Vedanta Opt. Excel in Mathematics Book 8 Final (2078)

Transformations

direction such that ‘AOA' = 90° about centre O, A' is the image of A.
(ii) Join OB, measure the length OB. Take OB as a radius and rotate the point B

is antickclockwise direction. Such that ‘BOB' = 90° about centre O, B' is the
image of B.
(iii) Join OC, measure the length OC. Take OC as a radius and rotate the point C in
anticlockwise direction such that ‘COC' = 90° about centre O, C' is the image
of C.
(iv) Join A', B' and C'
Hence, 'A'B'C' is the image of 'ABC under rotation of +90° about the origin.
Are 'ABC and 'A'B'C' congruent?

12.5 Use of Coordinates in Rotation

Here, we study about the rotation of figures through some special angles such as
90q and 180q in anti-clockwise and clockwise directions about the centre at origin.
The rotation through 90q is also called a quarter-turn and the rotation through 180q
is called a half-turn.

Rotation through 90° in anti-clockwise or +90° about the centre at
origin

The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.

Y Y Y Y
A'(-4,5) A(5,4)
A(-5,4) A'(4,4)

X' O X X' O X X' O X X' O X

Y' A'(-4,-5) A'(5,-3) A(4,-4)
Y' A(-3,-5) Y'
A(5, 4) o A'(–4, 5)
? P(x, y) o P'(–y, x) A(–5, 4) o A'(–4, –5) Y'
? P(–x, y) o P'(–y, –x) A(–3, –5) o A'(5, –3) A(4, –4) o A'(4, 4)

? P(–x, –y) o P'(y, – x) ? P(x, –y) o P'(y, x)

Thus, when a point is rotated through 90q in anti-clockwise direction (+ 90q) about
the origin as the centre of rotation, the x and y-coordinates are interchanged by
making the sign of y-coordinate just opposite.

i.e. P(x, y) R[0, + 90°] P' (– y, x)

A(4, 5) o A'(–5, 4)

vedanta Excel in Additional Mathematics - Book 8 201

Transformations

Rotation through 90° in clockwise or –90° about the centre at origin

The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.

Y Y A'(-3,5)Y Y
A(2,5)
A(-4,3) A'(3,4)

X' O X X' O X X' O X X' O X

A'(5,-2) Y' A(-5,-3) A'(-5,-5) A(5,-5)
A(–4, 3) o A'(3, 4) Y'
Y' Y'
A(2, 5) o A'(5, –2) ? P(–x, y) o P'(y, x) A(–5, –3) o A'(–3, 5) A(5, –5) o A'(–5, –5)
? P(x, y) o P'(y, –x) ?P(–x, –y) o P'(–y, x)
? P(x, –y) o P'(–y, –x)

Thus, when a point is rotated through 90q in clockwise direction (– 90q) about the
origin as the centre of rotation, and the x and y-coordinates are exchanged by making
the sign of x-coordinate just opposite.

i.e. P(x, y) R[0, – 90°] P' (y, –x)

Rotation through 180° in anti-clockwise and clockwise about origin

When a point is rotated through 180q in anti-clockwise or in clockwise direction
about origin as the centre of rotation, the coordinates of the image are the same.
Study the following illustration.

The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.

Y Y Y Y
A(4,5)
A (-4, 3) A(2,4) A'(-2,4)

X' O X X' O X X' O X X' OX

A'(-4,-5) A'(4,-3) A'(-2,-4) A(2,-4)

Y' Y' Y' Y'

A(4, 5) o A'(–4, –5) A(–4, 3) o A'(4, –3) A(–2, –4) o A'(2, 4) A(2, –4) o A'(–2, 4)
? P(x, y) o P'(–x, –y) ? P(–x, y) o P(x, –y) ? P (–x, –y) o P(x, y) ? P(x, –y) o P'(–x, y)

Thus, when a point is rotated through 180q in anti-clockwise (+180q) or in clockwise

(– 180q) direction, about the origin as the centre of rotation, the x and y-coordinates

of the image remain the same just by changing their signs.

i.e. P(x, y) R[0, + 180°] P' (–x, –y)

A(4, 5) o A'(–4, –5)

202 vedanta Excel in Additional Mathematics - Book 8

Transformations

Worked Out Examples

Example 1. P(3, 1), Q(4, – 2), and R(5, 3) are the vertices of ∆PQR. Find the coordinates
Solution: of its image under the rotation through 90° in anti-clockwise direction
about origin. Also, draw the graphs ∆PQR and its image.

Here, P(3, 1), Q(4, – 2), and R(5, 3) are the Y

vertices of 'PQR. When it is rotated R'(-3,5) Q'(2,4)
R(5,3)
through + 90q about origin, P'(-1,3) P(3,1)

we have, P(x, y) R[0, + 90°] P' (–y, x) X' O X

P(3, 1) o P'(– 1, 3) Q(4,-2)

Q(4, – 2) o Q'(2, 4)

R(5, 3) o R'(– 3, 5) Y'

? P'(– 1, 3), Q'(2, 4) and R'(– 3, 5) are the

vertices of the image of 'PQR.

'PQR and its image 'P'Q'R' are plotted in same graph.

Example 2. A(4, 5), B(–3, 5), and C(–3, –2) are the vertices of ∆ABC. Find the
Solution: coordinates of its image under the rotation through 180° in clockwise
direction about origin. Also draw the graphs of this transformation.

Here, A(4, 5), B(–3, 5), and C(–3, –2) are B(-3,5) Y
the vertices of 'ABC, when 'ABC is A(4,5)
rotated – 180° about origin.
C'(3, 2)
We have, P(x, y) R[0, –180°] P' (–x, –y) X' OX

A(4, 5) o A'(–4, –5) C(-3,-2)

B(–3, 5)) o B'(3, –5) A'(-4,-5) B'(3,-5)
C(–3, –2) o C'(3, 2)

? A'(–4, –5), B'(3, –5), and C(3, 2) are the Y'

images of A, B and C respectively

'ABC and its image 'A'B'C' are plotted in same graph.

Example 3. P(4, 4), Q(3, –4) and R(–5, –5) are the vertices of ∆PQR. Find the
Solution:
coordinates of the vertices of 'P'Q'R' under the rotation through –90°

about origin.

Here, P(4, 4), Q(3, –4), and R(–5, –5) are the vertices of 'PQR, when

it is rotated through –90° about origin.

We have, P(x, y) o P'(y, –x)

P(4, 4) o P'(4, –4)

Q(3, –4) o Q'(–4, –3)

R(–5, –5) o R'(–5, 5)

? P'(4, –4), Q'(–4, –3), and R'(–5, 5) are the vertices of 'P'Q'R'.

vedanta Excel in Additional Mathematics - Book 8 203

Transformations

Exercise 12.2

Short Questions :

1. Write images of the following points:

(a) Rotate the following points through 90° in anticlockwise direction and find
their images:

(i) P(x, y) (ii) P(–x, y) (iii) P(–x, –y) (iv) P(x, – y)

A(2, 3) A(–4, 5) A(–2, –3) A(4, –5)

B(4, 5) B(–6, 7) B(–4, –5) B(6, –7)

(b) Rotate the following points through 90° in clockwise direction and find
their images:.

(i) P(x, y) (ii) P(–x, y) (iii) P(–x, –y) (iv) P(x, – y)

A(2, 3) A(–2, 4) A(–2, –3) A(4, –5)

B(4, 5) B(–6, 7) B(–7, –8) B(6, –7)

(c) Rotate the following points through 180° in anticlockwise or clockwise
direction and find their images:

(i) P(x, y) (ii) P(–x, y) (iii) P(–x, –y) (iv) P(x, – y)

A(2, 3) A(–2, 5) A(–2, –3) A(4, –5)

B(4, 5) B(–5, 7) B(–5, 6) B(6, –7)

2. Draw the images of the following figures rotating through the given angles in
anti-clockwise and clockwise direction about the given centre of rotation.

(a) P (b) A C (c) M

O P

Q R O
O
B N
Rotation through 60° Rotation through 45° Rotation through 120°

(d) P OQ (e) M (f) A OC
D
NR
SR P B
Rotation through 100° Rotation through 180°
Rotation through 90°

3. Copy the following figures in your own graph papers. Draw their images rotating
through 90q in (i) anti-clockwise, (ii) clockwise directions about the centre at
origin. Also, write the coordinates of the vertices of images.

204 vedanta Excel in Additional Mathematics - Book 8

Transformations

(a) Y (b) (c)

Y Y

P
R

X' A O X X' QO X X’ MO X

B

B NP

Y' Y' Y’
4. Copy the following figures in your own graph papers. Draw their images

rotating through 180q in anti-clockwise direction. Also, write the coordinates

of the vertices of images.

(a) Y (b) Y (c) Y

A KI

B C X X' C O X X' J X
X' O B O

A

Y' Y' Y'

Long Questions :

5. (a) A(– 2, 2), B(3, 7), and C(4, 2) are the vertices of 'ABC. Find the coordinates
of its image under the rotation through 90q in

(i) anti-clockwise (ii) clockwise direction about origin.

(b) A(6, 7), B (2, 3), and C (8, 2) are the vertices of 'ABC. Find the coordinates
of its image under the rotation through 180q in anti-clockwise direction
about origin.

6. (a) P(2, 3), Q(5, 6), and R(8, 2) are the vertices of 'PQR. Find the coordinates
of 'P'Q'R' rotating through 180° about the origin. Draw graph of 'PQR and

its image 'P'Q'R' on the same graph paper.

(b) Let P(1, 5), Q(8, 5), R(5, 12), and S(1, 12) be the vertices of square PQRS.
Rotate the square through 90° in clockwise direction and plot squares
PQRS and its image P'Q'R'S' on the same graph.

(c) Let P(2, 6), Q(9, 6), R(9, 10), and S(2, 10) be the vertices of rectangle PQRS.
Rotate the rectangle PQRS through 90° in anti-clockwise direction about
the origin and obtain image P'Q'R'S'. Plot the rectangles PQRS and its image
P'Q'R'S' on the same graph.

vedanta Excel in Additional Mathematics - Book 8 205

Transformations

1., 2., 3. and 4. Show to your answer.
5. (a) (i) A'(–2, –2), B'(–7, 3), C'(–2, 4) (ii) A'(2, 2), B'(7, –3), C'(–2, 4)

(b) A'(–6, –7), B'(–2, –3) and C'(–8, –2)
6. (a) P'(–2, –3), Q'(–5, –6) and R'(–8, –2) (b) P'(5, –1), Q'(5, –8), R'(12, –5), S'(12, –1)

(c) P'(–6, 2), Q'(–6, 9), R'(–10, 9) and S'(–10, 2)

12.6 Translation - Introduction

It is the translation of geometrical figures in which each vertex of a figure is displaced
by equal distance to the same direction. Study the following illustrations:

D

P' AC
P

Q' R' B D'
Q A' C'

R

B'

'PQR is displaced to 'P'Q'R' Quadrilateral ABCD is displaced to quad. A'B'C'D'.
Here, PP' = QQ' = RR' Here, AA' = BB' = CC' = DD'
Also, PP' // QQ' // RR' Also, AA' // BB' // CC' // DD'

Thus, the displacement of a geometrical figure has magnitude as well as direction.
So, it is a vector quantity.

Now, let’s learn to displace a given geometrical figure.

12.7 Translation Using Translation Vector

Let A(2, 3) be a point. Plotting it in a grid we get,

Let’s shift point A by a translation vector T = 4 .
5
It simply means shifting (moving) point A is (2, 3), 4 units along x-axis, and 5

units along y-axis. So, the point A is shifted to A' as in the figure (2). So, A' is the

translated image of A:

Y Y A'(6, 8)

4

=

5

T

A(2, 3) A(2, 3)

X' O X X' O X
Y' Fig (1)
Y' Fig (2)

206 vedanta Excel in Additional Mathematics - Book 8

Transformations

Translated by a
b
Hence, P(x, y) P'(x + a, y + b)

Worked Out Examples

Example 1. Displace the given ∆PQR in the magnitude and P R
Solution: direction of (ot ). t
Q
Example 2. (i) From P, draw PP' // ot and cut off PP' = ot P
Solution:
(ii) From Q, draw QQ' // ot and the cut off

QQ' = ot ' QR
(iii) From R, draw RR' // ot and cut off P’ t

RR' = ot '. R’
Q’
(iv) Join P', Q' and R'

(v) 'P'Q'R' is the required image of 'PQR under the displacement

of ot .

Translate the point A(–2, 3) by a translation vector T = –2 .
4
Here, the given point is A(–2, 3)
a
Translation vector (T) = b
Now, we know,
a
T= b

P(x, y) P' (x + a, y + b)

T= –2
4
So, A(–2, 3) A'(–2 – 2, 3 + 4) = A'(– 4, 7)

In graph

Y

A'(–4,7)

A(-2, 3) X

X' O

Y' 207

Hence, A(–2, 3) is the translated to A'(–4, 7)

vedanta Excel in Additional Mathematics - Book 8

Transformations

by a translating vector T = –2
4

It means that A is translated through 2 units left along X-axis and 4

units upward along Y-axis.

Example 3. ABC is a triangle with vertices A(–4, 5), B(4, 4), and C(0, 2). Find the

image of 'ABC under translation T = 2 .
–1

Represent both the object and image in the same graph.

Solution: Let A(–4, 5), B(4, 4), and C(0, 2) be the vertices of 'ABC.

Translation vector, T = 2
–1

Under the translation by translating vector 2 ,
–1

we have,

T= a
b
P(x, y) P' (x + a, y + b)

T= 2
–1
Now, A(–4, 5) A'(–4 + 2, 5 – 1) = A'(–2, 4)

T= 2
–1
B(4, 4) B'(4 + 2, 4 –1) = B'(6, 3)

T= 2
–1
C(0, 2) C'(2 + 0, 2 –1) = C'(2, 1)

Hence, 'A'B'C' is the image of 'ABC.

Showing 'ABC and its image 'A'B'C' in graph.

Y

A(-4,5) B(4,4)
A'(-2,4) B'(6,3)

C(0,2)

X' O C'(2,1) X

Y'

208 vedanta Excel in Additional Mathematics - Book 8

Transformations

Exercise 12.3

Short Questions :

1. Write the coordinates of images of given points:

Translate the following points through a translating vector 2
3
(a) A(4, 5) (b) B(2, 5) (c) C(4, 2)

(d) D(–5, –6) (e) E(6, 4) (f) F(–4, –7)

2. Translate the following figures by a translating vector T = 3 :
2

(a) Y (b) Y

A

X’ A X X’ B
X

(c) Y’ (d) Y’
Y
X’ X’ Y
A D

B CX AC

BX

Y’ Y’

3. Displace the following geometrical figures in the magnitude and direction of ot .

(a) (b) F (c) F G
t
B tt E
H
C
E

D

vedanta Excel in Additional Mathematics - Book 8 209

Transformations

(d) L M (e) Q (f) N M
N R L
K Pt t
t O K
S
P

4. (a) Translate the point (2, 3) by a translating vector T = 3 . Show it in a graph.
4

(b) Translate the point (4, 5) by a translating vector T = 3 . Show it is a graph.
2
Long Questions :

5. (a) A(0, –6), B(–3, 2), and C(5, 6) are the vertices of 'ABC. Translate ABC by a

translating vector T = 1 to get 'A'B'C'. Find the coordinates of 'A'B'C'
2

and present both object and image in the graph.

(b) P(1, 2), Q(5, 3), and R(–2, –5) are the vertices of 'PQR. Translate 'PQR by
–2
at translating vector T = –1 to get 'P'Q'R'. Find the co-ordinates of P'Q'R

and present both object and image in the same graph.

6. If 'PQR is translated by a translation vector T = a , the image of P(4, 5) is
b
obtained as P'(7, 8). Find the values of a and b. Under this translation, what

happens to the images of Q(8, 5) and R(7, 7)? Plot 'PQR and 'P'Q'R' on the

same graph.

1. (a) A'(6, 8) (b) B'(4, 8) (c) C'(6, 5) (d) D'(–3, –3)

(e) E'(8, 7) (f) F'(–2, –4) 2. and 3. Show to your teacher.

4. (a) (5, 7) (b) (7, 7)

5. (a) A'(1, – 4), B'(–2, 4), C'(6, 8) (b) P'(–1, 1), Q'(3, 2), R'(–4, –6)

6. a = 3, b = 3

210 vedanta Excel in Additional Mathematics - Book 8

Statistics

13Statistics

13.1 Introduction

Statistics is a branch of applied mathematics concerned with collection, organizing
and interpreting data. The data are presented by means of graphs. The presentation
is the most important, convincing, and appealing and easily understanding method
of presenting the collected data by using diagram and graphs.

Statistics are used to make decision and prediction about future plans and policies.
Statistics is also the mathematical study of likelihood and probability of events
occurring based on known quantitative data or collection of data.

The word 'statistics' comes from the word 'state'. The state needs different types of
data to keep them as records and used for present and future plannings.

13.2 Collection of Data

The marks obtained by 30 students of class 7 in Mathematics are given below.
12, 14, 18, 16, 14, 17, 18, 20, 20, 15,
22, 11, 5, 8, 7, 10, 8, 16, 13, 20,
13, 19, 9, 11, 16, 7, 14, 15, 12, 10

Such numerical figures are called data.

Data should be presented in a proper order so that it becomes easier to get the
necessary information for which they are collected. When the data are arranged
either in ascending or in descending order, they are said to be in proper order. The
properly arranged data are called arrayed data.

13.3 Frequency Table

The data given below are the daily wages (in Rs) of 20 workers of a factory.
75, 60, 60, 80, 70, 60, 75, 70, 80, 70,
75, 80, 90, 60, 70, 80, 75, 75, 70, 75

This form of data is called raw data. It can be arranged in ascending as follows:
60, 60, 60, 60, 70, 70, 70, 70, 70, 75,
75, 75, 75, 75, 75, 80, 80, 80, 80, 90

vedanta Excel in Additional Mathematics - Book 8 211

Statistics

Here, Rs. 60 is repeated 4 times. So, its frequency is 4.
Rs. 70 is repeated 5 times. So, its frequency is 5.
Rs. 75 is repeated 6 times. So, its frequency is 6.
Rs. 80 is repeated 4 times. So, its frequency is 4.
Rs. 90 is repeated 1 time. So, its frequency is 1.

Thus, a frequency is the number of times a value occurs. Data and their frequencies
can be presented in a table called frequency table.

Now, let’s present the above wages of 20 workers in a frequency table.

Wages (in Rs.) Tally marks Frequency
60 |||| 4
70 |||| 5

75 |||| 6

80 |||| 4
90 | 1
Total 20

Tallying is a system of showing frequencies using diagonal lines grouped in fives.
Each time five is reached, a horizontal line is drawn through the tally marks to make
a group of five. The next line starts a new group. For example:

Frequency Tally marks Frequency Tally marks
1 | 6 |||| |
||||||
2 || 7 |||| |||

3 ||| 8 |||| ||||
|||| ||||
4 |||| 9

5 |||| 10

13.4 Grouped and Continuous Data

Let’s consider the following marks obtained by 30 students in a test of mathematics.

25, 28, 36, 24, 24, 17, 9, 25, 25, 48,

25, 18, 27, 35, 25, 23, 24, 45, 32, 16,

26, 39, 24, 20, 33, 40, 37, 25, 48, 35

The above mentioned data are called individual data. Another way of organizing
data is to present them in a grouped form. For grouping the given data, we should
first see the smallest value and the largest value, then we have to divide the data
into an appropriate class-interval. The numbers of values falling within each class–
interval give the frequency. For example:

212 vedanta Excel in Additional Mathematics - Book 8

Statistics

Marks Tally marks Frequency
0 - 10 | 1
10 - 20 ||| 3
20 - 30 15
|||| |||| ||||

30 - 40 |||| || 7
||||
40 - 50 4
Total N = 30

In the above series, 9 is the smallest value and 48 is the largest value. So, the data
are grouped into the interval of 0 - 10, 10 - 20, etc. so that the smallest and the largest
values should fall in the lowest and the highest class-interval respectively.

Let’s consider a class-interval 10 - 20.

Here, 10 is called the lower limit and 20 is the upper limit of the class-interval (10 - 20).
The difference between two limit is called the length or height of each class-interval.
For example, in 10 - 20 the length of the class-interval is 10.

Again, let’s take class-intervals, 0 - 10, 10 - 20, 20 - 30, …

Here, the upper limit of a pervious class–interval has repeated as the lower limit
of the consecutive next class–interval. Such an arrangement of data is known as
grouped and continuous data.

In the continuous class, the value of upper limit is excluded in the class. For example,
in class (10 - 20), 20 is excluded. The value 20 is kept in the class (20 - 30). Hence
continuous class is also called exclusive class.

13.5 Cumulative Frequency Table

The word ‘cumulative’ is related to the word ‘accumulated’, which means to ‘gathered
together’. The table given below shows the marks obtained by 30 students in a
mathematics test and the corresponding cumulative frequency of each class-interval

Marks Frequency Cumulative o 5 students obtained marks less than 10.
(f) frequency (cf) o 11 students obtained marks less than 20.
0 - 10 5 o 19 students obtained marks less than 30.
10 - 20 6 5 o 25 students obtained marks less than 40.
20 - 30 8 5 + 6 = 11 o 30 students obtained marks less than 50.
30 - 40 6 11 + 8 = 19
40 - 50 5 19 + 6 = 25
Total 25 + 5 = 30
N = 30

Thus, cumulative frequency corresponding to a class-interval is the sum of all
frequencies up to and including that class-interval.

vedanta Excel in Additional Mathematics - Book 8 213

Statistics

Exercise 13.1

Short Questions :
1. Complete the cumulative frequency table given below:

(a) x f cf (b) Marks No. of students (f) cf
50 3 3
15 2 2 60 5 8
70 20
20 4 6 80 12
90 8
25 6 100 2

30 10 N = 50

35 8

40 10

N = 40

2. (a) From the marks given below obtained by 30 students in Mathematics,
construct a frequency distribution table with tally marks.
10, 15, 18, 12, 15, 16, 18, 17, 10, 15,
18, 16, 15, 12, 20, 22, 17, 10, 12, 15,
12, 16, 18, 12, 15, 12, 16, 18, 23, 24

(b) The marks obtained by 30 students in mathematics in SEE examination
are given below. Group the data into the class intervals of length 10 and
construct a cumulative frequency distribution table.

42, 80, 45, 92, 36, 17, 49, 5, 98, 74,
65, 72, 28, 46, 86, 70, 62, 27, 16, 44,
85, 59, 51, 73, 78, 81, 97, 77, 45, 67

(c) From the data given in 2(a) group the data into class intervals of length 5
and construct a cumulative frequency table.

(d) The marks obtained by 40 students of class 7 in Mathematics are shown in
the table given below. Construct a cumulative frequency table to represent
the data.

Marks 10 20 30 40 50

No. of students 4 9 20 15 2

(e) The daily wages of 40 workers in a factory are shown in the table given
below. Show their wages in a cumulative frequency table.

Wages (in Rs.) 50 60 70 80 90
No. of workers
7 14 11 5 3

Show to your teacher vedanta Excel in Additional Mathematics - Book 8
214

Statistics

13.6 Histogram

Histogram is the most popular and widely used in practice for graphical presentation
of a continuous frequency distribution. It consists of a series of rectangles with no
gap between them. Width of each rectangle is based on the respective class size and
the height of each rectangle based on the frequency of each class.

Histograms may be with equal class intervals and unequal class intervals. Here, we
discuss only about histograms with equal class intervals.

Steps for Constructing Histograms

(i) On a graph paper draw X-axis and Y-axis.

(ii) With an appropriate scale, mark the class limits on the X-axis and the frequencies
on the Y-axis. The scales on both axes may be different.

(iii) Construct erect rectangles with class intervals as bases and the corresponding
frequencies as heights.

Example Construct a histogram from the following data :

Marks 0-20 20-40 40-60 60-80 80-100
No. of students 5 20 15 10 5

Solution: Here, class width = 20 for all the classes.
So the data can be presented in histogram as follows :

Y

No. of students 25 10
20 20 5
15 15

10
55

OX
20 40 60 80 100

Marks

vedanta Excel in Additional Mathematics - Book 8 215

Statistics

Exercise 13.2

1. Look the adjoining histogram and answer. No. of people Y
25
(a) Which age group people are maximum? 20 X
How many are they? 15

(b) How many people are there in age group 10
of (40 - 50) years? 5

(c) Find the total number of people. O 10 20 30 40 50
Age (years)
2. Draw a histogram from each of the following
data :

(a) Daily wage (Rs.) 400-500 500-600 600-700 700-800
No. of workers 4 12 16 14

(b) Marks 0-10 10-20 20-30 30-40 50-60 60-70
No. of family 4 5 20 15 10 5

(c) Income (Rs) 200-400 400-600 600-800 800-1000 1000-1200

Family 5 10 15 20 15

(d) Height (cm) 10-20 20-30 30-40 40-50 50-60
No. of plants 12 15 20 10 5

1. (a) 10 - 20, 25 (b) 15 (c) 75
2. Show to your teachers.

13.7 Measures of Central Tendency

The measure of central tendency gives a single central value of the given data. A
single central value is the best representative of the given data towards which the
values of all other data are approaching.

Average of the given data is the measure of central tendency. There are three types
of averages which are commonly used as the measure of central tendency. They are:
mean, median, and mode.

11.8 Arithmetic Mean

Arithmetic mean is the most common type of average. It is also called average. It

is the number obtained by dividing the sum of all the all the variate values by the

number of items. sum of all the variate values
the number of items
i.e. mean =

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Mean of Individual Data ¦X
n
If X represents all the items and n be the number of items, then mean (X) =

Worked Out Examples

Example 1. Calculate the average of the following marks obtained by 10 students
Solution: of a class in mathematics.
Example 2.
Solution: 18, 13, 12, 17, 15, 10, 20, 22, 23, 25

Here, 6X = 18 + 13 + 12 + 17 + 15 + 10 + 20 + 22 + 23 + 25 = 175

n = 10 ¦X 175
n 10
Now, mean (x) = = = 17.5.

If the average of the following wages received by 10 workers is Rs. 37.5,

find the value of p.

27, 28, 29, 30, 36, p, 40, 44, 45, 56

Here, 6X = 27 + 28 + 29 + 30 + 36 + p + 40 + 44 + 45 + 56

= p + 335

n = 10 ¦X
n
Now, average =

or, 37.5 = p + 335 or, 375 = p + 335
10

? p = 40

So, the required value of p is Rs. 40.

(ii) Mean of individual repeated data (Mean of a frequency distribution)
In the case of repeated data, follow the steps given below to calculate the mean:

(a) Draw a table with 3 columns.

(b) Write down the items (x) in ascending or descending order in the first column
and the corresponding frequencies in the second column.

(c) Find the product of each item and its frequency (fx) and write in the third column.

(d) Find the total of f column and fx column.

(e) Divide the sum of fx by the sum of f (total number of items), the quotient is the
required mean.

Example 3. From the table given below, calculate the mean mark.

Marks 12 16 23 28 30 36
No. of students 438942

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Solution: Calculation of mean marks:

Marks (x) No. of students (f) fx
48
12 4 48
184
16 3 252
120
23 8 72
¦fx = 724
28 9

30 4

36 2

Total N = 30

Now, mean mark (x) = ¦fx = 724 = 24.13
N 30

So, the required mean mark is 24.13.

Example 4. If the mean of the data given below be 18, find the value of m.

Solution: x 5 10 15 20 25 30
f34 9 m 4 3

Here,

xf fx

53 15

10 4 40

15 9 135

20 m 20m

25 4 100

30 3 90

Total N = 23 + m ¦fx = 380 + 20m

Now, mean (x ) = ¦fx
N
380 + 20m
or, 18 = 23 + m

or, 414 + 18m = 380 + 20m

or, 2m = 34

? m = 17

So, the required value of m is 17.

(iii) Mean of Continuous Data

In the case of continuous data, we should find the mid–values (m) of each class

interval and it is written in the second column. The mid–value of each class interval
lower limit + upper limit
is obtained as: Mid–value = 2

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Then, each mid–value is multiplied by the corresponding frequency and the product
fm is written in the fourth column. The process of calculation of mean is similar to
the above mentioned process.

Example 5. Calculate the mean from the table given below.

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

No. of students 3 6 11 8 4

Solution: Calculation of mean

Marks (x) mid-value (m) No. of student (f) fm
15
0 - 10 5 3 90
275
10 - 20 15 6 280
180
20 - 30 25 11 ¦fm = 840

30 - 40 35 8

40 - 50 45 4

Total N = 32

Now, mean marks (x ) = ¦fx = 840 = 26.25
N 32

Exercise 13.3

Short Questions :

1. (a) Write the formula to find mean of individual series.

(b) Write the formula to find mean of discrete data.

(c) Write the formula to find mean of continuous data.

2. Calculate the mean from the given data:

(a) 6x = 220, n = 10 (b) 6x = 540, n = 27

(c) 6fx = 1250, n = 30 (d) 6fx = 1500, n = 40

(e) 6fm = 1000, N = 50

3. (a) The ages of Hari, Mohan, Sudip, Jeevan, and Nirajan are 12, 18, 13, 16, and
6 years respectively. Find their average age.

(b) Find the mean value of the following:

5, 11, 14, 10, 8, 6

(c) Find the mean from the data given below.

57, 74, 83, 76, 60

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4. (a) The average age of 5 students is 9 years. Out of them, the ages of 4 students
are 5, 7, 8, and 15 years. What is the age of the remaining student?

(b) If 7 is the mean of 3, 6, a, 9, and 10, find the value of a.

(c) Find x if the mean of 2, 3, 4, 6, x, and 8 is 5.

Long Questions :

5. (a) Find the mean from the given table:

Marks obtained 15 25 35 45 55
No. of students 5 7 15 6 7

(b) The ages of the students of a school are given below. Find the average age.

Age (in years) 5 8 10 12 14 16

No. of students 2 4 6 8 4 1

(c) Compute the arithmetic mean from the following frequency distribution table.

Height (in cm) 58 60 62 64 66 68
No. of plants 12 14 20 13 8 5

6. (a) Find the mean of the following frequency distribution:

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency
56 7 10 6 4

(b) Find mean.

Daily wage (Rs) 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

No. of labours 2 3 6 5 4 3

(c) Find the mean from the following data:

Marks obtained 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of students 245321

(d) Compute the mean from the table given below.

Age (years) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Men 478541

(e) The table below gives the weekly earnings of 115 workers in a textile mill.

Daily earning (Rs) 200-300 300-400 400-500 500-600 600-700 700-800

No. of workers 20 30 40 15 10 5

Find the average weekly earning.

(f) The ages of workers in a factory are as follows:

Age in years 18 – 24 24 – 30 30 – 36 36 – 42 42 – 48 48 – 54

No. of people 5 6 8 6 5 4
Calculate the average age of the groups.

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7. (a) Find the mean by constructing a frequency table of class interval of 10
from the data given below.

7, 20, 47, 36, 39, 31, 45, 19, 41, 49,
9, 16, 51, 29, 22, 28, 59, 17, 49, 21,
24, 12, 31, 8, 25, 36, 18, 32, 16, 23
(b) Construct a frequency table of class interval of 10 from the given data and
find the mean:

10, 12 23, 5, 15, 17, 28, 39, 52, 16,
22, 69, 75, 17, 73, 41, 33, 9, 49, 34,
59, 72, 46, 65, 58, 18, 26, 60, 48, 64,
32, 50, 73, 57, 51, 63, 36, 20, 22, 21

2. (a) 22 (b) 20 (c) 41.66 (d) 37.5 (e) 20
3. (a) 13 (b) 9 (c) 70
4. (a) 10 (b) 8 (c) 7 (d) 35.34
5. (a) 35.75 (b) 10.8 (c) 62.16 cm (b) 39.5
6. (a) 29.73 (b) 71.52 (c) 46.17
(f) 35.11 7. (a) 28.33
(e) 433.33

The partition Values

The variate values which divide given data into equal parts are known as the
partition values. The given data may be divided into two, four, ten and hundred
equal parts. The most widely used partion values in the analysis of data are:

a) Quartiles b) Deciles c) Percentiles

In this section, we study medians and quartiles of individual and discrete series
only.

13.9 Median

Let us observe the given data.

4, 5, 7, 15, 17, 19, 24, 28, 30

4 items Middle item 4 items

In the above series, the numbers are arranged in ascending order. Here, the fifth
item 17 has four items before it and four items after it. So, 17 is the middle item in
the series. 17 is called the median of the series.

Thus, median is the value of the middle item when the data are arranged in ascending
or descending order of magnitudes.

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(a) Median of Individual Data

To find the median of an individual data, arrange them in ascending or descending

order. Let the total number of observation be n. n +1
2
(i) If n is odd, the median is the value of the th observation.

(ii) If n is even, the median is the average of the n th and n + 1 th observation.
2 2

Worked Out Examples

Example 1. The weights in kg of 9 students are given below. Find the median
Solution:
weight. 53, 50, 48, 43, 54, 62, 65, 45, 55

Arranging the weights in ascending order, we have,

43, 45, 48, 50, 53, 54, 55, 62, 65

Here, n = 9 n+1
2
Now, the position of median = th item

= 9+1 th item = 5th item
2
i.e. 5th item is the median.

? Median = 53.

Example 2. The marks obtained by 10 students are given below. Calculate the
median mark.

13, 12, 15, 25, 10, 30, 20, 35, 40, 18

Solution: Arranging the marks in ascending order, we have,

10, 12, 13, 15, 18, 20, 25, 30, 35, 40

Here, n = 10

Now, the position of median = n+1 th item
2

= 10 + 1 th item = 5.5th item
2
5.5th item is the average of 5th and 6th items.

? Median = 5th item + 6th item
(b) Median of Discrete series 2
18 + 20 38
= 2 = 2 = 19

To calculate the median of a discrete series of frequency distribution, we should
display the data in ascending or descending order in a cumulative frequency table.
Then, the median is obtained by using the formula:

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Median = value of N+1 th item
2
Note :

The variate value corresponding to c.f. just greater or equal to n+1 th item is
median for discrete series. 2

Example 3. Compute the median from the table given below.

Solution: x 10 12 14 17 18 19 22
f 2248753

Let us prepare a cumulative frequency table.

x f cf
10 2 2
12 2 4
14 4 8
17 8 16
18 7 23
19 5 28
22 3 31
Total N = 31

Now, position of median = N+1 th item
2

= 31 + 1 th item = 16th item
2

In c.f. column, the c.f. just greater than or equal to 16 is 16 and

its corresponding values is 17.

? Median = 17.

13.10 Quartiles

25% 75%

Q1 Q2 Q3

Quartiles are the variate values that divide the data arranged in ascending or
descending order into four equal parts. A distribution is divided into four equal
parts by three quartiles.

(a) The first quartile or lower quartile (Q1) is the point below which 25 % of the
items lie and above which 75 % of the items lie.

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(b) The second quartile (Q2) is the point below which 50 % of the items lie and
above which 50 % of the items lie. the second quartile is the median.

(c) The third quartile or upper quartile (Q3) is the point below which 75 % of the
items lie and above which 25 % of the items lie.

If N be the number of items in ascending (or descending) order of a distribution,

then in the case of discrete data

the position of the first quartile (Q1) = n+1 th item
4

the position of the second quartile (Q2) = 2(n + 1) th = n+1 th item
4 2

the position of the third quartile (Q3) = 3(n + 1) th item
4

Here we study to calculate the median of individual and discrete data only.

Example 4. Find the first quartile (Q1) and the third quartile (Q3) from the data
Solution: given below.

5, 7, 9, 17, 10, 20, 22, 13, 28, 23, 24

Arranging the data in ascending order,
Example 5.
Solution: 5, 7, 9, 10, 13, 17, 20, 22, 23, 24, 28

Here, n = 11

The position of the first quartile (Q1) = n+1 th item
4

= 11 + 1 th item = 3rd item
4
The value of 3rd item is 9.

? The first quartile (Q1) = 9.

Again, the position of the third quartile (Q3) = 3 n+1 th item
4

=3 11 + 1 th item
4
= 9th item

The value of 9th item is 23.

? The third quartile (Q3) = 23.

The marks obtained by 10 students of class 8 in mathematics are
given below. Compute Q1 and Q3.

20, 15, 25, 27, 40, 45, 37, 23, 17, 35

Arranging the marks in ascending order,

15, 17, 20, 23, 25, 27, 35, 37, 40, 45

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Here, n = 10

The position of the Q1 = n+1 th item = 10 + 1 th item
4 4
= 2.75th term

= 2nd item + (3rd item – 2nd item) × 0.75

Here, the 2nd item is 17 and 3rd item is 20.

? Q1 = 17 + (20 – 17) × 0.75 = 17 + 3 × 0.75 = 19.25

Again, the position of Q3 =3 n+1 th item = 3 × 2.75th item
Example 6. 4
= 8.25th item

Here, 8th item is 37 and 9th item is 40.

? Q3 = 37 + (40 – 37) × 0.25 = 37 + 3 × 0.25 = 37.75

Compute the first and the third quartiles from the table given below:

Marks 20 30 40 50 60 70 80 90
No. of students 3 4 6 10 12 5 2 2

Solution: Cumulative frequency distribution table

Solution: Marks (x) No. of students (f) cf
20 3 3
30 4 7
40 6 13
50 10 23
60 12 35
70 5 40
80 2 42
90 2 44

Total N = 44

Now, the position of the first quartile (Q1) = N+1 th item
4

= 44 + 1 th item
4
= 11.25th item

In cf column, cf just greater 11.25 is 13 whose corresponding value is 40.

? The first quartile (Q1) = 40 N+1
4
Again, the position of the third quartile (Q3) = 3 th item

= 33.75th item

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In c.f. column, the c.f. just greater than 33.75 is 35
and its corresponding value is 60.
? The third quartile (Q3) = 60

13.11 Mode

The most repeated value in the set of given data is called mode. A distribution that
has two modes is called bimodal. The mode of a set of data is denoted by Mo.

Here, we study to calculate the mode of individual and discrete data.

(a) Mode of Individual Series

Change the given individual series into frequency distribution. For the given data,
the variate value which is mostly repeated is mode of the data.

(b) Mode of Discrete Data

In the case of discrete data, mode can be found just by inspection, i.e. just by taking
an item with highest frequency.

Example 7. Find the mode for the following distribution.
Solution: 15, 25, 18, 12, 20, 18, 21, 22, 18, 20, 18, 20, 22, 27
Arranging the data in ascending order.
12, 15, 18, 18, 18, 18, 20, 20, 20, 21, 22, 22, 25, 27
Here, maximum repeated value is 18, which is repeated four times.
? Mode = 18

Exercise 13.4

Short Questions :

1. (a) Define median.

(b) Write formula to compute median for individual series.
(c) What is most repeated value in a set of data called?
2. (a) Find the medians of the following sets of data.

(i) 13, 27, 36, 16, 28
(ii) 20, 15, 25, 40, 30, 35, 40
(iii) 40, 50, 60, 30, 70, 80, 100

(b) The weights of five students are as follows. Find their median weight.

42 kg, 57 kg, 45 kg, 60 kg, 55 kg, 80 kg

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(c) Find the median age of a group of 10 people whose ages in years are as
follows:

37, 51, 23, 43, 55, 23, 33, 22, 70, 35

3. Find the first quartiles (Q1) and Q3 of the following sets of data.
(a) 13, 12, 18, 23, 22, 26, 11

(b) 13, 26, 15, 40, 38, 18, 17

(c) 40, 20, 30, 10, 60, 55, 25, 45, 28, 75

(d) 10, 19, 11, 23, 37, 40, 24, 55

(e) 18, 13, 14, 22, 25, 42, 34, 40

(f) 10, 5, 50, 30, 40, 60, 70, 90

4. Find the modes of the following distributions.

(a) 7, 9, 5, 7, 10, 9, 9, 7, 9, 12, 9

(b) 12 kg, 21 kg, 18 kg, 24 kg, 28 kg, 21 kg, 15 kg, 21 kg, 28 kg

5. (a) A shopkeeper sold 50 pairs of shoes of 5 number, 40 pairs of 4 number, 20
pairs of 6 number, 45 pairs of number 7, 70 pairs of 8 number. What is the
size of modal shoes?

(b) In a factory, number of labourers and their remuneration are as follows.
Find the modal class.

Remuneration (in Rs) 1000 – 2000 2000 – 3000 3000 – 4000

No. of labourers 100 200 70

(c) Find the modal size of the shoes from the data given below:

Size of shoe 5 6 7 8 9 10
No. of men 12 16 40 35 20 15
(d) Find the mode from the following data:

Daily wages (in Rs) 70 90 110 130 150 170

No. of workers 10 20 40 15 25 18

Long Questions :
6. Find the median marks from the data given below:

(a) Marks 24 36 50 65 78

No. of students 4 8 12 10 7

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(b) Marks 10 20 30 40 50 60
7 3
No. of students 3 7 15 12

7. Compute the median wage from the table given below:

Wage in Rs. 145 155 165 175 185 195

No. of workers 10 35 14 4 8 6

8. Calculate the first quartile (Q1) and Q3 from the data given below:

(a) Marks obtained 40 50 60 70 80 90

No. of students 3 6 7 6 5 4

(b) Wages in Rs. 50 60 70 80 90 100

No. of students 6 10 15 13 8 3

(c) Ages in years 14 24 34 44 54
No. of people 5 15 20 20 14

2. (a) (i) 27 (ii) 30 (iii) 60 (b) 56 kg (c) 36 yrs.
3. (a) 12, 23 (b) 15, 38 (c) 23.75, 56.25 (d) 13, 39.25
(f) 12.5, 67.5
(e) 15, 38.5 (b) 21 (c) 7 (d) Rs. 110
4. (a) 9 (b) 2000 - 3000
5. (a) 8 (b) 30
6. (a) 50
7. 155 (b) 60, 80 (c) 24, 44
8. (a) 50, 80

13.12 Range

Range is the difference between the largest and smallest item in the set of observation.
It is denoted by R and calculated as:

R=L–S
where, L = Largest item

S = Smallest item

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Worked Out Examples

Example 1. Find the range of given data : 5, 10, 15, 20, 22, 25, 30
Solution: Here, largest item (L) = 30

Example 2. smallest item (S) = 5
Now, Range (R) = L – S = 30 – 5 = 25

Find the range from the given data :

Age (years) 15 17 18 20 22 25
No. of people 2 8 10 15 10 5

Solution: Here, largest item (L) = 25 years

smallest item (S) = 15 years

Now, Range (R) = L – S = 25 – 15 = 10 years.

Example 3. Find the range from the following data :

Temperature (°C) 0-10 10-20 20-30 30-40

No. of days 2 10 5 3

Solution: There are two methods of calculating the range of continuous data set
(i) First Method

Here, highest class interval = 30 - 40

lowest class interval = 0 - 10

Range = Upper limit of highest – Lower limit of the
class interval smallest class interval

= 40 – 0 = 40

(ii) Second Method

Here, mid-value of lowest C.I. = 0 + 10 = 5
2
30 + 40
mid-value of highest C.I. = 2 = 35

Range = mid-value of height C.I. – mid-value of lowest C.I.

= 35 – 5 = 30

Exercise 13.5

Short Questions :
1. Find the range of the following data :

(a) 20 cm, 25 cm, 30 cm, 25 cm, 40 cm, 45 cm, 50 cm

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(b) 40 kg, 42 kg, 46 kg, 50 kg, 70 kg 35
(c) 70, 50, 80, 90, 20, 100, 92, 88 3
(d) 22, 23, 27, 28, 30, 25, 32, 47, 50
(e) 20 mm, 25 mm, 40 mm, 28 mm, 30 mm, 35 mm
2. Find the range of the following given data :

(a) x 15 23 25 28 30
f 2 5 10 7 2

(b) Weights (kg) 35 40 45 50 55 60

No. of students 2 4 6 10 3 2

(c) Heights (cm) 10 20 30 40 50 60
No. of plants 2 3 9 21 11 5

(d) Marks in Maths 20 30 40 50 60 70 80 90 10
No. of students 2 4 6 10 25 15 8 3 1

Long Questions :
3. Find the range of the following given data :

(a) x 0-10 10-20 20-30 30-40
f2 4 6 3

(b) Value (L) 5-10 10-15 15-20 20-25 25-30 30-35
No. of family
4 12 16 14 5 4

(c) Income (Rs.) 10-20 20-30 30-40 40-50 50-60 60-70
No. of workers 2 4 8 15 10 8

(d) Income (Rs.) 100-200 200-300 300-400 400-500 500-600 600-700

No. of workers 2 5 15 12 13 5

1. (a) 30 cm (b) 30 kg (c) 80 (d) 28 (e) 15 mm
2. (a) 20 (b) 25 (c) 50 (d) 80
3. (a) 40 or 30 (b) 30 or 25 (c) 60 or 50 (d) 600 or 500

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Model Questions

Class : 8 Annual Examination Time : 1.5 hours
Attempt all the questions. FM - 50 PM - 20

Group A [9×2 = 18]

1. Find the values of x and y if (2x + y, 2) = (1, x – y).

2. If P = 1 2 and Q = 4 3 , find P – Q and P + Q.
3 4 2 1

3. For what value of k does the point (2, 1) lie in the locus whose equation is

x2 + y2 – 4x + 3y + k = 13?

4. Find the equation of a straight line whose x-intercept and y-intercept are
respectively 4 and 5.

5. Find the interior and exterior angle of regular hexagon in degrees and grades.

6. Prove that : 1 – cos4 T = 1 + 2 cot2 T
sin4 T

7. Define a unit vector. Find the magnitude of vector op = 2 .
–3

8. If P(2, 3), Q(–2, –3), and R(0, 4) are the vertices of 'PQR, find the coordinates of

image vertices of the triangle when it is rotated through 180° about origin.

9. Compute Q1 and Q3 from the given data :
6, 10, 5, 7, 11, 12, 8

Group B [8×4 = 32]

10. If f(x) = 2x + 3, find the value of f(a + h) – f(a) .
h

11. If P = 1 2 , Q = –4 5 , and R = 2 3 , then verify that
3 7 2 1 4 1

(P + Q) + R = P + (Q + R)

12. Find the equation of a straight line whose perpendicular distance from the
origin is 4 units and angle made by the perpendicular with X-axis is 60°.

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13. If tan T = pq, then prove that

p cos T + q sin T = p2 + q2
p cos T– q sin T p2 – q2

14. From a top of building of 80 m height, the angle of depression of a bus on

the ground is to be found to be 60°. Find the distance between the foot of the

building and the bus.

15. If A(2, 7), B(5, 2), C(2, 5), and D(5, 0), prove that AoB = CoD.

16. A(2, –4), B(2, 4), and C(7, 3) are the vertices of 'ABC. Find coordinates of its
image under the reflection Y-axis. Plot 'ABC and its image on the same graph.

17. Find the mean from the given data:

Marks obtained 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of students 2 4 10 15 10 3 1

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