MCQ Questions for Class 12 Commerce Maths Matrices Quiz 2

$A = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 & 8\end{bmatrix},$

$B = \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 & 5\end{bmatrix}$ and

$A+B = \begin{bmatrix}1 & a & b \\ c & 2 & 13\end{bmatrix},$ then find the value of

$a+b+c.$

0%
$12$
0%
$21$
0%
$15$
0%
$5$

Explanation

Given,

$A = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 & 8\end{bmatrix},$

$B = \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 & 5\end{bmatrix}$ and

$\quad A+B = \begin{bmatrix}1 & a & b \\ c & 2 & 13\end{bmatrix}$ ....(1)

Therefore,

$\quad A+B = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 & 8\end{bmatrix} + \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 & 5\end{bmatrix}=\begin{bmatrix}2-1 & 3+4 & 4+7 \\ -3-3 & 4-2 & 8+5\end{bmatrix}= \begin{bmatrix}1 & 7 & 11 \\ -6 & 2 & 13\end{bmatrix}$ .....(2)

By equating (1) and (2), we get

$a+b+c=7+11-6=12$

$A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$ , Then

$A^2$ =

0%
$\begin{bmatrix}8 & -5 \\ -5 & 3\end{bmatrix}$
0%
$\begin{bmatrix}8 & -5 \\ 5 & 3\end{bmatrix}$
0%
$\begin{bmatrix}8 & -5 \\ -5 & -3\end{bmatrix}$
0%
$\begin{bmatrix}8 & 5 \\ -5 & 3\end{bmatrix}$

Explanation

Given

$A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$

$A^{2}= \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix} \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$

$=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

$\displaystyle 2\begin{bmatrix}3 &4 \\5 &x \end{bmatrix}+\begin{bmatrix}1 &y \\0 &2 \end{bmatrix}=\begin{bmatrix}7 &0 \\10 &5 \end{bmatrix}$ , then the values of

$x\ and\ y$ are :

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$\displaystyle x=0,y=-2$
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$\displaystyle x=\dfrac{3}{2},y=-8$
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$\displaystyle x=-2,y=-8$
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$\displaystyle x=2,y=8$

Explanation

$2\left[ \begin{matrix} 3 & 4 \\ 5 & x \end{matrix} \right] +\left[ \begin{matrix} 1 & y \\ 0 & 2 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 6 & 8 \\ 10 & 2x \end{matrix} \right] +\left[ \begin{matrix} 1 & y \\ 0 & 2 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 7 & 8+y \\ 10 & 2x+2 \end{matrix} \right] =\left[ \begin{matrix} 7 & 0 \\ 10 & 5 \end{matrix} \right]$

$\Rightarrow 8+y=0\;\&\;2x+2=5$

$\Rightarrow y=-8\;\&\;2x=3$

$\Rightarrow x=\dfrac { 3 }{ 2 }$

$\therefore x=\dfrac { 3 }{ 2 }\; \&\;y=-8$

Hence, the answer is

$x=\dfrac { 3 }{ 2 }\; \&\;y=-8.$

$\begin{bmatrix} x & y \\ u & v \end{bmatrix}$ is symmetric matrix, then

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$x+v=0$
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$x-v=0$
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$y+u=0$
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$y-u=0$

Explanation

$A =\begin{bmatrix} x & y \\ u & v \end{bmatrix}$

$A^T =\begin{bmatrix} x & u \\ y & v \end{bmatrix}$
So for

$A$ to be symmetric matrix,

$A=A^T$

$\Rightarrow y = u\Rightarrow y-u=0$

$\displaystyle A=\begin{bmatrix}x &y \\z &w \end{bmatrix},B=\begin{bmatrix}x &-y \\-z &w \end{bmatrix}$ and

$C=\begin{bmatrix}-2x &0 \\0 &-2w \end{bmatrix},$ then

$A+B+C$ is a:

0%
identity matrix
0%
null matrix
0%
row matrix
0%
column matrix

Explanation

Given:

$\displaystyle A=\begin{bmatrix}x &y \\z &w \end{bmatrix},B=\begin{bmatrix}x &-y \\-z &w \end{bmatrix}$ and

$C=\begin{bmatrix}-2x &0 \\0 &-2w \end{bmatrix}$

Consider,

$\displaystyle A+B+C=\begin{bmatrix}X &Y \\Z &W \end{bmatrix}+\begin{bmatrix}X &-Y \\-Z &W \end{bmatrix}+\begin{bmatrix}-2X &0 \\0 &-2W \end{bmatrix}$

$=\begin{bmatrix}X+X+\left ( -2X \right ) &Y+\left ( -Y \right )+0 \\Z+\left ( -Z \right )+0 &W+W+\left ( -2W \right ) \end{bmatrix}$

$=\begin{bmatrix}0 &0 \\0 &0 \end{bmatrix}$

Hence,

$A+B+C$ is a null matrix.

If matrix

$A$ is of order

$p\times q$ and matrix

$B$ is of order

$r\times s$ ,then

$A-B$ will exist if

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$p=q$
0%
$p=r, q=s$
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$p=q, r=s$
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$p=s, q=r$

Explanation

If matrix

$A$ is of order

$p\times q$ and matrix

$B$ is of order

$r\times s,$ then

$A-B$ will exist if order of

$A$ and

$B$ is same
Therefore,

$p=r,q=s$

Ans: B

$\displaystyle A=\left [ a_{ij} \right ]_{m\times\:n}, B=\left [ b_{ij} \right ]_{m\times\:n},$ then the element

$\displaystyle C_{23}$ of the matrix

$C=A+B$ is

0%
$\displaystyle C_{32}$
0%
$\displaystyle a_{23}+b_{32}$
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$\displaystyle a_{23}+b_{23}$
0%
$\displaystyle a_{32}+b_{23}$

Explanation

$\displaystyle A=\left [ a_{ij} \right ]_{m\times\:n}$ and

$B=\left [ b_{ij} \right ]_{m\times\:n}$

$\therefore C=A+B$

$\Rightarrow \left [ c_{ij} \right ]_{m\times\:n}=\left [ a_{ij}+b_{ij} \right ]_{m\times\:n}$

Hence,

$c_{23}=a_{23}+b_{23}$

$A-A'=0$ , then

$A'$ is

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orthogonal matrix
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symmetric matrix
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skew-symmetric matrix
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triangular matrix

Explanation

It is fundamental concept that,
If

$A = A'$ then

$A=A'$ is called symmetric matrix.

For any square matrix

$A, \ A+{A}^{T}$ is-

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unit matrix
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symmetric matrix
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skew symmetric matrix
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zero matrix

Explanation

Let,

$B = A+A^T$

$\Rightarrow B^T = (A+A^T)^T =A^T+(A^T)^T =A^T+A =B$
Hence

$B = A+A^T$ is a symmetric matrix.

$\displaystyle \begin{bmatrix} x & y \\ 1 & 6 \end{bmatrix}$ =

$\displaystyle \begin{bmatrix} 1 & 8 \\ 1 & 6 \end{bmatrix},$ then

$x+2y=$

0%
$13$
0%
$17$
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$19$
0%
None of these

Explanation

By equating given matrices, we have

$x=1$ and

$y=8$

$\displaystyle \therefore x+2y=17$

$A= \displaystyle \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$ and B=

$\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ then

$A+B=$

0%
$A$
0%
$B$
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$\displaystyle \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$
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$\displaystyle \begin{bmatrix} 0 & 2 \\ 2 & 2 \end{bmatrix}$

Explanation

Given,

$A= \displaystyle \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$ and B=

$\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Then,

$A+ B = \displaystyle \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix}$

If A is any square matrix, then

$(A\, +\, A^T)$ is a ............ matrix

0%
symmetric
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skew symmetric
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scalar
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identity

Explanation

Lets matrix

$A=\begin{bmatrix} a & b & c\\ d & f & g\\ e & h & i\end{bmatrix}$

Then, matrix

$A^T$ , after transforming rows with each other will be,

$A^T=\begin{bmatrix} a & d & e\\ b & f & h\\ c & g & i\end{bmatrix}$

on adding

$A+A^T$ , we get

$\begin{bmatrix} 2a & (b+d) & (c+e)\\ (b+d) & 2f & (h+g)\\ (e+c) & (h+g) & 2i\end{bmatrix}$

which is clearly symmetry about its diagonal.

$A$ is a square matrix, then

$A-{A}^{T}$ is-

0%
unit matrix
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null matrix
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$A$
0%
a skew symmetric matrix

Explanation

${ \left( A-{ A }^{ T } \right) }^{ T }={ A }^{ T }-{ \left( { A }^{ T } \right) }^{ T }={ A }^{ T }-A=-\left( A-{ A }^{ T } \right)$
Therefore, it is a skew symmetric matrix

Ans: D

$\displaystyle \begin{bmatrix} 2 & 3 \\ 4 & 4 \end{bmatrix}$ +

$\displaystyle \begin{bmatrix} x & 3 \\ y & 1 \end{bmatrix}$ =

$\displaystyle \begin{bmatrix} 10 & 6 \\ 8 & 5 \end{bmatrix},$ then

$(x,y)=$

0%
$(4,8)$
0%
$(8,4)$
0%
$(1,2)$
0%
$(2,4)$

Explanation

By equating given matrices, we have

$2+x=10 \Rightarrow x=8$

Also,

$4+y=8 \Rightarrow y=4$ .

For square matrix

$A$ ,

$A{A}^{T}$ is-

0%
unit matrix
0%
symmetric matrix
0%
skew symmetric matrix
0%
diagonal matrix

Explanation

Since,

${ \left( A{ A }^{ T } \right) }^{ T }={ \left( { A }^{ T } \right) }^{ T }{ A }^{ T }=A{ A }^{ T }$
Therefore,

$A{ A }^{ T }$ is symmetric matrix

Ans: B

$A=\begin{bmatrix} -1 & 0 & 2 \\ 3 & 1 & 2 \end{bmatrix}$ and

$B=\begin{bmatrix} -1 & 5 \\ 2 & 7 \\ 3 & 10 \end{bmatrix},$ then

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$AB$ and $BA$ both exist
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$AB$ exists but bot $BA$
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$BA$ exists but not $AB$
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but $AB$ and $BA$ do not exist

Explanation

Since, order of

$A$ is

$2\times3$ and order of

$B$ is

$3\times 2$
Thus,

$AB$ and

$BA$ exists.

$A$ and

$B$ are symmetric matrices of order

$\displaystyle n,\left( A\neq B \right)$ , then

0%
$A+B$ is skew-symmetric
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$A+B$ is symmetric
0%
$A+B$ is a diagonal matrix
0%
$A+B$ is a zero matrix

Explanation

Since

$A$ and

$B$ are symmetric matrices,

$A' = A$ and

$B'=B$

$\therefore (A+B)'=A'+B'=A+B$

$\therefore A+B$ is symmetric matrix

$A$ is a skew symmetric matrix then

$\displaystyle A^{T}$

0%
$-A$
0%
$A$
0%
$0$
0%
diagonal matrix

Explanation

In mathematics, a skew-symmetric (or anti-symmetric or antimetric) matrix is a square matrix whose transpose is its negation; that is, it satisfies the condition

$-A=A^{T}$

Hence,

$A^{T}=-A$ .

$A= \begin{bmatrix} 1 & 2\end{bmatrix}, B=\begin{bmatrix} 3 & 4\end{bmatrix}$ then

$A+B=$

0%
$\begin{bmatrix}1 & 4\end{bmatrix}$
0%
$\begin{bmatrix}4 & 4\end{bmatrix}$
0%
$\begin{bmatrix}4 & 6\end{bmatrix}$
0%
None of these

Explanation

Given

$A=\begin{bmatrix} 1 & 2 \end{bmatrix}$

$B=\begin{bmatrix} 3 & 4 \end{bmatrix}$

Now,

$A+B=\begin{bmatrix} 1+3 & 2+4 \end{bmatrix}$

$\Rightarrow A+B=\begin{bmatrix} 4 & 6 \end{bmatrix}$

$\therefore$ Option C is correct

$A$ is matrix of order

$\displaystyle m\times n$ and

$B$ is a matrix of order

$\displaystyle n\times p,$ then the order of

$AB$ is

0%
$\displaystyle p\times m$
0%
$\displaystyle p\times n$
0%
$\displaystyle n\times p$
0%
$\displaystyle m\times p$

Explanation

$A$ is matrix of order

$m \times n$ and

$B$ is a matrix of order

$n \times p,$ then the order of

$AB$ is

$m \times p.$

Two matrices

$A$ and

$B$ are added if

0%
both are rectangular
0%
both have same order
0%
no of columns of A is equal to columns of B
0%
no of rows of A is equal to no of columns of B

Explanation

While adding two matrices we add the numbers which belong to some row and column of each matrix or two matrices can be added if there are equal number of rows and columns in both.

Hence, both matrices should have same order.

$\therefore$ option B is correct

Find the output order for the following matrix multiplication

$A_{4 \times 2}\times B_{2\times4}$ ?

0%
$2 \times 4$
0%
$4 \times 4$
0%
$4 \times2$
0%
Multiplication not possible

Explanation

Note:

$m×n$ is the order of matrix

$AB$ if the order of matrix

$A$ is

$m×p$ and the order of

$B$ is

$p×n$ .

2. For matrix multiplication to work, the columns of the second matrix have to have the same number of entries as do the rows of the first matrix.

Here

$A,B$ are conformant matrices, Multiplication is possible.
Therefore,

$A$ is of size

$4 \times 2$ , and

$B$ is of size

$2 \times 4$ , in which case the output is of size

$4 \times 4$ and is the matrix product of

$A$ and

$B$ .

If the matrices

$A=\begin{bmatrix}2 & 1 & 3 \\4 & 1 & 0\end{bmatrix}$ and

$B=\begin{bmatrix}1 & -1\\ 0 & 2 \\5 & 0\end{bmatrix}$ , then AB will be

0%
$\begin{bmatrix}17 & 0 \\4 & -2\end{bmatrix}$
0%
$\begin{bmatrix}4 & 0 \\ 0 & 4\end{bmatrix}$
0%
$\begin{bmatrix}17 & 4 \\0 & -2\end{bmatrix}$
0%
$\begin{bmatrix}0 & 0 \\0 & 0\end{bmatrix}$

Explanation

Given,

$A=\begin{bmatrix}2 & 1 & 3 \\ 4 & 1 & 0\end{bmatrix}$ and

$B=\begin{bmatrix}1 & -1 \\ 0 & 2 \\ 5 & 0\end{bmatrix}$
Now,

$AB=\begin{bmatrix}2\times1+1\times0+3\times5 & 2\times(-1)+1\times2+3\times0 \\ 4\times1+1\times0+0\times5 & 4\times(-1)+1\times2+0\times0\end{bmatrix}$

$=\begin{bmatrix}17 & 0 \\ 4 & -2\end{bmatrix}$

Least number of changes for the expression

$ax^{2} + bxy + cy^{2} + dx + ey + f$ to be symmetric in x and y is

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$a = b, c = d$
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$b = c, e = f$
0%
$a = c, d = e$
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$a = f, b = e, c = d$

Explanation

$ax^2+bxy+cy^2+dx+ey+f$ .......(1)

Replace by

$x$ by

$y$ and

$y$ by

$x$

$\Rightarrow ay^2+byx+cx^2+dy+ex+f$ .......(2)

if (1) and (2) are equal then

$\boxed{a=c\,and\,d=e}$

What is the output for the following matrix multiplication

$A_{3 \times 2}\times B_{2\times 3}$ ?

0%
$(AB)_{3\times 2}$
0%
$(AB)_{3\times 3}$
0%
$(AB)_{2\times 3}$
0%
$(AB)_{2\times 2}$

Explanation

Here

$A,B$ are conformant matrices.
Therefore,

$A$ is of size

$3 \times 2$ , and

$B$ is of size

$2 \times 3$ , in which case the output is of size

$3 \times 3$ and is the matrix product of

$A$ and

$B$ .
The result will be

$(AB)_{3\times 3}.$

Find the output order for the following matrix multiplication

$X_{5 \times 3}\times Y_{3\times 5}$ ?

0%
$5 \times 5$
0%
$3 \times 5$
0%
$5 \times 3$
0%
$3 \times 3$

Explanation

Since,

$X$ is of size

$5 \times 3$ , and

$Y$ is of size

$3 \times 5$

$\therefore$ The matrix product of

$X$ and

$Y$ has order

$5 \times 5.$

Find the value in place of question mark in the following:

$A_{6 \times 2}\times B_{2\times 6} = C_{?\times6}$ ?

0%
$2$
0%
$6$
0%
$12$
0%
None of these

Explanation

$A_{6 \times 2}$ denotes a matrix having

$6$ rows and

$2$ columns.

Similarly,

$B_{2 \times 6}$ denotes a matrix having

$2$ rows and

$6$ columns.

When multiplied, we get

$6$ rows and

$6$ columns based on the matrix multiplication rule.

Thus, if the product of matrices

$A$ and

$B$ is

$C$ , we can write the answer as

$C_{6 \times 6}$

$A=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$ and

$AB=BA=I$ , then B is equal to

0%
$\begin{bmatrix} -\cos\theta & \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$
0%
$\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$
0%
$\begin{bmatrix} -\sin\theta & \cos\theta \\ \cos\theta & \sin\theta\end{bmatrix}$
0%
$\begin{bmatrix} \sin\theta & -\cos\theta \\ -\cos\theta & \sin\theta\end{bmatrix}$

Explanation

Given,

$A=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ and

$AB=BA=I$

$\Rightarrow B=A^{-1}I=A^{-1}$

$=\displaystyle\frac{1}{\cos^2\theta +\sin^2\theta}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$

$\Rightarrow B=\begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$

$A=\begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}$ is symmetric matrix then the value of

$x$ is

0%
$4$
0%
$3$
0%
$-4$
0%
$-3$

Explanation

Since

$A$ is symmetric matrix, then

$A=A'$

$\Rightarrow$

$\begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}=\begin{bmatrix} 3 & 2x+3 \\ x-1 & x+2 \end{bmatrix}$

$\Rightarrow$

$x-1=2x+3$

$\Rightarrow$

$x=-4$

What is the output order for the following matrix multiplication

$A_{2 \times 1}\times B_{1\times 2}$ ?

0%
$1 \times 1$
0%
$2 \times 3$
0%
$2 \times 2$
0%
$2 \times 1$

Explanation

We know that

$m\times n$ is the order of matrix

$AB$ if the order of matrix

$A$ is

$m\times p$ and the order of

$B$ is

$p\times n.$

Since,

$A$ is of size

$2 \times 1$ , and

$B$ is of size

$1 \times 2$ , in which case the output is of size

$2 \times 2$

and is the matrix product of

$A$ and

$B$ .

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Matrices Quiz 2 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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