Optional Math 8

iii) If a 2 = 3 n , find the direction of a – 2b .
= d n and b d
44

iv) If A(–3, –1) and B(2, 4), find the magnitude and direction of AB .

v) If magnitude of a = x is 5 units, find the value of ‘x’.
c4m

6. Project work
Collects the different activities in your daily life which can be

represented in vector notation. Also present such examples in your
classroom.

Answer

1. i) 19 units ii) 5 units iii) 10 units

iv) 4 2 units v) 5 units Vector Geometry

2. i) 30° ii) 60° iii) 0°

iv) 45° v) 0°

3. i) Va = KKKKJKKK 2 5 OOOOOONO ii) Ub = KLJKKKKKK–4355 OOOOOOPNO iii) KKJKKKKLK 12 55OPNOOOOOO
1 5 P

L

iv) LKKKKKKJK–3455 OOOOOOONP v) JKKKLKKKK34 55 OOOOOOONP

4. i) e 4 o ii) d20n iii) d76n
4

iv) d72n v) d1110n

5. i) 34 units ii) 10 units iii) 45°

iv) 5 2 units, 45° v) 3

PRIME Opt. Maths Book - VIII 147

Vector Geometry

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. What do you mean by unit vector?
2. a. If A(1, 2) and B(4, –2) are any two points find the magnitude of AB .

b. If a =e 3 o , find the direction of a .
3

c. If a = 1 and b = 3 n , find 2a + b .
dn d
2 –1

3. a. If A(3, 2), B(4, –3), P(1, 4) and Q(0, 9), prove that AB = – PQ .

Vector Geometry b. If a = 3 , b = –1 n , find the magnitude of 2a – b .
dn d
2
4

4. If A(3, –2) and B(7, m) find the value of m where magnitude of AB is
4 2 units. Also find the direction of AB .

148 PRIME Opt. Maths Book - VIII

Unit 7 Transformation

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – 1 – 1 2 7 10
Weight – 2 – 5

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can define the transformation.
• Students can identify the isometric transformations.
• Students can find the image of a point under isometric transformations

geometrically.
• Students can find the image using co-ordinate.
• Students can plot the object and image in graph.

Materials Required:
• Chart paper.
• Graph board.
• Sample of transformations.
• List of the formula of transformations.
• Geo-board
• Chart paper.
• Shape of objects

PRIME Opt. Maths Book - VIII 149

Introduction of transformation

Let us see,

A Ai)

Here, the image of A is seen into a mirror, where
• A is the object and
• A into the mirror is called the image of A.

ii)

object image
mirror
Transformation
Here,
• 1st fish is taken as the object.
• Second fish is taken as the image.

Above examples show that one object can be transferred from one place
to another place with same shape & size or different size which is called
transformation.

The process of changing the position or
size of an object under any geometrical
conditions is called transformation.

By using transformation patterns of objects can be drawn which is useful
to fill the pictures in clothes walls and in any kind of objects to make them
very attractive.

Here in grade VIII we discuss only three types of transformations
i) Reflection ii) Rotation iii) Translation

150 PRIME Opt. Maths Book - VIII

7.1 Reflection A P

The image of person AB is A‘B‘ which is formed Q A‘
m B‘
at equal distance from the mirror line m as the
distance of object where AP = PA‘ or BQ = QB‘.
The image A‘B‘ so formed is laterally inverted as B
the image formed in the looking glass but size is

same as the object.

Here, AB is called the object
m is called reflection axis (mirror line)
A‘B‘ is called the image.

The transformation of an object from one place to Transformation
another place by a mirror line is called reflection.
• Object and image so formed are always congruent

in reflection.

Reflection using co-ordinates axes.

i) Reflection about x-axis (y = 0).

Y

4

3 M(2, 3)

2

1

X‘ P X
-4 -3 -2 -1 O 1 2 3 4
-1

-2

-3 M’(2, –3)

-4

Y‘

PRIME Opt. Maths Book - VIII 151

Draw MP^OX and produce MP to M‘ such that MP = PM‘
Here, M(2, 3) is an object.
x-axis (XX‘) is the reflection axis.
M‘(2, –3) is the image.

It shows that: P(x, y) → P‘(x, –y)

ii) Reflection about y-axis (X = 0)

Y

P‘(–3, 4) M4 P(3, 4)

3

2

1

X’ -4 -3 -2 -1 O 1 2 3 4 X
-1

Transformation -2

-3

-4

Y’
Draw PM^OY & produce PM to MP‘ to make PM = MP‘.
Here, P(3, 4) is an object.
Y -axis (YY‘) is the reflection axis.
P‘(–3, 4) is the image.

It shows that : P(x, y) → P‘(–x, y)

152 PRIME Opt. Maths Book - VIII

Worked out Examples

1. Find the image of the plane figure given below under the given mirror
line M.
A
B
M

C

Solution: A Transformation
B

M

P

Q

C‘ B‘

R

C A‘
Here, DABC is the given object, mirror line M is the reflection axis.
Draw, BP^M, BP = PB‘
AQ^M, AQ = QA‘
CR^M, CR = RC‘

Then, Join A‘, B‘ & C‘ so DA‘B‘C‘ is the image of DABC after reflection
under a mirror line M.

PRIME Opt. Maths Book - VIII 153

2. Find the image of a point A(–2, 3) under reflection about x-axis.
Solution:
Under reflection about x-axis.
P(x, y) → P‘(x, –y)
A(–2, 3) → A‘(–2, –3)

3. Find the image of DABC having vertices A(2, 3), B(4, 6) & C(7, 1) under
reflection about x = 0. Also plot them in a graph.

Solution :
Under reflection about x = 0 (y-axis)
P(x, y) → P‘(–x, y)
A(2, 3) → A‘(–2, 3)
B(4, 6) → B‘(–4, 6)
C(7, 1) → C‘(–7, 1)

Y

B‘ B

Transformation X’ C‘ A‘ A C
O X

Y’

Here,
DABC is the given object.
Y = axis (x = 0) is the mirror line.
DA‘B‘C‘ is the image.

154 PRIME Opt. Maths Book - VIII

Exercise 7.1

1. Find the image of the following plane figures under the given reflection

axis ‘M’. B
M
A

B

C A

i) M ii)

R

Q

P Q M Transformation
S R

iii) M P
S
iv)

TR

P
Q

v) M vi) M

2. Find the image point of the following under the reflection about

x - axis.

i) A(2, –3) ii) P(–3, 5) iii) M(–3, –2)

iv) N(3, 7) v) Q(3, –7) vi) D(–4, 6)

PRIME Opt. Maths Book - VIII 155

3. Find the image of the following points under reflection about y-axis.

i) M(–3, –5) ii) N(–7, 2) iii) A(2, 5)

iv) B(5, –3) v) C(–8, –3) vi) D(–5, 4)

Transformation 4. PRIME more creative questions
i) Find the image of the points A(2, 2) and B(6, 7) under reflection

about x-axis. Also plot them in graph. Also join AB and A‘B‘.

ii) Find the image of DPQR having vertices P(–2, 1), Q(–5, 4) &
R(–6, –2) under reflection about y-axis. Also plot them in graph.

iii) Find the image of quadrilateral having vertices A(1, 2), B(3, 5),
C(6, 6) and D(7, 1) under reflection about y = 0. Also plot them in
graph.

iv) Find the image of quadrilateral having vertices P(3, 1), Q(4, 5),
R(7, 4) and S(8, –2) under reflection about X = 0. Also plot them
in graph.

v) Find the coordinate of image of triangle having vertices A(2, 1),
B(4, 5) & C(5, –3) under reflection on x–axis to DA‘B‘C‘. Also plot
DABC, DA‘B‘C‘ & DA‘‘B‘‘C‘‘ in graph. Also find the coordinates of
image of DABC under the reflection about y-axis and plot DABC,
DA‘B‘C‘ and DA‘‘B‘‘C‘‘ in graph.

Answer

1. Discuss with your subject teacher. iii) M‘(–3, 2)
2. i) A‘(2, 3) ii) P‘(–3, –5)
iv) N‘(3, –7) v) Q‘(3, 7) vi) D‘(–4, –6)
3. i) M‘(3, –5) ii) N‘(7, 2) iii) A‘(–2, 5)
iv) B‘(–5, –3) v) C‘(8, –3) vi) D‘(5, 4)
4. i) A‘(2 –2) & B‘(6, –7); graph
ii) P‘(2, 1), Q‘(5, 4) & R‘(6, –2); graph
iii) A‘(1, –2), B‘(3, –5), C‘(6, –6), D‘(7, –1); graph
iv) P‘(–3, 1), Q‘(–4, 5), R‘(–7, 4), S’(–8, –2)
v) A‘(2, –1), B‘(4, –5), C‘(5, 3) and A‘‘(–2, –1), B‘‘(–4, –5) & C‘‘(–5, 3)

Note: Shot to your subject teacher after plotting in graph.

156 PRIME Opt. Maths Book - VIII

7.2 Rotation: A A‘
B‘ C‘
B
C

O Transformation

Here, O is the center of rotation, object DABC is rotated in clockwise
direction with the angle of rotation 90° where,
\ AOA’ = \ BOB’ = \ COC’ = 90°
DABC is an object.
DA’B’C’ is the image.

The transformation of an object from one
place to another place according to centre,
direction and angle is called rotation.

Note : Anticlockwise direction = Positive direction
Clock wise direction = Negative direction
+ve quarter turn = +90° or – 270°

–ve quarter turn = – 90° or + 270°
Half turn = +180° or –180°
Full turn (complete rotation) = 360°

PRIME Opt. Maths Book - VIII 157

Rotation using co-ordinate having center origin.
1. Rotation about positive quarter turn.

Y

P’(–3, 2) P(2, 3)
X’ OX

Transformation Y’

Above example shows that A(2, 3) → A‘(–3, 2) under rotation about +90°.
It gives image of a point P(x, y) as
\ P(x, y) $ P‘(–y, x)

2. Rotation about negative quarter turn.

Y

P(4, 3)

X’ O X

P(3, –4)

Y’

Above example shows that A(4, 3) → A’(3, –4) under rotation about, – 90°.
It gives image of a point P(x, y) as,
\ P(x, y) → P‘(y, –x)

158 PRIME Opt. Maths Book - VIII

3. Rotation about half turn.

Y

A(4, 5)

X’ O X Transformation

A’(–4, –5)

Y’

Above example shows that A(4, 5) → A‘(–4, –5) under rotation about
180°.

It gives image of a point P(x, y) as,
\ P(x, y) → P‘(–x, –y)

4. Rotation about full turn.
Rotation about full turn (360°) is invariant transformation.
i.e. P(x, y) → P‘(x, y)

A‘, A

Here, 159
Image of A is A`.
after full turn.

PRIME Opt. Maths Book - VIII

Worked out Examples

1. Rotate the given triangle under positive quarter turn with center ‘O’.
A

C B
Solution:
O
A

Transformation A‘ B‘ B
C

O

C‘

Here, DABC is an object.
O is center of rotation.
\ AOA‘ = \ BOB‘ = \ COC‘ = 90° in positive direction (Anticlockwise).
\ DA‘B‘C‘ is the image of DABC.

160 PRIME Opt. Maths Book - VIII

2. Find the image of a point A(–3, 2) under rotation about –90° with Transformation
centre origin.

Solution :
Under rotation about –90°.
P(x, y) → P‘(y, –x)
\ A(–3, 2) → A‘(2, 3)

3. Find the image of DPQR under rotation about positive quarter turn.
Also plot the object and image in graph. Where P(1, 2), Q(3, –2) & R(5,
4) are the vertices.

Solution:
Under rotation about +90°
P(x, y) → P‘(–y, x)
\ P(1, 2) → P‘(–2, 1)
Q(3, –2) → Q‘(2, 3)
R(5, 4) → R‘(–4, 5)

Y

R‘
Q‘ R

P

X’ P‘ X
O

Q

Y’

Here,
DPQR is the object.
under rotation about +90°.
DP‘Q‘R‘ is the image.

PRIME Opt. Maths Book - VIII 161

Exercise 7.2

1. Draw the image of the following diagrams under the following

conditions given in question.

i) A ii) P

B RQ
C

O (+90°) O (–90°)

iii) Q iv) E

R AD

Transformation S P
BC

O (–90°) O (–270°)

v) G vi) A
W
A E Q T

HF V S
C U

BD R

O (+270°) O (–270°)

2. Find the image of the following points under rotation about negative

quarter turn taking center origin.

i) (3, –5) ii) (5, 2) iii) (–2, –7)

iv) (–6, 3) v) (–7, –1) vi) (5, –3)

162 PRIME Opt. Maths Book - VIII

3. Find the image of the following points under rotation about positive

quarter turn taking center origin.

i) (–7, 2) ii) (–5, –3) iii) (8, –2)

iv) (3, 7) v) (–6, 2) vi) (–2, –8)

4. Find the image of the following points under rotation about half turn

taking center origin.

i) (–2, –6) ii) (5, –3) iii) (3, 6)

iv) (–5, 4) v) (–4, –5) vi) (2, –5)

5. i) Find the co-ordinate of image of DABC having vertices A(–2, 3), Transformation
B(–3, –1) and C(–5, 4) under rotation about –270° taking center
origin. Also plot the object and image in graph.

ii) Find the co-ordinate of the image of DPQR having vertices P(2,
1), Q(3, 5) & Q(5, 0) under rotation about +270° taking center
origin. Also plot the DPQR and DP‘Q‘R‘ in graph.

iii) Which rotation gives the image of point A(2, –1) to A‘(1, 2)? Also
find the image of the points B(3, 2) and C(5, –4). Also plot DABC
and DA‘B‘C‘ in graph.

iv) Find centre of rotation, angle of rotation and direction of rotation
which gives A(2, –3) to A‘(–3, –2) and B(4, 1) to B‘(1, –4) by
plotting the points in graph.

v) Find the co-ordinate of the image of quadrilateral having vertices
A(–3, 2), B(–1, 5), C(2, 4) & D(3, –1) under rotation about positive
quarter turn taking center origin. Also plot them in graph.

PRIME Opt. Maths Book - VIII 163

Answer

1. Consult to your subject teacher.

2. i) (–5, –3) ii) (2, –5) iii) (–7, 2)
vi) (–3, –5)
iv) (3, 6) v) (–1, 7)

3. i) (–2, –7) ii) (3, –5) iii) (2, 8)
iv) (–7, 3) v) (–2, –6) vi) (8, –2)

4. i) (2, 6) ii) (–5, 3) iii) (–3, –6)
iv) (–5, –4) v) (4, 5) vi) (–2, 5)

Transformation 5. i) A‘(–3, –2), B‘(1, –3) & C‘(–4, –5)
ii) P‘(1, –2), Q‘(5, –3) & R‘(0, –5)
iii) Rotation about +90°, B‘(–2, 3) & C(4, 5)
iv) Centre origin, Angle 90° and direction is clock wise.
v) A‘(–2, –3), B‘(–5, –1), C‘(–4, 2) & D‘(1, 3)

Note : Show to your subject teacher after plotting in graph.

164 PRIME Opt. Maths Book - VIII

7.3 Translation P’

P

Q’ B R’

QA

R
Here, vector AB is the magnitude & direction of translation.
DPQR is an object.
PP‘ = QQ‘ = RR‘ = AB
PP‘ ' QQ‘ ' RR‘ ' AB
DP‘Q‘R‘ is the image of DPQR

The transformation of an object from one place to Transformation
another place according to the magnitude and direction
of the given vector is called translation.

i) Translation using co-ordinate:

Y

A’(7, 5)
A(4, 3)

X’ B’(6, 1) C’(9, 1)
X
O
B(3, -1) C(6, -1)

Y’

PRIME Opt. Maths Book - VIII 165

Here, A(4, 3) is translated to A’(7, 5)
B(3, –1) is translated to B’(6, 1)
C(6, –1) is translated to C’(9, 1)

i.e. All the points are translated with constant number 3 for
x-component and 2 for y - component.

i.e. A (4, 3) → A‘(4 + 3, 3 + 2) = A‘(7, 5)

3
i.e. Translation vector is T = <F
2

a
i.e. under translation T = <F
b

P(x, y) → P‘(x + a, y + b)

iii) Translation using vector:

Transformation Let us consider a translation vector is AB where A(1, 2) and B(3, 5)

are any two points.

Then, according to the concept of column vector AB

AB = x2 – x1 = 3–1 = 2
y2 – y1 5–2 3

2
\ Translation vector T = AB = 3

Worked out Examples

1. Find the image of a point A(3, –2) under a translation vector of T = <13F .
Solution:

1
Under translation vector T = <F
P(x, y) → P‘(x + a, y + b) 3

→ P‘(x + 1, y + 3)
A(3, –2) → A‘(3 + 1, –2 + 3)
→ A‘(4, 1)

1
T=< F

\ A(3, –2) 3 A’(4, 1)

166 PRIME Opt. Maths Book - VIII

2. If A (3, –1) and B(1, 2) are any two points, find the image of point A

under AB .

Solution :

Here,

Translation vector AB from the given points A(3, –1) & B(1, 2) is,

<x2 – x1F 1–3 –2
= y2 – y1 = < F = <F =T
AB 2 + 1
3

–2
Then, under translation vector T = < 3 F

p (x, y) → P‘(x + a, y + b) = P‘(x – 2, y + 3)
A(3, –1) → A‘(3 – 2, –1 + 3) = A‘(1, 2)

–2
T=< F

\ A(3, –1) 3 A’(1, 2)

3. Find the image of DABC having vertices A(3, 1), B(–1, 5) & C(1, –3) under

a translation vector T = <32F . Also draw the object and image in graph. Transformation
3
Solution: Under a translation vector T = <F
P(x, y) → P‘(x + a, y + b) 2

→ P‘(x + 3, y + 2)

Then,
A(3, 1) → A‘(3 + 3, 1 + 2) = A‘(6, 3)
B(–1, 5) → B‘(–1 + 3, 5 + 2) = B‘(2, 7)
C(1, –3) → C‘(1 + 3, –3 + 2) = C‘(4, –1)

Y
B’

B

X’ A A’
X
Here, DABC is the O
object. C’ 167

DA‘B‘C‘ is the image of C
DABC Y’

PRIME Opt. Maths Book - VIII

4. Draw the image of given tangle under the given vector AB .
Q

A B
P R

Solution: Q Q’

A B

Transformation P P’

R R’

Here, Draw PP‘, QQ‘, RR‘ || AB .
Then,
Taking PP‘ = QQ‘ = RR‘ = AB
DPQR is the object.
DP‘Q‘R‘ is the image.

168 PRIME Opt. Maths Book - VIII

Exercise 7.3

1. Find the image of the following figures under the given vector.
A
M

PB B

N

i) Q A R ii) C
S
BP

T R D Transformation
A B C

iii) Q iv) A Q

B AG F
v) H
N
M
D
CE

2. Find the image of the following points under the translation vector.

2
T = <F
1

i) A(3, 2) ii) P(–2, 5) iii) M(–3, –1)
iv) N(4, –6) v) O(–3, –4)

3. Find the image of the points P(3, –2) under the translation vector AB
for the following points A and B.

PRIME Opt. Maths Book - VIII 169

i) A(3, 2) and B(1, 0) ii) A(1, –2) & B(3, 1)
iii) A(2, 1) and B(4, 5) iv) A(–1, 2) & B(2, 1)
v) A(4, 1) and B(1, –2)

4. Find the image of the following triangles under the given vector. Also

plot the object and image in graph.

2
i) A(1, –2), B(–3, 4) and C(3, –3) under a vector T= <F
3

–2
ii) P(2, 5), Q(–2, 1) and R(5, 2) under a vector T = <F
–3

1
iii) A(–2, –4), B(–4, –1) and C(3, 0) under a vector T = <F
3

–2
iv) K(2, 3), L(4, 6) & M(6, 1) under a vector T = < 4 F

3
v) X(–2, 5), Y(3, 1) & Z(–4, –1) under a vector T = <F
Transformation –4

PRIME more creative questions

5. i) If a translation T gives image of an object A(1, 3) → A’(3, 4). Find

the value of translation vector T.
ii) If a point P(3, –1) is translated to P’(5, 2) under a vector T, find

the value of translation vector T.

iii) If a vector T = <1F translate a point P to P’(3, 7), find the
3

co-ordiante of the point P.

iv) Find the co-ordiantes of image of DABC having vertices P(1, 2),

Q(3, 5) and R(6, –2) under a translation vector PQ . Also plot
DPQR and DP‘Q‘R‘ on graph paper.

v) Find the co-ordinates of the image of DABC having A(3, 1),

B(7, 3), C(5, –2) under a translation vector AB . Also pot them in

graph.

Project work

6. Collects the formula used in transformations in a chart paper and

present in your classroom as the project work.

170 PRIME Opt. Maths Book - VIII

Answer

1. Show to your subject teacher.

2. i) A‘(5, 3) ii) P‘(0, 6) iii) M‘(–1, 0)
iv) N‘(6, –5) v) O‘(–1, –3) iii) P‘(5, 2)

3. i) P‘(1, –4) ii) P‘(5, 1)
iv) P‘(6, –3) v) P‘(0, –5)

4. i) A‘(3, 1), B‘(–1, 7) & C‘(5, 0)
ii) P‘(0, 2), Q‘(–4, –2) & R‘(3, –1)
iii) A‘(–1, –1), B‘(–3, 2), C‘(4, 3)
iv) K‘(0, 7), L‘(2, 10), M‘(4, 5)
v) X(1, 1), Y‘(6, –3), & Z‘(–1, –5)

2 2 Transformation
5. i) T = <F ii) T = <F iii) P(2, 4)
1 3

iv) P‘(3, 5), Q‘(5, 8) & R‘(8, 1)

v) AB = 4 n ; A‘(7, 3), B‘(11, 5) & C‘(9, 0)
d
2

Note : show to your subject teacher after plotting in graph.

PRIME Opt. Maths Book - VIII 171

Transformation

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. What is transformation? Name any two of them.

2. a. Find the image of a point A(3, –2) under reflection about x-axis.
b. Find the image of a point P(–2, 5) under rotation about +90° with

c. center origin.

Find the translation vector ‘T’ Which transform A(2, –1) to A’(4, 2).

Transformation 3. a. Find the image of DABC having vertices A(3, 4), B(1, 0) and
C(4, – 3) under reflection about y-axis. Also plot them in graph.

b. Find the co-ordinate of image of a triangle having vertices P(–2, 3),

Q(–4, 0) and R(0, – 3) under rotation about –90° with centre origin.

Also plot them in graph.

4. The vertices of DABC are A(1, 2), B(3, 5)and C(4, –2). Find the co-
ordinate of image of DABC under translation about AB . Also plot them
in graph.

172 PRIME Opt. Maths Book - VIII

Unit 8 Statistics

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – 1 – 1 2 7 6
Weight – 2 – 5

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can find the values of central tendency mean, median & mode.
• Students can find the partition values like quartiles and deciles.
• Students can find the measures of dispersion range and quartile deviation.
• Students can tabulate the collected data.
• Students can present the dat in diagrams.

Materials Required:
• Chart paper.
• Sample of data collection.
• Sample of tabulation of data.
• List of formula used in statistics.
• Graph paper.
• Scissors
• Geo-board

PRIME Opt. Maths Book - VIII 173

Introduction of statistics

Data Collection
The set of numerals collected for any fact for the purpose of investigation
or analysis during study is called statistical data. It can be collected by the
individuals directly for personal investigation by travelling to the related
area, through interviews and questionnaires etc, is called primary data.

It can be collected by taking the informations from related offices like
VDC, Municipality, sub metropolitan city, metropolitan city, district office,
province office, central office etc is called secondary data.

In statistics, the collected raw data should be tabulated, represented
diagrammatically and processed for different analysis. Among them here
we discuss about two central values called mean and median, partition
values quartiles and deciles as well as measure of dispersion range and
quartile deviation.

Statistics Tabulation of data
The collected data which is written roughly is called raw data. The categorical
presentation of data in a table using frequency is called the frequency
distribution table. There are three types of frequency distribution table viz.
individual, discrete and continuous frequency distribution tables. Discrete
frequency distribution table and continuous frequency distribution table.

i) Discrete frequency distribution (ungrouped) :
The raw data collected for the investigation is tabulated individually with

respect to their repeated numbers with tally marks in it. The number of
repeatation of any variable is called the frequency of the variable.
Example : The number of family members collected in a village is as 3,
2, 4, 4, 3, 2, 3, 3, 4, 4, 4, 5, 5, 6, 5, 6, 4, 3, 2, 5, 6, 4, 3, 4, 4, 3, 4, 5, 2, 5

No. of members Tally Marks Frequency

2 |||| 4

3 |||| || 7

4 |||| |||| 10

5 |||| | 6

6 ||| 3

Total 30

174 PRIME Opt. Maths Book - VIII

ii) Continuous frequency distribution (grouped):
When the collected raw data is too large having long range between

smallest and highest items, the discrete frequency distribution is
dif�icult to prepare and lead to a confusion. In such situation, we divide
the distribution in many subgroups of suitable sizes, called the class

intervals. Which makes our presentation and study more comfortable,

understanding and easier. This distribution is called the continuous

frequency distribution table. There are two types of continuous

frequency distribution viz.

• inclusive class interval

• exclusive class interval
Example: The marks obtained by 40 students in �irst term examination
of optional maths out of 50 marks is as :

(No. of students)

Marks Tally marks Frequency

0 - 10 || 2

10 - 20 |||| 5 Statistics

20 - 30 |||| |||| |||| | 16

30 - 40 |||| ||| 8

40 - 50 |||| |||| 9

This is an example of exclusive class interval as upper limits are
excluded in each class intervals.

8.1 Measure of Central Tendency:

The calculation of a central numerical value of the given statistical data
which suppose to represent the entire data is called the measure of central
tendency. They are mean, median and mode.

Arithmetic Mean

The average value of the statistical data which is the ratio of sum of

observations to the total number of observations is called the arthmetic

mean or simply mean. Total sum of observations
Total no. of observations
i.e. Arithmetic Mean =

PRIME Opt. Maths Book - VIII 175

i) For individual observations: Mean ( x ) = Rx
n
Where, x = observation (value of variable)

Rx = sum of the observations

n = no. of observations

ii) For discrete observation
Rfx
Mean ( x ) = N

Where,

x = observations (value of variable)

f = No of observations (No. of repeatation of particular variable)

fx = product of observations and respective no. of observations

R fx = sum of (f × x)

N = R f = sum of number of observations

iii) For continuous class interval
Rfm Rfx
Mean ( x ) = N or N

Statistics Where,

m = Mid value (mean) of each classes. (Also can be taken x)

f = No. of observations

fm = Product of mid-value and number of observation (fx)

R fm = sum of (fm) or (fx)

N = R f = sum of number of observations.

Worked out Examples

1. Find the arthmetic mean of the observations 7, 9, 11, 15, 20, 29, 24, 32.

Solution :

The observations are :

x : 7, 9, 11, 15, 20, 29, 24, 32

No. of observations (n) = 8

Then, = Rx
Mean ( x ) n

= 7 + 9 + 11 + 15 + 20 + 29 + 24 + 32
8
140
= 8

= 17.5

176 PRIME Opt. Maths Book - VIII

2. If mean of the observations 3, 7, x, 12, 15 and 14 is 10, �ind the value

of ‘x’.

Solution:

The observations are :

x : 3, 7, x, 12, 15,14

No. of obs (N) = 6

Mean ( x ) = 10

We have, Rx
N
Mean ( x ) =

or, 10 = 3 + 7 + x + 12 + 13 + 14
6

or, 60 = 49 + x

∴ x = 11

3. Find the mean of the marks obtained by the students given below.

Marks 12 18 24 30 36

F45894 Statistics

Solution :

x f f×x

12 4 48

18 5 90

24 8 192

30 9 270

36 4 144

N=30 R fx = 744

We have, = Rfx
Mean ( x ) N

= 744
30

= 24.8

PRIME Opt. Maths Book - VIII 177

4. Find the arithmetic mean of the given observations.

Class 0-10 10-20 20-30 30-40 40-50

f 45894

Solution:

Class f Mid value (x) f×x

0 - 10 4 0 + 10 =5 20
2

10 - 20 5 10 + 20 = 15 75
2

20 - 30 8 20 + 30 = 25 200
2

30 - 40 9 30 + 40 = 35 315
2

40 - 50 4 40 + 50 = 45 180
2

N = 30 R fx = 790

Statistics We have, Rfx
N
Mean ( x ) =

= 790
30

= 26.33

5. If mean of the observations given in table is 16, find ‘m’.

x 5 10 15 18 20 24
5
f 4 6 m 10 7

Solution:

xf f×x

54 20

10 6 60

15 m 15 m

18 10 180

20 7 140

24 5 120

N = 32 + m fx = 520 + 15m

178 PRIME Opt. Maths Book - VIII

We have, Rfx
N
Mean ( x ) =

16 = 520 + 15m
32 + m

or, 512 + 16m = 520 + 15m

or, 16m – 15m = 520 – 512

\ m = 8

Exercise 8.1 Statistics

1. Tabulate the following statistical data in frequency distribution table.
a. Discrete frequency distribution table.
i) Age Group of 30 students of grade VIII of a school.
12, 12, 13, 14, 12, 14, 13, 15, 12, 14, 14, 14, 13, 14, 13, 15, 12,

14, 12, 13, 16, 16, 15, 15, 12, 14, 13, 14, 15, 14
ii) Number of food item used in dinner of 40 families.
2, 3, 3, 3, 4, 2, 4, 4, 4, 4, 2, 5, 6, 5, 6, 5, 4, 4, 2, 3, 3, 4, 3, 3, 2, 6,

6, 5, 4, 4, 3, 5, 3, 6, 4, 3, 5, 6, 3, 4

b. Continuous class interval taking suitable classes.
i) Marks obtained by 40 students of grade VIII is optional

maths of Enlighten school out of 50 full marks.
33, 37, 49, 48, 40, 30, 12, 7, 24, 28, 47, 49, 30, 20, 10, 40, 16,

39, 38, 50, 32, 44, 19, 29, 28, 27, 33, 35, 42, 15, 43, 36, 24, 26,
28, 30, 30, 47, 1
ii) Systolic Blood pressure of 30 people found in Pulchowk .
60, 70, 110, 100, 100, 90, 80, 80, 120, 140, 100, 70, 70, 150,
140, 80, 80, 60, 70, 70, 80, 80, 90, 90, 70, 100, 120, 90, 80, 70
iii) Number of students found in 20 government schools of
Lalitpur district.
410, 250, 210, 370, 860, 340, 890, 420, 500, 300, 400, 450,
730, 640, 670, 520, 260, 380, 360, 430

PRIME Opt. Maths Book - VIII 179

2. Find the arithmetic mean for the followings.
i) R x = 200, N = 25, of a statistical data.
ii) R fx = 450, R f = 30 of the observations.
iii) 24, 16, 12, 28, 20, 10, 24, 26
iv) 7, 9, 15, 18, 21, 24, 30, 32, 34, 40
v) 102, 112, 120, 126, 132, 136, 231

3. Find the arithmetic mean for the following observations. 24
i) x 5 9 12 18 1

f 3583 50
6
ii) Marks 36 42 46 48
64
No. of students 5 8 12 9 9

iii) Age 25 32 42 58 32 – 40
5
No. of people 8 12 18 13
50 – 60
Statistics iv) Class 0 – 8 8 – 16 16 – 24 24 – 32 6

f 7 9 13 6

v) Marks 10 – 20 20 – 30 30 – 40 40 – 50

f 8 12 15 9

4. PRIME more creative questions:
i) If mean of the observations 5, 7, 9, 11, 13 and m is 10.5, find the

value of ‘m’.
ii) If R x = 208 + m, N = 5 + m and x = 30, find the value of ‘m’.
iii) If R fx = 240 + 6p, R f = 3p – 10 and x = 15, find the value of ‘p’.
iv) If mean of the observation given below is 15, find ‘k’.

x 5 10 15 20 25

f 2 6 5 k 3
v) If mean of the observation given below is 25, find ‘p’.

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50

f 3 4 5 p 2

180 PRIME Opt. Maths Book - VIII

5. Project work
Collect the marks obtained by the students of grade VIII of first

terminal examination and construct frequency distribution table by
taking suitable class interval. Also find the arithmetic mean.

Answer

1. Show to your subject teacher

2. i) 8 ii) 15 iii) 20
iii) 43.5
iv) 23 v) 137 iii) 10

3. i) 11.7 ii) 45

iv) 18.6 v) 33.6

4. i) 18 ii) 2

iv) 4 v) 6

Statistics

PRIME Opt. Maths Book - VIII 181

Y

50% 50%
OX

Median

Statistics ` N + 1 th
2
j

` N + 1 th
2
j

sum of two consecutive observation of size ` N + 1 th item
2
j

2

` N + 1 th
2
j

` N + 1 th
2
j

` N + 1 th
2
j

182 PRIME Opt. Maths Book - VIII

Worked out Examples

` N + 1 jth
2

a 5 + 1 th
2
k

` N + 1 jth Statistics
2
6 + 1 jth
` 2

3rd obs. + 4th obs
2

24 + 30
2

PRIME Opt. Maths Book - VIII 183

3. Find median of the data.

x 18 15 10 30 24

f 95326
Solution :
The cummulative frequency table taking the observation in ascending

order.

xf c.f.

10 3 3

15 5 3+5=8

18 9 8 + 9 = 17

24 6 17 + 6 = 23

30 2 23 + 2 = 25

N = 25

Here,

Median lies in,

= size of ` N + 1 th item
2
Statistics j

= size of a 25 + 1 th item
2
k

= size of 13th item

\ 17 is just greater than 13 in c.f. column.

\ Median (Md) = corresponding observation of 17 c.f. is 18.
\ Md = 18

2. Quartiles:
The numerical values which divide the statistical data in four equal parts
are called quartiles.

• They are Q1, Q2 and Q3.

Q1 Q2 Q3
• Q1 = 1st quartile (lower quartile)

= It divides the data 25% in left & 75% in right.
• Q2 = 2nd quartile (median)

= It divide the data 50% in left and 50% in right as the median.
• Q3 = 3rd quartile (upper quartile)

= It divide the data 75% in left & 25% in right.

184 PRIME Opt. Maths Book - VIII

• They are calculated as the median by calculating the size as follows.
N + 1 th
Q1 = size of ` 4 item
j

Q2 = size of 2` N + 1 th item
4
j

Q3 = size of 3` N + 1 th item
4
j

3. Deciles:
The numerical values of the statistical data which divide the data in ten

equal parts are called deciles.
• They are D1, D2, D3, ....... D9.

D1 D2 D3 ...................................... D9

• D1 = 1st decile which divides the data 10% in left and 90% in right. Statistics
• D2 = 2nd decile which divides the data 20% in left and 80% in right.
• D3 = 3rd decile which divides the data 30% in left and 70% in right.
• Likewise other deciles D4, D5, D6, D7, D8, D9 can be calculated.
• They are calculate as the median by calculating the sizes as,

D1 = size of a N+1 th item
10
k

D2 = size of 2^N + 1h th item
10

D3 = size of 3^N + 1h th item
10

4^N + 1h th
D4 = size of 10 item

D9 = size of 9^N + 1h th item
10

PRIME Opt. Maths Book - VIII 185

Worked out Examples

1. Find the first quartile of the observations : 2, 5, 15, 11, 8, 21, 18

Solution: The observations in ascending order are : 2, 5, 8, 11, 15, 18,

21

No. of observations (n) = 7

Now, = size of `n + 1 th item
1st quartile (Q1) 4
j

= size of `7 +1 th item
4
j

= size of 2nd item

= corresponding observation is 5

\ Q1 = 5

2. Find the third quartile of the data 14, 18, 22, 26, 32, 36, 40, 45, 48, 52.

Solution : The observations is ascending order are :

14, 18, 22, 26, 30, 36, 40, 44, 48, 52

Statistics No. of observation (n) = 10

Now, = size of 3`n + 1 th item
3rd quartile (Q3) 4
j

= size of 3 × 11 item
4

= size of 8.25th item

= 8th + 0.25(9th – 8th) obs.

= 44 + 0.25(48 – 44)

= 44 + 0.25 × 4

= 44 + 1

\ Q3 = 45

3. Find the 9th decile of the above observations given in example 2.

Solution : The observations is ascending order are :

14, 18, 22, 26, 30, 36, 40, 44, 48, 52

No. of observation (n) = 10

Now,

9th quartile (D9) = size of 9^n + 1h th item
10

= size of 9 × 11 item
10

186 PRIME Opt. Maths Book - VIII

= size of 9.9th item

= 9th + 0.9(10th – 9th) obs.

= 48 + 0.9(52 – 48)

= 48 + 0.9 × 4

\ D9 = 51.6

4. Find the third decile from the observations given in descrete frequency
distribution table.

x 12 18 24 28 36 42 48

f 2 4 9 15 8 5 2

Solution :

xf c.f.

12 2 2

18 4 2+4=6

24 9 6 + 9 = 15

28 15 15 + 15 = 30

36 8 30 + 8 = 38 Statistics

42 5 38 + 5 = 43

48 2 43 + 2 = 45

N = 45

For the third decile (D3)
item
= size of 3 a N+1 th
10
k

= size of 3 × 4.6th item

= size of 13.8th item

15 is just greater than 13.8 in c.f. column

\ D3 = corresponding observation of 15c.f. is 24
\ D3 = 24

PRIME Opt. Maths Book - VIII 187

Exercise 8.2

1. Find the median value of the followings:
i) 11, 15, 18, 21, 25, 30, 36

ii) 32, 8, 24, 12, 28, 20, 16, 4

iii) x 12 16 20 25 30
9 7
f 5 7 12 10
2
iv) Marks 50 40 20 30 55
f 6 7 8 10
4
v) x 15 25 45 35 65 75
f 25 13 763 36
6
Statistics 2. Find the first quartile of the followings. 20
i) 7, 11, 15, 18, 22, 25, 30, 34, 38, 40, 45 11

ii) 48, 60, 54, 42, 36, 30, 20, 24, 52 36
6
iii) 35, 25, 15, 5, 65, 55, 45, 95, 75, 85 20
11
iv) Marks 12 16 24 30
9
No. of Students 5 8 12
28
v) x 7 25 18 10 35 8

f 32952

3. Find the third quartile of the data given below.
i) 7, 11, 15, 18, 22, 25, 30, 34, 38, 40, 45

ii) 48, 60, 54, 42, 36, 30, 20, 24, 52
iii) 35, 25, 15, 5, 65, 55, 45, 95, 75, 85

iv) Marks 12 16 24 30

No. of Students 5 8 12 9

v) x 7 25 18 10 35 28
f 329528

188 PRIME Opt. Maths Book - VIII

4. Find the third decile of the data given below.
i) 11, 15, 18, 21, 25, 30, 36

ii) 32, 8, 24, 12, 28, 20, 16, 4

iii) x 12 16 20 25 30

f 5 7 12 9 7

iv) Marks 50 40 20 30 10
f 6 7 8 10 2

v) x 15 25 45 35 65 75 55
f 25 13 763 4

5. i) Find the 6th decile of the observations 108, 100, 116, 112, 124, Statistics
118, 130, 142, 136.

ii) Find the 5th decile of the observations.
24, 36, 30, 56, 48, 42, 68, 64, 60, 72, 80
iii) Find the 4th decile of the observations.

x 15 25 32 40 48 52 60

f 2 3 8 12 5 6 4
6. i) Find the 9th decile of the observations 10, 20, 60, 50, 40, 30, 90,

80, 70, 100.
ii) Find 8th decile of the observations given below.

Height of plants 80 75 72 100 86 90 99
No of plants 10 9 6 5 12 8 7

7. PRIME more creative questions.
i) If median of the observations taken in order of 4, 8, 12, 2x + 6, 20,

24 & 28 is 16. Find the value of ‘x’.
ii) If fifth decile of the observation taken in order of: 11, 15, 18, 3p +

4, 5p-2, 36, 42, 48 is 29. Find the value of ‘p’.
iii) If upper quartile of the observations taken in order of: 20, 24, 30,

36, 42, 48, 2m + 12, 56, 60 is 54, find the value of ‘m’.
iv) Find the fourth decile of :

x 54 48 40 20 24 30 36

f 4 6 10 3 6 9 12

PRIME Opt. Maths Book - VIII 189

v) Construct the cummulative frequency distribution table of the
following data and compute lower quartile.
12, 12, 13, 14, 12, 14, 13, 15, 12, 14, 14, 14, 13, 14, 13, 15, 12, 14,
12, 13, 16, 16, 15, 15, 12, 14, 13, 14, 15, 14

Answer

1. i) 21 ii) 18 iii) 20 iv) 30 v) 45
2. i) 15 ii) 27 iii) 22.5 iv) 16 v) 18
3. i) 38 ii) 53 iii) 77.5 iv) 30 v) 28
4. i) 16.2 ii) 10.8 iii) 20 iv) 40 v) 35
5. i) 124 ii) 56 iii) 40 6.i) 99 ii) 99
7. i) 5 ii) 7 iii) 20 iv) 36 v) 13

Statistics

190 PRIME Opt. Maths Book - VIII

8.3 Measure of dispersion Statistics

Let us consider the observations taken from the collection of age of 5
parsons of two different families.
Family A : 2, 12, 24, 30, 32.
Family B : 15, 18, 20, 22, 25.

The arithmetic mean of such observations is 20 in both families where the
observations of family A are very far from the mean where as of family B are
very close from mean.

• Discuss which type of data is better ?
• Discuss such type of scatterdness of the observations by calculating

median also.

That type of scatteredness of the observations of the data from the central
value (mean, median, mode, quartiles) is called the measure of dispersion.
The measures of dispersion are of different types which are range, quartile
deviation, mean deviation and standard deviation. The coefficient of such
measure of dispersions are also can be taken as the measure of dispersion.

The measurement of scatteredness of the observations
of the collected data from their central value is called
measure of dispersion.

Range:
The measurement of difference between highest and lowest observations

of the collected data is called range.

Where, Range = Highest observation – lowest observation.

i.e. R = H – L

Also,
Coefficient of range is calculated from such values as,
H–L
Coefficient of Range = H+L .

PRIME Opt. Maths Book - VIII 191

Note : Range measures the measurement of closeness of the observations
from highest and lowest observations. [R = H – L]
• All the observations are not included in it.

Example: Find range and its coefficient of the observations taken in order

of 20, 24, 30, 36, 44, 50, 60.

Here,

Highest observation (H) = 60

Lowest observation (L) = 20

Range (R) = H – L

= 60 – 20

= 40

Again,

Coefficient of Range = H–L
H+L

= 60 – 20
60 + 20

Statistics = 40
80

= 0.5

Quartile deviation:

The measurement of dispersion by calculating the quartiles of the
observations of collected data is called quartile deviation where only first
quartile (Q1) and third quartile (Q3) are use to find it.

Quartile deviation (Q. D.) = 1 (Q3 – Q1)
2

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

Note: Quartile deviation is the measurement of the
observations from highest and lowest quartiles of the
observations.
• All the observations are not included in it.
• It is slightly better than range for the measurment of

dispersion.

192 PRIME Opt. Maths Book - VIII

Worked out Examples

1. Find range and its coefficient of the observations 18, 12, 24, 28, 20.

Solution:

The given observations taken in order are, 12, 18, 20, 24, 28.

Here,

Highest observation on (H) = 28

Lowest observation (L) = 12

Range = H – L

= 28 – 12

= 16.

Coefficient of range = H–L
H+L

= 28 – 12
28 + 12

= 16
40

= 0.4

2. Find the range and its coefficient of the given discrete observations. Statistics

Age 12 14 15 16 18

No. of students 5 6 12 10 7

Solution:

The given frequency distribution table is,

Age 12 14 15 16 18

No. of students 5 6 12 10 7

Here,

Highest observation (H) = 18

Lowest observation (L) = 12

Range = H – L

= 18 – 12

= 6

Again,

Coefficient of range = H–L
H+L

= 18 – 12
18 + 12

= 6
30

= 0.2

PRIME Opt. Maths Book - VIII 193

3. Find quartile deviation and its coefficient of 20, 14, 12, 36, 24, 30, 42.

Solution:

The given observations taken in order are: 12, 14, 20, 24, 30, 36, 42.

Number of obs. (N) = 7

Then,

First quartile lies in = Size of ` N + 1 th item
4
j

= ` 7 + 1 th item
4
j

= 2nd item

= 14

Third quartile lies in = size of 3 ` N + 1 th item
4
j

= size of (3 × 2)th item

= size of 6th item
= 36

Statistics Then,

Quartile deviation (Q.D.) = 1 (Q3 – Q1)
2

= 1 (36 – 14)
2

= 11

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

= 36 – 14
36 + 14

= 22
50

= 0.44.

194 PRIME Opt. Maths Book - VIII

4. Find quartile deviation and its coefficient of:

In come in thousands 25 27 30 32 37 50

Family No. 3 4 762 1

Solution:
Cumulative frequency table:

In come Family No. (f) C.F.

x in thousand 3
3+4=7
25 3 7 + 7 = 14
14 + 6 = 20
27 4 20 + 2 = 22
22 + 1 = 23
30 7

32 6

37 2

50 1

N = 23

Here, Statistics

First quartile (Q1) = Size of ` N + 1 th item
4
j

= Size of ` 23 + 1 th item
4
j

= Size of 6th item

7 is just greater than 6 in c. f.

\ Q1 = 27 (thousands)

Third quartile (Q3) = size of 3 ` N + 1 th item
4
j

= Size of 3(6)th item

= Size of 18th item

20 is just greater than 18 in c.f.

\ Q3 = 32 (Thousands)

Then,

Quartile deviation (Q.D.) = 1 (Q3 – Q1)
2

= 1 ×5
2

= 2.5

PRIME Opt. Maths Book - VIII 195

Coefficient of Q.D. = Q3 – Q1
Q3 + Q1

= 32 – 27
32 + 27

= 5
59

= 0.084.

5. If quartile deviation and its coefficient are 11 and 0.44 respectively,
find the upper and lower quartiles.

Solution:

Quartile deviation (Q.D.) = 11
Coefficient of Q.D. = 0.44.

We have,

Q.D. = 1 (Q3 – Q1)
2

or, 11 = 1 (Q3 – Q1)
2

Statistics or, 22 + Q1 = Q3 ................... (i)

Again,
Coefficient of Q.D. = 0.44
Q3 – Q1
or, Q3 + Q1 = 0.44

or, Q3 – Q1 = 0.44Q3 + 0.44Q1
or, 0.56Q3 = 1.44Q1
or, 0.56(22 + Q1) = 1.44Q1
or, 12.32 + 0.56Q1 = 1.44Q1

or, 12.32 = Q1
0.88

\ Q1 = 14

Putting the value of Q1 in (i),
Q3 = Q1 + 22
or Q3 = 14 + 22
\ Q3 = 36



196 PRIME Opt. Maths Book - VIII