$ , so

$R$ is not transitive:
(c) Since

$(2, 3)$

$\in$

$R$ but

$(3, 2)$

$\notin$

$'R$ , so

$R$ is not symmetric.
(d) Since

$(1, 1)$

$\notin$

$R$ , so

$R$ is not reflexive.
Hence, option C is correct.

For any two real numbers

$\theta$

$\phi$ , we define

$\theta R \phi$ if and only if

$\sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \phi } =1$ . The relation

$R$ is

0%
Reflexive but not transitive
0%
Symmetric but not reflexive
0%
Both reflexive and symmetric but not transitive
0%
An equivalence relation

Explanation

Given ,

$\theta$ R

$\phi$ where

$\theta$ and

$\phi$ are two real numbers.

$\sec^{2} { \theta } -\tan ^{2} { \phi } =1$

$\theta=\phi$

R is reflexive and symmetric.

R is not transitive since there are only two numbers.

The relation

$R$ defined on set

$A = \left \{x :|x| < 3, x\epsilon I\right \}$ by

$R = \left \{(x, y) : y = |x|\right \}$ is

0%
$\left \{(-2, 2), (-1, 1), (0, 0), (1, 1), (2, 2)\right \}$
0%
$\left \{(-2, 2), (-2, 2), (-1, 1), (0, 0), (1, -2), (2, -1), (2, -2)\right \}$
0%
$\left \{(0, 0), (1, 1), (2, 2)\right \}$
0%
None of the above

Explanation

$R$ defined on set

$A = \left \{x :|x| < 3, x\epsilon I\right \}$ by

$R = \left \{(x, y) : y = |x|\right \}$

$A = \left \{x :|x| < 3, x\epsilon I\right \}$

$=\left\{x:-3<x<3, x\epsilon I\right\}$

$\therefore$

$R = \left \{(x, y) : y = |x|\right \}$

$=\left\{(-2,2),(-1,1),(0,0),(1,1),(2,2)\right\}$

Hence, option A is correct.

Let the number of elements of the sets

$A$ and

$B$ be

$p$ and

$q$ respectively. Then the number of relations from the set

$A$ to the set

$B$ is

0%
${ 2 }^{ p+q }$
0%
${ 2 }^{ pq }$
0%
$p+q$
0%
$pq$

Explanation

Given

$A$ and

$B$ are two sets with number of elements

$p$ and

$q$ respectively.

The cartesian product of

$A$ and

$B = A \times B = \{(a,b): (a\in A)$ and

$(b\in B)\}$

Number of elements in

$A \times B = |A\times B| =|A|.|B| = pq$

Any relation from

$A$ to

$B$ is a subset of

$A\times B$ .

Hence number of relations from

$A$ to

$B$ is the number of subsets of

$A\times B$

$= 2^{|A\times B|} = 2^{pq}$

A relation

$\rho$ on the set of real number

$R$ is defined as follows:

$x\rho y$ if any only if

$xy > 0$ . Then which of the following is/are true?

0%
$\rho$ is reflexive and symmetric
0%
$\rho$ is symmetric but not reflexive
0%
$\rho$ is symmetric and transitive
0%
$\rho$ is an equivalence relation

Explanation

$x\rho x : x^{2} > 0$ not true for

$x = 0$

$x\rho y\Rightarrow y\rho x$

$x\rho y$ so

$xy > 0, y\rho z$ so

$yz > 0$ .

$xzy^{2} > 0$ , Hence

$xz > 0$ so

$x\rho z$ .

$N$ denote the set of all natural numbers and

$R$ be the relation on

$N\times N$ defined by

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

$ad(b + c) = bc (a + d)$ , then

$R$ is

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

Explanation

For

$(a, b), (c, d)\epsilon N\times N$

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

$\Rightarrow ad(b + c) = bc(a + d)$
Reflexive:

$ab(b + a) = ba(a + b), \forall ab\epsilon N$

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

$R$ is reflexive,
Symmetric:

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

$\Rightarrow bc(a + d) = ad(b + c)$

$\Rightarrow cd(d + a) = da(c + b)$

$\Rightarrow (c, d)R(a, b)$
So,

$R$ is symmetric.
Transitive: For

$(a, b), (c, d), (e, f)\epsilon N\times N$
Let

$(a, b)R(c, d),\ (c, d) R(e, f)$

$\therefore ad(b + c) = bc(a+ d)\ and\ cf(d + e) = de(c + f)$

$\Rightarrow adb + adc = bca + bcd .... (i)$
and

$cfd + cfe = dec + def ... (ii)$
On multiplying eq. (i) by

$ef$ and eq. (ii) by

$ab$ and then adding, we have

$adbef + adcef + cfdab + cfeab$

$= bcaef + bcdef + decab + defab$

$\Rightarrow adcf (b + e) = bcde (a + f)$

$\Rightarrow af (b + e) = be (a + f)$

$\Rightarrow (a, b)R(e, f)$
So,

$R$ is transitive.
Hence,

$R$ is an equivalence relation.

$p - q > 0$ , which of the following is true?

0%
If q = 0, then p < 0
0%
If q < 0, then p < 0
0%
If p > 0, then q > 0
0%
If q = 0, then p > 0
0%
If q < 1, then p > 1

Explanation

Given $p-q>0$
If $q=0$ , then $p-0>0$ , which gives $p>0$
Therefore option $D$ is correct

Let

$R$ be a relation defined on the set

$Z$ of all integers and

$xRy$ when

$x + 2y$ is divisible by

$3$ . Then

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

Explanation

$aRa$ is positive as

$a+2a=3a$ is divisible by

$3$

Hence reflexive

$aRb$ is positive

$a+2b$ is divisible by

$3$

Hence

$3a+3b-(2a+b)$ is divisible by

$3$

$\Rightarrow 2a+b$ is divisible by

$3$

$aRb$ is positive

$\Rightarrow bRa$ is positive hence symmetric

$aRb,bRc$ is

$a+2b,b+2c$ divisible by

$3$

$\Rightarrow a+2b+b+2c$ divisible by

$3$

$\Rightarrow a+3b+2c$ divisible by

$3$

$a+2c$ divisible by

$3$ hence transitive

So it is an equivalence relation.

The relation R define on the set of natural numbers as {(a, b) : a differs from b by 3} is given.

0%
$\{(1, 4), (2, 5), (3, 6), ........\}$
0%
$\{(4, 1), (5, 2), (6, 3), ........\}$
0%
$\{(1, 3), (2, 6), (3, 9), ........\}$
0%
None of above

Explanation

Let

$R=\{(a,b): a, b \in N, a-b=3\}$

$=\{(n+3,n):n \in N\}$

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

Hence, option B is correct.

A relation R in N is defined such that

$xRy \Leftrightarrow x + 4y = 16 ,$ then the range of R is

0%
{ 1 , 2 , 4}
0%
{ 1 , 3 , 4}
0%
{ 1 , 2 , 3}
0%
{ 2 , 3 , 4}

Rule from of relation {(1 ,2) , ( 2 , 5) , (3 , 10) , ( 4 , 17), ......} in N

0%
${(x ,y) : x , y \in N = y = 2x + 1 }$
0%
${( x , y ) : x , y \in N , y = x^{2} + 1 }$
0%
${( x , y ) : x , y \in N , y = 3x - 1}$
0%
${( x , y ) : x , y \in N , y = x + 3}$

The relation 'has the same father as' over the set of children is:

0%
Only reflective
0%
Only symmetric
0%
Only transitive
0%
An equivalence relation

Explanation

Let this relation be called as

$R$ .
Then, if

$xRx$ then clearly

$xRx$ for all

$x$ . Also

$xRy$ it means

$yRx$

$xRy, yRz$ it means

$x$ has same father as

$y$ and

$y$ has same father as

$z$ .

So we can say that,

$x$ has same father as

$z$

Thus,

$xRz$ .
Thus, the given relation is reflexive, symmetric and transitive.

Hence, the relation is an equivalence relation.
Option D is correct.

Let

$X$ be the set of all persons living in a city. Persons

$x, y$ in

$X$ are said to be related as

$x < y$ if

$y$ is at least

$5$ years older than

$x$ . Which one of the following is correct?

0%
The relation is an equivalence relation on $X$
0%
The relation is transitive but neither reflexive nor symmetric
0%
The relation is reflexive but neither nor symmetric
0%
The relation is symmetric but neither transitive nor reflexive

Explanation

Given,

$X$ is the set of persons living ins a city.

Relation(R) = {(x,y): x,y

$\in$

$R$

$\&$ x<y if y is atleast

$5$ years older than x}

For Reflexive:

$(x,x)$ should

$\in R$ for all

$x \in X$

now,

$x$ cannot be be

$5$ years older than himself. So the relation is not reflexive.

For Symmetric: If

$(x,y) \in R \Rightarrow (y,x) \in R$

$(x,y) \in R \Rightarrow y$ is at least

$5$ years older than

$x$ .

$(y,x) \in R \Rightarrow x$ is at least

$5$ years older than

$y$ . This contradicts the above statement. Hence the relation is not symmetric

For Transitive: If

$(x,y) \in R$ and

$(y,z) \in R \Rightarrow (x,z) \in R$ .

$(x,y) \in R \Rightarrow y$ is at least

$5$ years older than

$x$ .

$(y,z) \in R \Rightarrow z$ is at least

$5$ years older than

$y$ .

Then,

$(x,z) \in R \Rightarrow z$ is at least

$5$ years older than

$x$ .

Since,

$z$ is at least

$10$ years older than

$x$ . The relation is transitive.

If A and B are two non-empty sets having n elements in common, then what is the number of common elements in the sets

$A\times B$ and

$B\times A$ ?

0%
$n$
0%
$n^2$
0%
$2n$
0%
Zero

Explanation

Say A has x elements and B has y elements in total.

Their cartesian product

$A\times B$ will have

$x\times y$ elements.

Hence if they have n elements in common.

$n^2$ common elements are present in the products

$A\times B$ and

$B\times A$

Let

$A=\left\{ x\in W,the\quad set\quad of\quad whole\quad numbers\quad and\quad x<3 \right\}$

$B=\left\{ x\in N,the\quad set\quad of\quad natural\quad numbers\quad and\quad 2\le x<4 \right\}$ and

$C=\left\{ 3,4 \right\}$ , then how many elements will

$\left( A\cup B \right) \times C$ conatin?

0%
$6$
0%
$8$
0%
$10$
0%
$12$

Explanation

$A=\{0,1,2\}, B=\{2,3\}$ and

$C=\{3,4\}$

$A \cup B=\{0,1,2,3\}$
No. of elements in

$(A\cup B)\times C$

$=$ No. of elements in

$(A \cup B)\times$ No. of elements in

$C$

$=4\times 2=8$

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

$B = \left\{ 2,3 \right\}$ and

$C = \left\{ 3,4 \right\}$ , then what is the cardinality of

$\left( A\times B \right) \cap \left( A\times C \right)$

0%
$8$
0%
$6$
0%
$2$
0%
$1$

Explanation

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

$B = \left\{ 2,3 \right\}$ and

$C = \left\{ 3,4 \right\}$

Now,

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

$A\times C=\{ (1,3),(1,4),(2,3),(2,4)\}$ And

$(A\times B)\cap (A\times C)=\{ (1,3),(2,3)\}$

So cardinality is

$2$ .

Hence, option C is correct.

Let

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

$B=\left\{ x,y,z \right\}$ . What is the number of elements in

$A\times B$ ?

0%
$6$
0%
$7$
0%
$12$
0%
$64$

Explanation

Given sets are

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

$\{x,y,z\}$

$A\times B=\{(a,x),(a,y),(a,z),(b,x),(b,y),(b,z),(c,x),(c,y),(c,z),(d,x),(d,y),(d,z)\}$

No. of elements in

$A\times B$ is

$3\times 4=12$

Let

$Z$ be the set of integers and

$aRb$ , where

$a, b\epsilon Z$ if an only if

$(a - b)$ is divisible by

$5$ .
Consider the following statements:

$1.$ The relation

$R$ partitions

$Z$ into five equivalent classes.

$2.$ Any two equivalent classes are either equal or disjoint.
Which of the above statements is/are correct?

0%
$1$ only
0%
$2$ only
0%
Both $1$ and $2$
0%
Neither $1$ nor $2$

Explanation

A relation is defined on

$Z$ such that

$aRb \Rightarrow (a-b)$ is divisible by

$5$ ,

For Reflexive:

$(a,a) \in R$ .

Since,

$(a-a) = 0$ is divisible by

$5$ . Therefore, the realtion is reflexive.

For symmetric: If

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

$(a,b) \in R \Rightarrow (a-b)$ is divisible by

$5$ .

Now,

$(b-a) =-(a-b)$ is also divisible by

$5$ . Therefore,

$(b,a) \in R$

Hence, the relation is symmetric.

For Transitive: If

$(a,b) \in R$ and

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

$(a,b) \in R \Rightarrow (a-b)$ is divisible by

$5$ .

$(b,c) \in R \Rightarrow (b-c)$ is divisible by

$5$ .

Then

$(a-c)=(a-b+b-c)= (a-b)+(b-c)$ is also divisible by

$5$ . Therefore,

$(a,c) \in R$ .

Hence, the relation is transitive.

There, the relation is equivalent.

Now, depending upon the remainder obtained when dividing

$(a-b)$ by

$5$ we can divide the set

$Z$ iinto

$5$ equivalent classes and they are disjoint i.e., there are no common elements between any two classes.

Therefore, the correct option is

$(C)$ .

$A$ is a finite set having

$n$ elements, then the number of relations which can be defined in

$A$ is

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

Explanation

A relation is simply a subset of cartesian product

$A\times A$ .

${A\times A}= [(a_1,a_1), (a_1,a_2),.....(a_1,a_n),$

$(a_2,a_1),(a_2,a_2),........(a_2,a_n)$

$......$

$(a_n,a_1), (a_n,a_2).........(a_n,a_n)]$

We can select first element of ordered pair in

$n$ ways and second element in

$n$ ways.

So, clearly this set of ordered pairs contain

$n^2$ pairs.

Now, each of these

$n^{2}$ ordered pairs can be present in the relation or can't be. So, there are

$2$ possibilities for each of the

$n^{2}$ ordered pairs.

Thus, the total no. of relations is

$2^{n^{2}}$ .

Hence, option C is correct.

A and B are two sets having

$3$ elements in common. If

$n(A)=5, n(B)=4$ , then what is

$n(A\times B)$ equal to?

0%
$0$
0%
$9$
0%
$15$
0%
$20$

Explanation

$n(A) =5$ and

$n(B) = 4$

For this type cases we know that the formula for the no. of elements in

$n(A\times B)$ =

$5\times4 = 20$

If A = { a ,b , c} , then numbers of possible non zero relations in A is

0%
$511$
0%
$512$
0%
$7$
0%
$8$

Let R be a relation from A = {1, 2, 3, 4} to B = {1, 3, 5} such that
R = [(a, b) : a < b, where a

$\varepsilon$ A and b

$\varepsilon$ B].
What is RoR

$^{-1}$ equal to?

0%
${(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}$
0%
${(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}$
0%
${(3, 3), (3, 5), (5, 3), (5, 5)}$
0%
${(3, 3), (3, 4), (4, 5)}$

Explanation

$R$ gives the set

$\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$

$R^{-1}$ gives the set

$\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4),(3,3)\}$

To compute

$RoR^{-1}$ , we pick 1 element from

$R^{-1}$ and its corresponding relation from

$R$ .

eg:

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

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

Similarly, computing for all such pairs, we have

$RoR^{-1}=\{(3,3),(3,5),(5,3),(5,5)\}$

Let

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

$R$ be a relation in

$A$ given by

$R = \{ (1,1) , (2,2) (3,3) , (4,4) , (1,2) , (3,1) , (1,3) \}$ then

$R$ is :

0%
Reflexive and transitive only
0%
Transitive and symmetric only
0%
An equivalence relation
0%
None of the above

Explanation

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

$R = \{ (1,1) , (2,2),(3,3),(4,4),(1,2),(3,1)(1,3) \}$
clearly

$(1,1) \epsilon R, (1,2) \epsilon (2,2) \epsilon R , (3,3) \epsilon R, (4,4) \epsilon R$

$\Rightarrow R$ is Reflective.
also

$(1,1) \epsilon R, (1,2) \epsilon R \Rightarrow (1,2) R$
and

$(1,2) \epsilon R, (1,3) \epsilon R \Rightarrow (1,3) \epsilon R \Rightarrow R$ is transitive
But for

$(1,2) \epsilon R (2,1) \notin R, \Rightarrow R$ is not symmetric

$\Rightarrow R$ is reflective and transitive only.

How many ordered pairs of (m, n) integers satisfy

$\displaystyle\frac{m}{12}=\frac{12}{n}$ ?

0%
$30$
0%
$15$
0%
$12$
0%
$10$

Explanation

$m \times n = 144$

144 can be written in multiplication of two integers in the following ways:

$1 \times 144 = 2 \times 72 = 3 \times 48 = 4 \times 36 = 6 \times 24 = 8 \times 18 = 9 \times 16 = 12 \times 12$

For all the factors except

$(12, 12)$ there would be 14 ordered pairs.

Adding one more, yields 15 possibilities

${ y }^{ 2 }={ x }^{ 2 }-x+1$ and

$\quad { I }_{ n }=\int { \cfrac { { x }^{ n } }{ y } } dx$ and

$A{ I }_{ 3 }+B{ I }_{ 2 }+C{ I }_{ 1 }={ x }^{ 2 }y$ then ordered triplet

$A,B,C$ is

0%
$\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right)$
0%
$\left( 3,1,0 \right)$
0%
$\left( 1,-1,2 \right)$
0%
$\left( 3,-\cfrac { 5 }{ 2 } ,2 \right)$

Explanation

We have given

${y^2} = {x^2} - x + 1$

${I_n} =\displaystyle \int {\dfrac{{{x^n}}}{y}dx}$

And,

$A{I_3} + B{I_2} + C{I_1} = {x^2}y$

Order triplet

$A,\, B,\,C$

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

Hence, the option

$(A)$ is correct.

The number of ordered pairs

$(x, y)$ of real numbers that satisfy the simultaneous equations.

$x + y^{2} = x^{2} + y = 12$ is.

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

Explanation

$x + y^{2} = x^{2} + y = 12$

$x + y^{2} = x^{2} + y$

$x - y = x^{2} - y^{2} \Rightarrow x = y, x + y = 1$
when

$x = y, x^{2} + x = 12$

$x^{2} + x - 12 = 0$

$x^{2} + 4x - 3x - 12 = 0$

$(x + 4)(x - 3) = 0$

$x = -4, 3$

$(3, 3),(-4, -4)$
When

$y = 1 - x$

$x + (1 - x)^{2} = 12$

$x^{2} - x - 11 = 0$

$x = \dfrac {1\pm \sqrt {1 + 44}}{2}$ for two value of

$x$ there are two value of

$y$ .
So four pair.

$R$ is a relation on

$N$ given by

$R = \left \{(x, y)|4x + 3y = 20\right \}$ . Which of the following doesnot belong to

$R$ ?

0%
$(-4, 12)$
0%
$(5, 0)$
0%
$(3, 4)$
0%
$(2, 4)$

Explanation

$R$ is given by

$\{(x,y)|4x+3y=20\}$

$4 \times (-4)+3 \times 12=-16+36=20$

$4 \times 5+3 \times 0=20+0=20$

$4 \times 3+3 \times 4=12+12=24\neq 20$

$4 \times 2+3 \times 4=8+12=20$

So, among the given options

$C$ does not satisfy the given condition.

All except

$C$ , does not belong to the set

$R.$

Hence, option C is correct.

$n (A) = 5$ and

$n (B) = 7 ,$ then the number of relations on

$A \times B$ is :

0%
$2^{35}$
0%
$2^{49}$
0%
$2^{25}$
0%
$2^{70}$
0%
$2^{35 \times35 }$

Explanation

Given,

$n (A) = 5$ and

$n (B) = 7$
Therefore, number of relations on

$A \times B = 2^{[ n(A) \times n(B) ]}$

$= 2^{(5 \times 7)} = 2^{(35)}$

The number of ordered pairs

$\left( m,n \right)$ , where

$m, n \in \left\{ 1,2,3,\dots ,50 \right\}$ , such that

${ 6 }^{ m }+{ 9 }^{ n }$ is a multiple of

$5$ is

0%
$1250$
0%
$2500$
0%
$625$
0%
$500$

Explanation

For a number to be divisible by

$5$ the last digit shall be a multiple of

$5$
Any power of 6 has its last digit(or unit's digit) equal to 6.
The power of

$9$ of the form

$4p+1$ has it's last digit as 9

$4p+2$ has it's last digit as 1

$4p+3$ has it's last digit as 9

$4p$ has it's last digit as 1
Thus

$6^{m}+9^{n}$ will be a multiple of

$5$ only if

$m$ is any number from the given set and

$n$ is a number of the form

$4p+1$ or

$4p+3$ .
Therefore the number of ways to select

$m$ is

$\binom{50}{1}$ and the number of ways to select

$n$ from the given set is

$\binom{25}{1}$
hence the total number of ordered pairs

$(m,n)$ is

$\binom{50}{1} \times \binom{25}{1}=50 \times 25=1250$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 9 - 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 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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