Vedanta Opt. Maths Book 10 Final (2078)

vedanta Excel In Opt. Mathematics - Book 10

(d) In the given circle, O is the centre and P is the point of contact. What is A
the relation between OP and AB ?
OP
(e) What are the meanings of similarity and line of symmetry ? B

11.7 Inversion Transformation

Transformation is a type of function. So, it may also have its inverse. The inverse of a
transformation is called its inverse. If T is a transformation on a set A to itself which carries
each element 'a' of A into a corresponding element 'b' of A. Then, the transformation that
reverses transformation T, by carrying each element 'b' of A back into its original element 'a'
of A is called inverse transformation. Then, it is denoted by T–1. We can write:

T:aob

T–1 : b o a

The inverse of reflection is the reflection on the same line of reflection. Similarly, the inverse

of translation is a transformation in the same magnitude but opposite in direction. The size
1
of transformation of magnitude p and p , having the same centre are inverse to each other.

Inversion transformation does not preserve collinearity and distance. It can be generalized

as reflection in a line.

Concept of Inversion Circle and Inversion Point

O P=P' OP P' O P' P

Fig (i) Fig (ii) Fig (iii)

In all figure O is the centre of circle, r is radius. In figure (i), the points P and P' are on the
circumference of the circle.

In figure (ii) P lies inside the circle and P' lies outside the circle.

In figure (iii) P' lies inside the circle and P lies outside the circle.

In each of above cases OP. OP' = r2 is true. We can observe that P o P', P' o P the points P

and P' are called relative inverse of each other.

Inversion in a circle is a method of transformation of A
geometric figure into other figure. It is like process of D

reflection on a line. Since the reflection on a line depends O P' P'
on the particular line chosen, inversion in a circle depends

on the particular circle. Let D be a circle with centre O and

radius r and P are any point other than origin. Then, P' is B
called inversion of P.

Inversion is the process of transformation of point P to corresponding set of points P' called
in the inverse points.

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Let D be a fixed circle. Then, in the above figure, O is the inversion centre, D is called
inversion circle, AB is called polar and P is the pole.

For every point P other than O, we define a point P' is the inverse point of P. It is to be noted
that if P is inside the circle P' is outside of the circle and vice-versa.

Features of Inversion Circle

The following are the some important features of inversion circle.

(a) To each point P of the plane other than O, there corresponds an inverse point P'.

(b) If P lies on the circumference of the inversion D, then P = P'

(c) If OP < r, i.e. if P lies inside there circle P' lies outside the circle EFG. Then
OP' > r. Conversely if OP > r, i.e. if P lies outside the circle, P' and inverse point
P' lies inside the circle EFG.

(d) The point P and the inverse point P' can always be interchanged.

(e) If P' is the image of P, P is the image of the P'. Hence, the point and its inverse point
can be always be interchanged. This property can be referred as symmetric feature
of inversion. i.e., (P')' = P

11.8 Construction of Inverse Point Geometrically

There may arise three cases.

(a) When the point P lies on the circumference of the circle then the point O
P' is the inverse point itself. P' = P

(b) When the point P is inside the circle D. Draw D U P'
o OP

a ray OP passing through O and P. A chord UV V
is drawn perpendicular to the ray OoP passing
through the point P. Then tagents UP' and VP'
are drawn to the circle at the points U and V. OU
and OV are joined. The point of intersection of
the tangents UP' and VP' is P' which lies on the
ray OoP. Join OU and OV. Then ‘OPU and ‘OUP'
will be right angles. In right angled triangles
∆OPU and ∆OP'U, we have,

‘OPU = ‘OUP' (Both are right angles)

‘UOP = ‘UOP' (Common angle)

‘OUP = ‘UP'O (the remaining angles)

Hence 'OPU and 'OP'U are similar.

Therefore, from similar triangles ∆OPU and ∆OP'U. we have,

OP = OU or. OP.OP' = OU2
OU OP'
? OP.OP' = r2(i.e. OU = r)

Hence, the point P' is the inverse of point P.

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P
(c) Let the P be outside the inversion circle O P'
D. Join OP. Let M be the mid point of V
OP. Draw another circle E about centre
M and radius OM = MP. Then the two
circles D and E intersect at the points U
and V. Join UP and VP. Then ‘OUP and
‘OVP are inscribed in semi-circle E. D
Hence, they are right angles. Again UP
and VP are tangents to the circle D. The
chord UV meets OP at P'. The point P' is
the inverse of point P. Hence, the point
P' is the inverse of the point P relative to
the circle D.

11.9 Use of Coordinates in Inversion Circle

(a) Centre at the origin (Standard form): P (x, y)
y
The inverse of a point P(x, y) with respect
to a circle with centre at the origin and the
radius r is a point P'(x', y') such that

x' = r2x , y = r2y P'(x',y')
x2 + y2 x2 + y2
Proof: Sx
x' OR

Let O (0, 0) be the centre and r be the radius y'

of inversion circle. Then equation of the
circle is x2 + y2 = r2. Let P'(x', y') be the

inverse point of P(x, y). Then, by definition,
OP.OP' = r2, where,

OP = x2 + y2

OP' = x'2 + y'2

From P and P' draw PSAOX and P'RAOX. Since the points O, P', P are collinear, the
triangles OPS and OP'R are similar. In similar triangles the corresponding sides are
proportional.

OR = P'R = OP'
OS PS OP
x' y' OP' OP
or, xx' = yy' = OP . OP
or, x = y =
OP.OP' = x2 r2 y2
OP2 +
Then,

and x' = r2x
y' = x2 r+2y y2
x2 + y2
Hence the equation of inverse of any point P(x, y) with respect to a circle with centre at

the origin and the radius r is a point P'(x', y) such that

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x' = r2x and y' = r2y
x2 + y2 x2 + y2

Note :

1. When r = 0, x2 + y2 = 0, is the equation of inversion circle. Then x' and y' do not have
any real values, then the center of inverse is a point at infinity.

2. For any point on the circumference of the inversion circle x2 + y2 = r2, its inverse
point P'(x', y') is given by x = x' and y = y'.

(b) Centre at any Point (Central form) y P(x, y)

The equations of inverse of P'(x', y')P'(x', y') S
any point P(x, y) with respect C(h, k) r x'
to a circle with centre at
C(h, k) and radius r is a point R W
P'(x', y') such that O

x' = h + (x – r2 (x – h) k)2 UV
h)2 + (y –
y'
and y' = k + r2 (y – k)
(x – h)2 + (y – k)2
Proof:
x
Let C(h, k) be the centre and r
be the radius of the circle. Then
equation of the inversion circle is
given by

(x – h)2 + (y – k)2 = r2

Let P'(x', y') be the inverse point of P(x, y)

Then by definition, CP.CP' = r2.

Now, CP = (x – h)2 + (y – k)2

CP' = (x' – h)2 + (y' – k)2

Draw CU A OX, P'V A OX, PW A OX. Then also draw CS A PW, CS cuts PV at R. Since
the points C, P' and P are collinear, 'CPS and 'CP'R are similar.

In similar triangles, the corresponding sides are proportional

CR = P'R = CP'
CS PS CP
x' – h y' – k CP'.CP
or, x–h = y–k = CP2

or, x' – h = y' – k = r2
x–h y–k (x – h)2 + (y – k)2
r2 (x – h) r2 (y – k)
Then, x' = h + (x – h)2 + (y – k)2 and y' = k+ (x – h)2 + (y – k)2

Hence, the equation of inverse of any point P(x, y) with respect to a circle with centre
at (h, k) and radius r is a point P'(x', y') such that

x' = h + r2 (x – h) and y' = k + r2 (y – k)
(x – h)2 + (y – k)2 (x – h)2 + (y – k)2

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Worked out Examples

Example 1: From the adjoining figure give the following concepts:
Solutions:
(a) Inversion circle (b) Inversion radius

(c) Centre of inversion (d) Relation between OP, OP' and OA.

From the above figure, N
D

(a) Circle D is the inversion circle A
(b) OA = r, radius of inversion circle
OP P'

R

(c) O is the centre of inversion circle M

(d) By definition of inversion circle

OP.OP' = OA2

or, OP.OP' = r2

Example 2: Find the inverse of the point P, Q, R which are at distances 2, 4, 8 units from
Solutions: the centre of O of the circle with radius 4 units.

Let OP = 2, OQ = 4, OR = 8 units.

Let P', Q', and R' the inverse points of P, Q, and R' respectively.

Now, for inverse point of P.

OP.OP' = r2, r = 4 R

or, 2.OP' = 42

or, OP' = 16 Q
2 Q'
? OP' = 8 units
R'
i.e., P' is at 8 units from O. OP

Similarly, for the inverse point of Q, P'

OQ.OQ' = r2

or, 4 × OQ' = 16

or, OQ'= 4 units

i.e., Q' is at 4 units from O.

Here, OQ' = OQ = r. The point Q itself is its inverse point.

For the inverse point of R.

OR.OR'= r2

or, OR' = 42 = 16 =2
OR 8
? R' is 2 units from O.

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Example 3: Find the inverse of the point (i) A(3, 5) (ii) B(2, – 5) with respect to the circle
Solutions: x2 + y2 = 1.

Example 4: Equation of given circle is x2 + y2 = 1.
Solutions:
radius (r) = 1 unit, centre of circle = O (0, 0)

(i) To find inverse of A(3, 5).

r = 1, x = 3, y = 5. Let A'(x', y') be inverse of A.

x' = r2x = 1×3 = 3 = 3
x2 + y2 32+ 52 9 + 25 34
r2y 12 × 5 5 5
y' = x2 + y2 = 32+ 52 = 9 + 25 = 34

Hence, the required inverse point is A'( 3 , 354 )
34

(ii) To find the inverse of B(2, –5)

r = 1, x = 2, y = 5

Let B'(x', y') be inverse image of B.

Then, x' = r2x
y' x2 + y2

= 12 × 2 = 2
22+ (–5)2 29

= r2y = 12 × (–5) = – 5
x2 + y2 22+ (–5)2 29

Hence, the required inverse of point B is B' 2 , – 5 .
29 29

Find the inverse of the point (i) P(3, 2), (ii) Q(3, 3) with respect to the circle
with centre (2, 3) and radius of inverse circle 2 units.

According question,

radius (r) = 2, centre (h, k) = (2, 3)

(i) To find inverse of point P(3, 2).

r = 2, (h, k) = (2, 3), (x, y) = (3, 2)

Let, P'(x', y') be required inverse of point P.

Then, =h+ r2 (x – h) =2+ 22 (3 – 2)
x' (x – h)2 + (y – k)2 (3 – 2)2 + (2 – 3)2

and y' = 2 + 22 × 1 = 2 + 4 = 2 + 2 = 4
(1)2 + (–1)2 2

=k+ r2 (y – k) =3+ 22 (2 – 3)
(x – h)2 + (y – k)2 (3 – 2)2 + (2 – 3)2

= 3 + 22 × 1 = 3 + –4 = 3 – 2 = 1.
(1)2 + (–1)2 2

Hence, the inverse point of P(3, 2) is P'(4, 1).

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(ii) To find inverse of point Q(3, 3)

r = 2, (h, k) = (2, 3), (x, y) = (3, 3)

x – h = 3 – 2 = 1, y – k = 3 – 3 = 0

Let Q' (x', y') be required inverse of point Q.

x' = h+ r2 (x – h) =2+ 22 × (3 – 2)
(x – h)2 + (y – k)2 (3 – 2)2 + (3 – 3)2
22 × 1
= 2 + (1)2 + (0)2

=2+4

=6

and y' = k+ (x – r2 (y – k) k)2 = 3 + 22 × (3 – 3)
h)2 + (y – (3 – 2)2 + (3 – 3)2

= 3 + 22 × 0 = 3 + 0 = 3.
(1)2 + (0)2

Hence, the inverse point of Q(3, 3) is Q'(6, 3).

Exercise 11.6

Very short questions A P'
1. From the given figure, give the following concepts:
OP
(a) inversion circle D
(b) inversion radius
(c) centre of inversion
(d) relation among OP, OP' and r.
2. Fill in the gaps in the following table:

Distance of the Distance of the
inverse point from
S.N. given point from the Radius of inversion centre of inversion
centre of inversion Circle
circle
circle ..........
..........
(a) 1 2
1
(b) 1 3 2
9
2
(c) 18 .......... 3
625
(d) 16 .......... 1
(e) 3 .......... 4
(f) .......... 12
(g) .......... 25
9

(h) .......... 24

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Short question

3. Find the inverse points of the following points with respect to the inversion circle with
centre at the origin.

S.N. Given points Radius of inversion circle

a P(2, 3) 5

b Q(1, 2) 2

c R(4, 5) 4

d S(3, 0) 3

4. (a) Find the distance of inverse of the points A, B and C which are at distances 3, 6, 9
units respectively from the centre O of the circle with radius 6 units.

(b) Find the inverse of the points P, Q and R which are at distances 2, 4, 8 respectively
units from the centre O of the circle with radius 8 units.

(c) Find the inverse of the points M, N and P which are at distances 2.5, 5 and 10 units
from the centre O of the circle with radius 5 units.

5. (a) Find the inverse of the point (2, –5) with respect to the circle x2 + y2 = 1.

(b) Find the inverse of the point (3, 4) with respect to the circle x2 + y2 = 4.

(c) Find the inverse of the point (4, 5) with respect to the circle x2 + y2 = 25.

6. (a) Find the inverse of the point (6, 2) with respect to a circle centre at the point (1, 2)
of radius 3 units.

(b) Find the inverse of the point (4, 2) with respect to a circle centre at the point (5, 6)
of radius 5 units.

(c) Find the inverse of the point (5, 10) with respect to a circle with centre at the point
(3, 4) of radius 6 units.

Long Questions

7. (a) Find inversion line segment of AB with points A(1, 2), B(3, 3) with respect to
inversion circle x2 + y2 + 8x + 8y + 24 = 0.

(b) Find the inverse line segment of PQ where P(4, 5) and Q(4, 3) with respect to the
circle x2 + y2 – 4x – 6y = 3, show both of the line segments on the same graph with
the inversion circle.

2. (a) 4 (b) 81 (c) 3 (d) 4 2

(e) 3 (f) 1, (g) 324 (h) 48
(d) (9, 0)
3. (a) 50 , 75 (b) 4 , 8 (c) 64 , 80
13 13 5 5 41 41

4. (a) 12, 6, 4 (b) 32, 16, 8 (c) 10, 5, 2.5

5. (a) 2 , – 5 (b) 12 , 16 (c) 100 , 125
29 29 25 25 41 41

6. (a) (2.8, 2) (b) 60 , 2 (c) 24 , 47
17 17 5 5

7. (a) A'B', A' - 204 , - 196 , B' –24 , 24 (b) P'(6, 7), Q'(10, 3), P'Q'
61 61 7 7

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11.10 Use of Matrices in Transformation

Matrices can be used in most of the elementary transformation.

Let us take a point P(x. y) on the plane.

( ) (Let it be pre-multiplied by ) ( ) ( )0

1
-1 0 then -1 x -x
0 1 0 y =y
Also, P(x, y) y - axis
P'(-x, y)

( )-x

Above two images y and (-x, y) are same image of the point P(x, y).

( )Hence, -1 0 represents a transformation matrix that represents a transformation
0 1
matrix that reflects a given point on Y-axis.

Definition : The matrix which is used to transform a point to get its image is called
transformation matrix.

Translation

Translation by using 2 × 1 matrix

aa

( ) ( )Let us consider T = b , a column vector. Here, T is a 2 × 1 matrix. This matrix b

represents a translation of 'a' unit in the direction of X-axis and 'b' units in the direction of

Y-axis. Y

Let P(x. y) be any point on the plane. P'(x + a, y + b)

a P(x, y) a b
X' O X
( )Let us translate P(x, y) by T = b .
Y'
P T P'

( )x ( a ) ( )x + a
y+ b
= y+b

( )We get that P'(x + a, y + b) is the image of P due to translation T = a .
b

Example: ( )Find the image of P(4, 5) under translation vector -2 .
Solution: 2

( )Given point P(4, 5) can be written as 4
5

( )and -2
translation vector T = 2 which is 2 × 1 matrices.

Now, P T P'

( ) ( ) ( )4 -2 2
5 + 2 =7

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Alternatively,
Let P(x', y') be image of P (4, 5) under translation T we get,

P T P'

( ) ( ) ( ) ( )x' 4 -2 2
y' = 5 + 2 = 7
Hence, P'(2, 7) is the required image of P(4, 5).
Transformation using 2 × 2 matrices

x

[ ]A point (x, y) can be written in the matrix form as y .

[ ]a b

Let it be pre-multiplied by a matrix c d of order 2 × 2.
Then, we get the resulting matrix of order 2 × 1.

[ ] [ ] [ ] [ ]a b

i.e. c d
x ax + by ax + by
y = cx + dy cx + dy is the image of P

[a ]b

under transformation matrix. c d is the transformation matrix
of order 2 × 2.

Example : Let P(3, 4), Q(0, 3), and R(3, 0) the vertices of ∆PQR. Find the image of ∆PQR
Solution:
[ ]0 -1

using 2 × 2 matrix -1 0 .
The vertices of given ∆PQR can be written in the matrix form as,

P QR

[ ]3 0 3
4 30

[ ]Transformation matrix is 0 -1

Now, Image = transformation-1mat0rix × object matrix

P QR [= P' Q' R'
[ ] [ ]0 -1
-1 0 0-4 ]0-3 0+0
3 03 -3+0
4 30 0 + 0 -3 + 0

P' Q' R'
[ ]=
-4 -3 0
-3 0 -3
Hence, the coordinates of image vertices are P'(-4, -3), Q'(-3, 0) and R'(0, -3).

∆ P'Q'R' is the image of ∆PQR.

Reflection using 2 × 2 matrices.

Let P(x, y) be a point and P'(x' y') its image under certain geometrical transformation.

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(a) Reflection on X-axis.

Since we have P(x, y) P'(x, -y)

Which shows that x' = x and y' = - y.

We can write,

x' = 1 . x + 0 . y

y' = 0 . x + (-1) . y

Writing above linear equations in matrix form.

[ ] [ ] [ ]1 0
0 -1
xx
y = -y

[-1 ]0

Hence, 0 1 is the transformation matrix of reflection on y-axis.

(b) Reflection in y - axis
Since P(x, y) y-axis P'(-x, y)

which shows that x' = -x and y' = y ,

We can write x' = x (-1).x + 0.y

y' = y = 0.x + 1.y

Writing above linear equations in matrix form.

[ ] [ ] [ ] [ ]x' -1 0
y' = 0 1
x -x
y= y

[-1 ]0

Hence 0 1 is the transformation matrix of reflection on y-axis.

(c) Reflection on the line y = x.
Since we have,
P(x, y) y= x P'(y, x)

i.e. x' = y and y' = x

x' = y = 0.x + 1.y

y' = x = 1.x + 0.y

Writing these linear equation in the matrix form,

[ ] [ ] [ ] [ ]x' 0 1 y
y' = 1 0 x.
x
y=

[0 ]1

The matrix 1 0 is the transformation matrix of reflection on the line y = x

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(d) Reflection on the line y = -x
Since we have,
P(x, y) y=-x P' (-y, -x)
i.e. x' = -y, y' = -x
x' = -y = 0. x + (-1) y
y' = -x = (-1)x + 0.y
Writing these equations in the matrix form

[ ] [ ] [ ] [ ]x' 0 -1
y' = -1 0
x -y
y = -x .

[ ]0 -1

Hence, the matrix -1 0 describes reflection on the line y = -x.

Rotations Using 2 × 2 Matrices

(a) Rotation of +90° about origin O.
Since we have,

P(x, y) P(-y, x)

i.e. x' = -y and y' = x

x' = 0.x + (-1)y

y' = x = 1.x + 0.y

Writing these equation in matrix form, we write

[ ] [ ] [ ] [ ]x' 0 -1
y' = 1 0
x -y
y= x.

[0 ]-1

Hence, the matrix 1 0 represents a transformation matrix of rotation of +90°
about origin.

(b) Rotation of 180° about origin.
Since we have,
P(x, y) R[0,180°] P'(-x, -y)
i.e. x' = -x and y' = -y
x' = - x = (-1) . x + o.y
y' = -y = o.x + (-1). y
Writing there equations in matrix form, we get,

[ ] [ ] [ ] [ ]x' -1 0
y' = 0 -1
x -x
y = -y .

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[ ]Hence, the matrix
-1 0 represents the rotation of 180° about the origin.
0 -1

(c) Rotation of +270° about origin
Since we have,
P(x, y) R[0,270°] P'(y, -x)

i.e. x' = y and y' = -x

x' = y = 0.x + 1. y

y' = - x = (-1).x + o.y

Writing these equations in the matrix form,

[ ] [ ] [ ] [ ]x' 0 1
y' = -1 0
xy
y = -x .

[ ]Hence, the matrix
01
-1 0 represents rotation of + 270° or -90° about the origin.

Enlargement Using 2 × 2 Matrices

Let O be the centre of enlargement and k the scal factor. Let it be denoted by E[(0, 0), k].
Now, P(x, y) E[(0,0),k] P'(kx, ky)

i.e x' = kx = k.x + 0.y

y' = ky = 0.x + k.y

There equations can be written in the matrix form

[ ] [ ] [ ] [ ]x' k
y' = 0
0 x kx
k= y= ky .

[k ]0

Hence, 0 k represents enlargement matrix with centre O and scale factor k.

Unit Square Y B(1, 1)

The square having vertices 0(0, 0), (1, 0), (1, 1) and (0, 1) is C
called unit square. (0, 1)

In the figure, OABC is a unit square. X' O
Y'
i.e. OA = OC = AB = CB = 1 unit X

In matrix form, unit square is represented by A(1, 0)

[ ]0 1 1 0
00 1 1

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Transformations in terms of matrices can be tabulated as follows:

S.N Transformations Objects Images Equations Matrices
1. Reflection on x-axis (x, y) (x, -y) x'=1.x + 0.y
[ ]1 0
2. Reflection on y-axis (x, y) (-x, y) y'=0.x+(-1)y 0 -1
x'=(-1)x+o.y
3. Reflection on y=x (x, y) (y, x) [ ]-1 0
y'=o.x+1.y 01
4. Reflection on line y=-x (x, y) (-y, -x) x'=o.x+1.y
[ ]0 1
5. Rotation of +90° or (x, y) (-y,x) y'=1.x+o.y 10
+270° about 0 (y, -x) x'=o.x+(-1)y
(-x, -y) [ ]0 -1
6. Rotation of -90° or (x, y) (kx, ky) y'=(-1)x +o.y -1 0
+270° about 0 (x, y) x'=o.x+(-1)y
[ ]0 -1
7. Rotation of 180° about (x, y) y'=1.x+o.y 10
origin x'=o.x+1.y
[ ]0 1
8. Enlargement with centre (x, y) y'=(-1)x+o.y -1 0
origin and scale factor k x'=(-1)x+o.y
[ ]-1 0
9. Identity transformation (x, y) y'=o.x+(-1y) 0 -1
x'=k.x+o.y
[ ]k 0
y'=o.x+k.y 0k
x'=1.x+o.y
[ ]1 0
o.x+1.y 01

Worked out Examples

Example 1. [ ]Find the image of P(2, 3) using matrix which is translated by the vector
Solution: -3
T= 1
Example 2.
Solution: [ ]-3

Here, the translation vector T 1

Translating P by translation vector T, we get,

[ ] [ ] [ ] [ ] [ ]x' 2 -1
-3 2 - 3
y' = 3 + 1 = 3 + 1 = 4 .

[0 ]-1

Which transformation is associated with matrix -1 0 ? Use it to
transform the point (6, 4).

Let P(x, y) be any point on the plane.

[ ] [ ] [ ] [ ]Then -y
0 -1 x 0 + (-y) .= -x .
-1 0 y= -x + 0

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P(x, y) P'(-y, -x)

This shows that P is reflected on line y = -x.

[ ]0 -1

Hence, -1 0 is associated with transformation of reflection on line y = -x.

[ ] [ ] [ ]Again,
0 -1 6 -4
-1 0 4 = -6 .

? (-4, -6) is the image of (6, 4).

[find the image of R(3, 2). ]0 ? Use it to
Example 3. Which transformation is associated with the matrix 3
Solution: 3
0

Let P(x, y) be any point on the plane.

3

[ ] [ ] [ ]Transformation matrix M = 0
0 x 3x
3 y = 3y .

? P(x. y) P'(3x, 3y)

Hence, the given matrix M is associated with enlargement about O with scale
factor 3.

R'
[ ][ ] [ ]Also,
30 39
03 2= 6

Example 4. Write down the 2 × 2 transformation matrix which represents enlargement
Solution: with centre (0, 0) by scale factor 2. Use it find the image of point (4, 3).

Transformation matrix representing enlargement about O and scale factor 2 is

[ ]2 0

M= 0 2 .

Let given point be P(4, 3)

M P P'
[ ] [ ] [ ]2 0
4 8
Now, 0 2 = 3 = 6

? The image of P(4, 3) is P'(8, 6)

Example 5. [4 ]0

To which transformation is the matrix P = 0 4 associated ? What
types of transformation do P2 and P3 represent?

Solution: [4 ]0

Here, P = 0 4 is associated with the enlargement about centre origin

O(0, 0) and scale factor 3.

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40
4
[ ] [ ] [ ]Now, P2 = P × P = 0
40 16 0
0 4 = 0 16

Hence P2 represents an enlargement about centre origin O(o,o) with scale
factor 16.

16 0

[ ][ ][ ]Again. P3 = P2. P = 0 16
40 64 0
04 0 64 .

Hence P3 represents an enlargement matrix about centre origin O(0, 0) with scale
factor 64.

Example 6. The vertices of a unit square are O(0, 0), A(1, 0), B(1, 1) and C(0, 1), The
Solution: square OABC is reflected on the line y = x. Find the coordinates of image
using 2 × 2 matrix.

Given unit square OABC can be written in the matrix form.

OA B C

[ ]0 1 1 0
00 11

01
0
[ ]2 × 2 matrix of reflection on line y = x is M = 1

OA B C O' A' B' C'
[ ][ ] [ ]Now, 1 0 0 0 1 1 = 0 1 1 0 .
01 0 1 1 0 00 11

Hence, the coordinates of image of unit square are O(0, 0), A(0, 1), B(1, 1), and
C'(1, 0).

Example 7. Find the image of ∆ABC with vertices A(2, 3), B(-1, 2), and C(2, -3) using

[ ] [ ]the matrix
0 -1 0 -1
-1 0 . Which transformation is the matrix -1 0
associated with ?

Solution: Writing the coordinates of vertices A(2, 3), B(-1, 2) and C(2, -3) of ∆ABC in the
AB C
[ ]2 -1 2
matrix form , 3 2 -3

0 -1

[ ]Given transformation matrix is (M) = -1 0

AB C A' B' C'
[ ] [ ] [0 -1
-3 -2 ]3
Now, -1 0
2 -1 2 -2
3 2 -3 = -2 1

Hence, the coordinates of image of the ∆ABC are

A'(-3, -2), B'(-2, 1) and C'(3, -2).

[ ]The matrix
0 -1 is associated with reflection on the line y = -x.
-1 0

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Example 8. [Transform a unit square under the matrix 3 ]2
Solution: coordinates of the image. 1
1 and write the

Let the vertices be O(0, 0), A(1, 0), B(1, 1) and C(0, 1) of a unit square.

OA B C
01 10

[ ]Writing the given unit square in the matrix form 0 0 1 1

[ ]3 2

Transformation matrix, M = 1 1 .
Now, M × object OABC = Image O'A'B'C'.

OA B C O' A' B' C'
[ ][ ] [ ]i.e., 1 1 0 0 1 1 = 0 1 2 1
32 0 1 1 0 03 52

Hence the coordinates of image vertices are O'(0, 0). A(3, 1), B'(5, 2), C(2, 1)

Example 9. ∆PQR whose vertices are P(3, 6), Q(4, 2), and R(1, 1) maps onto ∆P'Q'R'
Solution: with the vertices P'(-6, 3), Q'(-2, 4), and R'(-2, 2). Which is the single
transformation for this mapping? Also find a 2 × 2 matrix that represents
this transformation.

Here, ∆PQR with vertices P(3, 6), Q(4, 2), and R(1, 1) is mapped onto ∆P'Q'R'
with the vertices P'(-6, 3), Q(-2, 4), and R(-1, 1)

Here, P(3, 6) P'(-6, 3)

i.e. (x, y) (-y, x)

Hence, the point (x, y) is transformed to (-y, x) which represents the rotation
of +90° about the origin.

[a ]b

LetM = c d be the required 2×2 transformation matrix which transform

∆PQR to ∆P'Q'R' .

Then,

AB C P' Q' R'
[ ] [ ] [ ]a b
2 -1 2 -6 -2 -1
c d . 3 2 -3 = 3 4 1 .

[ ] [ ]or, 3c + 6d 4c + 2d c + d = 3 4 1 .
3a + 6b 4a + 2b a + b -6 -2 -1

Equating the corresponding elements, we get,

3a + 6b = -6 or, a + 2b = -2........ (i)

4a + 2b = -2 or, 2a + b = -1........ (ii)

a + b = -1 ........ (iii)

3c + 6d = 3 or, c + 2d = 1 ....... (iv)

4c + 2d = 4 or, 2c + d = 2....... (v)

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c + d = 1 ......... (vi)
Solving equation (i) and (iii), we get,

a + 2b = -2
a + b = -1
-- +
b=-1

Put the value of 'a' in equation (iii), we get,
a=0

Again solving equations (iv) and (vi), we get,
c + 2d = 1
c+ d = -1

-- +
d=0

Put the value of d in equation (vi), we get, c = 1

[ ]0 -1

? M = 1 0 which is the required matrix.

[Example 10. Find the 2 × 2 matrix which transforms the unit square 0 1 1 ]0
[ ]to a parallelogram03 4 1 0 0 1
0 1 3 2 1

.

Solution: [Let M = a ]b be the required matrix. Then by question
c
d

[ ][ ] [a b 0 1 1 0 = 0 3 4 ]1
cd 0 0 1 1 0 1 3
[ ] [ ]or, 2

0 a a+b b = 0 3 4 1
0 c c+d d 0 1 3 2

Equating the corresponding elements of equal matrices, we get,

a = 3, b = 1, d = 2, c = 1

[ ].Required transformation matrix is M =3 1
1 2

Example 11. Prove by matrix method that the reflection in x-axis followed by rotation
about (0, 0) through 180° is equivalent to the reflection on y-axis.

Solution: The 2 × 2 transformation matrix representing the reflection on x-axis

0 1 -1 0
-1 0 -1
[ ] [ ]0
is and rotation about O(0, 0) through 180° is

Now, the reflection followed by rotation about O(0, 0) through 180° is given

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by

[ ] [ ] [ ]-1 01 0 = -1 0 ............................. (i)
0 -1 0 -1 0 1

The 2 × 2 transformation matrix representing reflection in y-axis is

[ ]-10 .............. (ii)
0 1

From (i) and (ii), we conclude that the reflection on x-axis followed by rotation

about (0, 0) through 180° is equivalent to the reflection on y-axis.

Example 12. A unit square is rotated about origin through 180°. Find the image of unit
square using 2 × 2 matrix. Plot the graphs of unit square and its image on
the same graph paper.

Solution: Let O(0, 0). A(1, 0). B(1, 1) and C(0, 1) be the vertices of unit square OABC.

OA B C
[ ]Theunitsquarecanbeexpressedinthematrixformas.
0 1 1 0
0 0 1 1

-1 ]0

[The 2 × 2 matrix of rotation about origin through 180° is 0 -1 .
C'
OA B C O' A' B'
[ ][ ] [-1 0 0 1 1 0 ]0
0 -1 -1
Now, 0 -1 0 0 1 1 = 0 0 -1 -1

? The required image vertices of unit square are O'(0, 0), A'(-1, 0), B'(-1.
-1). C'(0, -1).

Graphs of unit square OABC and its image O'A'B'C' are drawn in the graph
given below.

Scale 20 Y

lines - 1 unit
B'

C(0, 1) B(1, 1)

X' A' X
O
A(1, 0)

C'

Y'

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Exercise 11.6

Very short Questions

1. Let P(x, y) be any point on the plane and P(x', y') be the image of P. Write the transformation
matrices under the following cases:

(a) (x', y') = (-y, - x) (b) (x', y') = (-x, -y) (c) (x', y') =(-x, -y)

(d) (x', y') = (-x, y) (e) (x', y') = (y, x) (f) (x' , y') =(kx, ky)

(g) (x', y') = (x + a, y + b) (h) (x', y') = (3x, 3y)

2. Write the type of transformation represented by each of the following transformation
matrices:

( )1 0 ( )1 0 ( )0 1

(a) 0 1 (b) 0 -1 (c) 1 0

( )0 -1 ( )0 -1 ( )0 1

(d) -1 0 (e) 1 0 (f) -1 0

( )-1 0 ( )2 0 (4 )0

(g) 0 -1 (h) d 2 (i) 0 4

Find the image of P(4, 5) using above matrices in each care.

Short Questions

3. Write down 2 × 2 transformation matrices in each of the following transformation and
using the matrix find image of given point.

(a) Reflection on y-axis, A(3, 4)

(b) Reflection on x-axis. B(6, 3)

(c) Reflection on line y = x , C(-4, -2)

(d) Reflection on line x + y = 0, D(2, 3)

(e) Rotation about origin through, +90°, E(-4, -4)

(f) Rotation about origin through 180°, F(6, 7)

(g) Enlargement with centre at the origin by scale factor 3, G(4, 7)

[ ]4. (a) If a matrix
value of a. a0
3 2 maps the point (3, 4) on to the point (15, 17), find the

(b) If a point (x, y) is transformed into (y, x) by a 2 × 2 matrix, find the 2 × 2
transformation matrix.

(c) Find a 2 × 2 matrix which transforms point (8, -4) to P'(4, 8).

2

[ ]5. (a) Find the image of A(6, 7), under the translation matrix 4 followed by

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[ ]3

translation 2 .

34

[ ] [ ](b) Find the image of M(2. 3) under translation 2 followed by translation 2 .

2 ]0

[6. (a) To which transformation is the matrix M= 0 2 associated with ? What
types of transformation do M2 and M3 represent ?

[1 ]0

(b) What does the matrix 0 -1 represent? Find the image of P (4, 5) using the
matrix.

Long Questions

7. (a) P(2, 1), Q(5, 1), R(5, 4) and S(2, 4) are the vertices of a square PQRS. The square

[1 ]2

is transformed by matrix 1 -2 into a parallelogram. Find the vertices the
parallelogram.

(b A triangle with vertices P(2, 1), Q(5, 3), and R(6, 7) is transformed by the matrix

[ ]0 -1
-1 0 . Find the image of the triangle.
(c) A square with vertices A(1, 1), B(6, 1), C(6, 6), and D(1, 6) is transformed by the

[transformation matrix 3 ]0
0
3 . Find the image of the square ABCD.

(d) A quadrilateral PQRS with vertices P(2, 1) Q(5, 2), R(6, 5) and S(3, 5) is transformed

[ ]-1 0

by matrix 0 -1 find the image of the quadrilateral PQRS.

8. (a) A line PQ having P(5, 1), Q(8, 6) map's onto P'Q' having P'(-5, 1), Q(-8, 6) so that
the image P'Q' is formed. What is the single transformation for the mapping ? Also
find the 2 × 2 matrix.

(b) A square ABCD with vertices A(2, 0), B(5, 1) C(4, 4) and D(1, 3) is mapped onto
parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of the parallelogram
are A'(2, 2), B'(7,3) C'(12, - 4) and D'(7, -5). Find the 2 × 2 matrix.

(c) ∆ABC with the vertices A(2, 7), B(2, 9), and C(6, 7) is mapped onto ∆A'B'C' whose
vertices are A'(7, 2), B(9, 2) and C'(7, 6). Which is the single transformation for the
mapping? Also find the 2 × 2 matrix of transformation ?

(d) ∆PQR with vertices P(5, 1), Q(12, 4) and R(4, 5) maps onto the ∆P'Q'R' with the
vertices P'(-5, -1) Q'(-12, -4) and R'(-4, -5). Which is the single transformation for
this mapping ? Also find the 2 × 2 matrix that represents the transformation.

9. (a) A unit square having vertices A(o. o), B(1, 0), C(1, 1) and D(0, 1) is mapped to the
parallelogram A'B'C'D' by a 2 × 2 matrix so that the vertices of parallelogram are
A'(o,o), B'(3, 0), C(4, 1) and D'(1, 1). Find the 2 × 2 transformation matrix.

(b) Find the 2 × 2 matrix which the unit square is transformed to parallelogram

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( )0 3 4 1 .
0 1 3 2

[ ](c) 0 1 1 0
Find the 2 × 2 matrix which transforms the unit square 0 0 1 1

[ ]onto a parallelogram 0 4 6 2 .
0 1 3 2

[ ](d) A rectangle
02 20 is mapped onto a rectangle
00 -1 -1
[ ] [ ]0
0 2 2 0 a b , find the matrix of
0 1 1 by the 2 × 2 matrix c d

transformation.

(e) A unit square OMNP having vertices O(o, o), M(1, 0), N(1, 1), and P(0, 1) is
transformed to unit, square O'M'N'P' through reflection y = -x. Write the vertices
of the image unit square so formed. Draw the graphs of both of the units square on
the same graph paper.

10. (a) Verify by the matrix method that the reflection on the line y = x followed by the
reflection on the y-axis is equivalent to rotation about origin through +90°.

(b) Verify by the matrix method that a reflection in the line X - axis followed by the
Y - axis is the rotation about the origin through 180°.

(c) Prove by the matrix method that the reflection on the line y = x followed by the
rotation about the origin +90° is the reflection on y-axis.

3. (a) A'(-3, 4) (b) B'(6, -3) (c) C'(-2, -4) (d) D'(-3, -2)

(e) E'(4, -4) (f) F'(-6, -7) (g) G'(12, 21)
4. (a) 5
5. (a) A'(11, 13) (b) 01 (c) 0 -1
10 10

(b) M'(9, 7)

6. (a) E[(0, 0), 2], M2 represents E[(0, 0), 4], M3 represents E[(0, 0), 8]

(b) Reflection in X-axis, P'(4, -5)

7. (a) P'(4, 0), Q'(7 3), R'(13, -3), S'(10, -5) (b) A'(-1, -2), Q'(-3, -5), R'(-7, -6)

(c) A'(3, 3), B'(18, 3), C'(18, 18), D'(3, 18) (d) P'(-2, -1), Q'(-5, -2) R'(-6, -5), S'(-3, -5)

8. (a) -1 0 (b) 12 (c) reflection on y = x, 0 1
0 1 1 -2 1 0

(d) Rotation of 180° about origin, -1 0
0 -1

9. (a) 3 1 (b) 31 (c) 42 (d) 10
0 1 12 12 0 -1

(e) O'(0, 0), M'(0, -1), N(-1, -1)

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12Statistics

12.0 Measure of Dispersion

Introduction

Measures of central tendency gives us an idea of the concentration of observations about the
central part of the distribution. It tells nothing about how each variate value of a set of data is
scattered from an average value of those items. Two or more two data may have equal mean,
median or mode but with more scattered than the other. It may be clear from the following
two set of data.

Averages X Md Mo

Set I 63 64 65 66 66 67 68 69 66 66 66

Set II 64 65 66 66 66 66 67 68 66 66 66

The above two data sets have the same means, the same medians and same mode, all equal to
66. They have the same number of observations, n = 8. But these two data set are different.
What is the main difference between them ?

The two data sets have the same measure of central tendency (as measured by any of the
three measures of central tendency - mean, median and model) but they have a different
variability. In particular, we can see that data set I is more scattered than data set II. The
values of data set I are more spread out: they lie farther away from their mean than do those
data set II.

Hence, the certain measures are evolved which reflect on the scattering of values of
numerical terms are known as measures of dispersion. In statistics, the term dispersion is
used commonly to mean scatteredness, variation, diversity, deviation, fluctuations, spread,
heterogeneity, etc.

Dispersion may be defined as the degree of scatteredness or variation of a variable about a
central value. Less the value of dispersion more data will be representative. Dispersion of the
data are measured in terms of the following measures of dispersion.

1. Range 2. Quartile Devication

3. Mean Deviation 4. Standard Deviation

In this chapter we study about quartile deviation, mean deviation, and standard deviation
for continuous series of data.

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12.1 Quartile Deviation (Q.D)

Quartile deviation is a measure of dispersion based on the upper quartile (Q3) and the lower
quartile (Q1) of a set of data. The difference between the upper quartile (Q3) and the lower
quartile (Q1) is called interquartile range.

Interquartile range = Q3 - Q1

Half of the interquartile range is called semi-interquartile range or quartile deviation. It is
denoted Q.D. and is given by the formula.

Quartile Deviation (Q.D.) = 1 (Q3 - Q1)
2

It is an absolute measure of dispersion. It has the same unit as that of original data.

For comparative study of two or more data sets, we calculate coefficient of quartile deviation.

It is the relative measure of dispersion based the quartiles Q3 and Q1

Coefficient of Quartile Deviation = Q3 - Q1
Q3 + Q1
The coefficient of Q.D. is unitless. It is used to compare variability of two or more data

sets. The data having less value of coefficient of quartile deviation is less variable or less

heterogeneous or more uniform or more stable than other and vice-versa.

Calculation of Quartile Deviation for continuous Series.

For continuous data in the form of frequency distribution, the following formulas are used.

For lower quartile or the first quartile (Q1)
N
Find the value of 4 for the first quartile class.

N - c.f
4
Q1 = L + ×i
f
Where, (i) L is the lower limit of the class in which the first quartile lies.

(ii) f is the frequency of the first quartile class.

(iii) c.f. is the cumulative frequency of the class preceding the first quartile class.

(iv) i is the length of class interval (or class - size).

For the upper quartile or the third quartile (Q3)

Find the value of 3N to find the third quartile class.
4
3N
4 - c.f

Q3 = L + f ×i

Where, (i) L is the lower limit of the class in which the third class lies.

(ii) f is the frequency of the third quartile class.

(iii) c.f. is the cumulative frequency preceding the third quartile class.

(iv) i is the length of class interval (or class size).

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Worked Out Examples

Example 1. The quartile deviation of a continuous data is 20. If the first quartile is 45,
Solution: find the third quartile.

Example 2. Given, Quartile Deviation (Q.D.) = 20
Solution:
The first quartile (Q1) = 45
Example 3.
Now, Q.D. = 1 (Q3 - Q1)
2
1
or, 20 = 2 (Q3 - Q1)

or, 20 = 1 (Q3 - 45) or, 40 = Q3 - 45
2
? Q3 = 85

The coefficient of quartile deviation of a grouped data is 0.25. If the upper
quartile is 75, find the lower quartile.

Coefficient of Q.D. = 0.25

The upper quartile (Q3) = 75

Now, coefficient of Q.D. = Q3 - Q1
Q3 + Q1
5041Q.215===27727755555++-- QQQQ1111
or, or, 75 + Q1 = 300 - 4Q1
? Q1 = 45
or,
or,

Calculate the quartile deviation and its coefficient from the following data :

Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of students 1 18 25 28 30 33 22 15 22

Solution: To calculate quartile deviation and its coefficient.

Marks obtained No. of students c. f.
0-10 11 11
10-20 18 29
20-30 25 54
30-40 28 82
40-50 30 112
50-60 33 145
60-70 22 167
70-80 15 182
80-90 22 204

N = 204

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To find Q1 = N = 204 = 51
4 4
c.f., just greater 51 is 54. So, Q1 lies in the class (20 - 30). i.e. (20-30) is the first
quartile class.

L = 20, c.f = 29, f = 25, i = 10

N - c.f
4
Now, Q1 =L+ ×i
f
51 - 29
= 20 + 25 × 10 = 20 + 8.8 = 28. 8 marks.

To find Q3 , 3N = 3 × 204 = 153
4 4
c.f. just greater 153 is 167.

Q3 lies in the class (60 - 70) i.e. (60 - 70) is the third quartile class.

L = 60, c.f. = 145, f = 22, i = 10

3N - c.f
4
Now, Q3 =L+ ×i
f
Dev=iat6io0n+(Q.1D5)3=2-21214(5Q3×- Q110)
= 60 + 3.64 = 63.64 marks

Quartile = 1 (63.64 - 28.8) =17.42
2
Q3 - Q1 63.64 - 28.8
coefficient Quartile Deviation (Q.D) = 2 = 63.64 + 28.8 = 0.38

Example 4. Calculate the quartile deviation and its coefficient.

x less than 10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80 and above

f3 4684783 5

Solution: The given data is open ended. We can calculate deviation.

x f cf
Less than 10 3 3
4 7
10-20 6 13
20-30 8 21
30-40 4 25
40-50 7 32
50-60 8 40
60-70 3 43
70-80 5 48
80 and above N = 48

To find the first quartile (Q1)

N = 48 = 12
4 4

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c.f., just greater than 12 is 13.

So, Q1 lies in (20 - 30)

L = 20, c.f. = 7, f = 6, i = 10

N - c.f 12 - 7 50
4 6 6
Q1 =L+ × i = 20 + × 10 = 20 +
f
= 20 + 8.33 = 28.33

To find the upper quartile (Q3), 3N = 3 × 48 = 36
4 4
c.f. just greater 36 is 40. So, Q3 lies in (60 - 70)

L = 60, c.f. = 32, f = 8, i = 10

3N - c.f 36 - 32 × 10
4 8 - 28.33) =
Q3 =L+ ×i = 60 +
Quartile f 65 Q1) =
= 60 + 4 × 10 = (Q3 -
8 1
Deviation (Q.D) = 2 1 (65 18.34
2
Q3 - Q1 65 - 28.33
coefficient of quartile diviation = Q3 + Q1 = 65 + 28.33 = 0.393

Example 5. Calculate the quartile deviation and its coefficient from the given data :

x 20-29 30-39 40-49 50-59 60-69 70-79 80 - 89
f 4 6 8 12 7 6 5

Solution: To calculate quartile deviation and its coefficient, given inclusive data is
changed into exclusive. The correction factor is given by using the formula.

Correction factor = Lower limit of the 2nd class – upper limit of first class

30 - 29 2
2
= = 0.5

So, the exclusive class of above data are given by

xf cf
4
19.5 - 29.5 4 10
18
29.5 - 39.5 6 30
37
39.5 - 49.5 8 43
48
49.50 - 59.5 12

59.5 - 69.5 7

69.5 - 79.5 6

79.5 - 89.5 5

N = 48

To find the lower quartile (Q1), N = 48 = 12
4 4

c.f., just greater 12 is 18.

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? Q1 lies in class (39.5 - 49.5)

L = 39.5, c.f. = 10, f = 8, i = 10

N - c.f 12 -10
4 8
Now, Q1 =L+ × i. = 39.5 + × 10
f
2
= 39.5 + 4 × 10 = 39.5 + 5 = 44.5

To find upper quartile (Q3) , 3N = 3 × 48 = 36
4 4

c.f., just greater 36 is 37.

Hence, Q3 lies in the class (59.5 - 69.5)

L = 59.5, c.f. = 30, f = 7, i = 10

3N - c.f 36 - 30
4 7
Now, Q3 =L+ × i. = 59.5 + × 10
f
= 59.5 + 8.57 =68.07

Q.D. = 1 (Q3 - Q1) = 1 (68.,07 - 44.5) = 11.785
2 2
Q3 - Q1
coefficient of Q.D. = Q3 + Q1

= 68.07 - 44.5 = 0.2094
68.07 + 44.5

Exercise 12.1

Short Questions

1. (a) Define measure of dispersion.

(b) List methods of measure of dispersion.

(c) Define quartile deviation.

(d) Write down formula to calculate Q.D.

(e) Write down formula to calculate coefficient of Q.D.

(f) Write a difference between Q.D. and its coefficient.

2. Calculate quartile deviation from following:

(a) Q1 = 25, Q3 = 75 (b) Q1 = 25, Q3 = 65

(c) Q1 = 40, Q3 = 80 (d) Q1 = 5, Q3 = 75

3. Calculate the coefficient of quartile deviation from the following:

(a) Q1 = 20, Q3 = 40 (b) Q1 = 40, Q3 = 80

(c) Q1 = 2, Q3 = 22 (d) Q1 = 2.5, Q3 = 22

4. (a) In a set of continuous data, if Q1 = 35, and quartile deviation is 20, find Q3.

(b) The coefficient of quartile diviation of a grouped data is 0.25 and the upper quartile
is 60. Find the value of lower quartile.

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(c) The coefficient of quartile deviation for continuous data is 0.39 and upper quartile
49.75. Find the first quartile.

Long Questions :

5. Calculate the quartile deviation and its coefficient from the following data :

(a) Marks 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No. of students 4 12 16 10 8 6 2

(b) Marks 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No. of students 3 5 6 8 4 4 3

(c) Volume of water (l) 5-10 10-15 15-20 20-25 25-30 30-40
No. of families 4 12 16 6 2 1

(d) Age (years) 0-5 5-10 10-15 15-20 20-25 25-30
No. of persons 5 20
15 20 10 2

(e) Age (years) 60-65 65-70 70-75 75-80 80-85 85-90
No. of students 7 5 8 4 3 3

(f) x 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 30-40
f 6 10 18 30 15 12 10 6 5

6. Calculate quartile deviation and its coefficient from the following data:

(a) Marks Less than 30 30-40 40-50 50-60 60-70 70 and above

No. of students 3 6954 2

(b) x Below 25 25-30 30-35 35-40 40-45 45 and above

f 5 12 22 25 20 9

(c) x Below 30 30-40 40-50 50-60 60-70 70 and above

f 2 4686 4

(d) Income (Rs.) Below 50 50-70 70-90 90-110 110- 130-150 150 above
130

Workers 5 10 20 25 18 12 10

7. Calculate the quartile deviation and its coefficient from the given data:

(a) Class interval 20-29 30-39 40-49 50-59 60-69 70-79
Frequency 100 80 75 95 70 40

(b) Marks 0-9 10-19 20-29 30-39 40-49 50-59

No. of students 1 4 8 10 5 3

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8. (a) Calculate the semi-interquartile range and its coefficient of the following data:
(prepare a frequency distribution table taking class interval 10)

70 55 51 42 57 45 60 47 63 53 33 65 39
82 55 64 58 61 65 42 50 52 54 45 45 25
36 59 63 39 65 45 49 54 64 75 42 41 52
35 30 35 15 48 26 20 40 55 46 18

(b) Prepare a frequency distribution table taking (0 - 4) as one of the class. The calculate
the quartile deviation. 7, 3, 10, 4, 1, 9, 11, 18, 8, 5, 6,4, 13, 17,6,8, 12,17,19, 5, 3,
17, 16, 3, 2, 14, 13, 4, 10

2. (a) 25 (b) 20 (c) 20 (d) 35
3. (a) 0.33 (b) 0.33 (c) 0.83 (d) 0.7959
4. (a) 75 (b) 36 (c) 21.83
5. (a) 11.56, 0.23 (b) 13.18, 0.25 (c) 3.51, 0.22 (d) 5.13, 0.38
(f) 5.5, 0.27
(e) 6.31, 0.088 (b) 5.008, 0.137 (c) 10.83, 0.20 (d) 23.33, 0.22
6. (a) 10.21, 0.215 (b) 8.035, 0.251
7. (a) 13.79, 0.30 (b) 3.97, 0.3683
8. (a) 9.62, 0.1924

12.2 Mean Deviation

Mean deviation is a measure of dispersion that is based on all the values of a set of data. It
is defined as the arithmetic mean of the absolute deviations taken from central value (mean,
median or mode). It shows the variation of the items from an average. Therefore, it is also
called average deviation. Mean deviation is denoted by MD.

While computing mean deviation, the deviation of items can be taken from mean, median
or mode. However, it gives the best results when deviations are taken from the median. In
calculation of mean deviation, negative sings are ignored and all the values are treated as
positive.

Calculation of Mean Deviation for Continuous Series :

Let m be the mid values of corresponding classes in continuous series. Then,

Mean Deviation from mean = ∑ f|m- x|
Mean Deviation from median =
∑ Nf|m-Md|
N
∑ f|m-Mo|
Mean Deviation from mode = N

Here, we study only mean deviation from mean or median

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Mean deviation is an absolute measure of dispersion. For computing two or more series
having different units, we calculate coefficient of mean deviation. The relative measure of
dispersion based on mean deviation is called coefficient of mean deviation.

Coefficient of M.D. from mean = M.D. from mean
mean
M.D. from median
Coefficient of M.D. from median = median

Worked out Examples

Example 1. In a grouped data if ∑f|m-x| =400, x = 20, N = 50, find M.D. and its
Solution: coefficient.

Example 2. Here, N = 50, x = 20, ∑f|m-x| =400

∑ f|m - x | 400
Now, Mean Deviation from mean = 50 = 50 = 8
M.D. from mean 8
coefficient of M.D. from mean = = 20 = 0.4
Mean

Calculate the mean deviation and it coefficient from the following data :

(i) deviations from mean

(ii) deviations from median.

Class 0 - 10 10 -20 20 -30 30 -40 40 -50
Frequency 5 8 15 10 6

Solution: (i) To calculate Mean Deviation from mean.

Class Mid-values Frequency (f) fm |m-x| f|m - x|
(m)
104.5
0-10 5 5 25 20.9 87.2
13.5
10-20 15 8 120 10.9 91.0
114.6
20-30 25 15 375 0.9 6f|m - x| =
410.9
30-40 35 10 350 9.1

40-50 45 6 270 19.1

N = 44 6fm = 1140

Mean ( x ) = ∑fm = 1140 = 25.9
N 44
∑f|m-x |
Mean deviation from mean = 410N.9
= =
44 9.34

Coefficient of M.D. from mean = M.D. from mean
mean
9.34
= 25.9 = 0.36

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(ii) Calculation of mean deviation from median:

Class Mid-values Frequency (f) cf |m-md| f|m -md|
(m)

0-10 5 5 5 21 105

10-20 15 8 13 11 88

20-30 25 15 28 1 15

30-40 35 10 38 9 90

40-50 45 6 44 19 114

N = 44 6f|m - md| =
412

To calculate median,

N = 44 = 22
2 2

c.f just greater than 22 is 28. Hence median class is (20 -30).

L = 20, c.f. = 13, f = 15, i = 10

N - c.f
2
Median (Md) =L+ ×i
f
22 - 13
= 20 + 15 × 10

= 20 + 6 = 26

Now, M.D. from median = ∑f |m-Md|
N
412
= 44 = 9.36

Exercise 12.2

Short Question

1. (a) Define Mean Deviation

(b) Write down the formula to compute mean deviation for continuos series of data:
(i) from mean

(ii) from median

(c) Write down the formula to compute the coefficient of mean deviation:

(i) from mean

(ii) from median
2. (a) In a grouped data, if ∑ f|m- x | = 400, N = 80, x = 40, find the M.D. and its

coefficient .

(b) In a continuous series of data, if ∑f |m-Md| = 250, Md = 22, N = 25, find the M.D.
and its coefficient.

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Long questions
3. Calculate the mean, deviation from (i) mean (ii) median for the following data :

(a) Class interval 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 2 3 4 9 2 4

(b) Class interval 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 10 12 25 35 40 50

(c) Marks 0-10 10-20 20-30 30-40 40-50

Number of students 9 6 4 12 9

(d) Class interval 0≤x≤10 10≤x≤20 20≤x≤30 30≤x≤40
Number of students 15 12 7 8

4. Calculate mean deviation from median and its coefficient for the following grouped
data :

(a) Marks 20-30 30-40 40-50 50-60 60-70

No. of students 5 7 8 6 4

(b) Marks 0-20 20-40 40-60 60-80 80-100
No. of students 5 6 7 10 12

(c) Marks 0-10 10-20 20-30 30-40 40-50

No. of students 5 8 15 16 6

(d) Profit (Rs) 0-10 10-20 20-30 30-40 40-50 50-60
No. of persons 10 12 25 35 40 50

5. Compute mean deviation and its coefficient from mean of the following date :

(a) Mid point 2 6 10 14 18 22 26
Frequency 7
13 10 5 8 4 3

(b) Mid-value 5 15 25 35 45 55
No. of students 2 46864

6. Construct a frequency distribution table taking a class interval of 10 and calculate the
M.D. from median.

28, 49 37, 5, 18, 14, 24, 7, 38, 46, 30, 21, 16, 31,

45, 27, 10, 4, 17, 29, 35, 36, 41, 47, 44, 33, 34, 17, 18, 20.

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7. Compute the mean deviation its coefficients, from (i) mean and (ii) median from the
given data :

Age (years) Number of people
Less then 10 10
Less than 20 25
Less than 30 40
Less than 40 45
Less than 50 70
Less than 60 85
Less than 70 100
Less than 80 110
Less than 90 120

2. (a) 5, 0.125 (b) 10, 0.45 (b) (i) 12.57, 0.33 (ii) 12.45, 0.28
3. (a) (i) 11.46, 0.35 (ii) 11.25, 0.34 (d) (i) 9.59, 0.56 (ii) 9.05, 0.60
(ii) 13.21, 0.43 (c) 9.56, 0.34 (d) 12.45, 0.30
(c) (i) 13.43. 0.51 (b) 23.60, 0.37
4. (a) 10.58, 0.24 (b) 11.73, 0.36 (ii) 19.75, 0.43
5. (a) 6.05, 0.36 7.(i) 16.32, 0.37
6. 12.67, 0.63

12.3 Standard Deviation

Standard deviation is the positive square root of the mean of the square of the deviations
about mean. It is also known as "Root mean square deviation" Standard deviation is denoted
by Greek letter V (read as sigma).

Among all the methods of finding out dispersion, standard deviation is regarded as the best
because of the following reasons.

1. Its value of based on all the variate values.
2. The deviation of each variate is taken from the mean.
3. Algebraic sign of each deviation is considered.

Calculation of Standard Deviation for Continuous Series :
To calculate standard deviation for a continuous series, the following methods can be applied.

(a) Direct Method

In this method, the formulae used is

( )V =
∑fm2 - ∑fm 2
N N ..............................(I)

where, m is mid-value of each class.

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To use the above formula, follow the following steps.
1. Find the mid value of each class which is denoted by m.
2. Find ∑f = N
3. Find the products fm and ∑fm
4. Multiply fm by m to get fm2 and ∑fm2,
5. Substitute all above values of ∑fm, ∑fm2, N in formula (I), we get standard deviation of

given data.

(b) Actual Mean Method

In this method, deviations of each mid point of class intervals are taken from the

arithmetic mean. The formula used in this method is V = ∑f(m - x )2 ...(II)
N
Note :
Above formulae (I) and (II) represent same direct method

To use this formula, follow the steps below:

1. Find the mid point of each class which is denoted by m

2. Find arithmetic mean, x = ∑fm where N= ∑f
N
3. Take deviations of each mid point from the arithmetic mean. i.e. (m - x )

4. Square each of (m - x ), to get (m - x )2

5. Multiply each of (m - x )2 with their corresponding frequencies to get f(m - x )2 and
obtain ∑f(m - x )2

6. Substitute the values of each of ∑f (m- x )2 and N in formulae (II), we get standard
deviation.

(c) Short-cut Method

In this method, deviation of each of mid-point the class is taken from assumed mean.

The fromula used in this method is

( )V =
∑fd2 - ∑fd 2
N N ......... (III)

Where, d = m - a

a = assumed mean

m = mid point of each class .

The following step is used in this method:
1. Find the mid-point of each class which is denoted by m.

2. Take middle figure of mid points as an assumed mean and denote it by a. Take deviations
of each mid point from the assumed mean. Denote it by d i.e. d = m - a

3. Multiply each of d with the corresponding frequency f, to get fd and obtain ∑fd.

4. Multiply each of fd with the corresponding d to get fd2 and then find ∑fd2.

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5. Substitute all the values of ∑fd2 , ∑fd, N in formula (III), we get standard deviation.

(d) Step Deviation Method :

This method is used mostly in the cases, when each of variety value has a common
factor. In case of continuous series, the common factor, is h, the class-size. The formula
used in this method is

( )V = ∑fd'2 - ∑fd' 2
N N
× h ........... (IV)

Where, d' = m -a
h
a = assumed mean

m = mid point of each class

h = class size

The following steps are used in this method:
1. Find the mid point of each class which is denoted by m.

2. Take an assumed mean 'a' from the mid figure of mid-points.

3. Find deviation of each mid point from the assumed mean i.e. (m - a)

4. Divide each of (m - a) by h, denote it by d' = m -a
h
5. Multiply each of d' by the corresponding frequency f to get fd' and find ∑fd'

6. Multiply each of fd' by d' and denote thus by fd'2 and find ∑fd'2

7. Substitute the values ∑fd' , ∑fd', ∑fd'2, N in formula (IV) to get standard deviation.

Coefficient of Standard Deviation

The relative measure of standard deviation is known as the coefficient of standard deviation.

? Coefficient of S.D. = standard deviation = V
mean x

If the coefficient of standard deviation is multiplied by 100%, then it is called coefficient of

variation (C.V.).

? Coefficient of variation (c.v) = V × 100%
x

Varience

The square of standard deviation is called varience. If standard deviation is V, the varience
is V

Example : If standard deviation of a set of data is 4,

then the varience is V2 = 16.

i,e, S.D. (V)= 4,

varience (V2) = 42 = 16

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Worked out Examples

Example 1. Calculate the standard deviation from given data by direct method.

Class 40-50 50-60 60-70 70-80 80-90 90-100
20 14
Frequency 6 10 20 30

Solution: To calculate standard deviation by direct method :

Example 2. Class Mid-value f fm fm2
Solution: 12150
40-50 45 6 270 30250
84500
50-60 55 10 550 168750
144500
60-70 65 20 1300 126350
6fm2 = 566500
70-80 75 30 2250

80-90 85 20 1700

90-100 95 14 1330

N = 100 6fm = 7400

( )∑fm2- ∑fm 2
N N
Standard deviation (V) =

( )=
566500 - 7400 2
100 100

= 5665 - 5476 = 13.74

Find the standard deviation from the following data by
(a) Actual mean (or direct Method) (b) Short -cut method
(c) Step deviation method
Also find the coefficient of S.D. and coefficient of variation.

x 0 -10 10-20 20-30 30-40 40-50 50-60
F 4 6 10 20 6 4

(a) Actual mean method (or direct method)
To calculate standard deviation by actual mean method (or direct method).

x Mid-value f fm |m-x| |m-x|2 f|m - x|2
4
0-10 5 6 20 -26 676 2704
10-20 15 10 90 -16 256 1536
20-30 25 20 250 -6 36 360
30-40 35 6 700 4 16 320
40-50 45 4 270 14 196 1176
50-60 55 220 24 576 2304
N = 50 6fm = 6fm = 6f|m - x|2
7400 1550 = 8400

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Mean (x ) = ∑fm = 1550 = 31
N 50
∑f(m - x )2
Standard deviation =
N

= 8400 168 = 12.96
50 =

coefficient of S.D. (V) = V × 100%
x

= 12.96 × 100% = 41.81%
31

(b) To calculate standard deviation by shortcut method.

Let = 35

x Mid-value f d=m–d fd fd2

0-10 5 4 -3 -120 3600
-120 2400
10-20 15 6 -20 -100 1000

20-30 25 10 -10

30-40 35 20 0 00
40-50 45 6 10 60 600

50-60 55 4 20 80 1600

N = 50 6fd = -200 6fm = 1550

( )∑fd2– ∑fd 2
N N
Standard deviation (V) =
= ( )9200– – 200 2
50 50

= 184 – 16 = 168 = 12.96

(c) To calculate standard deviation by step-deviation method

x Mid-value (m) f m – 35 fd' fd'2
10

0-10 5 4 -3 -12 36

10-20 15 6 -2 12 24

20-30 25 10 -1 -10 10

30-40 35 20 0 0 0

40-50 45 61 6 6

50-60 55 4 2 8 16

N = 50 6fd' = -20 6fd'2 = 92

We have from above table, N = 50, ∑fd' = –20, ∑fd'2

Standard deviation (V) = ( )∑fd'2–∑fd' 2
N N ×h

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( )= 92 – –20 2 × 10
50 50

= 46 – –4 × 10
25 25

= 42 × 10 = 1.296 × 10 = 12.96
25

Example 3. Calculate the standard deviation from the given data

Class 0 -10 10-20 20-30 30-40 40-50
Frequency 4 10 15 18 20

Solution: To calculate standard deviation, the given data can be written in the following table:

x Mid-value f d = m– a fd fd2
h
16
0-10 5 4 -2 -8 6
0
10-20 15 6 -1 -6 3
8
20-30 25 = a 5 0 0 6fd2 = 38

30-40 35 31 3

40-50 45 22 4

N = 50 6fd = -7

standard deviation (V) = ( )∑fd2 – ∑fd 2
NN ×h

where, h = 10, ∑fd'2 = 33, ∑fd' = –7

( )= 33 – –7 2
20 20 × 10

? V = 1.65 – 0.1225 × 10 = 12.36

Exercise 12.3

Very Short Question
1. (a) Define standard deviation

(b) Write the formulas to calculate standard devition for continuous series of data by
(i) direct method
(ii) actual mean method
(iii) Short-cut method
(iv) step-deviation method

(c) Define coefficient of variation with formulae
(d) Define varience

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Short questions

2. Calculate standard deviation and its coefficient for continuous series, if

(a) ∑fm2 = 566500, N = 100, ∑fm = 7400, x = 31
(b) ∑fd2 = 993, N = 60, x = 37, ∑fd = –130
(c) ∑fd' = –20, ∑fd'2 = 92, h = 10, x = 37, N = 100
(d) ∑fd' = –4, ∑fd'2 = 28, N = 29, x = 24.5, i = 10
Long Questions

3. Calculate the standard deviation and its coefficient from the following data by

(i) direct method (ii) short-cut method (iii) step-deviation

(a) Class Interval 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 10 8 32 40 22 18

(b) Marks 0-10 10-20 20-30 30-40 40-50

No. of students 7 12 24 10 7

(c) Wages (Rs.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No. of Workers 12 18 35 42 50 45 20 8

(d) Age (yrs.) 20-25 25-30 30-35 35-40 40-45 45-50

No. of persons 170 110 80 45 40 45

4. Calculate the standard deviation and coefficient of variation:

(a) Mark Secured 0-20 20-40 40-60 60-80 80-100

No. of Students 2 8 16 10 4

(b) Height (cm) 0-8 8-16 16-24 24-32 32-40
No. of plants 6
7 10 8 9

5. Calculate the standard deviation and coefficient of variation from the following:

(a) x 0≤x<10 10≤x<20 20≤x<30 30≤x<40 40≤x<50
f 7 10 14 12 6

(b) Mid-value 4 8 12 16 20 24

Frequency 10 15 11 16 14 5

(c) x less than 10 less than 20 less than 30 less than 40 less than 50
f 12 19 24 33 40

(d) x above 20 above 40 above 60 above 80 above 100 and less than 120

f 50 42 30 18 7

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(e) Marks 0-10 0-20 0-30 0-40 0-50
54
No. of students 7 18 30 42

6. From the data given below which series is more variable (inconsistent).

Variable 10-20 20-30 30-40 40-50 50-60 60-70
Section A 10 18 32 40 22 18

Frequency 18 22 40 32 20 10
Section B

Project work

List the marks obtained by the students in your class in optional mathematics and compulsory
mathematics. Find the standard deviation, coefficient of standard deviation and coefficient
of variation (C.V.). Compare the marks obtained in these two subjects in terms of C.V.

2. (a) 13.74, 0.4432 (b) 3.44, 0.093 (c) 9.34, 0.2535 (d) 9.7, 0.3959
3. (a) 13.72, 0.3156 (b) 11.39, 0.4617 (c) 17.25, 0.426 (d) 8.22, 0.269
4. (a) 20.27, 38.24 (b) 10.86, 50.75
5. (a) 12.29, 49.16% (b) 6.07, 45.47% (c) 15.03, 75.15% (d) 25.79, 37.41%

(e) 12.87, 50.27% CV(A') = 33.34%, CV(B) = 37.02%
6. V (A) = 14.05, V (B) = 14.10

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Curriculum Development Centre, Sanothimi, Bhaktapur

Syllabus of Optional Mathematics - Grade 10

S.N. Content Course Title Tentative
1. Algebra periods

2. Continuity Functions: Algebraic and trigonometrical functions with 35
3. Matrix
4. Coordinate their graphs of the types y = mx + c, y = ax2, a ≠ 0; y = ax3,

Geometry a ≠ 0; y = sinA, y = cosA, y = tanA(–2S ≤ A ≤ 2S).

Composite function of two functions and inverse
functions with their presentation in arrow diagrams.

Polynomials: Simple operation of polynomials,
Synthetic division, Remainder and Factor theorem and
their applications; Solutions of polynomial equations
upto degree three using Remainder and Factor theorems.

Sequence and Series: Arithmetic sequence - Introduction,
General term, Mean, Sum of the terms of an arithmetic
sequence, Sum of the first n natural numbers including
odd and even natural numbers.

Geometric Sequence: Introduction, General term, Mean,
Sum of the finite terms of a geometric sequence.

Linear Programming: Introduction, Linear inequalities,
Detriment of inequality by graph, Maximization and
minimization of linear programming problems.

Quadratic Equations and Graphs: Graphs of quadratic 10
and cubic functions, Solution of quadratic equations
using graph, Solution of simultaneous linear and 15
quadratic equations by graphical and substitution 30
method.

Simple Concept of Continuity: Investigation of
continuity in set of real numbers, Investigation of
continuity and discontinuity in different types of set of
numbers, Investigation of the continuity of the function
using graph, Symbolic representation of the continuity
of the function.

Determinant of matrix of order 2 × 2, Inverse of a
2 × 2 matrix, Solution of simultaneous linear equations
using matrix method, Cramer's rule, and its application
upto a determinant of order 2 × 2.

Straight Line: Angle between two straight lines,
Condition for two lines to be parallel and perpendicular.

Pair of Straight Lines: Equations of the lines given
by the homogeneous equation of degree two and the
angle between them. Conditions for the two lines to be
coincident and perpendicular.

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Conic Section: Introduction, Different types of conic

section formed by the intersection of a cone and a plane.

Circle: Definition, Equation of the circle of the types:

x2 + y2 = r2, (x – h)2 + (y – k)2 = r2, (x – x1) (x – x2) +
(y – y1) (y – y2) = 0 and x2 + y2 + 2gx + 2fy + c = 0 and
the related problems.

5. Trigonometry Trigonometric Identities: Trigonometric ratios of 35
18
multiple and submultiple angles (sine, cosine and
15
tangent only), Transformation of identities into the sum, 12

difference and the product form (in sine and cosine only)

Conditional Identities: Solution of the trigonometric

identities under the condition A + B + C = Sc

Trigonometric Equations: Solution of trigonometric

equation (upto degree two) (0 ≤ T < 2Sc)

Height and Distance: Word problems of height and

distance consisting of two angle - angle of elevation and

angle of depression.

6. Vectors Scalar product of two vectors (dot product), Condition

for two vectors to be perpendicular.

Vector Geometry: Mid-point theorem, Section formula.

Theorems:

(i) The straight line joining the middle points of two

sides of a triangle is parallel to and half of the third

side.

(ii) The straight line joining the vertex and the middle

point of the base is perpendicular to the base.

(iii) The straight lines joining the middle points of

the sides of a quadrilateral taken in order is a

parallelogram.

(iv) The diagonals of a parallelogram bisect each other.

(v) The diagonals of a rectangle are equal.

(vi) The diagonals of a rhombus bisect each other at

right angles.

(vii) The angle in the semi-circle is a right angle.

(viii)The middle point of the hypotenuse of a right

angled triangle is equidistant from the vertices.

7. Transformation Coposite transformation of any of two transformations.

Reflection, Rotation, Translation and Enlargement,

Inversion transformation, Inverse circle, Use of matrix

in transformation.

8. Statistics Dispersion: Quartile deviation and its coefficient

(continuous series only), Mean deviation from mean

and median and its coefficient (continuous series only),

Standard deviation and its coefficient and analysis

(continuous series only), coefficient of variation (C.V.).

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vedanta Excel In Opt. Mathematics - Book 10

SEE Specification Grid 2076

Issued by CDC

K U A HA

SN Contents Topics Each Each Each Each TQ TM
of 1 of 2 of 4 of 5

Mark Marks Marks Marks

1. Algebra Function

Polynominals 2 3 2 1 8 21
Sequence and Series

Quadratic Equation and Graph

2. Limit and Numbers and Continuity

Continuity Discontinuity in Graph 1 - 1 - 25
Notational Representation of

Continuity

3. Matrix Determinant and Inverse of

Matrix

Solving Equations by Matrix 12 1 - 49

Method

Cramer's Rule

4. Coordinate Angle between two lines

Geometry Pair of straight lines 2 2 1 1 6 15
Conic Sections

Circle

5. Trigonometry Multiple and sub-multiple angles

Transformation of
Trigonometric Identies

Conditional Trigonometric 23 3 - 8 20

Identities

Trigonometric Equations

Height and Distance

6. Vectors Scalar Product 1 2 - 1 4 10
Vector Geometry

7. Transformation Combined Transformation

Inversion Transformation and 1 - 1 1 3 10
Inversion Circle

Matrix Transformation

8. Statistics Quartile Deviation

Mean Deviation -12 - 3 10
Standard Deviation and

Coefficient of Variation

Total 10 13 11 4 38 100

INDEX U = Understanding A = Application
K = Knowledge

HA = Higher Ability TQ = Total no. of Questions TM = Total Marks

368 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10

SEE New Model Questions 2076

Issued by CDC-2076 (2019)

Group A (10 × 1 = 10)

1. (a) Define trigonometric function.

(b) What is the arithmetic mean between two numbers 'a' and 'b'?

2. (a) Write a set of numbers which is continuous in number line.

(b) If matrix A = p q , what is the value of |A|?
r s
3. (a) If the slopes of two straight lines are m1 and m2 respectively and T be the angle
between them, write the formula for tanT.

(b) Which geometric figure is formed if a plane intersects a cone parallel to its base?

4. (a) Express sin2A in terms of tanA.

(b) Define angle of elevation.

5. (a) What is the scalar product of two vectors oa and ob if the angle between them is T?

(b) In an inversion transformation if P' is image of P and r is radius of inversion circle
with centre O, write the relation of OP.OP' and r.

Group B (13 × 2 = 26)

6. (a) Find f-1(x) if f(x) = 4x + 5.

(b) If g(x) = 2x – 1 and f(x) = 4x, find the value of gof(x).

(c) What are the points of intersection of the curve f(x) = x2 – 1 and f(x) = 3?

7. (a) If A = 2 –1 , find |A| and write if A–1 is defined.
3 1
(b) According to Cramer's rule, find the values of D1 and D2 for ax + by = c and
px + qy = r.

8. (a) Find the slopes of two straight lines 3x + 4y + 5 = 0 and 6x + 8y + 7 = 0 and
write the relationship between them.

(b) Find the single equation for the pair of lines represented by 3x + 2y = 0 and
2x – 3y = 0.

9. (a) Convert sin6A . cos4A into sum or difference of sine or cosine.

(b) Express 1 sinA in terms of sub-multiple angle of tangent.
+ cosA

(c) If 2sin2T = 3, find the value of T. (0° ≤ T ≤ 180°)

10. (a) Find the angle between two vectors oa and ob if |oa | = 2, |ob | = 12 and oa .ob = 12.
A
(b) From the adjoining figure, find AoP and express op in terms
of oa and ob . oa op P

(c) If the standard deviation of a set of data is 0.25, find its O ob B
variance.

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vedanta Excel In Opt. Mathematics - Book 10

Group C (11 × 4 = 44)

11. Solve : x3 – 3x2 – 4x + 12 = 0

12. Optimize P = 5x + 4y under the given constraints : x – 2y ≤ 1, x + y ≤ 4, x ≥ 0, y ≥ 0

13. For a real valued function f(x) = 2x + 3

(a) Find the values of f(2.95), f(2.99), f(3.01), f(3.05), and f(3).

(b) Is this function continuous at x = 3?

14. By using matrix method, solve the following system of equations: 3x + 5y = 11, 2x – 3y = 1

15. Find the single equation of pair of straight lines passing through the origin and
perpendicular to the lines represented by 2x2 – 5xy + 2y2 = 0.

16. Find the value of : sin20° . sin30° . sin40° . sin80°

17. If A + B + C = Sc, prove that sin2A – sin2B + sin2C = 2sinA.cosB.sinC

18. From a point at the ground level infront of a tower, the angle of elevations of the top
and bottom of flagstaff 6 m high situated at the top of a tower are observed 60° and
45° respectively. Find the height of the tower and the distance between the base of the
tower and point of observation.

19. Find the 2 × 2 matrix which transforms a unit square to a parallelogram 0 34 1 .
0 01 1
20. Find the mean deviation from mean and its coefficient from given data:

Marks obtained 0-10 10-20 20-30 30-40 40-50
No. of students 2 3 6 5 4

21. Find the standard deviation and coefficient of variation from given data:

Age 0-4 4-8 8-12 12-16 16-20 20-24
No. of students 7
7 10 15 7 6

Group D (4 × 5 = 20)

22. A contractor on construction job specifies a penalty for delay of completion beyond
a certain date as: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the
third day and so on. The penalty for each succeeding day is Rs. 50 more than that of
the preceding day. How much money does the contractor have to pay as penalty, if he
delays the work by 30 days?

23. On a wheel, there are three points (5, 7), (–1, 7) and (5, –1) located such that the distance
from a fixed point to these points is always equal. Find the coordinate of the fixed point
and then derive the equation, representing the locus that contains all three points.

24. By using vector method, prove that the quadrilateral formed by joining the mid-points
of adjacent sides of a quadrilateral is a parallelogram.

25. The coordinates of vertices of a quadrilateral ABCD are A(1, 1), B(2, 3), C(4, 2),
and D(3, –2). Rotate this quadrilateral about origin through 180°. Reflect this image
of quadrilateral about y = –x. Write the name of transformation which denotes the
combined transformation of above two transformations.

370 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur