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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 7 - MCQExams.com
CBSE
Class 12 Commerce Maths
Matrices
Quiz 7
Say true or false:
If
A
,
B
be two matrices such that they commute, then
A
2
−
B
2
=
(
A
−
B
)
(
A
+
B
)
.
Report Question
0%
True
0%
False
Explanation
A
and
B
commute when
A
B
=
B
A
Hence
(
A
−
B
)
(
A
+
B
)
=
A
2
+
A
B
−
B
A
−
B
2
=
A
2
−
B
2
If
A
=
[
4
2
−
1
1
]
, then
(
A
−
2
I
)
(
A
−
3
I
)
=
Report Question
0%
0
0%
A
0%
I
0%
5
I
Explanation
A
=
[
4
2
−
1
1
]
⇒
(
A
−
2
I
)
(
A
−
3
I
)
=
A
2
−
5
A
I
+
6
I
2
A
2
−
5
A
+
6
I
by Hemilton rule
|
4
−
λ
2
−
1
1
−
λ
|
=
0
after expanding determinant we got;
⇒
4
−
5
λ
+
λ
2
+
2
=
0
⇒
λ
2
−
5
λ
+
6
=
0
⇒
A
2
−
5
A
+
6
I
=
0
Ans: 0
Or
It can be done using scalar multiplication and followed by matrix multiplication.
If
A
×
[
1
2
3
4
5
6
]
=
[
1
2
3
3
2
1
3
1
2
]
then the order of A is _______
Report Question
0%
2
×
3
0%
3
×
3
0%
3
×
2
0%
2
×
2
Explanation
When a matrix of order
m
×
a
is multiplied with a matrix of order
a
×
n
, the resulting matrix is of order
m
×
n
As the second matrix is of order
2
×
3
and the product matrix is
3
×
3
, we get
a
=
2
;
m
=
3
;
n
=
3
Thus, the matrix A should have order
3
×
2
Inverse of a diagonal matrix is
Report Question
0%
Symmetric
0%
A skew-symmetric
0%
A diagonal matrix
0%
None of these
Explanation
Consider a diagonal
[
2
0
0
0
1
0
0
0
3
]
Minor of the matrix
[
4
0
0
0
6
0
0
0
2
]
Co factor of matrix
[
4
0
0
0
6
0
0
0
2
]
[
+
−
+
−
+
−
+
−
+
]
[
4
0
0
0
6
0
0
0
2
]
Adjoint of matrix
=
Transpose of co factor
=
[
4
0
0
0
6
0
0
0
2
]
Set of matrix considered
=
6
A
−
1
=
A
d
j
o
i
n
t
(
A
)
d
e
t
(
A
)
=
1
6
[
4
0
0
0
6
0
0
0
2
]
Clearly
A
−
1
is also diagonal matrix
∴
Inverse of a diagonal matrix is a diagonal matrix
Option C is correct
If
A
=
[
1
2
3
4
]
, then
A
2
−
5
A
is equal to
Report Question
0%
2
I
0%
−
2
I
0%
3
I
0%
Null matrix
Explanation
A
=
[
1
2
3
4
]
,
Also,
A
2
=
[
1
2
3
4
]
[
1
2
3
4
]
=
[
1
×
1
+
2
×
3
1
×
2
+
2
×
4
3
×
1
+
4
×
3
3
×
2
+
4
×
4
]
⇒
A
2
=
[
7
10
15
22
]
5
A
=
5
×
[
1
2
3
4
]
=
[
5
10
15
20
]
A
2
−
5
A
=
[
7
10
15
22
]
−
[
5
10
15
20
]
=
[
2
0
0
+
2
]
=
2
[
1
0
0
1
]
∴
A
2
−
5
A
=
2
I
If
A
+
I
=
[
3
−
2
4
1
]
, then
(
A
+
I
)
⋅
(
A
−
I
)
is equal to
Report Question
0%
[
−
5
−
4
8
−
9
]
0%
[
−
5
4
−
8
9
]
0%
[
5
4
8
9
]
0%
[
−
5
−
4
−
8
−
9
]
Explanation
Given,
A
+
I
=
[
3
−
2
4
1
]
∴
A
−
I
=
A
+
I
−
2
I
=
[
3
−
2
4
1
]
−
[
2
0
0
2
]
=
[
1
−
2
4
−
1
]
∴
(
A
+
I
)
⋅
(
A
−
I
)
=
[
3
−
2
4
1
]
[
1
−
2
4
−
1
]
=
[
3
−
8
−
6
+
2
4
+
4
−
8
−
1
]
=
[
−
5
−
4
8
−
9
]
Two matrices
A
and
B
are multiplied to get
A
B
if
Report Question
0%
both are rectangular
0%
both have same order
0%
number of columns of
A
is equal to rows of
B
0%
number of rows of
A
is equal to no of columns of
B
Explanation
According to condition of multiplication of two matrices:
The columns in first matrix should be same in number as the rows in the second matrix [For
A
B
to be defined, we must have columns in
A
=
rows in
B
]
Hence option C is true.
If
A
=
[
2
−
1
1
−
2
3
−
2
−
4
4
−
3
]
, then
A
2
is equal to
Report Question
0%
Null matrix
0%
Itself
A
0%
Unit matrix
0%
Scalar matrix
Explanation
Given,
A
=
[
2
−
1
1
−
2
3
−
2
−
4
4
−
3
]
∴
A
2
=
[
2
−
1
1
−
2
3
−
2
−
4
4
−
3
]
[
2
−
1
1
−
2
3
−
2
−
4
4
−
3
]
=
[
4
+
2
−
4
−
2
−
3
+
4
2
+
2
−
3
−
4
−
6
+
8
2
+
9
−
8
−
2
−
6
+
6
−
8
−
8
+
12
4
+
12
−
12
−
4
−
8
+
9
]
=
[
2
−
1
1
−
2
3
−
2
−
4
4
−
3
]
=
A
If
A
=
[
2
0
0
0
2
0
0
0
2
]
then
A
5
=
Report Question
0%
5
A
0%
10
A
0%
16
A
0%
32
A
Explanation
Given,
A
=
[
2
0
0
0
2
0
0
0
2
]
⇒
A
=
2
[
1
0
0
0
1
0
0
0
1
]
⇒
A
=
2
I
⇒
A
5
=
2
5
I
5
⇒
A
5
=
2
5
I
⇒
A
5
=
2
4
.2
I
⇒
A
5
=
2
4
.
2
[
1
0
0
0
1
0
0
0
1
]
⇒
A
5
=
16
[
2
0
0
0
2
0
0
0
2
]
.
⇒
A
5
=
16
A
Option
C
is correct.
If
A
and
B
are symmetric matrices, then
A
B
A
is
Report Question
0%
Symmetric
0%
Skew-symmetric
0%
Diagonal
0%
Triangular
Explanation
Given
A
and
B
are symmetric matrices.
⇒
A
T
=
A
and
B
T
=
B
Now, take
(
A
B
A
)
T
⇒
(
A
B
A
)
T
=
A
T
B
T
A
T
⇒
(
A
B
A
)
T
=
A
B
A
Hence,
A
B
A
is also a symmetric matrix.
If
A
is an
m
×
n
matrix such that
A
B
and
B
A
are both defined, then order of
B
is
Report Question
0%
m
×
n
0%
n
×
m
0%
n
×
n
0%
m
×
m
Explanation
Since
A
B
exists, so the number of column in
A
=
Number of column in
B
.
So
B
has
n
row.
Since
B
A
exists, so the number of column in
B
=
Number of column in
A
.
So
B
has
m
column.
∴
B is an
n
×
m
matrix.
If
A
=
[
0
0
1
0
1
0
1
0
0
]
, then
A
−
1
is.
Report Question
0%
−
A
0%
A
0%
1
0%
None of these
Explanation
We have,
A
=
[
0
0
1
0
1
0
1
0
0
]
⇒
|
A
|
=
0
(
0
−
0
)
−
0
(
0
−
0
)
+
1
(
0
−
1
)
⇒
|
A
|
=
−
1
and cofactors of A are
A
11
=
0
,
A
12
=
0
,
A
13
=
−
1
,
A
21
=
0
,
A
22
=
−
1
,
A
23
=
0
,
A
31
=
−
1
,
A
32
=
0
,
A
33
=
0
∴
A
−
1
=
a
d
j
(
A
)
|
A
|
=
−
1
1
[
0
0
−
1
0
−
1
0
−
1
0
0
]
If
A
=
[
1
1
1
1
]
, then
A
100
is equal to.
Report Question
0%
2
100
A
0%
2
99
A
0%
100
A
0%
299
A
Explanation
Given,
A
=
[
1
1
1
1
]
∴
A
2
=
A
⋅
A
=
[
1
1
1
1
]
[
1
1
1
1
]
=
2
[
1
1
1
1
]
=
2
A
Now,
A
4
=
A
2
⋅
A
2
=
2
A
⋅
2
A
=
4
A
2
=
4
×
2
A
=
8
A
=
2
3
A
Similarly,
A
8
=
2
7
A
∴
A
100
=
2
99
A
If
[
1
2
3
−
5
]
, then
A
−
1
is equal to
Report Question
0%
[
5
11
2
11
3
11
−
1
11
]
0%
[
−
5
11
−
2
11
−
3
11
−
1
11
]
0%
[
5
11
2
11
3
11
1
11
]
0%
[
5
2
3
−
1
]
Explanation
Since
A
=
[
1
2
3
−
5
]
∴
|
A
|
=
[
1
2
3
−
5
]
=
−
5
−
6
=
−
11
and
a
d
j
(
A
)
=
[
−
5
−
2
−
3
1
]
∴
A
−
1
=
1
|
A
|
a
d
j
(
A
)
=
−
1
11
[
−
5
−
2
−
3
1
]
=
1
11
[
5
2
3
−
1
]
=
[
5
11
2
11
3
11
−
1
11
]
Let
A
=
[
1
−
1
−
1
2
1
−
3
1
1
1
]
and
10
B
=
[
4
2
2
−
5
0
α
1
−
2
3
]
, if
B
is the inverse of matrix
A
, then
α
is
Report Question
0%
−
2
0%
1
0%
2
0%
5
Explanation
Since,
B
is the inverse of
A
.
ie,
B
=
10
A
−
1
∴
(
10
)
A
−
1
=
[
4
2
2
−
5
0
α
1
−
2
3
]
∴
(
10
)
A
−
1
⋅
A
=
[
4
2
2
−
5
0
α
1
−
2
3
]
A
⇒
10
I
=
[
4
2
2
−
5
0
α
1
−
2
3
]
[
1
−
1
1
2
1
−
3
1
1
1
]
⇒
[
10
0
0
0
10
0
0
0
10
]
=
[
10
0
0
−
5
+
α
5
+
α
−
5
+
α
0
0
10
]
⇒
5
+
α
=
10
⇒
α
=
5
Inverse of the matrix
[
cos
2
θ
−
sin
2
θ
sin
2
θ
cos
2
θ
]
is.
Report Question
0%
[
cos
2
θ
−
sin
2
θ
sin
2
θ
cos
2
θ
]
0%
[
cos
2
θ
sin
2
θ
sin
2
θ
−
cos
2
θ
]
0%
[
cos
2
θ
sin
2
θ
sin
2
θ
cos
2
θ
]
0%
[
cos
2
θ
sin
2
θ
−
sin
2
θ
cos
2
θ
]
Explanation
Let
A
=
[
cos
2
θ
−
sin
2
θ
sin
2
θ
cos
2
θ
]
∴
|
A
|
=
cos
2
2
θ
+
sin
2
2
θ
=
1
and
a
d
j
(
A
)
=
[
cos
2
θ
sin
2
θ
−
sin
2
θ
cos
2
θ
]
∴
A
−
1
=
1
1
[
cos
2
θ
sin
2
θ
−
sin
2
θ
cos
2
θ
]
=
[
cos
2
θ
sin
2
θ
−
sin
2
θ
cos
2
θ
]
If
A
=
[
1
−
2
1
2
1
3
]
and
B
=
[
2
1
3
2
1
1
]
, then
(
A
B
)
T
is equal to
Report Question
0%
[
−
3
−
2
10
7
]
0%
[
−
3
10
−
2
7
]
0%
[
−
3
7
10
2
]
0%
None of these
Explanation
Given,
A
=
[
1
−
2
1
2
1
3
]
,
B
=
[
2
1
3
2
1
1
]
∴
A
B
=
[
2
−
6
+
1
1
−
4
+
1
4
+
3
+
3
2
+
2
+
3
]
=
[
−
3
−
2
10
7
]
Now,
(
A
B
)
T
=
[
−
3
10
−
2
7
]
If
[
1
1
1
1
−
2
−
2
1
3
1
]
[
x
y
z
]
=
[
0
3
4
]
, then
[
x
y
z
]
is equal to.
Report Question
0%
[
0
1
1
]
0%
[
1
2
−
3
]
0%
[
5
−
2
1
]
0%
[
1
−
2
3
]
Explanation
Given,
[
1
1
1
1
−
2
−
2
1
2
1
]
[
x
y
z
]
=
[
0
3
4
]
∴
x
+
y
+
z
=
0
,
x
−
2
y
−
2
z
=
3
,
x
+
3
y
=
z
=
4
On solving these equations, we get
x
=
1
,
y
=
2
and
z
=
−
3
∴
[
x
y
z
]
=
[
1
2
−
3
]
A
=
[
−
2
4
−
1
2
]
, then
A
2
is equal to
Report Question
0%
Null matrix
0%
Unit matrix
0%
[
1
0
0
1
]
0%
[
0
0
0
1
]
Explanation
A
=
[
−
2
4
−
1
2
]
A
.
A
.
=
A
2
=
[
−
2
4
−
1
2
]
[
−
2
4
−
1
2
]
=
[
4
−
4
−
2
×
4
+
4
×
2
−
1
x
−
2
+
2
x
−
1
−
1
×
4
+
2
×
2
]
A
2
=
[
0
0
0
0
]
=
O
=
null matrix
.
If
A
=
[
x
y
z
]
,
B
=
[
a
h
g
h
b
f
g
f
c
]
and
C
=
[
x
y
z
]
Then,
A
B
C
=
0
, if
Report Question
0%
[
a
x
2
+
b
y
2
+
c
z
2
+
2
g
x
y
+
2
f
y
z
+
2
c
z
x
]
=
0
0%
[
a
x
2
+
c
y
2
+
b
z
2
+
x
y
+
y
z
+
z
x
]
=
0
0%
[
a
x
2
+
b
y
2
+
c
z
2
+
2
h
x
y
+
2
b
y
+
2
c
z
]
=
0
0%
[
a
x
2
+
b
y
2
+
c
z
2
+
2
g
z
x
+
2
f
y
z
+
2
h
x
y
]
=
0
Explanation
A
=
[
x
y
z
]
B
=
[
a
h
g
h
b
f
g
f
c
]
c
=
[
x
y
z
]
A
.
B
=
[
x
y
z
]
[
a
h
g
h
b
f
g
f
c
]
=
[
a
x
+
h
y
+
g
z
h
x
+
b
y
+
f
z
g
x
+
f
y
+
c
z
]
A
.
B
.
C
=
(
A
B
)
×
[
x
y
z
]
=
[
a
x
2
+
h
y
x
+
g
y
x
+
h
x
y
+
b
y
2
+
f
z
y
+
g
x
z
+
f
y
z
+
c
z
2
]
A
B
C
=
[
a
x
2
+
b
y
2
+
c
z
2
+
2
h
x
y
+
2
g
z
x
+
2
f
z
y
]
A
B
C
=
0
when,
a
x
2
+
b
y
2
+
c
z
2
+
2
h
x
y
+
2
g
z
x
+
2
f
z
y
=
0
If
A
and
B
are any two matrices, then
Report Question
0%
A
B
=
B
A
0%
A
B
=
I
0%
A
B
=
0
0%
A
B
may or may not be defined
Explanation
Step 1: Finding the order
Matrix multiplication is only defined for two matrices whose order is of the form
a
×
b
and
b
×
c
respectively
Since we cant say anything about the order of the given matrices it may or may not be defined
Hence, AB may or may not be defined
If
P
=
(
2
−
2
−
4
−
1
3
4
1
−
2
−
3
)
, then
P
5
is equal to
Report Question
0%
P
0%
2
P
0%
−
P
0%
−
2
P
Explanation
Given
P
=
(
2
−
2
−
4
−
1
3
4
1
−
2
−
3
)
P
2
=
P
.
P
=
(
2
−
2
−
4
−
1
3
4
1
−
2
−
3
)
(
2
−
2
−
4
−
1
3
4
1
−
2
−
3
)
=
(
4
+
2
−
4
−
4
−
6
+
8
−
8
−
8
+
12
−
2
−
3
+
4
2
+
9
−
8
4
+
12
−
12
2
+
2
−
3
−
2
−
6
+
6
−
4
−
8
+
9
)
=
(
2
−
2
−
4
−
1
3
4
1
−
2
−
3
)
=
P
∴
P
4
=
P
2
=
P
⇒
P
5
=
P
2
=
P
If
A
=
[
2
−
1
−
1
2
]
and
I
is the unit matrix of order
2
, then
A
2
equals
Report Question
0%
4
A
−
3
I
0%
3
A
−
4
I
0%
A
−
I
0%
A
+
I
Explanation
A
=
[
2
−
1
−
1
2
]
A
2
=
A
×
A
=
[
2
−
1
−
1
2
]
×
[
2
−
1
−
1
2
]
=
[
4
+
1
−
2
−
2
−
2
−
2
1
+
4
]
=
[
5
−
4
−
4
5
]
From the given option to get first entry i.e
5
of
A
2
, only option
(
A
)
is satisfied. Or we can check as below
4
A
−
3
I
=
[
8
−
4
−
4
8
]
−
3
×
[
1
0
0
1
]
=
[
8
−
4
−
4
8
]
−
[
3
0
0
3
]
=
[
5
−
4
−
4
5
]
=
A
2
If
A
is a matrix such that
(
2
1
3
2
)
A
(
1
1
)
=
(
1
1
0
0
)
then
A
=
Report Question
0%
(
1
1
0
1
)
0%
(
2
1
)
0%
(
1
0
−
1
1
)
0%
(
2
−
3
)
Explanation
(
2
1
3
2
)
A
(
1
1
)
=
(
1
1
0
0
)
To maintain the rules of matrix multiplication,
A
has to be of the order
2
×
1
(
2
1
3
2
)
(
x
y
)
(
1
1
)
=
(
1
1
0
0
)
⇒
(
2
x
+
y
3
x
+
2
y
)
(
1
1
)
=
(
1
1
0
0
)
⇒
(
2
x
+
y
2
x
+
y
3
x
+
2
y
3
x
+
2
y
)
=
(
1
1
0
0
)
⇒
2
x
+
y
=
1
and
3
x
+
2
y
=
0
⇒
4
x
+
2
y
−
(
3
x
+
2
y
)
=
2
−
0
⇒
x
=
2
and
y
=
−
3
For the matrices
A
=
[
1
3
0
−
1
]
B
=
[
2
2
−
1
4
]
What is
A
B
?
Report Question
0%
[
3
5
−
1
3
]
0%
[
1
−
1
−
1
5
]
0%
[
−
1
14
1
−
4
]
0%
[
2
4
−
1
−
7
]
0%
none of the above
Explanation
A
and
B
are given by
A
=
[
1
3
0
−
1
]
,
B
=
[
2
2
−
1
4
]
Thus,
A
B
=
[
(
1
×
2
)
+
(
3
×
−
1
)
(
1
×
2
)
+
(
3
×
4
)
(
0
×
2
)
+
(
−
1
×
−
1
)
(
0
×
2
)
+
(
−
1
×
4
)
]
=
[
2
−
3
2
+
12
0
+
1
0
−
4
]
=
[
−
1
14
1
−
4
]
The symmetric part of the matrix A=
[
1
2
4
6
8
2
2
−
2
7
]
.
Report Question
0%
[
1
4
3
2
8
0
3
0
7
]
0%
[
1
4
3
4
8
0
3
0
7
]
0%
[
0
−
2
−
1
−
2
0
−
2
−
1
−
2
0
]
0%
[
0
−
2
−
1
2
0
2
−
1
2
0
]
Explanation
Symmetric part of any matrix A is given by
A
+
A
′
2
, where
A
′
transpose of A
So symmetric part of given matrix
[
1
2
4
6
8
2
2
−
2
7
]
is
=
1
2
(
[
1
2
4
6
8
2
2
−
2
7
]
+
[
1
6
2
2
8
−
2
4
2
7
]
)
=
1
2
[
2
8
6
8
16
0
6
0
14
]
=
[
1
4
3
4
8
0
3
0
7
]
If
A
=
[
0
1
1
0
]
, then
A
2
is equal to ______
Report Question
0%
[
0
1
1
0
]
0%
[
1
0
1
0
]
0%
[
1
0
0
1
]
0%
[
0
1
0
1
]
Explanation
We have,
A
=
[
0
1
1
0
]
Hence
A
2
=
A
A
=
[
0
1
1
0
]
[
0
1
1
0
]
=
[
0
×
0
+
1
×
1
0
×
1
+
1
×
0
1
×
0
+
0
×
1
1
×
1
+
0
×
0
]
=
[
1
0
0
1
]
[
3
2
−
1
−
2
4
5
]
[
0
a
b
c
]
=
[
−
4
9
4
−
7
−
10
19
]
Find
a
,
b
and
c
.
Report Question
0%
2
,
−
1
,
3
0%
1
,
−
2
,
3
0%
3
,
−
2
,
1
0%
1
,
2
,
3
Explanation
The product is
[
3
2
−
1
−
2
4
5
]
[
0
a
b
c
]
=
[
2
b
3
a
+
2
c
−
2
b
−
2
c
−
a
5
b
5
c
+
4
a
]
By given, we have
[
2
b
3
a
+
2
c
−
2
b
−
2
c
−
a
5
b
5
c
+
4
a
]
=
[
−
4
9
4
−
7
−
10
19
]
By comparing, we get
b
=
−
2
,
a
=
1
,
c
=
3
If
P
=
(
2
−
2
−
4
−
1
3
4
1
−
2
−
3
)
then
P
5
equals
Report Question
0%
P
0%
2
P
0%
−
P
0%
−
2
P
Explanation
P
=
[
2
−
2
−
4
−
1
3
4
1
−
2
−
3
]
P
2
=
[
2
−
2
−
4
−
1
3
4
1
−
2
−
3
]
[
2
−
2
−
4
−
1
3
4
1
−
2
−
3
]
=
[
2
−
2
−
4
−
1
3
4
1
−
2
−
3
]
=
P
Now,
P
2
=
P
P
5
=
P
2
×
P
3
⇒
P
5
=
P
4
(
∵
P
2
=
P
)
⇒
P
5
=
P
2
×
P
2
P
2
=
P
×
P
⇒
P
5
=
P
2
=
P
∴
P
5
=
P
∴
Option A is correct
For a matrix
A
(
1
0
0
2
1
0
3
2
1
)
, if
U
1
,
U
2
and
U
3
are
3
×
1
column matrices satisfying
A
U
1
=
(
1
0
0
)
,
A
U
2
(
2
3
0
)
,
A
U
3
=
(
2
3
1
)
and
U
is
3
×
3
matrix whose columns are
U
1
,
U
2
and
U
3
Then sum of the elements of
U
−
1
is
Report Question
0%
6
0%
0
(
z
e
r
o
)
0%
1
0%
2
/
3
Explanation
Let
U
i
=
(
a
i
b
i
c
i
)
i
=
1
,
2
,
3
A
U
1
=
(
a
1
2
a
1
+
b
1
3
a
1
+
2
b
1
+
c
1
)
=
(
1
0
0
)
, So
a
1
=
1
,
b
1
=
−
2
,
c
1
=
1
A
U
2
=
(
a
2
2
a
2
+
b
2
3
a
2
+
2
b
2
+
c
2
)
=
(
2
3
0
)
,
So,
a
2
=
2
,
b
2
=
−
1
,
c
2
=
−
4
. Similarly,
a
3
=
2
,
b
3
=
−
1
,
c
3
=
−
3
So,
U
=
(
1
2
2
−
2
−
1
−
1
1
−
4
−
3
)
. So, sum of elements of
U
−
1
is zero.
0:0:2
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2
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5
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23
24
25
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2
3
4
5
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7
8
9
10
11
12
13
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16
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30
0
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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